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3 years ago

ASAP ASAP ASAP ASAP

Chemistry

1 answer:

IrinaVladis [17]3 years ago

5 0

Answer:

Molar mass of NaN3,

MM = 1*MM(Na) + 3*MM(N)

= 1*22.99 + 3*14.01

= 65.02 g/mol

mass of NaN3 = 1.3*10^2 g

mol of NaN3 = (mass)/(molar mass)

= 1.3*10^2/65.02

= 2.0 mol

Balanced chemical equation is:

2NaN3(s) ---> 2Na(s) + 3N2(g)

According to balanced equation

mol of N2 formed = (3/2)* moles of NaN3

= (3/2)*2.0

= 3.0 mol

Molar mass of N2 = 28.02 g/mol

mass of N2 = number of mol * molar mass

= 3.0*28.02

= 84.0 g

Now use:

density = mass / volume

0.916 g/L = 84.0 g / volume

volume = 91.7 g

Answer: 91.7 g

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2 years ago

A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal

goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂:

$1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC$

All of the hydrogens atoms in the compound ended up becoming H₂O, so the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O:

$0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH$

Because the compound is composed only by C, H and O, the mass of O in the compound can be calculated by substraction:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C

mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H

mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

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The answer is that exact locations within either cannot be determined at any given moment in time.

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Answer:

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