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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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3 years ago
One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silv
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Answer:

5.0x10⁻⁵ M

Explanation:

It seems the question is incomplete, however this is the data that has been found in a web search:

" One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose a EPA chemist tests a 250 mL sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:

NiCl₂ + 2AgNO₃ →  2AgCl + Ni(NO₃)₂

The chemist adds 50 mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 3.6 mg of silver chloride. Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits. "

Keep in mind that while the process is the same, if the values in your question are different, then your answer will be different as well.

First we <u>calculate the moles of nickel chloride found in the 250 mL sample</u>:

  • 3.6 mg AgCl ÷ 143.32 mg/mmol * \frac{1mmolNiCl_{2}}{2mmolAgCl} = 0.0126 mmol NiCl₂

Now we <u>divide the moles by the volume to calculate the molarity</u>:

  • 0.0126 mmol / 250 mL = 5.0x10⁻⁵M
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