A chemist compresses a 1.8 L sample of carbon dioxide gas at 2.4 atm to a final volume of 1.5 L. Assuming the temperature is constant, what is the final pressure, in arm, exerted by the gas

THE PRESSURE EXERTED BY THE GAS IS THEREFORE 2.88 atm.

Explanation:

Boyle's law states that at constant temperature, the volume of a given mass of gas is inversely proportional to the pressure of the gas.

Mathematically, P1 V1 = P2 V2

Write out the values of the variables given:

P1 = 2.4 atm

V1 = 1.8 L

V2 = 1.5 L

P2 = unknown

Re-arranging the variables by making P2 the subject of the equation, we have:

P2 = P1 V1 / V2

P2 = 2.4 * 1.8 / 1.5

P2 = 2.9=88 atm

Hence, the pressure exerted by the gas is therefore 2.88 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

20 drops hope this