You take a sample of water that is at room temperature and in contact with air and put it under a vacuum. right away, you see bubbles leave the water, but after a little while, the bubbles stop. as you keep applying the vacuum , more bubbles appear. a friend tells you that the first bubbles were water vapor, and the low pressure had reduced the boiling point of water, causing the water to boil. another friend tells you that the first bubbles were gas molecules from the air (oxygen, nitrogen , and so forth) that were dissolved in the water. which friend is mostly likely to be correct? what, then, is responsible for the second batch of bubbles?

The second friend is mostly correct.

Explanation:

When pressure decreases, gas solubility decreases. The bubbles are gas molecules that were dissolved in the water at atmospheric pressure. The bubbles started coming out as a result of the decrease pressure. The second bubbles would be water vapor because as the pressure continue to decrease, the amount of vapor pressure  ( vapor leaving the surface of the liquid ) equals the external pressure and water begin to boil. So the second bubbles must be water vapor.

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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