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3 years ago

Need help asap!!!!! Plz

Chemistry

1 answer:

Nostrana [21]3 years ago

6 0

Answer:

153.8 g/mol.

Explanation:

The molar mass of a compound = ∑ atomic masses of atoms.

∴ The molar mass of CCl₄ = 1(atomic masss of C) + 4 (atomic mass of Cl) = (12.0 g/mol) + 4 (35.45 g/mol) = 153.8 g/mol.

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What two atoms are held together by a covalent bond in NaHCO3

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Explanation:

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Any time an atom transfers or shares an electron a is formed

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Answer: Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

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$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

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Putting values in above equation, we get:

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For the given chemical equation:

$O_3+NO\rightarrow O_2+NO_2$

By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

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As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ozone is considered as a limiting reagent because it limits the formation of product.

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By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitrogen dioxide

Now, calculating the mass of nitrogen dioxide from equation 1, we get:

Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

$0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g$

Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

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