a gas occupies 600.0 ml at 20.00 degrees C and 70.00kPa what will be the pressure at 40.00 degrees C and 150.0mL?

P₂ = 299.11 KPa

Explanation:

Given data:

Initial volume = 600 mL

Initial pressure = 70.00 KPa

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

Final temperature = 40°C (40+273 = 313 K)

Final volume = 150.0 mL

Final pressure = ?

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₂ = P₁V₁ T₂/ T₁ V₂  

P₂ =  70 KPa × 600 mL × 313 K / 293K ×150 mL  

P₂ = 13146000 KPa .mL. K /43950 K.mL

P₂ = 299.11 KPa

ANSWER:

If 2f= -12 then,

F= -12/2

F= -6

Hope it helps!!

Step-by-step explanation:

ANSWER:

If 2f= -12 then,

F= -12/2

F= -6

Hope it helps!!

Step-by-step explanation:

ANSWER:

If 2f= -12 then,

F= -12/2

F= -6

Hope it helps!!

Step-by-step explanation:

Explanation: