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Related Questions
Hydrogen bonding activity drawings
Answers
Answer:UIHIU
A small amount of chemical splashes in Frank’s eye. What should Frank do immediately?
Explanation:A small amount of chemical splashes in Frank’s eye. What should Frank do immediately?
PROBLEM 19.12 Draw the structure of a triacylglycerol that fits each description: a. a saturated triacylglycerol formed from three 12-carbon fatty acids b. an unsaturated triacylglycerol that contains three cis double bonds c. a trans triacylglycerol that contains a trans double bond in each hydrocarbon chain
Answers
b. An unsaturated triacylglycerol that contains three cis double bonds would have three different unsaturated fatty acids attached to a glycerol backbone. Each fatty acid would contain a cis double bond.
c. A trans triacylglycerol that contains a trans double bond in each hydrocarbon chain would have three different trans fatty acids attached to a glycerol backbone. Each fatty acid would contain a trans double bond.
PLEASE HELP WITH CONVERSIONS
Answers
The grams of pigment needed is 9.20 grams
The grams of water needed is 0.199 grams.
The final molarity of the binder in the paint mixture is 20.04 M.
What are the masses of substances required?The masses of substances required are reived from the mole ratios given.
The grams of pigment needed:
mole ratios of CaCO3 and Cr2O3 is 1 : 1.52
Using the conversion factors:
molar mass of CaCO3 = 100.1 g
molar mass of Cr2O3 = 151.99 g
grams of pigment needed = (0.400 g CaCO3 x 1 mol CaCO3/100.1 g CaCO3) x (1.52 mol Cr2O3/1 mol CaCO3) x (151.99 g Cr2O3/1 mol Cr2O3) grams of pigment needed = 9.20 g Cr2O3
the grams of water needed:
mole ratio of CaCO3 and H2O is 1 : 27.5
molar mass of H2O = 18.0 g
the grams of water needed = 0.400 g CaCO3 x (1 mol CaCO3/100.1 g CaCO3) x (27.5 mol H2O/1 mol CaCO3) x (18.0 g H2O/1 mol H2O)
the grams of water needed = 0.199 g H2O
From the density of water, the volume of water needed is 0.199 mL of water.
Molarity = moles of solute/volume of solution in liters
Volume of solution = 0.000199 L
Number of moles of CaCO3 = 0.400/100.1 g
Number of moles of CaCO3 = 0.003998 mol
Molarity = 0.003998/0.000199
Molarity = 20.04 M
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The gas pressure in a can is 2.5 atm at 25 °C. Assuming that the gas obeys the ideal-gas equation, what is the pressure (in atm) when the can is heated to 525 °C?
Answers
The concept combined gas law is used here to determine the new pressure of the gas. This law states that the ratio between the product of pressure-volume and temperature of a system remains constant.
The combined gas law is the combination of Boyle's law, Charles's law and the Avogadro's law. These laws relate one thermodynamic variable to another holding everything else constant.
Here volume is constant, so the equation is:
P₁ / T₁ = P₂ / T₂
T₁ = 298 K
T₂ = 798 K
Pressure is:
P₂ = P₁ T₂/T₁
P₂= 2.5 × 798 / 298
P₂ = 6.69 atm
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in diagram A, what is the value of ^H
Answers
The value of h in the right triangle is 7.7 cm.
Why is this so?Trigonometric ratio is used to show the relationship between the sides and angles of a right angled triangle.
In the right triangle, using trigonometric ratio:
a) sin(31) = h/15
⇒ 0.515 = h/15
Making H the subject of the formula we have:
0.515 x 15 = h/15 x 15
h = 7.725
h [tex]\approx[/tex] 7,7
Hence, is is correct to state that the value of h in the diagram is 7.7 cm approximately. See the attached image.
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Full Question:
Although part of your question is missing, you might be referring to this full question:
the diagram shows a right-angled triangle. what is the value of h? give your answer correct to 1 decimal place.
Given the thermochemical equation:
4 AlCl3 (s) + 3 O2 (g) ⇒ 2 Al2 03 (s) + 6 Cl2 (g) ; ΔH = -529 kJ
Find ΔH for the following reaction:
1) 3 Al2O3 (s) + Cl2 (g) ⇒ 2/3 AlCl3 (s) + 1/2 O2 (g) ΔH= ?kJ
2) 88.2 kJ
b) 264.5 kJ
c) 529 kJ
d) 176.3 kJ
e) - 176.3 kJ
Answers
A thermochemical equation can be written by expressing the heat evolved or absorbed in terms of the enthalpy change ΔH. Here ΔH for the following reaction +88.2 kJ. The correct option is A.
