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kolezko [41]
3 years ago
6

What is the percent composition of dinotrogen pentoxide? With work shown please

Chemistry
1 answer:
iris [78.8K]3 years ago
7 0

Answer:

%N = 25.94%

%O = 74.06%

Explanation:

Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅

We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.

m(N): 2 × 14.01 g = 28.02 g

Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅

We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.

m(O): 5 × 16.00 g = 80.00 g

Step 3: Calculate the mass of 1 mole of N₂O₅

We will sum the masses of N and O.

m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g

Step 4: Calculate the percent composition  of N₂O₅

We will use the following expression.

%Element = m(Element)/m(Compound) × 100%

%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%

%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%

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Calculate the mass percent (m/m) of a solution prepared by dissolving 45.09 g of NaCl in 174.9 g of H2O.
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20.3 %  NaCl

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Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of
Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

  • We need to calculate the vapor pressure of each component after the stopcocks are opened.
  • Volume after the stopcocks are opened = 3.0 L.

<u><em>1) For N₂:</em></u>

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

<u><em>2) For H₂O:</em></u>

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: <em>(1.0 atm = 760.0 mmHg)</em>.

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

<em><u>3) For C₂H₅OH:</u></em>

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

<em>It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.</em>

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

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