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aleksley [76]
3 years ago
5

12. In a gas mixture consisting of carbon dioxide (CO2) and nitrogen (N2), the carbon dioxide molecules have a root-mean-square

speed of 550 m/s. What is the root-mean-square speed of the nitrogen molecules in the sample? CO2 has a molar mass of 44 grams, and N2 has a molar mass of 28 grams. a) 627 m/s b) 752 m/s c) 564 m/s d) 821 m/s e) 689 m/s
Chemistry
1 answer:
zaharov [31]3 years ago
7 0

Answer:

The correct answer is option e.

Explanation:

The formula used for root mean square speed is:

\nu_{rms}=\sqrt{\frac{3kN_AT}{M}}

where,

\nu_{rms} = root mean square speed

k = Boltzmann’s constant = 1.38\times 10^{-23}J/K

T = temperature

M = Molar mass

N_A = Avogadro’s number = 6.02\times 10^{23}mol^{-1}

Root mean square speed of  carbon dioxide molecule:

\nu_{rms}= 550 m/s

Temperature of the mixture = T =?

Molar mass of carbon dioxide = 44 g/mol = 0.044 kg/mol

\nu_{rms}=550 m/s=\sqrt{\frac{3\times 1.38\times 10^{-23}J/K\times 6.022\times 10^{23}mol^{-1}T}{0.044 kg/mol}}

T = 533.87 K

Root mean square speed of nitrogen  molecule:

\nu'_{rms}= ?s

Molar mass of nitrogen = 28 g/mol = 0.028 kg/mol

\nu'_{rms}=\sqrt{\frac{3\times 1.38\times 10^{-23}J/K\times 6.022\times 10^{23}mol^{-1}\times 533.87 K}{0.028 kg/mol}}

\nu'_{rms}=689.46 m/s\approx 689 m/s

689 m/s is the root-mean-square speed of the nitrogen molecules in the sample.

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Answer:

Mass of NaBr produced  = 23.67 g

Explanation:

Given data:

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Mass of NaBr produced = ?

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Answer:

Explanation:

Explanation:

As you know, the empirical formula tells you what the smallest whole number ratio that exists between the atoms that make up a compound is.

In your case, you know that the empirical formula is

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, which means that the regardles of how many atoms of each element you get in the actual compound, the ratio that exists between them will always be

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What you actually need to determine is how many empirical formulas are needed to get to the molecular formula.

Notice that the problem provides you with the molar mass of the compound. This means that you can use the molar mass of the empirical formula to determine exactly how many atoms you need to form the compound's molecule.

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