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Georgia [21]
3 years ago
8

A gas sample occupies 4.2 L at a pressure of 101 kPa.

Chemistry
2 answers:
Bond [772]3 years ago
8 0

Answer:

1.8 L

Explanation:

Given data:

Initial volume of gas = 4.2 L

Initial pressure of gas = 101 kpa

Final volume of gas = ?

Final pressure of gas = 235 kpa

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

101 kpa × 4.2 L = 235 kpa × V₂

V₂ = 424.2 kpa . L/ 235 kpa

V₂ = 1.8 L

So when pressure is increased the volume of gas becomes 1,8 L.

tekilochka [14]3 years ago
3 0

Answer:

1.8l

Explanation: is the sum

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3 years ago
Two systems with heat capacities 19.9 J mol-1 K-1 and 28.2 ] mol 1 K-1 respectively interact thermally and come to an equilibriu
MAVERICK [17]

Answer : The initial temperature of system 2 is, 19.415^oC

Explanation :

In this problem we assumed that the total energy of the combined systems remains constant.

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The mass remains same.

where,

C_1 = heat capacity of system 1 = 19.9 J/mole.K

C_2 = heat capacity of system 2 = 28.2 J/mole.K

T_f = final temperature of system = 30^oC=273+30=303K

T_1 = initial temperature of system 1 = 45^oC=273+45=318K

T_2 = initial temperature of system 2 = ?

Now put all the given values in the above formula, we get

-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K

T_2=292.415K

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8 0
4 years ago
What is the total amount of heat absorbed by 100.0
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Answer:

Option (B) 6270J

Explanation:

The following were obtained from the question:

M = 100g

T1 = 30°C

T2 = 45°C

ΔT = 45 —30 = 15°C

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Q = 6270J

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If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti
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Answer:

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Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

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The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

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Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

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Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

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The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

6 0
3 years ago
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