500. ml of gas is contained at stp. the volume changes to 560 ml at 20.°c
what is the pressure (in atm)?

P₂ = 0.96 atm

Explanation:

Given data:

Initial volume = 500 mL

Initial pressure = 1 atm

Initial temperature = 273 K

Final volume = 560 mL

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

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:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁ T₂/ T₁ V₂  

P₂ = 1 atm × 500 mL × 293 K / 273 K × 560 mL  

P₂ = 146500 atm .mL. K /  152880 k.mL

P₂ = 0.96 atm

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

the answer is c i took the test