(a) In Module 0, one of the demonstrations included two copper rods where one was unprocessed and the other had been bent and then heated to around 700°C for approximately 10 minutes. During bending, the type of processing that occurs is called cold working or cold forming.
Cold working involves bending, hammering, rolling, or pressing the metal at temperatures below its recrystallization temperature. This type of processing effects the strength of the copper rod, making it stronger and harder.(b) The technical name for the type of processing that was performed on the rod heated to 700°C is annealing.
The annealing process involves heating the metal to a high temperature, holding it at that temperature for some time, and then allowing it to cool slowly. This type of processing can reduce the strength of the copper rod, making it softer. Therefore, annealing is usually done to relieve stress or to improve machinability.
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2. Enthalpy Stoichiometry a) How much heat will be released when 6.44g of sulfur reacts with excess O, according to the following equation? 25+30,- 250, AH = -791.4 KJ b) How much heat will be released when 4.72g of carbon reacts with excess O, according to the following equation? C+0₁ - CO₂ AH=-393.5 kJ c) How much heat will be released when 4.77g of ethanol reacts with excess O, according to the following equation? AH = -1366.7 KJ C₂H,OH+30→2 CO₂ + 3 H₂O
a) The reaction of 6.44g of sulfur with excess oxygen will release approximately 791.4 kJ of heat. b) The reaction of 4.72g of carbon with excess oxygen will release approximately 393.5 kJ of heat. c) The reaction of 4.77g of ethanol with excess oxygen will release approximately 1366.7 kJ of heat.
To calculate the heat released in a chemical reaction, we need to use the enthalpy change (∆H) of the reaction and the amount of the reactant involved. The enthalpy change (∆H) represents the heat released or absorbed during a reaction.
In each case, we are given the ∆H value for the reaction. By multiplying the ∆H value by the moles of the reactant, we can determine the heat released or absorbed. To find the moles of the reactant, we divide the given mass by the molar mass of the substance. After calculating the moles, we multiply it by the ∆H value to obtain the heat released or absorbed. In this case, since all reactions involve excess oxygen, we assume that the given mass of the reactant is completely consumed.
Therefore, for each reaction, we can multiply the moles of the reactant by the ∆H value to find the heat released. The results are approximately 791.4 kJ, 393.5 kJ, and 1366.7 kJ for sulfur, carbon, and ethanol reactions, respectively.
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The following complex gas phase reactions follow elementary rate laws: (1) A+B→C, -TIA KIA CA CB, (2) 2A + C⇒ D, -r2c= k2c CA² Cc. The reactions are carried out isothermally in a PBR. The feed is equimolar in A and B with FAO-10 mol/min and the volumetric flow rate is 100 dm³/min. The catalyst weight is 50kg, the pressure drop parameter α-0.0019 kg¹, and the total entering concentration is CTO =0.2 mol/dm³. KIA=100 dm³/(mol kg cat min), k2e-1500 dm/(mol² kg cat min). Plot FA, FB, Fc, FD, dimensionless pressure, and overall selectivity ŠC/D as a function of catalyst weight W.
The overall selectivity of the reaction is ŠC/D = (FDO - FD) / (FA0 - FA)
The plot of FA, FB, FC, FD, the dimensionless pressure, and the overall selectivity SC/D as a function of catalyst weight W is shown below:
FA Remember that FAO is the inlet molar flow rate of species A.
FA is the molar flow rate of species A at any given catalyst weight W.
FA0 = 10 mol/min
FA = FA0 (1 - X)
where X is the conversion of A at any given catalyst weight W.
FBThe molar flow rate of species B at any given catalyst weight W is:
FB0 = 10 mol/min
FB = FB0 (1 - X)
where X is the conversion of A at any given catalyst weight W.
FCThe molar flow rate of species C at any given catalyst weight W is:
FC0 = 0 mol/minFC = FA0 * (1 - X)
where X is the conversion of A at any given catalyst weight W.
FD The molar flow rate of species D at any given catalyst weight W is:
FD0 = 0 mol/min
FD = FA0 * (1 - X)
where X is the conversion of A at any given catalyst weight W.
The dimensionless pressure The dimensionless pressure drop (P/PTO) as a function of the catalyst weight W is:
PTO = CTO * R * T / α * FTO
Where FTO is the volumetric flow rate and R is the gas constant.
Overall selectivity The overall selectivity of the reaction is given by:
ŠC/D = (FDO - FD) / (FA0 - FA)
where FDO is the molar flow rate of D at the outlet of the reactor.
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At −78∘C and 5.2 atm, carbon dioxide is in which phase?
OPTIONS:
solid
liquid
gas
supercritical fluid
solid-liquid equilibrium
liquid-gas equilibrium
solid-gas equilibrium
solid-liquid-gas equilibrium
At -78°C and 5.2 atm, carbon dioxide is in the solid phase according to its phase diagram. The temperature and pressure conditions fall below the sublimation curve, indicating the presence of solid carbon dioxide (dry ice).
At different combinations of temperature and pressure, substances can exist in different phases, such as solid, liquid, or gas. The phase of a substance is determined by the temperature and pressure conditions it is subjected to.
In the case of carbon dioxide (CO2), at atmospheric pressure (1 atm), it undergoes sublimation, transitioning directly from a solid to a gas without passing through the liquid phase. This is why we often refer to dry ice (solid carbon dioxide) "vaporizing" instead of melting.
At -78°C and 5.2 atm, the conditions fall below the triple point of carbon dioxide, which is the temperature and pressure at which the solid, liquid, and gas phases of a substance coexist in equilibrium. Therefore, carbon dioxide at these conditions is in the solid phase.
To provide a more detailed explanation, we can refer to the phase diagram of carbon dioxide. The phase diagram shows the relationship between temperature, pressure, and the phases (solid, liquid, and gas) of a substance.
Carbon dioxide has a phase diagram where the solid-gas equilibrium line slopes slightly downward from left to right, indicating that the solid phase of carbon dioxide (dry ice) can exist at low temperatures and moderate to high pressures. The phase diagram also shows a sublimation curve, separating the solid and gas phases.
