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arlik [135]
2 years ago
5

9. Suppose that 25.0 mL of a gas at 725 mm Hg and 20°C is converted to standard

Chemistry
1 answer:
storchak [24]2 years ago
8 0

Answer:

V₂ = 22.23 mL

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 25 mL

Initial pressure = 725 mmHg (725/760 =0.954 atm)

Initial temperature = 20 °C (20 +273 = 293 K)

Final pressure = standard = 1 atm

Final temperature = standard = 273.15 K

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁  P₂

V₂ = 0.954 atm × 25 mL × 273.15 K / 293 K × 1 atm

V₂ =  6514.63 mL . atm . K  / 293 K . atm

V₂ = 22.23 mL

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Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is C2H2(g)+2H2(g)?C2H6(g)
Alecsey [184]

Answer:

Keq =1.50108

Explanation:

The given reactionis

 C₂H₂(g) +2H₂(g) -------------> C₂H₂(g)

   ΔG0 f=ΔG0f n (products) - ΔG0f n (reactants )

              = -32.89 kJ/mol - (209.2 kJ/mol+2*0.0 kJ/mol)

              = - 242.09kJ/mol

   ΔG= -RTlnKeq

   ln Keq = -ΔG/RT

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    Keq =e0.406

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3 years ago
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lisabon 2012 [21]
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mass number = 5 + 7
mass number = 12
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Why doesnt science offer conclusive proof of phenomena
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3 years ago
Please help quick
Rama09 [41]

Answer:

c = 0.898 J/g.°C

Explanation:

1) Given data:

Mass of water = 23.0 g

Initial temperature = 25.4°C

Final temperature = 42.8° C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Specific heat capacity of water is 4.18 J/g°C

ΔT = 42.8°C - 25.4°C

ΔT = 17.4°C

Q = 23.0 g ×  × 4.18 J/g°C × 17.4°C

Q = 1672.84 j

2) Given data:

Mass of metal = 120.7 g

Initial temperature = 90.5°C

Final temperature = 25.7 ° C

Heat released = 7020 J

Specific heat capacity of metal = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 25.7°C - 90.5°C

ΔT = -64.8°C

7020 J = 120.7 g ×  c ×  -64.8°C

7020 J = -7821.36 g.°C ×  c

c = 7020 J / -7821.36 g.°C

c = 0.898 J/g.°C

Negative sign shows heat is released.

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3 years ago
How are flagella and cilia alike?
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Answer: Both store of celluar materials

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