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bekas [8.4K]
2 years ago
6

You need to make 10.0 L of 1.2 M KNO3. What molar ( concentration) would the potassium nitrate solution need to be if you were t

o use only 2.5 L of it? *​
Chemistry
1 answer:
solniwko [45]2 years ago
6 0

Answer:

2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.

Explanation:

Generally, moles of solute in solution before dilution must equal moles of solute after dilution.

By definition Molarity = moles solute/volume of solution in Liters

=> moles solute = Molarity x Volume (L)

Apply moles before dilution = moles after dilution ...

=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution

=> (M)(2.5L)before = (1.2M)(10.0L)after

=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate

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Concerning Boyle's Law, if you had a gas at a pressure of 101 kPa and decreased the volume of the container holding the gas to o
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Answer:

P_2=404 kPa

Explanation:

Hello,

In this case, the Boyle's is mathematically defined via:

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V_2=\frac{1}{4} V_1

We can compute the new pressure:

P_2=\frac{P_1V_1}{V_2} =\frac{P_1V_1}{\frac{1}{4} V_1} =\frac{101kPa*V_1}{\frac{1}{4} V_1} \\\\P_2=4*101kPa\\\\\\P_2=404 kPa

Which means the pressure is increased by a factor of four.

Regards.

7 0
2 years ago
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aleksley [76]

Answer:

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3 years ago
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Answer:

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