A chemical equation which indicates the heat change occuring during the reaction is defined as the thermochemical equation. In thermochemical equations, physical states of the reactants and products should be specified.
Here the given reaction 4 AlCl₃ (s) + 3 O₂ (g) ⇒ 2 Al₂O₃ (s) + 6 Cl₂ (g) is reversed as 1 /3 Al₂O₃ (s) + Cl₂ (g) ⇒ 2/3 AlCl₃ (s) + 1/2 O₂ (g) and multiplied by 1/6.
So the new enthalpy is +88.16 ≈ 88.2 kJ
Thus the correct option is A.
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Substance A decomposes at a rate proportional to the amount of A present.
Write an equation relating A to the amount left of an initial amount Ao after time t.
It is found that 8 g of A will reduce to 4 g in 3 hr. After how long will there be only 1 g left?
Answers
It will take approximately 2.079 hours for there to be only 1 g of substance A left.
The rate of decomposition of substance A is proportional to the amount of A present, which means that we can use the following differential equation to describe the decay;
dA/dt = -kA
where A is the amount of substance A at time t, k is the rate constant of the reaction, and the negative sign indicates that A is decreasing over time.
To solve differential equation, we separate the variables and integrate;
[tex]d_{A}[/tex]/A = -k [tex]d_{t}[/tex]
Integrating both sides gives;
ln(A) = [tex]-k_{t}[/tex] + C
where C is the constant of integration. To find the value of C, we can use the initial condition that 8 g of A reduces to 4 g in 3 hours. At t=0, A=Ao=8 g, and at t=3, A=4 g. Substituting values into the equation above, we have;
ln(8) = -3k + C
ln(4) = -6k + C
Subtracting first equation from second, we get;
ln(4/8) = -3k
Simplifying, we get;
k = ln(2)/3
Now, we can use the equation we derived earlier to find how long it will take for there to be only 1 g of A left;
ln(A) =[tex]-k_{t}[/tex] + C
ln(1) = -(ln(2)/3)t + C
Simplifying, we get:
t = 3 ln(2)
t ≈ 2.079 hours
Therefore, it will take 2.079 hours.
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Sometimes a fossil is formed as a result of the movement of an organism in soft sediment. Which of these are two kinds of trace fossils?
*
1 point
shells and bones
tracks and burrows
a bee and a beetle in amber
petrified and mummified fossils
Answers
Answer:The two kinds of trace fossils mentioned in the options are:
tracks
burrowsTracks and burrows are both examples of trace fossils, which are fossils thatprovide evidence of an organism's activity, rather than the organism itself.Tracks are impressions left by an organism's feet or other body parts as itmoved across soft sediment, while burrows are tunnels or other structurescreated by an organism as it burrowed into the sediment. Both types of tracefossils can provide valuable information about an organism's behavior, habitat, and interactions with other organisms.
How much energy is required to heat 192 grams of lead from 18 degrees Celsius to 44
degrees Celsius?
Answers
The amount of energy required to heat the given amount of lead from 18°C to 44°C is 638.976 J.
The specific heat capacity of lead is 0.128 J/g°C.
To calculate the amount of heat energy required to heat 192 grams of lead from 18°C to 44°C, we can use the formula:
Q = m × c × ΔT
where:
Q = heat energy (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
Plugging in the given values:
m = 192 g
c = 0.128 J/g°C
ΔT = (44°C - 18°C) = 26°C
Q = 192 g × 0.128 J/g°C × 26°C
Q = 638.976 J
So, the amount of heat energy required to heat 192 grams of lead from 18°C to 44°C is 638.976 Joules.
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5.4g of aluminum reacts with sulfuric acid (H₂SO4) to form aluminum sulfate and hydrogen.
a. Write the chemical equation.
b. Find mass of required sulfuric acid.
C. Find volume of the obtained gas.
(AI=23, S = 32, O=16, H =1, 2g of H2 has 22.4L).
