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3 years ago

How many moles of n are in 0.163 g of n2o?

Chemistry

2 answers:

slava [35]3 years ago

7 0

Answer:

$n_N=7.41x10^{-3}molN$

Explanation:

Hello,

In this case, one develops the following mass-mole relationship, taking into account that $N_2O$ has two nitrogens inside, thus:

$n_N=0.163gN_2O*\frac{1molN_2O}{44gN_2O}*\frac{2molN}{1molN_2O}\\n_N=7.41x10^{-3}molN$

Best regards.

zimovet [89]3 years ago

3 0

First convert 0.163 grams of N2O to mol by dividing it with the molecular weight. The molecular weight of N2O is 44 grams/mol. The answer would be 3,79x10^-3. Then multiply it with 2 since there are 2 Nitrogen in one mole of N2O. Therefore, there are 7.41x10^-3 moles of Nitrogen.

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If 0.00872 mol neon gas at a particular temperature and pressure occupies a volume of 211 mL, what volume would 0.00701 mol neon

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Answer:

0.00701 moles of neon gas will occupy 170 ml of volume under the same conditions.

Explanation:

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

where,

$V_1\text{ and }n_1$ are the initial volume and number of moles

$V_2\text{ and }n_2$ are the final volume and number of moles

We are given:

$V_1=211 mL\\n_1=0.00872 mol\\V_2=?\\n_2=0.00701 mol$

Putting values in above equation, we get:

$\frac{211 mL}{0.00872 mol}=\frac{V_2}{0.00701}$

$V_2=\frac{211 mL\tomes 0.00701 mol}{0.00872 mol}=169.62 mL\approx 170 mL$

0.00701 moles of neon gas will occupy 170 ml of volume under the same conditions.

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