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3 years ago

!!!Need answer for chem homework ASAP PLS !!!!

Chemistry

1 answer:

stealth61 [152]3 years ago

7 0

Answer:

The answer to your question is: kc = 6.48

Explanation:

Data

Given Molecular weight

CaO = 44.6 g 56 g

CO₂ = 26 g 44 g

CaCO₃ = 42.3 g 100 g

Find moles

CaO 56 g ---------------- 1 mol

44.6 g -------------- x

x = (44.6 x 1) / 56 = 0.8 mol

CO₂ 44 g ----------------- 1 mol

26 g ---------------- x

x = (26 x 1 ) / 44 = 0.6 moles

CaCO₃ 100 g --------------- 1 mol

42.3g -------------- x

x = (42.3 x 1) / 100 = 0.423 moles

Concentrations

CaO = 0.8 / 6.5 = 0.12 M

CO₂ = 0.6 / 6.5 = 0.09 M

CaCO₃ = 0.423 / 6.5 = 0.07 M

Equilibrium constant = $\frac{[products]}{[reactants]}$

Kc = [0.07] / [[0.12][0.09]

Kc = 0.07 / 0.0108

kc = 6.48

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Explanation:

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3 years ago

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Mass of Co(NO₃)₂ = 1.95 g
V KOH = 0.350 L
[KOH] = 0.220 M
Kf = 5.0 x 10⁹
molar mass of Co(NO₃)₂ = 182.943 g/mol
so [Co(NO₃)₂] = 1.95 / (0.350 * 182.943) = 0.03045 M
[Co²⁺] = 0.03045 M
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chemical reaction:
Co²⁺(aq) + 4 OH⁻ ⇄ Co(OH)₄²⁻
I (M) 0.03045 0.22 0
C (M) - 0.03045 - 4 (0.03045) 0.03045
E (M) - x 0.22 - 4(0.03045) 0.03045
= 0.0982
Kf = [Co(OH)₄²⁻] / [Co⁺²][OH⁻]⁴
5.0 x 10⁹ = (0.03045) / x (0.0982)⁴
x = 6.5489 x 10⁻⁸
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[Co²⁺] = 6.54 x 10⁻⁸
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