Answer:
A.Physical property
Explanation:
Physical properties such as mass,volume,density and colour can be observed without changing the identity of that matter.Examples of extensive properties include mass,volume and length.
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A sample of element X contains 100 atoms with a mass of 12.00 and 10 atoms with a mass of 14.00
Calculate the average atomic mass (in amu) of element X
Answer: 12.18 u
Explanation: The average atomic mass of an element is calculated by taking the weighted average of the atomic masses of its stable isotopes.
In other words, each stable isotope will contribute to the average mass of the element proportionally to its abundance.
What is the name of the ionic compound RbCI?
O rubidium chloride
O chlorine rubidium
O rubidium chlorine
O chlorine rubidide
Answer:A
Explanation:taking the test now
Answer:
A. rubidium chloride on edge 2020
Explanation:
i took the test and got it right
.For which of the following reactions is delta S at 25 degrees Celsius closest to zero?
A) C2H4 (g) + Br2 (l) -> C2H4Br2 (l)
B) H2 (g) + I2 (s) -> 2HI (g)
C) N2 (g) + O2 (g) -> 2NO (g)
D) CH3CHO (g) + 5/2O2 (g) -> 2CO2 (g) + 2H2O (g)
E) 2NO (g) + O2 (g) -> 2NO2 (g)
Among the given reactions, the reaction closest to having a ΔS (change in entropy) of zero at 25 degrees Celsius is reaction B: H2 (g) + I2 (s) -> 2HI (g).
Entropy is a measure of the disorder or randomness of a system. A reaction with a ΔS value close to zero indicates minimal change in entropy. To determine the relative ΔS values for the reactions, we consider the following factors:
A) C2H4 (g) + Br2 (l) -> C2H4Br2 (l)
This reaction involves the formation of a liquid from a gas and a liquid, which generally increases entropy.
C) N2 (g) + O2 (g) -> 2NO (g)
This reaction involves the formation of gas molecules from gas molecules, resulting in an increase in entropy.
D) CH3CHO (g) + 5/2O2 (g) -> 2CO2 (g) + 2H2O (g)
This reaction involves the formation of gas molecules from gas molecules, resulting in an increase in entropy.
E) 2NO (g) + O2 (g) -> 2NO2 (g)
This reaction involves the formation of gas molecules from gas molecules, resulting in an increase in entropy.
B) H2 (g) + I2 (s) -> 2HI (g)
In this reaction, a solid reacts with a gas to form a gas, resulting in a decrease in entropy.
Since reaction B involves a decrease in entropy, it is the closest to having a ΔS value of zero at 25 degrees Celsius.
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suppose that 3.14 g of a silver salt ( agx ) is dissolved in 720.0 ml of water. a current of 2.81 a , applied for 805 s , is required to plate out all of the silver in solution. what is the mass percentage of silver in the salt?
The mass percentage of silver in the salt is 8.18%.
Given that 3.14 g of a silver salt (AgX) is dissolved in 720.0 mL of water. A current of 2.81 A, applied for 805 s, is required to plate out all of the silver in the solution.
We are required to determine the mass percentage of silver in the salt.Main answer:Let's first calculate the amount of silver deposited using Faraday's laws of electrolysis:
From Faraday's first law of electrolysis,mass of silver deposited = ItM / 96500where I = current = 2.81 A, t = time for which current is passed = 805 s, M = molar mass of silver = 107.87 g/mol and 96500 C = 1 mol of electrons.
So, mass of silver deposited = (2.81 A × 805 s × 107.87 g/mol) / 96500= 0.256 g.
Now, let's determine the mass percentage of silver in the salt.
Mass percentage of silver in the salt can be calculated as:mass percentage = (mass of silver / mass of AgX) × 100where mass of silver = 0.256 g and mass of AgX = 3.14 g.
Therefore, mass percentage = (0.256 g / 3.14 g) × 100= 8.18%
The given question is based on Faraday's laws of electrolysis, which helps us to determine the amount of metal deposited during electrolysis.