At -78°C and 5.2 atm, we can locate this point on the phase diagram. The temperature (-78°C) falls well below the sublimation curve, indicating that carbon dioxide is in the solid phase at this temperature. The pressure (5.2 atm) is above the vapor pressure curve for the solid phase, further confirming that carbon dioxide is in the solid phase.
Therefore, at -78°C and 5.2 atm, carbon dioxide is in the solid phase.
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Consider an Irreversible, liquid-phase, elementary reaction 2A → B carried out (+25 in two continuous stirred tank reactors (CSTRS) in series. The two reactors have the same volume. The intermediate conversion of A exiting the first reactor prior to entering the second reactor is 50%. Assume that the volumetric flow rate remains constant throughout the process. (A) Write down the rate law describing the rate of disappearance of A (i.e., -ra) in terms of the rate constant (k), the conversion of A (X), and the initial concentration of A (CAO). (B) Calculate the value of the Damköhler's number for the operation. : (C) Determine the final conversion of A after the second reactor.
(A) Rate law: \(-rA = k \cdot CAO \cdot (1 - X)^2\) (B) \(Da = \frac{k \cdot V}{Q}\) (C) The final conversion of A after the second reactor is 25%.
(A) The rate law describing the rate of disappearance of A, denoted as -rA, in terms of the rate constant (k), the conversion of A (X), and the initial concentration of A (CAO), can be written as follows:
\(-rA = k \cdot CAO \cdot (1 - X)^2\)
(B) The Damköhler's number, Da, for the operation can be calculated using the formula:
\(Da = \frac{k \cdot V}{Q}\)
where k is the rate constant, V is the volume of each reactor, and Q is the volumetric flow rate.
In this given system, the reaction is irreversible and liquid-phase, with the reaction 2A → B occurring in two continuous stirred tank reactors (CSTRs) in series. The reactors have the same volume, and the intermediate conversion of A exiting the first reactor prior to entering the second reactor is 50%. The volumetric flow rate remains constant throughout the process.
To determine the final conversion of A after the second reactor, we can follow these steps:
1. We start with the rate law for the reaction, which is a second-order reaction:
\(-rA = k \cdot CAO \cdot (1 - X)^2\)
where -rA represents the rate of disappearance of A, k is the rate constant, CAO is the initial concentration of A, and X is the conversion of A.
2. In the first reactor, the conversion of A is 50%, meaning X = 0.5. We substitute this value into the rate law:
\(-rA_1 = k \cdot CAO \cdot (1 - 0.5)^2\)
\(-rA_1 = \frac{1}{4} \cdot k \cdot CAO\)
3. The concentration of A entering the second reactor is CAO * (1 - X) = CAO * (1 - 0.5) = CAO * 0.5.
4. In the second reactor, we have the same rate law, but with the concentration of A being CAO * 0.5:
\(-rA_2 = k \cdot (CAO \cdot 0.5) \cdot (1 - X)^2\)
\(-rA_2 = \frac{1}{4} \cdot k \cdot CAO \cdot (1 - 0.5)^2\)
\(-rA_2 = \frac{1}{16} \cdot k \cdot CAO\)
5. The total conversion of A after both reactors can be calculated by multiplying the individual conversions:
Total Conversion = X_1 * X_2 = 0.5 * (1 - 0.5) = 0.25
Therefore, the final conversion of A after the second reactor is 25%.
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A stream of o-Xylene vapor at its normal boiling point and 1 atm flowing at a rate of 250 kg/min is to be condensed at constant pressure. The product stream from the condenser is liquid o-Xylene at the condensation temperature (3 pts). a) Using data in Table B.1, calculate the rate (kW) at which heat must be transferred from the condenser.
b) If heat were transferred at a lower rate than that calculated in Part a), what would the state of the product stream be? (Deduce as much as you can about the phase and the temperature of the stream.) c) If heat were transferred at a higher rate than that calculated in Part a), what could you deduce about the state of the product stream?
a) The rate at which heat must be transferred from the condenser can be calculated using the enthalpy of vaporization and the mass flow rate of o-Xylene.
b) If heat is transferred at a lower rate than calculated in Part a), the product stream would remain in the vapor phase.
c) If heat is transferred at a higher rate than calculated in Part a), the product stream would be in the two-phase region, consisting of a mixture of liquid and vapor.
a) To calculate the heat transfer rate, we need to determine the enthalpy of vaporization and the mass flow rate of o-Xylene. The enthalpy of vaporization for o-Xylene can be found in Table B.1. Let's assume it is ΔH_vap = 320 kJ/kg. The heat transfer rate (Q) can be calculated using the equation Q = ΔH_vap * mass flow rate. Given that the mass flow rate is 250 kg/min, we can substitute these values into the equation to find the heat transfer rate.
b) If heat is transferred at a lower rate than calculated in Part a), the product stream would not have enough heat energy to condense completely. Therefore, it would remain in the vapor phase.
c) If heat is transferred at a higher rate than calculated in Part a), the product stream would receive more heat energy than required for complete condensation. This would result in the product stream being in the two-phase region, consisting of a mixture of liquid and vapor.
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which of the following is a lewis acid? group of answer choices more than one of these choices oh- h3o bf3 nf3
Answer:
BF_3 is a Lewis acid
Explanation:
Lewis Acid is a type of molecule that takes an electron pair from an electrophile and has empty orbitals. In the choices given, BF_3 is a strong electrophile and has an empty orbital that can accept a lone pair of electrons. It is therefore a Lewis acid. The other three OH- H_3O NF_3 are Lewis bases because they have a single pair of electrons.
A steam turbine is supplied with steam at a pressure of 5.8 MPa and a temperature of 450 °C. The steam is exhausted from the turbine at a pressure of 1.0 MPa. Determine the work output from the turbine per unit mass of steam, assuming that the turbine operates isentropically. You may assume negligable changes in kinetic and potential energy. Hint, use steam properties (online or tables) to determine enthalpy and entropy at the inlet and exit conditions. Enter the answer in units of kJ/kg to 1 dp. [Do not include the unit symbol]
To determine the work output from the steam turbine per unit mass of steam, we need to calculate the change in specific enthalpy (Δh) between the inlet and exit conditions. Since the turbine operates isentropically, the change in specific enthalpy can be calculated using the isentropic efficiency (η) of the turbine.
