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Blababa [14]
3 years ago
14

Would a precipitate form if a solution contained 8.0 x 10-4 m silver nitrate and 1.8 x 10-4 m potassium chromate? you will have

to find ksp values from your text or from another source.
Chemistry
1 answer:
Alina [70]3 years ago
6 0
Chemical reaction: 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄<span>(s).
</span>Ksp(Ag₂CrO₄) = 1,1·10⁻<span>¹².
</span>[Ag⁺] = c(AgNO₃) = 8.0·10⁻⁴ M.
[CrO₄²⁻] = c(K₂CrO₄) = 1.8·10⁻⁴ M.
Q(Ag₂CrO₄) = [Ag⁺]²·[CrO₄²⁻]. 
Q = (8.0·10⁻⁴ M)² · 1.8·10⁻⁴ M.
Q = 1.152·10⁻¹⁰.
Q<span> > Ksp, </span><span>a precipitate of silver chromate will form.
</span>Ksp is the solubility product constant for a solid substance dissolving in an aqueous solution.

<span> [Ag</span>⁺<span>] is equilibrium concentration of silver cations.
[CrO</span>₄²⁻] is equilibrium concentration of chromate anions.


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Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare. List from lowest freez
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Answer:

The solutions are listed from the lowest freezing point to highest freezing point.

Mg₃(PO₄)₂ -  AlBr₃ - BeBr₂ -  KBr

Explanation:

Colligative property of freezing point, is the freezing point depression.

Formula is:

ΔT = Kf . m . i

Where ΔT = T° freezing pure solvent - T° freezing solution

Kf is the cryoscopic constant

m is molality and i is the Van't Hoff factor (number of ions, dissolved in solution)

In this case, T° freezing pure solvent is 0° because it's water, so the Kf is 1.86 °C/m, and m is 0.1 molal.

The i modifies the T° freezing solution. The four salts, are ionic compounds, so for each case:

BeBr₂  → Be²⁺  +  2Br⁻   i = 3

AlBr₃ →  Al³⁺  + 3Br⁻   i = 4

Mg₃(PO₄)₂  →  3Mg²⁺  +  2PO₄⁻³    i = 5

KBr →  K⁺  +  Br⁻   i = 2

Solution of Mg₃(PO₄)₂ will have the lowest temperature of freezing and the solution of KBr, the highest (always lower, than 0°).

8 0
3 years ago
An industrial chemist is studying the rate of the haber synthesis: n2 + 3h2 → 2nh3. starting with a closed reactor containing 1.
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A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62
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Answer: The concentration of H_2SO_4 is 0.234 M

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity H_2SO_4 = 2

M_1 = molarity of H_2SO_4 solution = ?

V_1 = volume of  H_2SO_4 solution = 50.0 ml

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Putting in the values we get:

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M_1=0.234M

Therefore concentration of H_2SO_4 is 0.234 M

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