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Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all charges, stoichiometric coefficients, and phase subscripts.

Answers

Answer 1

Answer:

Answer:

2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓

Ksp = [2s]² . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺ + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s) ⇄ 2Ag⁺ (aq) + CrO₄⁻² (aq) Ksp

2 s s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]² . [s] → 4s³

Related Questions

Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you the substance is a pure substance or a mixture, if you can. • Sample A is a solid yellow cube with a total mass of 50.0 g. The cube is put into a beaker filled with 250. mL of water. The cube collapses into a small pile of orange powder at the bottom of the beaker. When this powder is filtered out, dried and weighed, it has a total mass of 29.9 g. If the experiment is repeated with 500. mL of water, the powder that's left over has a mass of 10.0 g. Sample B is 100. g of a coarse grey powder with a faint unpleasant smell. 15. mg of the powder are put into a very thin tube and heated. The powder begins melting at 66.2 °C.The temperature stays constant as the powder slowly melts. After the last of the powder melts, the temperature starts to rise again, eventually reaching 76.0 °C. O pure substance Is sample A made from a pure substance or a mixture? x ? mixture If the description of the substance and the outcome of the experiment isn't enough to decide, choose "can't decide. (can't decide) O pure substance Is sample B made from a pure substance or a mixture? If the description of the substance and the outcome of bstance and the outcome of the experiment isn't enough to decide, choose "can't decide." mixture (can't decide)
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A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.956 g lead (II) bromide (PbBr2 solid at a temperature of 25.0°C. Under these conditions, calculate the reaction free energy AG for the following chemical reaction: Pb2+ (aq) + 2Br (aq) = PbBr2 (s) Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

A compound is found to contain 18.28 % phosphorus , 18.93 % sulfur , and 62.78 % chlorine by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 169.4 g/mol. The molecular formula for this compound is

Answers

Answer:

1. EF = PSCl₃; 2. MF = PSCl₃

Explanation:

1. Empirical formula

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our first job is to calculate the molar ratio of P:S:Cl.

Assume 100 g of the compound.

(a) Calculate the mass of each element.

Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.

(b) Calculate the moles of each element

$\text{Moles of P} = \text{18.28 g C} * \frac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\n\n\text{Moles of S} = \text{18.93 g S} * \frac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\n\n\text{Moles of Cl} = \text{62.78 g Cl} * \frac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}$

(c) Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3

(d) Write the empirical formula

EF = PSCl₃

The empirical formula for this compound is PSCl₃.

2. Molecular formula

(a) Calculate the ratio of the molecular and empirical formula masses

n = (169.4 u)/(169.40 u) = 1.000 ≈ 1

(b) Calculate the molecular formula

MF = (EF)ₙ = (EF)₁ = PSCl₃

The molecular formula for this compound is PSCl₃.

The empirical formula of the compound is PSCl₃.

To find the empirical formula, we first need to find the moles of each element in the compound. We can do this by dividing the mass of each element by its molar mass. The molar masses of the elements are:

P = 30.97 g/mol

S = 32.06 g/mol

Cl = 35.45 g/mol

The mass percentages given are for 100 g of the compound. So, the mass of each element in 100 g of the compound is:

P = 18.28 g

S = 18.93 g

Cl = 62.78 g

The moles of each element are then:

P = 18.28 g / 30.97 g/mol = 0.590 mol

S = 18.93 g / 32.06 g/mol = 0.590 mol

Cl = 62.78 g / 35.45 g/mol = 1.770 mol

The smallest whole number ratio of the moles of each element is 1:1:3. So, the empirical formula of the compound is PSCl3.

The molecular formula of the compound can be the same as the empirical formula, or it can be a multiple of the empirical formula. The molecular formula is not given, so we cannot say for sure what it is. However, we can say that the molecular formula must be a whole number multiple of the empirical formula PSCl3.

