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3 years ago

which law is used to determine the relationship between the volume and the number of moles in the equation

Chemistry

1 answer:

Lana71 [14]3 years ago

8 0

MC=n÷V where MC is mole concentration and n is number of moles and V is volume. please give brainliest

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Hello! Can you please help me do this question? I have to pick two but I’m not really sure what to pick..

In-s [12.5K]

Answer:

B and D

Explanation:

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7 0

3 years ago

Read 2 more answers

Need help !!!!! ASAP

Nat2105 [25]

<h2>Hello!</h2>

The answer is:

The temperature will be the same, 37°C.

Since from the statemet we know the first temperature, pressure and volumen of a gas, and we need to calculate the new temperature after the pressure and the volume changed, we need to use the Combined Gas Law.

The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using Boyle's Law, Gay-Lussac's Law and Charles's Law.

The law establishes the following equation:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}$

Where,

$P_{1}$ is the first pressure.

$V_{1}$ is the first volume.

$T_{1}$ is the first temperature.

$P_{2}$ is the second pressure.

$V_{2}$ is the second volume.

$T_{2}$ is the second temperature.

Then, we are given the following information:

$V_{1}=200mL\\P_{1}=4atm\\T_{1}=37\°C\\V_{2}=400mL\\P_{2}=2atm$

So, isolating the new temperature and substituting the given information, we have:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}\\\\T_{2}=P_{2}V{2}*\frac{T_{1}}{P_{1}V_{1}} \\\\T_{2}=2atm*400mL*\frac{37\°C}{4atm*200mL}=37\°C$

Hence, we have that the temperature will not change because both pressure and volume decreased and increased proportionally, creating the same relationship that we had before the experiment started.

The temperature will be the same, 37°C

Have a nice day!

4 0

3 years ago

Heeeeeeeeeeeeeeeeeeeeeelp

Genrish500 [490]

Answer:

The correct answer is 2.33g of $H_{2}O$ , A

Explanation:

We can use basic stoich to get this question.

To start, lets convert our grams to moles in order to use our equation. 5.38g of $O_{2} = (\frac{5.38}{2*15.999}) moles = 0.168136$ moles.

We then can use our molar ratio of 13:10. Simply divide by our molar ratio in order to go from moles of $O_{2}$ to moles of $H_{2}O$ . This gets us 0.129335 moles of $H_{2}O$ .