Given:
Inlet pressure (P1) = 5.8 MPa
Inlet temperature (T1) = 450 °C
Exit pressure (P2) = 1.0 MPa
Step 1: Determine the specific enthalpy at the inlet condition (h1) using steam tables or steam properties.
Step 2: Determine the specific entropy at the inlet condition (s1) using steam tables or steam properties.
Step 3: Determine the specific entropy at the exit condition (s2) using steam tables or steam properties.
Step 4: Determine the specific enthalpy at the exit condition (h2) using the isentropic relation: s2 = s1.
Step 5: Calculate the change in specific enthalpy (Δh) as Δh = h2 - h1.
Step 6: Calculate the work output per unit mass of steam (W) as W = Δh.
Now let's perform the calculations:
Step 1: Using steam tables or steam properties, determine the specific enthalpy at the inlet condition:
h1 = [specific enthalpy at P1 and T1]
Step 2: Using steam tables or steam properties, determine the specific entropy at the inlet condition:
s1 = [specific entropy at P1 and T1]
Step 3: Using steam tables or steam properties, determine the specific entropy at the exit condition:
s2 = [specific entropy at P2]
Step 4: Using the isentropic relation, determine the specific enthalpy at the exit condition:
h2 = [specific enthalpy at P2 and s2]
Step 5: Calculate the change in specific enthalpy:
Δh = h2 - h1
Step 6: Calculate the work output per unit mass of steam:
W = Δh
Make sure to refer to steam tables or steam properties to obtain the specific enthalpy and entropy values at the given conditions.
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Identify 3 mistakes (not mentioned in the questions) in the previous method section part
Consider the Method part below in answering questions 3-4: Materials: 1. S. tamariscina (batch no: 20180325) 2. Benzoxazole 3. 50 mL of absolute ethanol 4. three-neck flask 5. 100 mesh, 21-73 μm 6. 1260 high-performance liquid chromatography (HPLC) system 7. ultrasonic device Preparation of Benzoxazole Ionic Liquids Firstly, place the solution in a three-neck flask, and cool it 5 °C after you put 50 mL of absolute ethanol in a 100 mL beaker. Then, dissolve Benzoxazole (0.1 mol). Then, mix the sample with 20 mL of 0.1 mol methanesulfonic acid. Then, let to react at room temperature. Then, recrystallize in absolute ethanol, and dry to obtain orange-red crystals (melting point, 104 °C; the yield will be, 75.6%).
The concentration of the solution is not mentioned in the method provided. Mistake 1: The size of the three-neck flask used is not mentioned. The size of the three-neck flask should have been mentioned in the Method part for the preparation of Benzoxazole Ionic Liquids.
It is a critical piece of information that needs to be included in the Method part for the preparation of Benzoxazole Ionic Liquids.Mistake 2: The mesh size of 100 is not enough to be mentioned, and the range is also not given. The mesh size of 100 is not sufficient to be mentioned in the Method part for the preparation of Benzoxazole Ionic Liquids. The range of the mesh size should have been mentioned, along with the size of the mesh used. The range of the mesh size is generally between 21-73 μm, but it should have been mentioned in the Method part for the preparation of Benzoxazole Ionic Liquids.
Mistake 3: The concentration of the solution is not mentioned in the method provided. The concentration of the solution should have been mentioned in the Method part for the preparation of Benzoxazole Ionic Liquids. It is important information that needs to be included in the Method part.
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Ethane is burned with 75% excess air. The percentage conversion of ethane is 88% of the ethane fed. 35% reacts to form CO and the balance reacts to form CO₂. Calculate the molar composition of the stack gas on a dry basis and the mole ratio of water to dry stack gas. 100 mol C₂H (mol C₂H₂) n₂(mol O₂) ny(mol N₂) 75% excess air (mol CO) ng(mol CO₂) ng(mol) ne(mol H₂O) 0.21 mol Oy/mol 0.79 mol Ny/mol C₂H6+0₂-200₂ + 3H₂O 200+ 3H₂O 3. A gas having a composition of CO₂ = 10.2 %, CO =2.5 %, O₂ = 6.3 %,N₂ = 81.0%, is heated from 91°C to 290°C.
Molar composition of the stack gas on a dry basis is approximately: CO: 30.8 / 112.5 ≈ 27.4%, CO₂: 57.2 / 112.5 ≈ 50.9%, N₂: 12.25 / 112.5 ≈ 10.9%, O₂: 12.25 / 112.5 ≈ 10.9%. Mole ratio : +approximately 0.826.
To calculate the molar composition of the stack gas on a dry basis and the mole ratio of water to dry stack gas, we need to determine the number of moles of each component in the gas mixture. This can be done using the given percentages and molar ratios of the reactants and products. Given: Ethane (C₂H₆) is burned with 75% excess air.
The percentage conversion of ethane is 88% of the ethane fed.
35% of the ethane reacts to form CO (carbon monoxide), and the remaining 65% reacts to form CO₂ (carbon dioxide).
The molar composition of the stack gas on a dry basis is required, along with the mole ratio of water to dry stack gas.
First, let's consider 100 moles of ethane (C₂H₆) as the basis. With 88% conversion, 88 moles of ethane will react.
From the given data, we can determine the moles of CO and CO₂ produced:
35% of 88 moles of ethane will react to form CO, so the moles of CO produced is 0.35 * 88 = 30.8 moles.
The remaining 65% of 88 moles of ethane will react to form CO₂, so the moles of CO₂ produced is 0.65 * 88 = 57.2 moles.
To determine the moles of nitrogen (N₂) and oxygen (O₂) in the stack gas, we need to consider the molar ratios:
From the balanced equation, for every 2 moles of ethane, we need 7 moles of O₂.
Given that there is 75% excess air, the moles of O₂ will be 1.75 times the stoichiometric requirement. Thus, the moles of O₂ are 1.75 * 7 = 12.25 moles.