According to Faraday's laws of electrolysis, the amount of metal deposited is proportional to the quantity of electricity passed through the electrolyte.
Hence, we have used Faraday's laws to determine the amount of silver deposited by applying a current of 2.81 A for 805 s.
From the calculation, we have obtained that 0.256 g of silver has been deposited on the cathode. Now, we are required to determine the mass percentage of silver in the salt.
The mass percentage can be calculated as the ratio of the mass of silver deposited to the mass of AgX taken initially multiplied by 100.
From the calculation, we have obtained that the mass percentage of silver in the salt is 8.18%.Hence, the mass percentage of silver in the salt is 8.18%.
The mass percentage of silver in the salt is 8.18%. This answer has been calculated using Faraday's laws of electrolysis, which helps in determining the amount of metal deposited during electrolysis.
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25. List the phases of water in order, from the phase with the GREATEST kinetic energy (heat) to the phase with the LOWEST kinetic energy (heat), A. Gas-Liquid-Solid I B. Liquid-Solid-Gas C. Solid-Liquid-Gas D. Gas-Solid-Liquid
Answer:
A. Gas-Liquid-Solid.
Explanation:
Classify each property according to whether it is displayed by metals or by nonmetals
a. low melting point b. shiny dull c. poor conductor d. ductile e. malleable f. brittle g. good conductor h. high melting point
a. low melting point: Nonmetals. b. shiny: Metals. c. poor conductor: Nonmetals. d. ductile: Metals. e. malleable: Metals. f. brittle: Nonmetals. g. good conductor: Metals. h. high melting point: Metals
a. Low melting point is generally displayed by nonmetals. Metals tend to have high melting points.
b. Metals are typically shiny due to their ability to reflect light.
c. Nonmetals are generally poor conductors of heat and electricity.
d. Metals are ductile, meaning they can be drawn into thin wires without breaking.
e. Metals are malleable, meaning they can be hammered into thin sheets or shapes without shattering.
f. Nonmetals are usually brittle, meaning they are prone to breaking or shattering when subjected to stress.
g. Metals are good conductors of heat and electricity.
h. Metals typically have high melting points compared to nonmetals.
It's important to note that these are general trends and not absolute characteristics for all metals or nonmetals. Some exceptions or variations may exist within specific elements or compounds.
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True or False: A rocket ship is a good example of Newton's first law of motion.
Answer:
True.
Explanation:
(a) Calculate the pressure exerted by 1.00 mol of CO2 in a 1.00 L vessel at 300 K, assuming that the gas behaves ideally. (b) Repeat the calculation using the van der Waals equation. (c) Explain the difference between your answers to parts (a) and (b) using kinetic molecular theory.
(a) The pressure exerted by 1.00 mol of CO2 in a 1.00 L vessel at 300 K, assuming ideal gas behavior, is approximately 24.8 atm.
(b) Using the van der Waals equation, the pressure is calculated to be approximately 23.2 atm.
(c) The difference between the answers in parts (a) and (b) can be explained by the kinetic molecular theory, which takes into account the volume occupied by gas molecules and intermolecular forces.
(a) To calculate the pressure exerted by 1.00 mol of CO2 in a 1.00 L vessel at 300 K using the ideal gas law, we can use the formula:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin. Substituting the given values:
P = (1.00 mol) * (0.0821 L·atm/(mol·K)) * (300 K) / (1.00 L)
P ≈ 24.8 atm
(b) The van der Waals equation corrects for the volume occupied by gas molecules and intermolecular forces. It is given by:
(P + a(n/V)^2) * (V - nb) = nRT
Where a and b are constants specific to the gas. For CO2, a = 3.59 L^2·atm/(mol^2) and b = 0.0427 L/mol. Substituting the values and solving for P:
(P + 3.59 atm * (1.00 mol / 1.00 L)^2) * (1.00 L - 0.0427 L/mol) = (1.00 mol) * (0.0821 L·atm/(mol·K)) * (300 K)
P ≈ 23.2 atm
(c) The difference between the answers in parts (a) and (b) can be explained by the kinetic molecular theory. The ideal gas law assumes that gas molecules occupy negligible volume and experience no intermolecular forces. However, in reality, gas molecules have a finite volume and experience attractive intermolecular forces. The van der Waals equation incorporates corrections for these factors, resulting in a slightly lower pressure compared to the ideal gas law prediction. This difference highlights the importance of considering molecular properties and intermolecular interactions when calculating the behavior of real gases.
