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Klio2033 [76]
3 years ago
15

Select the best answer for the question.

Chemistry
1 answer:
Delicious77 [7]3 years ago
5 0

Answer:

B

Explanation:

The rate constant is equal to forward rate divided by reverse rate.

K = f/r

-> r = f/K

0.62/0.75 = 0.8267

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Suppose 215 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?
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Answer:

The answer is "41.23 \ L\  N_2"

Explanation:

2 NO_3^{-} + 10 e^{-} + 12 H^{+} \longrightarrow N_2 + 6 H_2O\\\\= \frac{( 215 \ g \ NO_3^{-})}{(62.0049  \frac{\ g NO_3^{-}}{mol})} \times  \frac{(1 \ mol \ N_2}{ 2 \ mol \ NO_3^{-})}\\\\

=3.46746789 \times 0.5\\\\= 1.733 \ mol \ N_2 \\\\\to V = \frac{nRT}{P} \\\\= (1.733 \ mol) \times (0.08205746 \frac{L\ atm}{Kmol}) \times \frac{ (17 + 273) K}{(1.00 atm)}\\\\= 41.23

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2 years ago
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Why is it important to write a lab report once the experiment is finished?
cluponka [151]

Answer:

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3 years ago
If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0
3 years ago
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