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Marina86 [1]
3 years ago
9

Assuming the same temperature and pressure for each gas, how many milliliters of carbon dioxide are produced from 16.0 mL of CO?

Chemistry
1 answer:
Vlada [557]3 years ago
4 0

Answer:

V_{CO_2}=16.0mL

Explanation:

Hello,

In this case, given that the same temperature and pressure is given for all the gases, we can notice that 16.0 mL are related with two moles of carbon monoxide by means of the Avogadro's law which allows us to understand the volume-moles relationship as a directly proportional relationship. In such a way, since in the chemical reaction:

2CO(g)+O_2(g)\rightarrow 2CO_2(g)

We notice two moles of carbon monoxide yield two moles of carbon dioxide, therefore we have the relationship:

n_{CO}V_{CO}=n_{CO_2}V_{CO_2}

Thus, solving for the yielded volume of carbon dioxide we obtain:

V_{CO_2}=\frac{n_{CO}V_{CO}}{n_{CO_2}} =\frac{2mol*16.0mL}{2mol}\\ \\V_{CO_2}=16.0mL

Best regards.

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Answer:

0.256 L  

Explanation:

We should use the following formula:

concentration (1) × volume (1) =  concentration (2) × volume (2)

concentration (1) = 0.82 M NaOCl

volume (1) = ?

concentration (2) = 0.21 M NaOCl

volume (2) = 1 L

volume (1) = [concentration (2) × volume (2)] / concentration (1)

volume (1) = [0.21 / 1] / 0.82 = 0.256 L

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3 years ago
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How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
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Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

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3 years ago
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A sample of oxygen gas has a volume of 3.24 L at 29°C. What volume will it occupy at 104°C if the pressure and number of mol are
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<u>Answer:</u> The final volume of the oxygen gas is 4.04 L

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=3.24L\\T_1=29^oC=(29+273)K=302K\\V_2=?\\T_2=104^oC=(104+273)K=377K

Putting values in above equation, we get:

\frac{3.24L}{302K}=\frac{V_2}{377K}\\\\V_2=4.04L

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3 years ago
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Answer:

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<em />

La ley de Avogadro dice que el volumen de un gas bajo temperatura y presión constantes es proporcional a las moles de este gas. Así, como 2 moles de acetileno reaccionan con 5 moles de oxígeno, los litros de O₂ necesarios para quemar 9340L de acetileno son:

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Si un tanque contiene 7x10³ L de O₂ serán necesarios:

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