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Lee heats up a saucepan of water to make some macaroni and cheese. He places the saucepan on the stove, turns on the burner, walks away and is distracted by a phone call. He returns one hour later, turns off the stove, and finds the saucepan is now empty. Which statement below explains why the saucepan is empty?
Answer:
the water heats up and evaporates after surpassing its boiling temperature
Answer:
THe water heats up
Explanation:
Consider a weak acid hx. If a. 10m solution of this acid has a ph of 5. 84 at 25c what is the deta g 298 for the acid
The problem is related to thermodynamics and acid dissociation constant. The expression for acid dissociation constant (Ka) of the weak acid can be given as the following: hx + H2O ↔ H3O+ + x-Ka is the equilibrium constant for this reaction which can be given as, Ka = ([H3O+][x-])/[hx] pH of a weak acid solution can be given as pH = pKa + log([x-]/[hx]) The above expression is also called the Henderson-Hasselbalch equation.
According to the question, pH of the weak acid solution is given as 5.84 at 25°C. Hence, [H3O+] = 10^(-5.84) = 1.554 x 10^(-6) M. The weak acid concentration is also given as 10 M. Hence, [hx] = 10 M. The concentration of the conjugate base of the weak acid can be calculated as follows:[x-] = (Ka[hx])/[H3O+][x-] = (Ka[hx])/[H3O+]So, [x-] = (Ka[hx])/[H3O+] = (Ka)/[H3O+][hx] = 10(-5.84 - pKa). From the above two expressions, it can be observed that Ka and [hx] are related to pH.
Therefore, the pH of a weak acid solution can be used to calculate the Ka value of the weak acid. Using the above equations, we can calculate the value of Ka. However, the question requires the value of ΔG°298 of the weak acid. ΔG°298 can be calculated as follows:ΔG°298 = -RT. ln(Ka)Where R is the gas constant (8.314 J/K mol)T is the temperature (298 K)So, the expression for ΔG°298 can be written as:ΔG°298 = -8.314 x 298 x ln(Ka)ΔG°298 = -8.314 x 298 x ln(2.06 x 10^(-6))ΔG°298 = 3277 J/mol (approximately 3.3 kJ/mol). Therefore, the value of ΔG°298 of the weak acid hx is approximately 3.3 kJ/mol.
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calculate the amount of energy required to melt a 48.9 g sample of cobalt at its normal melting point. boiling point
The amount of energy required to melt a 48.9 g sample of cobalt at its normal melting point and boiling point is given below:
the amount of heat required to melt a certain amount of a solid is called the heat of fusion. It is usually expressed in Joules per gram (J/g). Cobalt's heat of fusion is 16.06 J/g.
Thus, to determine the quantity of heat needed to melt 48.9 grams of cobalt at its normal melting point, we may employ the formula
Q=m x ΔHf. (48.9 g) × (16.06 J/g)
= 784.14 J
Ans: 784.14 J or 7.84 x 10^2J.
The amount of heat required to transform a liquid into a gas is known as the heat of vaporization. It is frequently stated in J/g. Cobalt's heat of vaporization is 375 J/g.
Thus, to determine the amount of heat required to boil 48.9 grams of cobalt at its normal boiling point, we can use the formula
Q=m x Hv. (48.9 g) × (375 J/g)
= 18,337.5 J or 1.83 x 10^4 J
Ans: 1.83 x 10^4 J.