The moles of nitrogen (N₂) will be the same as the moles of O₂, since air consists mainly of nitrogen. Therefore, the moles of N₂ are also 12.25 moles.
Now, let's calculate the molar composition of the stack gas on a dry basis:
The total moles of dry stack gas (excluding water vapor) is the sum of CO, CO₂, N₂, and O₂: 30.8 + 57.2 + 12.25 + 12.25 = 112.5 moles.
The mole ratio of water to dry stack gas can be determined by dividing the moles of water (which is 3 times the moles of CO produced) by the moles of dry stack gas: (3 * 30.8) / 112.5 ≈ 0.826.
Therefore, the molar composition of the stack gas on a dry basis is approximately:
CO: 30.8 / 112.5 ≈ 27.4%
CO₂: 57.2 / 112.5 ≈ 50.9%
N₂: 12.25 / 112.5 ≈ 10.9%
O₂: 12.25 / 112.5 ≈ 10.9%
The mole ratio of water to dry stack gas is approximately 0.826. Understanding the molar composition and ratios in a gas mixture is important in various fields, including combustion analysis, air pollution control, and environmental engineering. It allows us to assess the composition and properties of the gas mixture, which is essential for understanding its behavior and potential impacts.
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zinc metal reacts with aqueous hydrochloric acid to form zinc chloride in solution and hydrogen gas. identify the oxidizing agent, the reducing agent, and the substances being oxidized and reduced.
The oxidation and reduction reactions in a redox reaction are known as redox half-reactions. The oxidation reaction is a half-reaction in which a substance loses electrons, while the reduction reaction is a half-reaction in which a substance gains electrons.
In this case,Oxidizing agent: H+ ions in aqueous HCl are the oxidizing agent. Reducing agent: Zinc metal is the reducing agent. Substances being oxidized: Zinc atoms in the zinc metal are oxidized to Zn2+ ions. Substances being reduced: H+ ions in aqueous hydrochloric acid are reduced to H2 gas.Explanation:Consider the following chemical reaction:Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)In this reaction, the oxidation state of zinc goes from zero in the solid state to +2 in the aqueous solution.
This means that zinc has lost electrons and has been oxidized. This implies that zinc is the reducing agent in this reaction. The oxidation state of H+ ions in aqueous hydrochloric acid is +1. When H+ ions in hydrochloric acid reduce to H2 gas, their oxidation state drops to zero. This implies that H+ ions are being oxidized and hence they are the oxidizing agent in this reaction. Therefore, the oxidizing agent is H+ ions, the reducing agent is zinc, the substance being oxidized is H+ ions in aqueous hydrochloric acid, and the substance being reduced is zinc metal.
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which of the following liquids do you expect to be most miscible with water mercury hexanes benzene's ammonia bromine
Ammonia is the liquid that is most miscible with water.
Out of the given options, ammonia is the liquid that is most miscible with water. Miscibility is the ability of two liquids to mix together without separating into layers. It is measured on a scale from complete immiscibility to complete miscibility.
When two liquids are miscible, they dissolve in each other in any proportion forming a homogeneous solution. In the case of ammonia, it is highly miscible with water due to the presence of a hydrogen bond between the nitrogen of ammonia and the hydrogen of water.
This leads to the formation of a strong intermolecular force between them, making ammonia soluble in water. Mercury is the least miscible with water since mercury is an elemental metal, and water is a polar solvent. Benzene and hexane are also non-polar liquids, and since water is polar, they are not miscible with water.
Bromine is slightly soluble in water, but its solubility decreases with temperature.
Therefore, ammonia is the liquid that is most miscible with water.
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A seawater sample contains the following components: NaCl-2.8%, MgCl₂-0.5% and NaBr-0.0085% by weight. The average specific gravity of the seawater is 1.03. What mass of magnesium, sodium and chlorine can be obtained from 100 m³ of this seawater?
The mass of magnesium, sodium, and chlorine can be obtained from 100 m³ of seawater is 3308.5 kg.
Given,
Percentage of NaCl (Sodium chloride) = 2.8%
Percentage of MgCl2 (Magnesium chloride) = 0.5%
Percentage of NaBr (Sodium bromide) = 0.0085%
Average specific gravity of seawater = 1.03
Volume of seawater = 100 m³
To calculate the mass of magnesium, sodium, and chlorine, we need to calculate their concentration first.
Concentration of NaCl:
It is given that the percentage of NaCl in seawater is 2.8%.
Therefore, for 100 m³ of seawater, the mass of NaCl can be calculated as follows:
Mass of NaCl
= (2.8 / 100) × (100 m³ × 1000 kg/m³)
= 2.8 × 1000 kg
= 2800 kg
Concentration of MgCl2:
It is given that the percentage of MgCl2 in seawater is 0.5%.
Therefore, for 100 m³ of seawater, the mass of MgCl2 can be calculated as follows:
Mass of MgCl2
= (0.5 / 100) × (100 m³ × 1000 kg/m³)
= 0.5 × 1000 kg
= 500 kg
Concentration of NaBr:
It is given that the percentage of NaBr in seawater is 0.0085%.
Therefore, for 100 m³ of seawater,
the mass of NaBr can be calculated as follows:
Mass of NaBr
= (0.0085 / 100) × (100 m³ × 1000 kg/m³)
= 0.0085 × 1000 kg = 8.5 kg
Average Specific Gravity of seawater:
It is given that the average specific gravity of seawater is 1.03.
Therefore, for 100 m³ of seawater, the total mass of the seawater can be calculated as follows:
Mass of seawater = Volume of seawater × Density of seawater
= 100 m³ × 1030 kg/m³
= 103,000 kg
Thus, the mass of chlorine in 100 m³ of seawater can be calculated as follows:
Mass of chlorine = Mass of NaCl + Mass of MgCl2 + Mass of NaBr
= 2800 + 500 + 8.5= 3308.5 kg
Hence, the mass of magnesium, sodium, and chlorine can be obtained from 100 m³ of seawater is 3308.5 kg.
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Look up the structures of the four standards which are caffeine, ibuprofen, acetaminophen, and aspirin. Which of these compounds would you expect to be visible under the UV lamp? Explain why the compounds are visible.
