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ira [324]
3 years ago
12

How many grams of water vapor (H2O) are in a 10.2-liter sample at 0.98 atmospheres and 26°C?

Chemistry
1 answer:
diamong [38]3 years ago
5 0
Hey there! 

<span>Use the equation of Clapeyron:
</span>
T in kelvin :

26 + 273.15 => 299.15 K

R = 0.082

V = 10.2 L

P = 0.98 atm

number of moles :

P *V = n * R * T

0.98 * 10.2 =  n * 0.082 * 299.15

9.996 = n * 24.5303

n = 9.996 / 24.5303

n = 0.4074 moles

Therefore:

Molar mass H2O = 18.01 g/mol

1 mole H2O ------------- 18.01 g
0.4074  moles ----------- m

m = 0.4074 * 18.01 / 1

m = 7.339 g of H2O
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What are the two elements sand is made of? *
olga55 [171]

Answer:

tiny crystals of the mineral quartz, which is made out of silica and oxygen

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3 years ago
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At 9°C a gas has a volume of 6.17 L. What is its volume when the gas is at standard temperature?
Alex17521 [72]

Answer:

V₂ = 5.97 L

Explanation:

Given data:

Initial temperature = 9°C (9+273 = 282 K)

Initial volume of gas  = 6.17 L

Final volume of gas = ?

Final temperature = standard = 273 K

Solution:

Formula:

The Charles Law will be apply to solve the given problem.

According to this law, 'the volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure'

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 6.17 L ×  273K /  282  k

V₂ = 1684.41 L.K / 282 K

V₂ = 5.97 L

5 0
3 years ago
What is it called when a gas is converted into a liquid
alex41 [277]
Your answer is probably
Vaporization point
3 0
3 years ago
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If 30.943 g of a liquid occupy a space of 35.0 ml. What is the density of the liquid in g/cm3?
timofeeve [1]
Mass / volume = density
30.943g / 35ml = 0.88408571g/ml 

7 0
3 years ago
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What mass of barium sulfate (233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of
Rzqust [24]

Answer:

4.37 g of barium sulphate

Explanation:

The reaction equation is;

3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)

From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles

To find the limiting reactant;

3 moles of barium chloride yields 3 moles of barium sulphate

0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate

1 mole of iron III sulphate yields 3 moles of barium sulphate

0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate

Hence,barium chloride is the limiting reactant

Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate

4 0
3 years ago
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