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Alex_Xolod [135]
3 years ago
13

Let P and V represent the pressure and volume of the Xe(g) in the container in diagram 3. If a piston is used to reduce the volu

me of the gas to V2 at a constant temperature, what is the new pressure in the container in terms of the original pressure, P ?
Chemistry
1 answer:
Reil [10]3 years ago
7 0

Answer:

2p

Explanation:

To solve this question, we can use Boyle's Law, which states that:

"For a fixed mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume"

Mathematically:

p\propto \frac{1}{V}

where

p is the pressure of the gas

V is its volume

The equation can be rewritten as

p_1 V_1 = p_2 V_2

where in this problem we have:

p_1 = p is the initial pressure of the Xe(g) gas

V_1=V is the initial volume of the Xe(g) gas

V_2=\frac{V}{2}  is the final volume of the Xe(g) gas

Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 V_1}{V_2}=\frac{pV}{V/2}=2p

So, the final pressure is twice the initial pressure.

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A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.
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<u>Answer:</u> The p-function of Zn^{2+} and NO_3^{-} ions are 2.51 and 2.14 respectively.

<u>Explanation:</u>

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

  • Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, M_1=(2\times 0.0031)=0.0062M

  • Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, M_2=(2\times 0.0042)=0.0084M

To calculate the concentration of nitrate ions in the solution, we use the equation:

M=\frac{M_1V_1+M_2V_2}{V_1+V_2}

Putting values in above equation, we get:

M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M

Calculating the p-function of zinc ions and nitrate ions in the solution:

  • <u>For zinc ions:</u>

\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]

\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51

  • <u>For nitrate ions:</u>

\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]

\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14

Hence, the p-function of Zn^{2+} and NO_3^{-} ions are 2.51 and 2.14 respectively.

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What relationship do you see between an electronic configuration of an element to its most common ion formed?
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20.0 g of bromic acid, HBrO3, is reacted with excess HBr.
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After Rounding off The percentage yield is 64%

<h3>What is Percentage Yield ?</h3>

It is the ratio of actual yield to theoretical yield multiplied by 100% .

It is given in the question

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

The reaction is

HBrO₃ (aq) + 5 HBr (aq) → 3 H₂O (l) + 3 Br₂ (aq)

Actual yield = 47.3 grams

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After Rounding off The percentage yield is 64% .

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