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A sample of aluminum foil contains 8.00 x 10^23 atoms. What is the mass of the foil? mass of foil ______ g
The mass of the foil containing 8.00 x 10^23 atoms is 35.88 g
The given number of atoms, i.e., 8.00 x 10^23, of aluminum foil will be used to find the mass of the foil.
Molar mass of aluminum (Al) = 26.982 g/mol ; Avogadro number = 6.022 x 10²³.
Step 1. To find the mass of aluminum foil, first, we must convert the number of atoms of aluminum foil to moles, and then we can calculate the mass of the foil using the molar mass of aluminum.
No. of moles = No of atoms / Avogadro number
Moles of Al = 8.00 x 10^23 Al atoms x 1 mol Al/6.022 x 10²³ Al atoms = 1.33 moles
Step 2. To determine the mass of Al foil
Mass = moles × molar mass
Mass = 1.33 x 10 mol Al × 26.982 g/mol Al = 35.88 g
Thus, the mass of the foil containing 8.00 x 10^23 atoms is 35.88 g.
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if a certain isotope of an element is 2.24846 times more massive than carbon-12, what mass does it have on the amu scale?
An atomic mass unit (amu) is defined as one twelfth of the mass of one carbon-12 atom. Therefore, a certain isotope of an element that is 2.24846 times more massive than carbon-12 has an atomic mass of 2.24846 amu. This means that its mass is 2.24846 times that of a carbon-12 atom, which has a mass of 12 amu.
Isotopes of an element have the same atomic number (number of protons), but different mass numbers (number of protons and neutrons combined). The atomic mass of an isotope is the weighted average of the masses of all its naturally occurring isotopes, taking into account the relative abundance of each isotope.
To calculate the atomic mass of an isotope, the mass of each isotope is multiplied by its relative abundance, then added together. For example, if an element has two isotopes, one with a mass of 10 amu and a relative abundance of 25%, and another with a mass of 12 amu and a relative abundance of 75%, the atomic mass of the element would be:
(10 amu x 0.25) + (12 amu x 0.75) = 11 amu
Therefore, if a certain isotope of an element is 2.24846 times more massive than carbon-12, it has a mass of 2.24846 amu on the amu scale.
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Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3NO2(g)+H2O(l)→2HNO3(l)+NO(g).
Suppose that 5.0 mol NO2 and 0.20 mol H2O combine and react completely. Which reactant is in excess?
How many moles of the reactant in excess are present after the reaction has completed?j
NO₂ is the reactant in excess, and 4.4 moles of NO₂ remain after the reaction has completed.
To determine which reactant is in excess, we need to compare the stoichiometric ratios of the reactants. According to the balanced equation, the ratio between NO₂ and H₂O is 3:1.
To determine the limiting reactant, we can calculate the moles of each reactant required for the reaction. Since the ratio is 3:1, we need 3 times more moles of NO₂ than H₂O.
Moles of NO₂ required = 3 * 0.20 mol = 0.60 mol
Comparing this with the actual amount of NO₂ provided (5.0 mol), we see that there is an excess of NO₂.
Therefore, NO₂ is the reactant in excess.
To find the number of moles of the excess reactant remaining after the reaction has completed, we can subtract the moles of the limiting reactant used from the initial moles of the excess reactant.
Moles of excess reactant remaining = Moles of excess reactant initially - Moles of excess reactant used
Moles of excess reactant remaining = 5.0 mol - 0.60 mol
Moles of excess reactant remaining = 4.4 mol
After the reaction has completed, there are 4.4 moles of NO₂ remaining as the excess reactant.
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what elements could be in the gas when sodium bicarbonate and citric acid react
Which of the following monosaccharides are used to synthesize nucleotide building blocks? Select all that apply. 2-deoxyribulose ribose 2-deoxyribose ribulose
Monosaccharides are used to synthesize nucleotide building blocks Ribose 2.2-deoxyribose. The correct options 1.
