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lora16 [44]
2 years ago
11

How many moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Cel

sius?
**I NEED ANSWERS STEP BY STEP PLEASE**
Chemistry
1 answer:
o-na [289]2 years ago
7 0

1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

<h3>What is an ideal gas equation?</h3>

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa). V = volume, in m^3.

We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol:

PV= nRT

Given data:

P=100.0 kPa =0.986923 atm

T=100 degree celcius= 100 + 273 =373 K

V=35.5 L

Substituting the values in the equation.

n= \frac{\;0,98 \;atm \;X \;35,5 \;L }{\;0,082\;atm / \;K mol \;X \;373 K}

n= 1.137448506 mol

Hence, 1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

Learn more about ideal gas here:

brainly.com/question/16552394

#SPJ1

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Match the following with their correct molecular weight. 2-butanone Propyl acetate 4-methyl-2-pentanone Butyl acetate Methanol E
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Answer:

2-butanone = 72.11 g/mol (option F)

Propyl acetate  = 102.13 g/mol (option C)

4-methyl-2-pentanone = 100.16 g/mol (option D)

Butyl acetate = 116.16 g/mol (option B)

Methanol = 32.04 g/mol (option E)

Ethanol  = 46.07 g/mol (option A)

Explanation:

Step 1: Data given

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Step 2:

2-butanone = C4H8O

⇒ 4*12.01 + 8*1.01 + 16.00 = 72.11 g/mol (option F)

Propyl acetate = C5H10O2

⇒ 5*12.01 + 10*1.01 + 2*16.00 = 102.13 g/mol (option C)

4-methyl-2-pentanone = C6H12O

⇒ 6*12.01 + 12*1.01 + 16.00 = 100.16 g/mol (option D)

Butyl acetate = C6H12O2

⇒ 6*12.01 + 12*1.01 + 2*16.00 = 116.16 g/mol (option B)

Methanol = CH3OH = CH4O

⇒ 12.01 + 4*1.01 + 16.00 = 32.04 g/mol (option E)

Ethanol = C2H5OH = C2H6O

⇒ 2*12.01 + 6*1.01 + 16.00 = 46.07 g/mol (option A)

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8 0
3 years ago
How many grams of K2O will be produced from 0.50 g of K<br> and 0.10 g of O2?
Rudik [331]

Answer:

0.6g

Explanation:

Given parameters:

Mass of K = 0.5g

Mass of O₂  = 0.10g

Unknown:

Mass of K₂O  = ?

Solution:

To solve this problem, let us write the reaction equation first;

                   4K   +   O₂     →     2K₂O

The reaction above delineates the balanced chemical reaction.

To solve this problem, we need to know the limiting reactant. This reactant is the one that determines the amount and extent of the reaction because it is given in short supply. The other reactant is the one in excess.

Start off by find the number of moles of the reactant;

     Number of moles =  \frac{mass}{molar mass}

         Molar mas of K  = 39g/mol

          Molar mass of O₂   = 2(16) = 32g/mol

 Number of moles of K  = \frac{0.5}{39}   = 0.013moles

 Number of moles of O₂    = \frac{0.1}{32}   = 0.031moles

From the balanced reaction;

          4 moles of K reacted with 1 mole of O₂

         0.013 moles of K will react with \frac{0.013}{4}   = 0.0078 moles of O₂

We see that oxygen gas is in excess. We were given 0.031moles of the gas but only require 0.0078moles of oxygen gas.

The limiting reactant is potassium.

    therefore;

              4 moles of K produced 2 moles of K₂O

             0.013 moles of K will produce \frac{0.013 x 2}{4}   = 0.0065‬moles of K₂O

to find the mass of K₂O;

   Mass of K₂O  = number of moles x molar mass

                Molar mass of K₂O  = 2(39) + 16  = 94g/mol

  Mass of K₂O = 0.0065 x 94  = 0.6g

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3 years ago
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