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jek_recluse [69]
3 years ago
11

Which is the molar mass of BF3?

Chemistry
1 answer:
Eduardwww [97]3 years ago
3 0
The answer is 67.82 g/mol
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The analysis of a hydrocarbon revealed that it was 85.7% C and 14.3% H by mass. When 1.77 g of the gas was stored in a 1.500-L f
gtnhenbr [62]

Answer:

The formula of hydrocarbon = C_3H_6

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 85.7

Molar mass of C = 12.0107 g/mol

% moles of C = 85.7 / 12.0107 = 7.14

% of H = 14.3

Molar mass of H = 1.00784 g/mol

% moles of H = 14.3 / 1.00784 = 14.19

Taking the simplest ratio for C and H as:

7.14 : 14.19 = 1 : 2

The empirical formula is = CH_2

Also, Given that:

Pressure = 508 Torr

Temperature = 17 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (17 + 273.15) K = 290.15 K  

Volume = 1.500 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637 L.torr/K.mol

Applying the equation as:

508 Torr × 1.500 L = n × 62.3637 L.torr/K.mol × 290.15 K  

⇒n = 0.0421 moles

Given that :  

Amount  = 1.77 g

Molar mass = ?

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0421\ moles= \frac{1.77\ g}{Molar\ mass}

Molar mass of the hydrocarbon = 42.04 g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12 + 2×1= 14 g/mol

Molar mass = 42.04 g/mol

So,  

Molecular mass = n × Empirical mass

42.04 = n × 14

⇒ n = 3

<u>The formula of hydrocarbon = C_3H_6</u>

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Crystals most often form in igneous and metamorphic rocks. According to the rock cycle, which processes and features do these ro
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The features that are shared by the rocks include the melting and cooling of rocks.

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