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SVETLANKA909090 [29]
2 years ago
12

Consider the reaction: A(aq) + 2B (aq) === C (aq). Initially 1.00 mol A and 1.80 mol B

Chemistry
1 answer:
liraira [26]2 years ago
6 0

Answer:

17

Explanation:

Step 1: Calculate the needed concentrations

[A]i = 1.00 mol/5.00 L = 0.200 M

[B]i = 1.80 mol/5.00 L = 0.360 M

[B]e = 1.00 mol/5.00 L = 0.200 M

Step 2: Make an ICE chart

        A(aq) + 2 B(aq) ⇄ C(aq)

I       0.200    0.360        0

C        -x           -2x         +x

E     0.200-x  0.360-2x   x

Then,

[B]e = 0.360-2x = 0.200

x = 0.0800

The concentrations at equilibrium are:

[A]e = 0.200-0.0800 = 0.120 M

[B]e = 0.200 M

[C]e = 0.0800 M

Step 3: Calculate the concentration equilibrium constant (K)

K = [C] / [A] × [B]²

K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17

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17)c: the reactivity of a metal.

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At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume
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Answer: Volume would be 196.15 mL if the temperature were changed to 22^{o}C and the pressure to 1.25 atmospheres.

Explanation:

Given: T_{1} = 35^{o}C = (35 + 273) K = 308 K,     V_{1} = 256 mL,    

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T_{1} = 22^{o}C = (22 + 273) K = 295 K,       P_{2} = 1.25 atm  

Formula used to calculate volume is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL

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4 0
2 years ago
In the laboratory you are asked to make a 0.565 m sodium bromide solution using 315 grams of water. How many grams of sodium bro
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Answer : The mass of sodium bromide added should be, 18.3 grams.

Explanation :

Molality : It is defined as the number of moles of solute present in kilograms of solvent.

Formula used :

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}

Solute is, NaBr and solvent is, water.

Given:

Molality of NaBr = 0.565 mol/kg

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Now put all the given values in the above formula, we get:

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Thus, the mass of sodium bromide added should be, 18.3 grams.

6 0
3 years ago
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