Answers
Answer:
a. The chemical equation for the reaction is:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
b. To find the mass of required sulfuric acid, we need to use stoichiometry. We can start by finding the number of moles of aluminum used in the reaction:
Molar mass of Al = 27 g/mol
Number of moles of Al = 5.4 g / 27 g/mol = 0.2 mol
According to the balanced equation, 3 moles of H₂SO₄ are required to react with 2 moles of Al. Therefore, the number of moles of H₂SO₄ required is:
Number of moles of H₂SO₄ = 3/2 x 0.2 mol = 0.3 mol
Molar mass of H₂SO₄ = 2 x 1 g/mol + 32 g/mol + 4 x 16 g/mol = 98 g/mol
Mass of H₂SO₄ required = 0.3 mol x 98 g/mol = 29.4 g
Therefore, 29.4 g of sulfuric acid is required to react with 5.4 g of aluminum.
c. To find the volume of hydrogen gas obtained, we need to use the ideal gas law:
PV = nRT
where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, R is the universal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.
We can start by finding the number of moles of hydrogen gas produced in the reaction. According to the balanced equation, 3 moles of H₂ are produced for every 2 moles of Al. Therefore, the number of moles of H₂ produced is:
Number of moles of H₂ = 3/2 x 0.2 mol = 0.3 mol
Assuming the reaction occurs at standard temperature and pressure (STP), which is 0°C (273 K) and 1 atm, we can use the molar volume of a gas at STP, which is 22.4 L/mol. Therefore:
V = nRT/P = 0.3 mol x 0.0821 L atm/mol K x 273 K / 1 atm = 6.58 L
Therefore, the volume of hydrogen gas produced at STP is 6.58 L.
Explanation:
The diagram shows sound and light waves from an emergency vehicle traveling toward a brick wall. The brick wall has both smooth and rough surfaces.
Select the correct answer from each drop-down menu to complete the sentences about how each wave is affected by the brick wall.
The sound waves from the siren will
the smooth surface of the wall. The light waves from the emergency vehicle will
the smooth surface of the wall. Rougher sections of the wall surface will cause the
from the emergency vehicle to scatter.
Answers
The sound waves from the siren will reflect off the smooth surface of the wall. The light waves from the emergency vehicle will reflect off the smooth surface of the wall. Rougher sections of the wall surface will cause the light waves from the emergency vehicle to scatter.
When sound waves hit a smooth surface, they reflect off the surface in a predictable way called the law of reflection. So, the sound waves from the siren will reflect off the smooth surface of the wall.
Similarly, light waves also follow the law of reflection when they hit a smooth surface. Therefore, the light waves from the emergency vehicle will also reflect off the smooth surface of the wall.
However, when light waves encounter a rough surface, they scatter in all directions due to the irregularities on the surface. Therefore, rougher sections of the wall surface will cause the light waves from the emergency vehicle to scatter.
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When you write the formula for sodium hydroxide, you do not have to put parentheses around the hydroxide polyatomic ion. However, when writing the formula for aluminum hydroxide, you must put parentheses around the hydroxide polyatomic ion. a) Write each formula. b) Explain why the parentheses are necessary for aluminum hydroxide.
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(a). Sodium hydroxide: NaOH, aluminum hydroxide: [tex]Al(OH)_3[/tex]
(b). The parentheses are necessary for aluminum hydroxide because the hydroxide polyatomic ion has a subscript of 3, indicating that there are three hydroxide ions for every one aluminum ion.
a) The formula for sodium hydroxide is NaOH, and the formula for aluminum hydroxide is [tex]Al(OH)_3[/tex]
b) Aluminum hydroxide requires brackets because there are three hydroxide ions for every one aluminum ion, according to the hydroxide polyatomic ion's subscript of 3. Without the parentheses, it would be unclear whether the subscript of 3 applies to only the oxygen or to the entire hydroxide ion. By enclosing the entire hydroxide ion in parentheses and placing the subscript outside the parentheses.
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What is the density (in g/L) of CO2 in a 5.20 L tank at 760.0 torr and 39.0°C .
Answers
The tank's CO₂ density is 1.84 g/L.