Caffeine, ibuprofen, acetaminophen, and aspirin are organic compounds with different chemical structures. Out of these four compounds, caffeine is the most likely to be visible under a UV lamp.
Caffeine has a conjugated system of double bonds in its structure, which allows it to absorb UV light. Conjugated systems consist of alternating single and multiple bonds, creating a delocalized system of electrons. This delocalization of electrons enables caffeine to absorb light in the UV range. As a result, when exposed to a UV lamp, caffeine will absorb UV radiation and become visible.
On the other hand, ibuprofen, acetaminophen, and aspirin do not possess extensive conjugated systems in their structures. These compounds do not absorb UV light strongly and are less likely to be visible under a UV lamp.
It's worth noting that the visibility of compounds under a UV lamp depends on their specific chemical structures and the presence of conjugated systems, which determine their ability to absorb UV light.
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there are ________ σ bonds and ________ π bonds in h 3c-ch 2-ch═ch-ch 2-c≡ch.
There are 6 σ bonds and 3 π bonds in H₃C-CH₂-CH═CH-CH₂-C≡CH.
In the given chemical structure, σ (sigma) bonds are formed by the overlap of atomic orbitals in a head-to-head fashion. These bonds allow for the sharing of electrons between the atoms involved. Each single bond, whether between carbon and hydrogen or carbon and carbon, is a σ bond. Therefore, we count the number of single bonds to determine the number of σ bonds.
In this molecule, there are six single bonds: three between carbon and hydrogen and three between carbon and carbon.Hence, there are 6 σ bonds in total.
In the given structure, there are three double bonds: one between carbon atoms (═), one between carbon and carbon (ch═ch), and one between carbon and carbon (c≡ch). Each double bond consists of one σ bond and one π bond.
Therefore, there are 6 σ bonds and 3 π bonds in the given chemical structure.
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Question 5 Flotation is best described as: a. the removal of suspended matter to clarify wastewater b. a process where air attached to dissolved particles which then float to the surface c. the use of floaters to coagulate smaller particles d. a process to clarify water by the removal of nitrogen.
Flotation is best described as a process where air attaches to dissolved particles, which then float to the surface. This corresponds to option b.
In flotation, the objective is to separate suspended or dissolved particles from a liquid or wastewater. The process involves introducing air bubbles into the liquid, typically using a flotation cell or tank. The air bubbles attach to the particles, forming aggregates or flocs. These aggregates are lighter than the surrounding liquid and float to the surface, creating a foam or froth layer. The foam layer can then be skimmed off to remove the floated particles.
Flotation is widely used in various industries for wastewater treatment, mineral processing, and water clarification. By selectively removing particles, such as oils, solids, or minerals, flotation helps to improve the quality and purity of the liquid or wastewater. The process relies on the attachment of air bubbles to the particles, allowing their separation based on their buoyancy properties.
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Urea dissolved in aqueous medium is separated into CO2 and ammonia by immobilized urease enzyme on the nonporous polymer bead surface. Given this system constant:
kL=0.2cm/s
Km=200mg/L
Vm'=0.1mg/urea/cm^2/s
Sb=2000 mg urea/L
(a) Determine the reaction rate.
(b) When increasing the concentration of immobilized enzymes by 4,000 times, Determine the reaction rate.
In a system where urea is dissolved in an aqueous medium and separated into CO2 and ammonia by immobilized urease enzyme, the initial reaction rate is approximately 0.0909 mg/urea/cm^2/s. When the concentration of immobilized enzymes is increased by 4,000 times, the new reaction rate becomes approximately 363.6 mg/urea/cm^2/s.
(a) The reaction rate can be determined using the Michaelis-Menten equation for enzyme kinetics. The reaction rate (v) is given by the equation:
v = Vm' * [Sb] / (Km + [Sb])
Where Vm' is the maximum rate of the reaction per unit area of the enzyme surface, [Sb] is the concentration of substrate (urea), and Km is the Michaelis constant.
Plugging in the given values: Vm' = 0.1 mg/urea/cm^2/s, [Sb] = 2000 mg urea/L, and Km = 200 mg/L, we can calculate the reaction rate (v):
v = (0.1 mg/urea/cm^2/s) * (2000 mg urea/L) / (200 mg/L + 2000 mg urea/L)
= 0.1 mg/urea/cm^2/s * 2000 mg urea/L / 2200 mg urea/L
= 0.0909 mg/urea/cm^2/s
Therefore, the reaction rate is approximately 0.0909 mg/urea/cm^2/s.
(b) When the concentration of immobilized enzymes is increased by 4,000 times, the new maximum rate of the reaction per unit area of the enzyme surface (Vm'') becomes:
Vm'' = Vm' * 4000
= 0.1 mg/urea/cm^2/s * 4000
= 400 mg/urea/cm^2/s
Using the same substrate concentration ([Sb] = 2000 mg urea/L) and Km value (Km = 200 mg/L), we can calculate the new reaction rate (v'):
v' = (400 mg/urea/cm^2/s) * (2000 mg urea/L) / (200 mg/L + 2000 mg urea/L)
= 400 mg/urea/cm^2/s * 2000 mg urea/L / 2200 mg urea/L
= 363.6 mg/urea/cm^2/s
Therefore, when the concentration of immobilized enzymes is increased by 4,000 times, the new reaction rate is approximately 363.6 mg/urea/cm^2/s.
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5. Why green leafy vegetables losses its color when subjected to thermal treatment? A. Due to inactivation of chloropyllase B. Due to inactivation of pheophytinase C. Due to release of organic acid D. Due to increase of pH 11. Flavors can come from different sources through different reactions. * 1p Which of these pairs are false example of taste substance and the resultant taste effect. A. Branched chain amino acids for ripe apple flavor B. Trimethylamine oxide for fishy aroma of seafood C. Methoxy alkyl pyrazines for green earthy aroma of vegetables D. Curcumin for the pungent and bitter flavor of cumin 12. Statements below are true, except 16 A. Fermentation of lactose by lactic acid bacteria contributes to cheese flavor and aroma B. Butyric acid in milk contributes to dairy goods flavor O C. Trimethylamine oxide (TMAO) is indicator of freshness in fish D. The longer storage period of fish will decrease the TMAO level
When we prepare green leafy vegetables , they sometimes change color and become less green. The best answer to explain this event is:
Due to inactivation of chloropyllase .
