Please help! Chemistry 113 smartworks question. Thanks !!!

Chemistry

1 answer:

Dominik [7]3 years ago

3 0

Kc' =Kc^1/3
=3√0.0061
=0.182716013

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Question 15 need this answered asap (chemistry)

Tcecarenko [31]

Answer:

80.7 L

Explanation:

PV = nRT

P = 1520 mmHg = 2 atm

n = 5 mol

R = 0.08206 (L * atm)/(mol * K)

T = 393.15 K

2 (V) = 5 (0.08206) (393.15)

V ≈ 80.7 L

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3 years ago

When a solution of barium nitrate and a solution of copper (II) sulfate are mixed, a chemical reaction produces solid barium sul

Artemon [7]

The total mass of the products is 10.76 g + 204.44 g = 215.20 g.
The masses of all the reactants but one are known so,

215.20 g - 120.00 g - 8.15 g - 75.00 g = 12.05 g

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Read 2 more answers

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$