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erastovalidia [21]
3 years ago
9

A 5.0L container is filled with H2 gas at STP. Predict what happens with the conditions of the gas change as described below (in

crease, decrease or stays the same).
What happens to average kinetic energy as temperature increases?
What happens to the average velocity if H2 is replaced with N2?
What happens to the average velocity as temperature increases?
What happens to average kinetic energy if H2 is replaced with N2?
Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
3 0

Answer:

(a)  As temperature increases, the average kinetic energy increases

(b) If H₂ is replaced with N₂, there will be increase in average velocity

(c) An increase in temperature will cause increase in average velocity

(d) The increase in entropy as a result of an increase in temperature and average velocity will cause the kinetic energy to increase.

Explanation:

Given;

5.0L container filled with H₂ at stp

Part (a) What happens to average kinetic energy as temperature increases

As temperature increases, the average kinetic energy increases

Part(b) What happens to the average velocity if H₂ is replaced with N₂

Entropy is directly proportional to average velocity of gases and entropy increases with an increase in molecular mass.The molecular mass of nitrogen gas N₂ is greater than molecular mass of hydrogen gas H₂.

Thus, if H₂ is replaced with N₂, there will be increase in average velocity.

Part(c) What happens to the average velocity as temperature increases

Kinetic energy increases with an increase in temperature and average velocity.

An increase in temperature will cause increase in average velocity

Part (d) What happens to average kinetic energy if H₂ is replaced with N₂

The increase in entropy as a result of an increase in average velocity will cause the kinetic energy to increase.

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a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

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b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

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= 0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}

= 0.00166 mole of H₃PO₄

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Molar Concentration =  \frac{mole \ \ of \ soulte }{ Volume \ of \ solution }

Molar Concentration = \frac{0.00166 \ mole \ of \  H_3PO_4 }{10}*1000

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

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