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2 years ago

Determine the molar concentration of cl- ions in a 1. 00 m solution of agcl2- with no excess silver ions

1 answer:

5 0

The molar concentration of Cl⁻ ions in a 1. 00 m solution of AgCl₂ with no excess silver ions is 0.020M.

<h3>How do we determine the molar concentration?</h3>

Molar concentration of Chloride ions present in the silver chloride solution will be calculated by using the formation constant equation as:

Kf = [AgCl₂] / [Ag²⁺][Cl⁻]²

For AgCl₂ value of Kf is 2.5×10⁵.

Chemical reaction with ICE table is:

Ag²⁺(aq) + 2Cl⁻(aq) → AgCl₂

Initial: 0 0 1

Change: +x +2x -x

Equilibrium: x 2x 1-x

On putting all the values on the Kf equation, we get

2.5×10⁵ = (1-x) / (x)(2x)²

x is negligible as compared to 1 so the term 1-x becomes 1,

2.5×10⁵ = 1 / 4x³

x = 9.97 × 10⁻³ M

Concentration of chloride ions = 2x = 2(9.97 × 10⁻³) = 0.020M.

Hence the molar concentration of chloride ions is 0.020M.

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State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and expl

Strike441 [17]

Answer:

a) ΔS is negative

b) ΔS is positive

c) ΔS is positive

d) ΔS is negative

Explanation:

Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.

The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.

S(solid) < S(liquid) < S(gas)

The entropy change for a given reaction is:

$\Delta S = S(products)-S(reactants)----(1)$

a) $PCl3(l) + Cl2(g) \rightarrow PCl5(s)$

Here the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

b) $2HgO(s) \rightarrow 2Hg(l) + O2(g)$

Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

c) $H2(g) \rightarrow 2H(g)$

Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

d) $U(s) + 3F2(g) \rightarrow UF6(s)$

Here one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

8 0

3 years ago

Which statement best describes thermodynamics?

hjlf

That person was right it is A

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3 years ago

Read 2 more answers

1. Determine the acid type for HBr.<br> O Brønsted-Lowry<br> O Ahrrenius<br> O Lewis<br> not a base

Licemer1 [7]

Answer:idk