Green plants explained.When we prepare green plants, they sometimes change color and become less green. The best answer to explain this event is:
Due to inactivation of chloropyllase .
Chloropyii is the pigment in plants that makes it green.
" This means that the green color of the vegetables becomes less bright and they don't look as fresh.
Option A says that when we prepare green plants , they lose their green color because a chemical called chlorophyllase stops working. This is why the plants lose their green color.
The false example about how something tastes and what happens after you taste it is called a false example of taste substance and taste effect.
Curcumin is what makes cumin taste strong and unpleasant.
Curcumin is a natural substance that is in turmeric and is what makes it yellow. Cumin doesn't directly make things taste like cumin. Cumin has a unique taste that is warm and earthy.
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A sample of oxygen is collected over water at a total pressure of 678.4 mmHg at 25°C. The vapor pressure of water at 25°C is 23.8 mmHg. The partial pressure of the O2 is
0.9239 atm.
0.9161 atm.
0.8926 atm
1.092 atm.
0.8613 atm.
To calculate the partial pressure of oxygen (O2) in the given scenario, we need to subtract the vapor pressure of water from the total pressure.
Given:
Total pressure = 678.4 mmHg
Vapor pressure of water = 23.8 mmHg
Partial pressure of O2 = Total pressure - Vapor pressure of water
Partial pressure of O2 = 678.4 mmHg - 23.8 mmHg
Partial pressure of O2 = 654.6 mmHg
To convert mmHg to atm, divide by the conversion factor of 760 mmHg = 1 atm:
Partial pressure of O2 = 654.6 mmHg / 760 mmHg/atm
Partial pressure of O2 ≈ 0.8613 atm
Therefore, the partial pressure of O2 in the given scenario is approximately 0.8613 atm.
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2M HCI solution at 25°C? stolnotupo b) 5x10-¹5 c) 3.4x10-¹2 d) 5.2x10-19 10) The pH of a buffer solution contains 0.25 M HF and 0.32 M NaF (K, for HF is 7.2 x10). a) 0.60 b) 2.93 c) 3.04 d) 3.25 11) Equilibrium is reached in chemical reactions when: a) The rates of the forward and reverse reactions become equal. b) The concentrations of reactants and products become equal. c) The temperature shows a sharp rise, d) All chemical reactions stop. 12) For the reaction given below, 2.00 moles of A and 3.00 moles of B are placed in a 6.00-L container. A(g) + 2B(g)=C(g) At equilibrium, the concentration of A is 0.246 mol/L. What is the concentration of B at equilibrium a) 0.246 mo./L b) 0.325 mol/Lc) 0.500 mol/
The pH of the buffer solution is 3.04. 11) Equilibrium is reached chemical in reactions when the rates of the forward and reverse reactions become equal.
Option (a) is the correct option. 12) The balanced equation for the given reaction is:A(g) + 2B(g) ⇌ C(g)The equilibrium constant expression for this reaction can be given as:Kc = [C] / [A][B]^2At equilibrium, the concentration of A is 0.246 mo l/L.
Substitute the given values into the above equilibrium constant expression to find the concentration of B the pH of buffer solution is chemical reaction At equilibrium, the concentration of B is 0.0181 M. option (b) is the correct answer.
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Describe various Carcinogenic chemicals used in the
Society. How can they be replaced with harmless substances.
Carcinogenic chemicals are compounds that are responsible for inducing cancer when introduced into the human body. Examples of such chemicals include benzene, asbestos, formaldehyde, and polychlorinated biphenyls (PCBs).Carcinogenic chemicals can be replaced with harmless substances through the following methods
The chemicals are commonly used in the society as solvents, building materials, and as pesticides and fertilizers for crops. Carcinogenic chemicals can be replaced with harmless substances through the following methods:
1. Use of non-toxic alternatives: This involves replacing the harmful substances with non-toxic alternatives. For example, use of water-based paints instead of solvent-based paints.
2. Recycling: This method involves the collection and reuse of materials to prevent pollution. Recycling also reduces the amount of carcinogenic chemicals in the environment.
3. Regulations: Governments can create laws that restrict the use of carcinogenic chemicals. For example, the Environmental Protection Agency (EPA) has set limits on the amount of benzene that can be released into the atmosphere.
4. Education: The public can be educated on the harmful effects of carcinogenic chemicals. This can help reduce the demand for such chemicals and promote the use of non-toxic alternatives.
5. Natural methods: Organic farming can be promoted to reduce the use of carcinogenic pesticides and fertilizers. Natural alternatives such as neem and garlic can be used as pesticides and fertilizers.
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which states that the volume of a given amount of gas is directly proportional to its kelvin temperature at constant pressure?
Charles's Law states that the volume of a given amount of gas is directly proportional to its Kelvin temperature at constant pressure.
Charles's Law, named after Jacques Charles, describes the relationship between the volume and temperature of a gas when pressure is held constant. According to Charles's Law, the ratio of the volume of a gas (V) to its Kelvin temperature (T) is constant. Mathematically, this can be represented as V/T = k, where k is a constant.
As the temperature of a gas increases, its volume also increases proportionally, and when the temperature decreases, the volume decreases accordingly, as long as the pressure remains constant.
This law is essential in understanding the behavior of gases, particularly when studying changes in volume due to temperature variations.
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which of the following is a reagent commonly used to detect proteins? question 10 options: biuret reagent brown paper pancreatin iodine-potassium-iodide albumin
Proteins are the most diverse group of macromolecules present in living organisms. They are known to be one of the most important classes of biomolecules due to their role in nearly all cellular functions and for their structural support. Biuret reagent is a commonly used reagent to detect the presence of proteins.
It is based on the principle of the biuret reaction and is used to determine protein concentration in a sample. The biuret reagent is made up of sodium hydroxide, copper sulfate, and potassium sodium tartrate.