Ribose and 2-deoxyribose are both monosaccharides that play important roles in the synthesis of nucleotide building blocks.
Ribose is a 5-carbon sugar and a key component of RNA (ribonucleic acid). It forms the backbone of RNA molecules and is involved in the synthesis of ribonucleotides, which are the building blocks of RNA.
2-deoxyribose is a modified form of ribose with a hydrogen atom replacing the hydroxyl group at the 2' carbon position. It is a crucial component of DNA (deoxyribonucleic acid). 2-deoxyribose forms the backbone of DNA molecules and is involved in the synthesis of deoxyribonucleotides, which are the building blocks of DNA.
Both ribose and 2-deoxyribose are important sugars in nucleotide synthesis, with ribose being used in RNA synthesis and 2-deoxyribose being used in DNA synthesis.
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write a balanced chemical equation that shows the formation of pentanoic acid from the elements carbon (c), hydrogen (h2 ) and oxygen (o2 ).
The balanced chemical equation for the formation of pentanoic acid from the elements carbon (C), hydrogen (H2), and oxygen (O2) is as follows: C5H10O2 + 7O2 → 5CO2 + 5H2O
Pentanoic acid is a carboxylic acid with the chemical formula C5H10O2. It is also known as valeric acid. The formation of pentanoic acid from carbon, hydrogen, and oxygen requires the presence of a catalyst such as manganese (IV) oxide.The equation above shows that five molecules of carbon dioxide (CO2) and five molecules of water (H2O) are formed as products when pentanoic acid is formed from carbon, hydrogen, and oxygen.
This means that the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side.The equation is balanced because the number of atoms of each element is equal on both sides. The balanced equation is useful because it allows chemists to calculate the amount of reactants required to produce a certain amount of products, or to determine the amount of products that will be produced from a given amount of reactants.
The conclusion is that pentanoic acid can be formed from the elements carbon, hydrogen, and oxygen by reacting them in the presence of a catalyst such as manganese (IV) oxide. The balanced equation for this reaction is
C5H10O2 + 7O2 → 5CO2 + 5H2O.
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Which of the following substances has the highest entropy value? a.CH3OH (g) b. CH3OH (I)
c. CH3CH2CH2OH (I)
d. CH3CH2CH2OH (g)
e. CH3CH2OH (I) f. CH3CH2OH (8)
The substance with the highest entropy value among the given options is CH3CH2CH2OH (g).
Entropy is a measure of the disorder or randomness in a system. In general, the entropy of a substance increases with increasing molecular complexity and freedom of motion.
Among the given options, CH3CH2CH2OH (g) has the highest entropy value because it is a gas. Gaseous substances generally have higher entropy compared to liquids or solids due to the increased freedom of motion of the molecules. The gaseous state allows molecules to move more freely and occupy a larger volume, leading to a higher degree of disorder and higher entropy.
In contrast, CH3OH (I) and CH3CH2OH (I) refer to liquid states, which have lower entropy compared to the gaseous state. CH3CH2OH (s) refers to the solid state, which generally has the lowest entropy among the three states of matter.
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Is sea water close to being a saturated sodium chloride solution? Explain your answer.
How could you tell that the original salt solution was saturated.
Sea water is not a saturated sodium chloride solution. A saturated solution is one where no more solute can dissolve, but sea water can dissolve more salt.
Determining if a solution is saturated involves observing if additional solute dissolves or if solid residue accumulates.
Sea water is not close to being a saturated sodium chloride solution. While sea water does contain dissolved sodium chloride, it is not at the point of saturation. A saturated solution is one in which no more solute (in this case, sodium chloride) can be dissolved at a given temperature. Sea water still has the capacity to dissolve more salt, so it is not saturated.
To determine whether a solution is saturated or not, one method is to observe if any undissolved solute remains in the solution under specific conditions. If, upon adding more solute to the solution, no further dissolution occurs and a solid residue accumulates, it indicates that the solution is saturated.
