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lianna [129]
3 years ago
5

g For the reaction Ag2S(s)⇌2Ag+(aq)+S2−(aq)Ag2S(s)⇌2Ag+(aq)+S2−(aq), Keq=2.4×10−4Keq=2.4×10−4, and the equilibrium concentration

of sulfide ion is [S2−]=0.0023M[S2−]=0.0023M. What is [Ag+]atequilibrium?
Chemistry
1 answer:
Roman55 [17]3 years ago
8 0

Answer:

0.32 M

Explanation:

Step 1: Write the balanced reaction at equilibrium

Ag₂S(s) ⇌ 2 Ag⁺(aq) + S²⁻(aq)

Step 2: Calculate the concentration of Ag⁺ at equilibrium

We will use the formula for the concentration equilibrium constant (Keq), which is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.

Keq = [Ag⁺]² × [S²⁻]

[Ag⁺] = √{Keq / [S²⁻]}

[Ag⁺] = √{2.4 × 10⁻⁴ / 0.0023} = 0.32 M

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What volume would 75.0g of oxygen gas occupy
seropon [69]

Answer:

Explanation: It is already known that 1 mole of the gas( or 32g of O2) is equivalent to 22.4 Litres of the oxygen gas. So, 8g is equivalent to = (22.4/32) × 8 = 5.6 L of the gas.

7 0
3 years ago
An ion is an atom that has has gained or lost:
Elena L [17]
Electrons, specifically valence electrons
5 0
3 years ago
Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4
inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

M=\frac{n_{solute}}{V_{solution}}

So you can solve for the moles of the solute:

n_{solute}=M*V_{solution}

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

n_{solute}=1mol/L*1.5L=1.5mol

But this is just a supposition.

Regards.

4 0
3 years ago
Assuming 1 mol of Fe3+ and 2 mol of SCN- were allowed to react and reach equilibrium. 0.5 mol of product was formed. The total v
zaharov [31]

Answer:

a. 0.5 mol

b. 1.5 mol

c. 0.67

Explanation:

Fe3+ + SCN- -----> [FeSCN]2+

a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium

b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium

c. KC =  0.5/(0.5*1.5) =  0.67

3 0
3 years ago
Why do the gases on the periodic table tend to form negative ions?
boyakko [2]
Because the valence shell of gases wants to become full
8 0
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