3. A 50.0 L sample of gas collected in the upper atmosphere at a pressure of 18.3 mmHg is compressed into a 150.0 mL container at the same temperature.

a. What is the new pressure, in atm?

b. To what volume would the original sample have had to be compressed to exert a pressure of 10.0 atm?

a.

b.

Explanation:

Hello,

a. In this case, we use the Boyle's law as en inversely proportional relationship between pressure and volume:

Thus, for the given conditions, one computes the new pressure as shown below:

b. Now, we should find the final volume for a new pressure of 10 atm:

Best regards.

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

it would be d. henry moseley

hope this !

and sorry if i'm late. have a good rest of your day! : d