the volume of a gas is 550 ml at 960 mm hg and 200.0°C. What volume would the pressure of the gas be 830 mm Hg if the temperature is reduced to 150°C?

V₂ = 568.9 mL

Explanation:

Given data:

Initial volume of gas = 550 mL

Initial pressure of a gas = 960 mmHg

Initial temperature = 200.0°C

Final volume = ?

Final pressure = 830 mmHg

Final temperature = 150°C

Solution:

Initial temperature = 200.0°C ( 200+273 = 473 K)

Final temperature = 150°C (150+273 = 423 K)

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂  

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 960 mmHg × 550 mL × 423 K / 473 K ×830 mmHg

V₂ = 223344000 mL / 392590

V₂ = 568.9 mL

the correct answer is answer a. a mixture

no other options make sense

hope this

have a great day : )

  - sprinkles

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer