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ale4655 [162]
3 years ago
9

An ideal gas is allowed to expand from 5.40 L to 35.1 L at constant temperature. By what factor does the volume increase?

Chemistry
1 answer:
horrorfan [7]3 years ago
6 0

Answer:

a) 6.5

b) pressure decreases

c)22 atm

Explanation:

Since the initial volume V1 is 5.40 L and the final volume V2 is 35.1 L, the ratio of V2:V1= 35.1/5.40= 6.5 hence the volume increases by a factor of 6.5.

When the volume increases, the pressure decreases accordingly in accordance with Boyle's law. Boyle's law states that the volume of a given mass of gas is inversely proportional to its pressure at constant temperature.

c) From Boyle's law

Initial volume V1= 5.40 L

Final volume V2= 35.1 L

Initial pressure P1= 143 ATM

Final pressure P2 = the unknown

P1V1= P2V2

P2= P1V1/V2

P2 = 143 × 5.40/ 35.1

P2= 22 atm

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If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at a constant temperature. What is the new volume​
wel

Answer :

=748

⋅22.5

725

=23.2L

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4 years ago
a 100g sample of material requires 5104.5 joules of heat energy to raise its temperature from 20 degrees Celsius to 80 degrees C
levacccp [35]

Answer: 0.85075J/g.K

Explanation:

Mass of the material = 100g

Energy (Q) = 5104.5J

T1 = 20°C = 293K

T2 = 80°C = 353K

Formula for heat energy (Q) = mc ∇T

Q = mc∇T

∇T = T2 - T1

∇T = 353K - 293K = 60K

Q = mc∇T

C = Q / m∇T

C = (5104.5) / (100 * 60)

C = 0.85075 J/gK

The specific heat capacity of the material is 0.85075J/gK.

5 0
4 years ago
Read 2 more answers
If 9.2 g of calcium react completely with excess aluminum chloride how many grams of aluminum would be produced?
Gekata [30.6K]

Explanation:

3Ca(s) + 2AlCl3(aq) -> 3CaCl2(aq) + 2Al(s)

According to the question, Ca is the limiting reactant.

Therefore, we equate Ca to Aluminium which is the product whose mass we want to find

Molar mass of Ca- 40g/mol

". ". of Al- 27g/mol

3Ca --> 2Al

3×40 --> 2×27

9.2 --> x

x = 9.2×2×27= 496.8÷120=4.14

8 0
3 years ago
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4 0
4 years ago
Use the following balanced reaction to solve:
Naily [24]

Answer:  60.7 g of PH_3 will be formed.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}    

\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles

The balanced chemical reaction is

P_4(s)+6H_2(g)\rightarrow 4PH_3(g)

H_2 is the limiting reagent as it limits the formation of product and P_4 is the excess reagent.

According to stoichiometry :

6 moles of H_2 produce = 4 moles of PH_3

Thus 2.68 moles of H_2 will produce=\frac{4}{6}\times 2.68=1.79moles  of PH_3

Mass of PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g

Thus 60.7 g of PH_3 will be formed by reactiong 60 L of hydrogen gas with an excess of P_4

3 0
3 years ago
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