Can someone answer this:

At 25 °C and 100 kPa a gas occupies a volume of 20 dm. Calculate the new
temperature of the gas if

a) the volume is decreased to 10 dm at constant pressure.

b) the pressure is decreased to 50 kPa at constant volume.

Changes specified for both cases => T(final) = 149K

Explanation:

 (Given)                           (a)                        (b)            

P₁ = 100KPa          P₂ = 100 KPa      P₃ =  50 KPa

V₁ =  20 dm³         V₂ =  10dm³       V₃ =  20dm³

T₁ = 25°C(298K)    T₂ =     (?)a         T₃ =     (?)b

a) Pressure is constant; V ∝ T => Charles Law problem

V₁/T₁ = V₂/T₂ => T₂ = T₁(V₂/V₁) = 298k(10dm³/20dm³) = 149K

b) Volume is constant; T ∝ P => Gay-Lussac Law problem

P₁/T₁ =  P₃/T₃ => T₂ = T₁(P₃/P₁) = 298k(50KPa/100KPa) = 149K

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

i wish i could but this one is weird give me abc and i can solve it.