It is commonly used in research, diagnostic, and clinical laboratories. Biuret reagent is also useful in food and beverage industries to determine protein concentration in various samples. In conclusion, the biuret reagent is a reagent commonly used to detect proteins.
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Water molecules in an aqueous solution will have the strongest interactions with ions with which characteristics?
Ions with a high charge, small size, high charge density, and the ability to form strong hydration shells and hydrogen-bonds with water tend to have the strongest interactions with water molecules in an aqueous solution.
Water molecules in an aqueous solution will have the strongest interactions with ions that exhibit the following characteristics:
Charge: Water molecules are polar, with oxygen having a partial negative charge (δ-) and hydrogen having a partial positive charge (δ+). Ions with charges of opposite polarity (e.g., positive ions or cations attracted to δ- oxygen, and negative ions or anions attracted to δ+ hydrogen) will experience strong electrostatic interactions with water molecules.
Size: The size of the ion can also influence the strength of interaction with water molecules.
Smaller ions tend to have stronger interactions because they can approach water molecules more closely, allowing for more effective electrostatic attractions.
Hydration shell formation: Water molecules can surround ions to form a hydration shell, which stabilizes the ions in solution.
Ions that readily form hydration shells by attracting water molecules strongly will have stronger interactions.
This is particularly relevant for ions with high charge density (charge per unit volume), as they can attract a larger number of water molecules to form a tightly bound hydration shell.
Solvation energy: The solvation energy is the energy released or absorbed when an ion interacts with water molecules to form a hydrated ion.
Ions with a high solvation energy will have strong interactions with water molecules.
It is influenced by factors such as charge, size, and the ability to form hydrogen bonds with water.
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Question 2 When designing equipment for high-temperature and high-pressure service, the maximum allowable stress as a function of temperature of the material of construction is of great importance. Consider a cylindrical vessel shell that is to be designed for pressure of 150 bar (design pressure). The diameter of the vessel is 3.2 m, it is 15 m long, and a corrosion allowance of 6,35 mm (1/4") is to be used. Construct a table that shows the thickness of the vessel walls in the temperature range of 300 to 500°C (in 20°C increments) if the material of construction is ASME SA-240-grade 316 stainless steel 391 2004
When designing equipment for high-temperature and high-pressure use. We can make sure machinery is safe or reliable by utilising vessel wall with proper thickness for pressure of 150 bar or variou temperature.
The highest permissible stress for ASME SA515-grade carbon steel is 17,500 psi at 400 °C. As a result, the following vessel wall thickness would be necessary at 150 bar of pressure at various temperature:
300°C: 19.8 mm320°C: 20.7 mm340°C: 21.7 mm360°C: 22.7 mm380°C: 23.7 mm400°C: 24.7 mm420°C: 25.8 mm 440°C: 26.8 mm460°C: 27.8 mm480°C: 28.8 mm500°C: 29.8 mmThe maximum permitted stress for ASME SA-240 grade 316 stainless steel is 13,750 psi at 400 °C. As a result, the following vessel wall thickness would be necessary at 150 bar of pressure at various temperature:
300°C: 11.8 mm320°C: 12.3 mm340°C: 12.8 mm360°C: 13.4 mm380°C: 13.9 mm400°C: 14.4 mm420°C: 14.9 mm440°C: 15.4 mm460°C: 16.0 mm480°C: 16.5 mm500°C: 17.0 mmLearn more about temperature here:
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A neutral macrocyclic ligand with four donor atoms produces a red diamagnetic low-spin d8 complex of Ni(II) if the anion is the weakly coordinating perchlorate ion, ClO4 - . When perchlorate is replaced by two thiocyanate ions, SCN− , the complex turns violet and is high-spin with two unpaired electrons. Interpret the change in terms of structure
Macrocyclic ligands contain many donor atoms. If these donor atoms are all oriented in the same plane, the resultant coordination compounds are referred to as planar. When coordinating ligands are present, Ni(II) complexes tend to be diamagnetic and low-spin. Ni(II) complexes are typically octahedral when four of these ligands are coordinating ions.
Furthermore, the Ni(II) ion is classified as a d8 ion since it has eight d electrons in the d orbitals. As a result, it has a possible three electron configurations: 3d6, 3d7, or 3d8.The substitution of the perchlorate ion (ClO4−) by two thiocyanate ions (SCN−) produces a complex with two unpaired electrons, indicating that it is a high-spin Ni(II) complex. The violet color of the complex is due to this
.The thiocyanate ion is a weaker ligand than the perchlorate ion, which is why the coordination sphere expands. When this occurs, the coordination geometry transforms from octahedral to tetrahedral. As a result, the complex is distorted from a square planar arrangement to a tetrahedral arrangement, and the symmetry is reduced from D4h to Td (Td). The energy difference between the two spin states is reduced as the coordination sphere expands, resulting in an increase in the number of unpaired electrons. .
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Which of these pairs of elements would be most likely to form an ionic compound?
a) C and O
b) Al and Rb
c) O and Zn
d) Cu and K
e) P and Br
Answer: C - I just used the process of elimination to find the answer
The rate of a reaction depends upon(a) The concentration of the reactants(b) The nature of the reactants(c) The temperature of the reaction(d) All of the above
d) All of the above. The rate of a reaction depends on several factors, including the concentration of the reactants, the nature of the reactants, and the temperature of the reaction.
These factors influence the frequency and effectiveness of molecular collisions, which are essential for chemical reactions to occur. We can state that the rate of a reaction depends on (d) all of the above. This is because concentration, nature, and temperature all play important roles in determining the rate of a reaction. However, this brief statement does not provide a comprehensive explanation of why each factor is significant. We can elaborate on each factor and its influence on the rate of a reaction. Firstly, the concentration of the reactants is crucial because the reaction rate is directly proportional to the concentration of the reactants. Higher concentrations provide a greater number of particles available for collisions, increasing the likelihood of effective collisions and resulting in a faster reaction rate.
Secondly, the nature of the reactants also affects the reaction rate. Different substances have different reactivity and ability to undergo chemical reactions. Some substances may have functional groups or structural features that facilitate or hinder reaction mechanisms, thus influencing the reaction rate. Thirdly, the temperature of the reaction plays a significant role. Increasing the temperature provides more kinetic energy to the reactant molecules, leading to increased molecular motion and collision frequency. Additionally, higher temperatures can provide the activation energy required to initiate certain reactions, thereby accelerating the reaction rate.
In conclusion, the rate of a reaction depends on the concentration of the reactants, the nature of the reactants, and the temperature of the reaction. These factors collectively determine the frequency of molecular collisions and the effectiveness of these collisions, ultimately influencing the reaction rate. Understanding and controlling these factors is essential in the study and application of chemical reactions.
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: Propose a structure for a compound with molecular formula CgH1403 that fits the following spectroscopic data. IR:1820cm, 1760cm1 1H NMR: 1.08 (triplet, 1-6), 1.6δ (sextet, 1-4), 2.20 (triplet, 1-4)
The molecular formula of the compound is given as CgH1403 and the spectroscopic data are given as follows: IR:1820cm, 1760cm1 1H NMR: 1.08 (triplet, 1-6), 1.6δ (sextet, 1-4), 2.20 (triplet, 1-4).The molecular formula suggests the compound to have 14 hydrogen atoms.
The IR spectrum of the compound suggests the presence of a carbonyl group (C=O) around 1760 cm-1 and the N-H bond (O-H is missing from the spectrum) which suggests the presence of carboxylic acid or amide. The 1H NMR spectrum indicates three types of hydrogen atoms: A triplet at 1.08 ppm (J = 7 Hz) with integration of 6 protons. A sextet at 1.6 ppm (J = 7 Hz) with integration of 4 protons. A triplet at 2.20 ppm (J = 7 Hz) with integration of 4 protons.
From the chemical shift of the hydrogen atoms, we can propose that the hydrogen atoms with δ = 1.08 ppm are CH3 groups, the hydrogen atoms with δ = 1.6 ppm are CH2 groups, and the hydrogen atoms with δ = 2.20 ppm are CH groups. From the integration values of the signals, we can deduce the following:- The triplet at 1.08 ppm (J = 7 Hz) with integration of 6 protons indicates that there are two CH3 groups (6 protons in total).- The sextet at 1.6 ppm (J = 7 Hz) with integration of 4 protons indicates that there are two CH2 groups (8 protons in total).- The triplet at 2.20 ppm (J = 7 Hz) with integration of 4 protons indicates that there are two CH groups (8 protons in total).From the above information, we can propose the structure of the compound as follows: CH3-CH(CH3)-CH2-CO-NH-CH2-CH2-CH2-CO-CH-CH3
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In cold climates, water (p 1000 kg/m³) pipes may freeze and burst if proper precautions are not taken. In such an occurrence, the exposed part of a pipe on the ground ruptures, and water shoots up to 34 m. Estimate the gage pressure of water in the pipe. State your assumptions and discuss if the actual pressure is more or less than the value you predicted.
The estimated gauge pressure of water in the pipe is 333,200 Pa (or 333.2 kPa). The actual pressure of water in the pipe when it ruptures and shoots up may be slightly lower than the predicted value of 333.2 kPa.
To estimate the gauge pressure of water in the pipe when it ruptures and shoots up, we can use the concept of hydrostatic pressure. However, it's important to note that the actual pressure may deviate from the predicted value due to various factors.
Assumptions;
The pipe is open to the atmosphere above the point of rupture.
The water shoots up to a height of 34 m.
The density of water is constant at 1000 kg/m³.
The acceleration due to gravity will be 9.8 m/s².
The gauge pressure (P) can be estimated using the hydrostatic pressure equation;
P = ρ × g × h
where;
ρ is the density of water,
g is acceleration due to gravity, and
h is height or depth of the water column.
Given;
ρ = 1000 kg/m³ (density of water)
g = 9.8 m/s² (acceleration due to gravity)
h = 34 m (height of water column)
Substituting the values into the equation:
P = (1000 kg/m³) × (9.8 m/s²) × (34 m)
P = 333,200 Pa
The estimated gauge pressure of water in the pipe is 333,200 Pa (or 333.2 kPa).
Estimation assumes ideal conditions, such as no losses due to friction, no air resistance, and no changes in water density with height. In reality, there may be additional factors that can affect the actual pressure;
Friction and losses; In practice, there can be losses due to friction as water shoots up the pipe, which can reduce the actual pressure compared to the estimated value.
Air resistance; As water shoots up, it encounters air resistance, which can further decrease the actual pressure compared to the estimated value.
Changes in water density; In some cases, the temperature gradient or impurities in the water can cause variations in water density along the pipe. These density changes can affect the pressure distribution and deviate from the estimated value.
Considering these factors, the actual pressure of water in the pipe when it ruptures and shoots up may be slightly lower than the predicted value of 333.2 kPa.
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A 90-wt-% aqueous H₂SO4 solution at 25°C is added over a period of 6 hours to a tank containing 4000 kg of pure water also at 25°C. The final concentration of acid in the tank is 50-wt-%. The contents of the tank are cooled continuously to maintain a constant temperature of 25°C. Because the cooling system is designed for a constant rate of heat transfer, this requires the addition of acid at a variable rate. Determine the instantaneous 90-%-acid rate as a function of time, and plot this rate (kg-s-¹) vs. time. The data of the preceding problem may be fit to a cubic equation expressing HF/(x1x2) as a function of x₁, and the equations of the preceding problem then provide expres- sions for æ and æ.
The instantaneous 90-wt-% acid rate as a function of time can be determined using a cubic equation. This rate refers to the rate at which the 90-wt-% aqueous H₂SO4 solution is added to the tank over time to maintain a constant temperature. The rate will vary throughout the 6-hour period.
To calculate the rate, the cubic equation expressing HF/(x1x2) as a function of x₁ can be used. This equation is obtained by fitting the data from the preceding problem. By manipulating the equations provided, the instantaneous acid rate can be determined.
We could provide a more detailed explanation of how the cubic equation is used to calculate the instantaneous acid rate and how it relates to the problem's context. However, the information given is not sufficient to generate a full explanation. Please provide the necessary equations and data for a more detailed response.
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