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asambeis [7]
2 years ago
11

How many grams of KNO3 are required to prepare 0.250 L of 0.70 M solution?​

Chemistry
1 answer:
drek231 [11]2 years ago
4 0

Answer:

about 3

Explanation:

just did this for the points just being honiest

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Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f
pishuonlain [190]

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>

<em />

<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>

T₂ = 0.5 T₁. According to Gay-Lussac's law,

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}

P₂ in terms of the ideal gas equation is:

P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}

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