How to calculate density?Use the ideal gas law to solve for the density of CO₂:
PV = nRT
where:
P = pressure = 760.0 torr
V = volume = 5.20 L
n = moles of CO2 (we don't know this yet)
R = gas constant = 0.08206 L·atm/K·mol
T = temperature = 39.0°C + 273.15 = 312.15 K
First, convert torr to atm:
760.0 torr ÷ 760 torr/atm = 1 atm
Rearrange the ideal gas law to solve for n:
n = PV/RT
n = (1 atm)(5.20 L)/(0.08206 L·atm/K·mol)(312.15 K)
n = 0.217 mol
Use the mass of CO₂ and the volume of the tank to find the density:
mass = n × molar mass
mass = 0.217 mol × 44.01 g/mol
mass = 9.57 g
density = mass/volume
density = 9.57 g/5.20 L
density = 1.84 g/L
Therefore, the density of CO₂ in the tank is 1.84 g/L.
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ties there khat is Chemical Compound?
Answers
Answer:
In chemistry, a compound is a substance made up of two or more different chemical elements combined in a fixed ratio. When the elements come together, they react with each other and form chemical bonds that are difficult to break.
Explanation:
brainlist
Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the
inner transition metals are also called the lanthanide and actinide series.
Main-group metal
(representative metal)
Ru
Ta
Main-group nonmetal
(representative nonmetal)
As
Rn
Main-group
semimetal
(metalloid)
Answer Bank
In
Transition metal
Eu
Se
Inner transition metal.
(lanthanide/actinide)
Answers
Main-group metal (representative metal): Ru, Ta. Main-group nonmetal (representative nonmetal): As, Rn. Main-group semimetal (metalloid): In. Transition metal: Eu. Inner transition metal (lanthanide/actinide): Se is not an inner transition metal. Therefore, the correct answer for Se should be replaced with an inner transition metal (lanthanide or actinide) .Possible replacements for Se: La (lanthanide), Am (actinide).
Main group elements can be further classified as metals, nonmetals, or metalloids (also known as semimetals) based on their physical and chemical properties. Metals are typically shiny, good conductors of heat and electricity, and malleable. Nonmetals are typically brittle, poor conductors of heat and electricity, and have low melting and boiling points. Metalloids have properties of both metals and nonmetals and are semiconductors.
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How many mL of a 5.00% (m/v) glucose solution will be needed to deliver 8.5 grams of glucose?
Answers
mass = volume x concentration x density
where mass is the amount of glucose needed, volume is the volume of the glucose solution we need to prepare, concentration is the percentage of glucose in the solution, and density is the density of the solution.
We can rearrange the formula to solve for volume:
volume = mass / (concentration x density)
The density of the 5.00% (m/v) glucose solution can be assumed to be 1.00 g/mL.
Plugging in the values, we get:
volume = 8.5 g / (5.00 g/100 mL x 1.00 g/mL) = 170 mL
Therefore, we need 170 mL of the 5.00% (m/v) glucose solution to deliver 8.5 grams of glucose.
S8 + 12O2 -------> 8SO3
If you start with 873.2 g of S8 and 859.3 g of O2, what mass of SO3 will be produced?
Answers
The balanced chemical equation is:
S8 + 12O2 → 8SO3
The molar mass of S8 is 256.52 g/mol. Therefore, the number of moles of S8 is:
n(S8) = mass / molar mass = 873.2 g / 256.52 g/mol = 3.4 mol
The molar mass of O2 is 32.00 g/mol. Therefore, the number of moles of O2 is:
n(O2) = mass / molar mass = 859.3 g / 32.00 g/mol = 26.85 mol
According to the balanced chemical equation, the stoichiometric ratio of S8 to O2 is 1:12. Therefore, the number of moles of O2 needed to react with all of the S8 is:
n(O2) needed = n(S8) × 12 = 3.4 mol × 12 = 40.8 mol
Since the number of moles of O2 available (26.85 mol) is less than the number of moles of O2 needed (40.8 mol), O2 is the limiting reactant.
The theoretical yield of SO3 is based on the number of moles of O2 used. Since 12 moles of O2 are required to produce 8 moles of SO3, the number of moles of SO3 produced is:
n(SO3) = n(O2) × (8/12) = 26.85 mol × (8/12) = 17.90 mol
The molar mass of SO3 is 80.06 g/mol. Therefore, the mass of SO3 produced is:
mass(SO3) = n(SO3) × molar mass = 17.90 mol × 80.06 g/mol = 1433 g
Therefore, the mass of SO3 produced is 1433 g.