In the case of sea water, if you were to collect a sample and attempt to dissolve additional sodium chloride in it, you would find that the additional salt would continue to dissolve. The water's ability to dissolve more salt indicates that the original salt solution (sea water) is not saturated.
It's worth noting that the composition of sea water is more complex than just sodium chloride. It contains various dissolved salts, minerals, and other substances, giving it a different composition from a pure sodium chloride solution.
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Describe why natural resources are distributed unevenly around the planet?
Answer:
Earth has different topography, climate and altitude at different places. Difference in these factors has resulted in unequal distribution of resources over the earth. Also, all these factors are different from place to place on the earth.
Explanation:
is centigrams or grams bigger.
example: which unit is larger.
35cg or 35 g
Answer:
35g
Explanation:
Answer:
35 grams is bigger
Explanation:
grams= 1
centigrams= .01
a sample of iron absorbs 81.0 j of heat, upon which the temperature of the sample increases from to if the specific heat of iron is 0.450 j/g-k, what is the mass (in grams) of the sample? group of answer choices -21.2 g 4.29 g 3.83 g 21.2 g 0.0472 g
The mass of the sample is 3.83 g.
The specific heat of a substance is defined as the amount of heat required to raise the temperature of a unit mass of the substance by one degree Celsius. The formula for calculating the heat absorbed by an object can be expressed as follows:Heat Absorbed = Mass × Specific Heat Capacity × Change in TemperatureThe change in temperature in the above equation is usually given in degrees Celsius or Kelvin. The specific heat capacity of iron is given as 0.450 J/g-K. Using this information, we can calculate the mass of the sample as follows:Heat Absorbed = Mass × Specific Heat Capacity × Change in Temperature81.0 = Mass × 0.450 × (760 - 20)81.0 = Mass × 0.450 × 740Mass = 81.0 / (0.450 × 740)Mass = 3.83 gTherefore, the mass of the sample is 3.83 g. This is the final answer.
The mass of the iron sample is 3.83 g.
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please need asap
A neutral atom with 4 electrons shells and 7 valence electrons must be
Bromine
Iodine
Manganese
Chlorine
Answer:
Bromine
Explanation:
We know that all halogens have seven valence electrons. So form given options bromine , iodine and chlorine could be the given atom but it should be the one which have 4 electronic shells. It will confirm from electronic configuration.
Electronic configuration of Br:
Br₃₅ = [Ar] 4s² 3d¹⁰ 4p⁵
Bromine have 4 electronic shell.
Electronic configuration of Iodine:
I₅₃ = [Kr] 4d¹⁰ 5s² 5p⁵
Iodine have 5 electronic shells.
Electronic configuration of Chlorine:
Cl₁₇ = [Ne] 3s² 3p⁵
Thus correct option is bromine.
Arrange these species from fastest SN 2 reaction rate to slowest SN2 reaction rate. Fastest Slowest Answer Bank
The order of SN2 reaction rate for the given species is : CH3CH2I > CH3CH2Cl > CH3CH2OH
SN2 reactions are nucleophilic substitution reactions. These reactions are important in organic chemistry since they involve substitution of a nucleophile at a saturated carbon atom.
Factors that affect the rate of the reaction include the nature of the leaving group, the solvent, and the nature of the nucleophile.
Primary substrates are more reactive in SN2 reactions than tertiary substrates.The best leaving groups are weak bases.The reaction rate is faster when there is no steric hindrance at the site of reaction.CH3CH2I (ethyl iodide)Ethyl iodide has a good leaving group (Iodide ion) and relatively low steric hindrance. Additionally, iodide is a weak base, making it a better leaving group than the hydroxide ion (from ethanol) or chloride ion (from ethyl chloride). Therefore, ethyl iodide is expected to have the fastest SN2 reaction rate.
