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dalvyx [7]
2 years ago
5

A gas is contained in a tick walled balloon. When the pressure changes from 2. 95atm to ___atm, the volume changes from 7. 456L

to 4. 782L and the temperature changes from 379k to 212k
Chemistry
1 answer:
tia_tia [17]2 years ago
7 0

A gas is contained in a tick-walled balloon. When the pressure changes from 2. 95atm to <u>2.57</u> atm. p.

<h3>What is pressure?</h3>

Pressure is a force perpendicular to the unit area on which it is applied.

Given that, P1, the initial pressure is 2.95 atm

The initial volume, V1 = 7.456 L

The final volume, V2 = 4.782 L

The initial temperature, T1 =  379k

The final temperature, T2 = 212k

The equation will be

\dfrac{P_1V_1}{T_1 } =\dfrac{P_2V_2}{T_2}

Putting the values in the equation

\dfrac{2.95 \times 7.456 L }{379k } =\dfrac{P_2\times 4.782 L}{212k}\\\\\\P_2 = \dfrac{2.95 \times 7.456 L \times 212k }{379k \times 4.782 L} = 2.57\;atm.

Thus, the pressure changes from 2. 95atm to <u>2.57</u> atm p.

Learn more about pressure

brainly.com/question/12971272

#SPJ1

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Answer:

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3 years ago
What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o
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Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

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The conversion of mL into L is shown below:

1 mL= 10^{-3} L

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Molarity of a solution is the number of moles of solute present in 1 L of the solution.

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

The formula can be written for the calculation of moles as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Thus,  

Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}

Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles

Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles

The chemical reaction taking place:

AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)

According to reaction stoichiometry:

<u>1 mole</u> of AgNO₃ reacts with <u>1 mole</u> of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}

Molarity\ of\ NaCl= 0.3374 M

<u>Thus, the concentration of NaCl = 0.3374 M</u>

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