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2 years ago

A compound contains 81.1 boron with the remainder being hydrogen what is the empirical formula of the compound

1 answer:

4 0

Empirical formula is the simplest ratio of whole numbers of components making up the compound
Mass percentage of boron - 81.1 %
Mass percentage of hydrogen - 18.9 %
For 100 g of the compound
B H
Mass   81.1 g 18.9 g
Number of moles 81.1 g /11 g/mol   18.9 g/1 g/mol
  = 7.37 mol = 18.9 mol
Divide by the least number of moles
7.37/7.37 =1.00   18.9/7.37 =2.56
Since we get 2.56 that cannot be rounded off we have to multiply both boron and hydrogen values by 2
B = 1.00 x2 = 2.00
H = 2.56 x 2 = 5.12 rounded off to 5
Therefore empirical formula is
B₂H₅

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vladimir2022 [97]

Answer:

The volume of the container is 59.112 L

Explanation:

Given that,

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Temperature of the gas, T = 300 K

Pressure of the gas, P = 1.25 atm

We need to find the volume of the container. For a gas, we know that,

PV = nRT

V is volume

R is gas constant, R = 0.0821 atm-L/mol-K

So,

$V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L$

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How many moles of tungsten (W,183.85 g/lol are in 415 grams of tungsten?

vladimir1956 [14]

Given mass of tungsten, W = 415 g

Molar mass of tungsten, W = 183.85 g/mol

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1 year ago

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

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After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

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2 years ago

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2 years ago

A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2

4vir4ik [10]

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

$C_3H_7+5O_2\rightarrow 3CO_2+4H_2O$

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

$\frac{3}{1}\times 3.85 mol=11.55 mol$ of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

$\frac{4}{1}\times 3.85 mol=15.4 mol$ of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

$\frac{4}{5}\times 0.647 mol=0.5176 mol$ of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

Then 0.647 moles of oxygen will give:

$\frac{1}{5}\times 0.647 mol=0.1294 mol$ of propane.

0.1294 moles of propane are consumed.

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