Learn more about empirical formula,here:

brainly.com/question/32125056

#SPJ3

The rate constant doubles when the temperature is increased from 45.0 C to 73.0 C. What is the activation energy for this reaction? (R=8.314 J/Kmol)?

Answers

Answer:

Ea =22542.6

Explanation:

The rate constant k is affected by the temperature and this dependence may be represented by the Arrhenius equation:

$k=Ae^{-(E_a)/(RT) }$

where the pre-exponential factor A is assumed to be independent of temperature, R is the gas constant, and T the temperature in K. Taking the natural logarithm of this equation gives:

ln k = ln A - Ea/(RT)

ln k = -Ea/(RT) + constant

ln k = -(Ea/R)(1/T) + constant

These equations indicate that the plot of ln k vs. 1/T is a straight line, with a slope of -Ea/R. These equations provide the basis for the experimental determination of Ea.

now applying the above equation in the problem

we can write that

$ln(k_2)/(k_1) = (E_a)/(R)[(1)/(T_1)-(1)/(T_2) ]$

solve for Ea:

Ea = R[Ln(k2/k1)] / [(1/T1) - (1/T2)]

but k_2 = 2 k_1, hence:

Ea = (8.314 J/moleK)[ln(2)] / [(1/273+45) - (1/273+73)]

Ea =22542.6

Answer:

The activation energy for this reaction is 22.6 kJ/ mol

Explanation:

Step 1: Data given

Rate constant doubles when Temperature goes from 45.0 °C to 73.0 °C

R = 8.314 J/K*mol

Step 2: Calculate the activation energy

Log (k2/k1) = Ea / 2.303R *((1/T1) - (1/T2))

⇒ with k1 = initial rate constant

⇒ with k2 = rate constant after doubled = 2k1

⇒ T1 = initial temperature = 45.0 °C = 318 Kelvin

⇒ T2 = Final temperature = 73.0 °C = 346 Kelvin

log (2) = Ea / (2.303*8.314) *((1/318) - (1/346))

log(2) = Ea / (2.303*8.314) * 0.00025448

Ea = 22649 J/mol = 22.6 kJ/mol

The activation energy for this reaction is 22.6 kJ/ mol

If sodium arsenite is Na3AsO3, the formula for calcium arsenite would be

Answers

Answer:

Ca₃(AsO₃)₂

Explanation:

Sodium arsenite, with the chemical formula Na₃AsO₃, is formed by the cation Na⁺ and the anion AsO₃³⁻. For the molecule to be neutral, 3 cations Na⁺ and 1 anion AsO₃³⁻ are required.

Calcium arsenite would be formed by the cation Ca²⁺ and the anion AsO₃³⁻. For the molecule to be neutral, we require 3 cations Ca²⁺ and 2 anions AsO₃³⁻. The resulting chemical formula is Ca₃(AsO₃)₂.

Which best illustrates the way in which radiation transfers thermal energy?

#edge2021

Answers

Answer:

it's b.

Explanation:

thank u so much for this. i appreciate it. lol.

First one
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Which step would help a student find the molecular formula of a compound from the empirical formula? Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Subtract the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound from the subscripts of the empirical formula. Divide the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Add the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound to the subscripts of the empirical formula..

Answers

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

got it right on edge 2020 :)

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen, if the rate of the formation of carbon monoxide is 0.35 M/s ? g

Answers

Answer : The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-(1)/(a)(d[A])/(dt)$

$\text{Rate of disappearance of B}=-(1)/(b)(d[B])/(dt)$

$\text{Rate of formation of C}=+(1)/(c)(d[C])/(dt)$

$\text{Rate of formation of D}=+(1)/(d)(d[D])/(dt)$

$Rate=-(1)/(a)(d[A])/(dt)=-(1)/(b)(d[B])/(dt)=+(1)/(c)(d[C])/(dt)=+(1)/(d)(d[D])/(dt)$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$CH_4+H_2O\rightarrow CO+3H_2$