3 years ago

How many molecules of N204 are in 85.0 g of N2O4?

Chemistry

1 answer:

alukav5142 [94]3 years ago

4 0

Answer:

5.56 × 10^23 molecules

Explanation:

The number of molecules in a molecule can be calculated by multiplying the number of moles in that molecule by Avagadro's number (6.02 × 10^23)

Using mole = mass/molar mass

Molar mass of N2O4 = 14(2) + 16(4)

= 28 + 64

= 92g/mol

mole = 85.0/92

= 0.9239

= 0.924mol

number of molecules of N2O4 (nA) = 0.924 × 6.02 × 10^23

= 5.56 × 10^23 molecules

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A 254.5 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 116.5 g of this mixture is dissolved i

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Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of $KNO_3$ in original 254.5 mixture.

moles of $BaSO_4 = \frac{mass}{Molecular\ Weight}$

moles of $BaSO_4 = \frac{68.3}{233.38}$

= 0.2926 mol of BaSO4

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0.2926 mol of BaCl2,

mass of $BaCl_2 = mol\times molecular weight$

$= 0.2926\times 208.23$

= 60.92 g

the AgCl moles $= \frac{mass}{Molecular\ Weight}$

$= \frac{199.1}{143.32}$

= 1.3891 mol of AgCl

note that, the Cl- derive from both, $BACl_2 and NaCl$

mole of Cl- f NaCl $= (1.3891) - (0.2926\times 2) = 0.8039$ mol of Cl-

mol of NaCl = 0.8039 moles

$mass = mol\times Molecular\ Weight = 0.8039 \times 58 = 46.626\ g \ of \ NaCl$

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Mass of KNO3 in the original mix is 146.954 g

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