CH3CH2Cl (ethyl chloride)Ethyl chloride has a weaker leaving group compared to ethyl iodide but is still better than ethanol. It also has moderate steric hindrance. Although it is slower than ethyl iodide, it is expected to have a faster SN2 reaction rate than ethanol.
CH3CH2OH (ethanol)Ethanol has a hydroxide ion as a leaving group, which is a weaker leaving group compared to both iodide and chloride ions. Additionally, ethanol has a higher steric hindrance due to the presence of the hydroxyl group. These factors contribute to a slower SN2 reaction rate compared to ethyl iodide and ethyl chloride.
Thus, the order of SN2 reaction rate for the given species is : CH3CH2I > CH3CH2Cl > CH3CH2OH.
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150.0 ml of a 0.100 m ba(ch3coo)2 solution is diluted to 4.00 l . what is the molarity of the diluted solution?
To determine the molarity of the diluted solution, you need to use the dilution formula which is:M1V1 = M2V2where M1 and V1 are the initial concentration and volume, respectively, and M2 and V2 are the final concentration and volume, respectively. Given: M1 = 0.100 M (initial concentration) V1 = 150.0 mL (initial volume) V2 = 4.00 L (final volume)To solve for M2, you need to convert the initial volume to liters: M1V1 = M2V2(0.100 M)(150.0 mL) = M2(4.00 L)0.015 = M2(4.00 L)M2 = 0.015 M. So, the molarity of the diluted solution is 0.015 M.
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How many atoms of hydrogen are in 160 g of hydrogen peroxide (H2O2)?
N= ___ atoms
There are approximately 9.03 x 10^23 atoms of hydrogen in 160 g of hydrogen peroxide (H2O2).
To determine the number of atoms of hydrogen in 160 g of hydrogen peroxide (H2O2), we need to use Avogadro's number and the molar mass of H2O2.
The molar mass of H2O2 is calculated as follows:
2 atoms of hydrogen (H) = 2 x 1.008 g/mol = 2.016 g/mol
2 atoms of oxygen (O) = 2 x 16.00 g/mol = 32.00 g/mol
Total molar mass of H2O2 = 34.016 g/mol
Next, we calculate the number of moles of H2O2:
Moles of H2O2 = mass of H2O2 / molar mass of H2O2
Moles of H2O2 = 160 g / 34.016 g/mol ≈ 4.703 mol
Since each mole of H2O2 contains 2 moles of hydrogen atoms, we can calculate the number of hydrogen atoms:
Number of hydrogen atoms = 2 moles of H2O2 x 6.022 x 10^23 atoms/mol
Number of hydrogen atoms ≈ 9.03 x 10^23 atoms
Therefore, there are approximately 9.03 x 10^23 atoms of hydrogen in 160 g of hydrogen peroxide (H2O2).
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What mass of iron(III) chloride contains 2.35×1023 chloride ions?
The mass of iron(III) chloride that contains 2.35×10²³ chloride ions can be calculated using the molar mass and stoichiometry of iron(III) chloride.
To calculate the mass of iron(III) chloride, we need to use the molar mass and stoichiometry. The molar mass of iron(III) chloride (FeCl₃) can be calculated by adding the atomic masses of iron (Fe) and three times the atomic mass of chlorine (Cl). The atomic mass of iron is 55.845 g/mol, and the atomic mass of chlorine is 35.453 g/mol.
Molar mass of FeCl₃ = 55.845 g/mol + 3 * 35.453 g/mol = 162.204 g/mol
From the balanced chemical equation of iron(III) chloride, we know that each formula unit of FeCl₃ contains three chloride ions (Cl⁻). Therefore, the ratio of chloride ions to iron(III) chloride is 3:1.
Given that there are 2.35×10²³ chloride ions, we can calculate the moles of iron(III) chloride using the mole ratio:
Moles of FeCl₃ = (2.35×10²³ chloride ions) / (3 chloride ions/1 FeCl₃)
= (2.35×10²³) / 3 moles
Finally, we can calculate the mass of iron(III) chloride using the moles and molar mass:
Mass of FeCl₃ = Moles of FeCl₃ * Molar mass of FeCl₃
Substituting the values, we can find the mass of iron(III) chloride containing 2.35×10²³ chloride ions.
