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Anastaziya [24]
2 years ago
12

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

uation P V = C , where C is a constant. Suppose that at a certain instant the volume is 500 cm 3 , the pressure is 200 kPa , and the pressure is increasing at a rate of 30 kPa/min . At what rate is the volume decreasing at this instant?
Chemistry
1 answer:
ad-work [718]2 years ago
5 0

Answer:

-75 cm^3/min

Explanation:

Given from Boyle's law;

PV=C

From product rule;

VdP/dt + PdV/dt = dC/dt

but dC/dt = 0, V= 500 cm^3, P= 200kPa, dP= 30kPa/min

PdV/dt = dC/dt - VdP/dt

dV/dt = dC/dt - VdP/dt/ P

substituting values;

dV/dt = 0 - (500 * 30)/200

dV/dt = -75 cm^3/min

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Answer:

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Answer:

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

b) 0.0035 mole

c)  0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

1 mole         3 mole

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

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

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

10 ml            17.50 ml

(x) M              0.200 M

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c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

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

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

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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