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krek1111 [17]
3 years ago
5

A student reported the following composition of an unknown mixture, 25% NH4Cl, 30% NaCl, and 50% SiO2. Assuming the student's ca

lculations are correct, explain the error.
Chemistry
2 answers:
marin [14]3 years ago
8 0
The three substances add up to 105% instead of 100%. Also, <span>the amount of either one of those compounds is more than according to the Stoichiometry of the reaction, meaning one of those compounds is in excess. </span>
saveliy_v [14]3 years ago
3 0

<u>Answer:</u> The composition of the unknown mixture is more than 100 %, which is not possible.

<u>Explanation:</u>

Composition of a mixture is defined as the sum of the amount of substances that are present in the mixture. The maximum composition of a mixture can be 100 %

We are given:

Composition of ammonium chloride in the mixture = 25 %

Composition of sodium chloride in the mixture = 30 %

Composition of silicon oxide in the mixture = 50 %

Composition of mixture = (25 + 30 + 50) = 105 %

As, the measured value of unknown mixture is more than the true value which is 100 %. Thus, the error lies here only.

Hence, the composition of the unknown mixture is more than 100 %, which is not possible.

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C. All electron carriers are mobile and hydrophobic

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How many atoms does 2NaOH have
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3. Which of the following most likely describes an isotope? *
Nimfa-mama [501]
I think the answer might be B but i’m not positive
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3 years ago
5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
3 years ago
The number of moles of solute in 200.ml of a 0.500 m solution is
IrinaK [193]

Answer:

I believe the answer is 0.100.

Explanation:

Hope my answer has helped you!

8 0
3 years ago
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