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Stells [14]
3 years ago
11

How many molecules of nitrogen dioxide would form from 2 molecules of no?

Chemistry
1 answer:
Nataly_w [17]3 years ago
3 0

2 molecules of nitrogen dioxide would form from 2 molecules of NO.

Explanation:

Data given:

molecule of NO given = 2

molecules of nitrogen dioxide formed = ?

Avagadro number = 6.022 X 10^{23} molecules

atomic mass of NO = 30 gram/mole

atomic mass of NO_{2} = 46 grams/mole

balanced chemical reaction is given by:

2NO + O_{2} ⇒ 2N0_{2}

to convert the molecules into moles, we use the formula:

moles = \frac{number of molecules}{Avagadro number}

the number of moles are same in the reaction for both reactant and product which means that 2 moles of NO reacts to form 2 moles of NO2.

thus when moles remain same, number of molecules are also same.

1 molecule = one mole x Avagadro's number

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When of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing poi
Nesterboy [21]

The given question is incomplete. The complete question is:

When 282. g of glycine (C2H5NO2) are dissolved in 950. g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 282. g of ammonium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for ammonium chloride in X.

Answer:  the van't Hoff factor for ammonium chloride is 1.74

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point  

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8.2^0C=1\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

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8.2^0C=1\times K_f\times \frac{282g}{75.07g/mol\times 0.95kg}

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ii) 20.0^0C=i\times \times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

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Mass of solute (ammonium chloride) = 282 g

20.0^0C=i\times 2.07\times \frac{282g}{53.49g/mol\times 0.95kg}

i=1.74

Thus the van't Hoff factor for ammonium chloride is 1.74

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