6. Calculate the pH of a buffer prepared by mixing 0.10 mol-L-' acetic acid (CH;COOH, HAc) and
0.10 mol L NaOH solution with the volume ratio of 3:1. (Answer pH = 4.45
Answers
CH3COOH + NaOH → CH3COONa + H2O
The resulting solution will contain both the weak acid (HAc) and its conjugate base (Ac-). The pH of the buffer can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([Ac-]/[HAc])
where pKa is the dissociation constant of HAc, [Ac-] is the concentration of the acetate ion, and [HAc] is the concentration of acetic acid.
The pKa of acetic acid is 4.76.
To calculate the concentrations of HAc and Ac-, we can use the volume ratio of 3:1 to find the total volume of the solution:
Total volume = 0.10 L HAc + 0.03 L NaOH = 0.13 L
The concentration of HAc is:
[HAc] = 0.10 mol / 0.13 L = 0.769 mol/L
The concentration of Ac- is:
[Ac-] = 0.03 mol / 0.13 L = 0.231 mol/L
Now we can substitute these values into the Henderson-Hasselbalch equation:
pH = 4.76 + log([0.231]/[0.769]) = 4.45
Therefore, the pH of the buffer is 4.45.
pH = pKa + log([A-]/[HA])
where [A-]/[HA] is the ratio of the concentration of the conjugate base to the concentration of the weak acid. In this case, acetic acid is the weak acid and its conjugate base is acetate (CH3COO-).
The volume ratio of the acetic acid and NaOH solutions is 3:1. Therefore, we can assume that we have 0.075 L of acetic acid solution and 0.025 L of NaOH solution.
The concentration of acetic acid is 0.10 mol-L¹. Therefore, the number of moles of acetic acid is:
moles of HAc = concentration × volume = 0.10 mol-L¹ × 0.075 L = 0.0075 mol
Since the volume of the NaOH solution is 0.025 L and its concentration is 0.10 mol-L¹, the number of moles of NaOH is:
moles of NaOH = concentration × volume = 0.10 mol-L¹ × 0.025 L = 0.0025 mol
The NaOH reacts with the HAc to form water and acetate:
NaOH + HAc → NaAc + H2O
Since the number of moles of NaOH is less than the number of moles of HAc, all of the NaOH will react with the HAc. Therefore, the number of moles of acetate formed is:
moles of acetate = moles of NaOH = 0.0025 mol
The number of moles of HAc remaining after the reaction is:
moles of HAc remaining = moles of HAc - moles of acetate = 0.0075 mol - 0.0025 mol = 0.0050 mol
The total volume of the buffer is 0.075 L + 0.025 L = 0.1 L. Therefore, the concentration of acetate is:
concentration of acetate = moles of acetate / volume of buffer = 0.0025 mol / 0.1 L = 0.025 mol-L¹
The concentration of HAc is:
concentration of HAc = moles of HAc remaining / volume of
Calculate the percentage by mass of oxygen in copper (II) tetraoxosulphate (VI) (Cu = 64.0, S = 32.0, 0 = 16.0)
Answers
The percentage by mass of oxygen in CuSO4 is 40%.
What is the percentage by mass of an atom in a compound?Finding the compound's molar mass is the first step in calculating the percentage of an atom by mass. The atomic masses of all the atoms in the compound are added up to determine the molar mass, which is the mass of one mole of the molecule.
We have the molar mass of the CuSO4 is given as;
64 + 32 + 4(16) = 160 g/mol
Then we have the percentage by mass of oxygen as;
4(16)/160 * 100/1
40%
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I just need # 6,8 and 12 pls
Answers
6. 1 [tex]SO_{3}[/tex] + 1 [tex]H_{2}O[/tex] --> 1 [tex]H_{2} SO_{4}[/tex]
8. 1 [tex]K_{2}O[/tex] + 1 [tex]H_{2}O[/tex] --> 2 KOH
12. 1 [tex]CdSO_{4}[/tex] + 1 [tex]H_{2} S[/tex] --> 1 CdS + 1 [tex]H_{2} SO_{4}[/tex]
find the ΔG°rxn for a reaction that has a K = 17.23 at 371 K. Is the reaction spontaneous. Note: R = 8.314 J/mol.K
Answers
The value of ΔG°rxn is - 6979 J/mol, and the reaction is spontaneous.
