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omeli [17]
3 years ago
5

The density of an unknown gas is found to be 4.12 g/L at standard temperature and pressure (STP). What is the molar mass of this

gas?
Chemistry
2 answers:
elena-s [515]3 years ago
7 0
The   molar  mass of this gas  is  92.3 g/mol

  Calculation

By  use  ideal gas  equation PV =nRT  where
n=mole  p=pressure V= volume R =  gas constant  T=  temperature

n =  mass /molar  mass(MM)
substitute in the equation

PV =(mass/MM)RT
mass = density  x  volume(V)

Therefore  PV =(density xV/ MM) xRT

divide both side by  by V

P=  (density/Mm) xRT

making MM  the subject  of the formula

MM = densityPRT

At  STP = P= 1 atm,  R=  0.0821 L.atm/Mol.k  T =  273 K

MM is therefore = 4.12 g/l x   1 atm x 0.081  L.atm/mol.k  x    273 K = 92.3 g/mol

hichkok12 [17]3 years ago
7 0
At standard temperature and pressure, 1 mole of any gas occupies 22.4 litres. We need to find the number of grams in 22.4 litres of the unknown gas.

4.12 g/L * 22.4 = <span>92.288. 

</span><span>92.288 or 92 grams are in one mole of the gas making the molar mass 92 g/mole. </span>
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But-1-ene can be prepared from butan-1-ol. which type of reaction takes place?
slega [8]

Answer:

              Dehydration Reaction

Explanation:

                  Alcohols can be converted into Alkenes using catalytic amounts of acids as a catalysts.

                 The water molecule is removed hence, it is called as dehydration reaction.

                  The reaction is attached.

8 0
2 years ago
How many moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas?
tino4ka555 [31]

Taking into account the reaction stoichiometry, 340.0 moles  of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CO₂ + 4 H₄  → CH₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CO₂: 1 mole
  • H₄: 4 moles  
  • CH₄: 1 mole
  • H₂O: 2 moles

<h3>Moles of CH₄ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

moles of CH_{4} =\frac{85.1 moles of CO_{2}x4 moles of CH_{4} }{1 moles of CO_{2}}

<u><em>moles of CH₄= 340.4 moles</em></u>

Then, 340.0 moles  of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

6 0
2 years ago
A 10.0 gram sample of Fe contains how many miles of Fe
Dafna1 [17]

Answer:

1.7857 moles

Explanation:

moles=mass/Mr

10/56=1.7857

moles of iron =1.7857

4 0
2 years ago
The combustion of 135 mg of a hydrocarbon produces 440 mg of CO2 and 135 mg H2O. The molar mass of the hydrocarbon is 270 g/mol.
NeX [460]

Answer:

Molecular formula = C20H30

Explanation:

NB 440mg = 0.44g, 135mg= 0.135g

From the question, moles of CO2= 0.44/44= 0.01mol

Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01

Also from the question, moles of H2O = 0.135/18= 0.0075mole

Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H

To get the empirical formula, divide by smallest number of mole

Mol of C = 0.01/0.01=1

Mol of H = 0.015/0.01= 1.5

Multiply both by 2 to obtain a whole number

Mol of C =1×2 = 2

Mol of H= 1.5×2 = 3

Empirical formula= C2H3

[C2H3] not = 270

[ (2×12) + 3]n = 270

27n = 270

n=10

Molecular formula= [C2H3]10= C20H30

5 0
3 years ago
How much does the air weigh within a column that is 1 square inch in area that extends from sea level all the way to the top of
fredd [130]

Answer:

14.7 lbs

Explanation:

Air pressure is the weight of the air above us. It is approximately 14.7 pounds or lbs per square inch at sea level. It means that an air column weights 14.7 lbs, 1 square inch in diameter, reaching all the way up to the top of the atmosphere.

5 0
3 years ago
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