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what is the number of atoms of various elements present in the following compound. 2Na2SO4,2PbO,(NH4)2SO4,4KMO4,2NH4CI,3AI2(SO4)3,Ca(HCO3)2,NH4NO2
Answer:
Detail is given below.
Explanation:
2Na₂SO₄
There are two sodium, one sulfur and four oxygen atoms are present in Na₂SO₄.
In 2 mole of Na₂SO₄ = four sodium, two sulfur and 8 oxygen atoms are present.
2PbO
There are one oxygen and one lead atoms are present in PbO.
In 2 mole of PbO = 2 lead and 2 oxygen atoms
(NH₄)₂SO₄
There are two nitrogen, eight hydrogen, one sulfur and four oxygen atoms are present in (NH₄)₂SO₄.
4KMnO₄
There are one potassium, one manganese and four oxygen atoms are present in KMnO₄.
In 4 mole of KMnO₄ = four potassium, four manganese and sixteen oxygen atoms
2NH₄Cl
There are one nitrogen, four hydrogen and one chlorine atoms are present in NH₄Cl.
In 2 mole of NH₄Cl = two nitrogen, eight hydrogen and two chlorine atoms
3Al₂(SO₄)₃
There are two aluminum, three sulfur and twelve oxygen atoms are present in Al₂(SO₄)₃.
In 3 mole of Al₂(SO₄)₃ = six aluminum, nine sulfur and thirty six oxygen atoms are present.
Ca(HCO₃)₂
There are one calcium, two hydrogen, two carbon and six oxygen atoms are present.
NH₄NO₂
There are two nitrogen, four hydrogen and two oxygen atoms are present in NH₄NO₂.
what condition is necessary for a sp3-hybridized atom to display asymmetry?
Asymmetric sp3-hybridized carbon atoms are present in the majority of chiral molecules.
An sp3-hybridized atom will display asymmetry when it has at least four different substituent atoms or groups in their valence shell. This condition is known as chirality.
Chirality is a term used to refer to molecules or atoms that are not superimposable on their mirror image. This term comes from the Greek word "hand," which indicates that chiral molecules or atoms, like hands, can exist in both left-handed and right-handed forms.
As a result, these chiral molecules and atoms have unique physical and chemical properties when compared to their non-chiral counterparts
A chiral molecule contains at least one atom that is sp3-hybridized to display asymmetry. When an sp3-hybridized atom has at least four different substituent atoms or groups in their valence shell, it is considered chiral and has non-superimposable mirror images (enantiomers).
Asymmetric sp3-hybridized carbon atoms are present in the majority of chiral molecules.
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calculate the molarity of 34.1 g of mgs in 945 ml of solution.
The molarity of the solution containing 34.1 g of MgS in 945 mL is approximately 0.639 M.
To calculate the molarity of a solution, we need to know the number of moles of the solute and the volume of the solution in liters. Let's calculate the molarity of 34.1 g of MgS in 945 mL of solution.
First, we need to convert the volume of the solution from milliliters to liters. Since 1 L is equal to 1000 mL, we divide 945 mL by 1000 to get 0.945 L.
Next, we need to calculate the number of moles of MgS. To do this, we divide the mass of MgS (34.1 g) by its molar mass. The molar mass of MgS is the sum of the atomic masses of magnesium (24.31 g/mol) and sulfur (32.07 g/mol), which gives us 56.38 g/mol. Therefore, the number of moles of MgS is 34.1 g / 56.38 g/mol = 0.604 mol.
Finally, we can calculate the molarity by dividing the number of moles of MgS by the volume of the solution in liters. The molarity (M) is equal to 0.604 mol / 0.945 L ≈ 0.639 M.
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