The change in Gibbs free energy (ΔG°rxn) for a reaction can be calculated using the equation;
ΔG°rxn = -RT ln(K)
where; R = gas constant = 8.314 J/mol.K
T = temperature in Kelvin (K)
K = equilibrium constant for the reaction
Given;
K = 17.23
T = 371 K
R = 8.314 J/mol.K
Plugging in the given values into the equation;
ΔG°rxn = - (8.314 J/mol.K) × (371 K) * ln(17.23)
Using a calculator, we can calculate the natural logarithm of 17.23:
ln(17.23) ≈ 2.848
Plugging this value back into the equation;
ΔG°rxn = - (8.314 J/mol.K) × (371 K) × 2.848
ΔG°rxn ≈ - 6979 J/mol
The negative value of ΔG°rxn indicates that the reaction is spontaneous, as ΔG°rxn < 0.
Therefore, the reaction will occur spontaneously in the forward direction at 371 K when the equilibrium constant (K) is 17.23.
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Can anyone please name this compound?
Answers
Answer:
Fluorobenzene
Explanation: Fluorobenzene is an organic compound, which is a derivative of benzene. It has a fluorine atom attached to one of the carbon atoms in the benzene ring. It appears as a colorless liquid and has a slightly sweet odor. Fluorobenzene is used as a solvent and catalyst in various chemical reactions. It is also used in the production of agrochemicals and pharmaceuticals. Due to its high solubility in water, it can contaminate groundwater and pose a risk to human health and the environment.
CO, (9) +2NH_(9) - CO(NH,) (s) +H, O(1)
a. What is the maximum mass of urea, CO(NH), that can be manufactured from the reaction of 2.20 moles of CO2 with sufficient amount of ammonia.
Answers
The mass of the ammonia that is required is 258 g.
What is the stoichiometry of the reaction?
The quantitative correlations between the reactants and products in a chemical reaction are the focus of the chemistry subfield known as stoichiometry.
We have to know that;
1 mole of CO2 produces 1 mole of urea
2.2 moles of CO2 produces 2.2 urea
Given that the number of moles of urea = 455 g/60 g/mol
= 7.58 moles
Now;
2 moles of NH3 produces 1 mole of urea
x moles of NH3 produces 7.58 moles of urea
x = 7.58 * 2/1
= 15.16 moles
Mass of the ammonia = 15.16 moles * 17 g/mol
= 258 g
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at 300K, a sample of gas occupies 1.5 dm3.
Assuming the pressure remains constant, the volume of this gas at 600K would be?
Answers
So, to find the volume of the gas at 600K, we can use the formula:
(Volume at 600K) = (Volume at 300K) x (Temperature at 600K / Temperature at 300K)
(Temperature at 600K) = (600°C + 273.15) = 873.15K
(Temperature at 300K) = (300°C + 273.15) = 573.15K
So, the volume of the gas at 600K would be:
(Volume at 600K) = (1.5 dm³) x (873.15K / 573.15K) = 2.28 dm³
Therefore, the volume of the gas at 600K would be 2.28 dm³.
How much energy is required to heat 186 grams of water from 32 degrees Celsius to 90
degrees Celsius?
Answers
q = m × c × ΔT
where q is the amount of energy required, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is 4.18 J/g°C.
The change in temperature is:
ΔT = 90°C - 32°C = 58°C
Substituting these values into the equation gives:
q = (186 g) × (4.18 J/g°C) × (58°C)
q = 44,222.56 J
Therefore, the amount of energy required to heat 186 grams of water from 32°C to 90°C is 44,222.56 J.
Can u mark my answer as the Brainlyest if it work Ty
Write short note on the
-physical and chemical methods
of monitoring the rate of
chemical reaction
Answers
Answer:
Physical and chemical methods can be used to monitor the rate of a chemical reaction. Physical methods measure changes in properties like temperature, pressure, or volume. Chemical methods track reactant consumption or product formation using techniques like titration or spectrophotometry. The choice of method depends on the reaction being studied, and scientists use these methods to gain insight into reaction kinetics and optimize conditions for better efficiency and selectivity.
A saturated solution of a salt was made by adding 36.00 g to 150.0 g water. There was 5.00 g of salt on the bottom that didn't dissolve. What is the % solubility of the salt ?
