Both hydrogen sulfide (H2S) and ammonia (NH3)
have strong, unpleasant odors. Which gas has the
higher effusion rate? If you opened a container of
each gas in a corner of a large room, which odor
would you detect first on the other side of the
room? Assume the temperature is constant.
Explain your answer.

The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster. Effusion rate is inversely proportional to molar mass. NH3 will have a higher average molecule velocity, so it will diffuse faster and will reach the other side of the room more quickly.

Explanation:

The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster.

Effusion rate is inversely proportional to molar mass.

NH3 will have a higher average molecule velocity, so it will diffuse faster and will reach the other side of the room more quickly.

Explanation:

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