Answers
Answer:
The mass of the salt that dissolved is:
36.00 g - 5.00 g = 31.00 g
The % solubility of the salt can be calculated using the formula:
% solubility = (mass of solute dissolved ÷ mass of saturated solution) × 100%
The mass of the saturated solution is:
36.00 g + 150.0 g = 186.00 g
Therefore, the % solubility of the salt is:
(31.00 g ÷ 186.00 g) × 100% = 16.67%
The % solubility of the salt is 16.67%.
Explanation:
how mony moles of each ion present in 100mL of0.12M Ca3(Po4)2 solution.
Answers
In 100 mL of 0.12 M Ca3(PO4)2 solution, there are 0.036 moles of calcium ions and 0.024 moles of phosphate ions.
To find the number of moles of each ion present in 100 mL of 0.12 M Ca3(PO4)2 solution, we first need to determine the formula weight of Ca3(PO4)2.
Ca3(PO4)2 contains three calcium ions (Ca2+) and two phosphate ions (PO43-).
The atomic weight of calcium (Ca) is 40.08 g/mol, and there are three calcium ions in Ca3(PO4)2, so the total weight of calcium in Ca3(PO4)2 is 3 x 40.08 = 120.24 g/mol.
The atomic weight of phosphorus (P) is 30.97 g/mol, and the atomic weight of oxygen (O) is 16.00 g/mol.
The molecular weight of the phosphate ion (PO43-) is therefore 30.97 + (4 x 16.00) = 94.97 g/mol. Since there are two phosphate ions in Ca3(PO4)2, the total weight of phosphate in Ca3(PO4)2 is 2 x 94.97 = 189.94 g/mol.
The total formula weight of Ca3(PO4)2 is therefore 120.24 + 189.94 = 310.18 g/mol.
To find the number of moles of each ion present in 100 mL of 0.12 M Ca3(PO4)2 solution, we can use the formula:
moles = concentration x volume
For calcium ions:
moles of Ca2+ = concentration of Ca3(PO4)2 x (3 moles of Ca2+ / 1 mole of Ca3(PO4)2) x volume
moles of Ca2+ = 0.12 M x (3 / 1) x 0.1 L
moles of Ca2+ = 0.036 mol
For phosphate ions:
moles of PO43- = concentration of Ca3(PO4)2 x (2 moles of PO43- / 1 mole of Ca3(PO4)2) x volume
moles of PO43- = 0.12 M x (2 / 1) x 0.1 L
moles of PO43- = 0.024 mol
Therefore, in 100 mL of 0.12 M Ca3(PO4)2 solution, there are 0.036 moles of calcium ions and 0.024 moles of phosphate ions.
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What is the rate of the reaction in L/min
Answers
The rate of the reaction is obtained from the calculation as 0.022 L/min.
What is the rate of reaction?We know that the rate of reaction has to do with the change in the volume or the concentration of one of the reactant which the volume of the oxygen have been chosen for the particular question that we are dealing with here.
Thus;
Rate of reaction = Change in the volume of oxygen/Time
= v2 - v1/t
= 0.35 L - 0.02 L/15 min
=0.022 L/min
Thus, we can see that the rate of the reaction can be obtained as 0.022 L/min.
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2. Balance the equation below for the reaction between acetylene and oxygen, using the smallest
whole-number coefficients.
C₂H₂(g) +
O₂(g) →→
_CO₂(g) +
H₂O(g) + heat
Answers
C₂H₂ + 5/2 O[tex]_2[/tex](g) → 2CO[tex]_2[/tex](g) +H [tex]_2[/tex]O(l) is the balanced equation for the given unbalanced chemical equation.
An equation per a chemical reaction is said to be balanced if both the reactants plus the products have the same number of atoms and total charge for each component of the reaction. In a nutshell, the two components of the reaction have an equal balance of mass and charge. preservation of charge and mass, equation and reaction balance, etc.
C₂H₂ + O[tex]_2[/tex](g) → CO[tex]_2[/tex](g) +H [tex]_2[/tex]O(l)
Firstly balance the number of carbon atoms on product side by multiplying by 2
C₂H₂ + O[tex]_2[/tex](g) → 2CO[tex]_2[/tex](g) +H [tex]_2[/tex]O(l)
Now balance hydrogen and oxygen
C₂H₂ + 5/2 O[tex]_2[/tex](g) → 2CO[tex]_2[/tex](g) +H [tex]_2[/tex]O(l)
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