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LenaWriter [7]
2 years ago
12

A 2 L bottle containing only air is sealed at a temperature of 22°C and a pressure of 0.982 atm. The bottle is placed in a freez

er and allowed to cool to -3°C. What is the pressure in the bottle if the volume changes to 1.8 L?
Chemistry
1 answer:
Degger [83]2 years ago
5 0

Answer:

.997 atm

Explanation:

1. Find the combined gas law formula...

(P1V1/T1 = P2V2/T2)

2. Find our numbers...

P1= .982 atm

P2= ? (trying to find)

V1= 2 L

V2= 1.8 L

T1= 22 C = 295 K

T2= -3 C = 270 K

- Note: always use Kelvin. To find Kelving add 273 to ___C.

3. Rearrange formula to fit problem...

(P2=P1V1T2/V2T1)

4. Fill in our values...

P2= .982 atm x 2 L x 270 K / 1.8 L x 295 K

5. Do the math and your answer should be...

.997 atm

- If you need more help or still do not understand please let me know and I would be glad to help!

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Give the atomic symbol for an element that does not exist as a molecule or extended structure. that is, an element that exists o
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An element that exist as a discreet atom has only one atom which can stand alone on its own. Example of this is Neon, which is a noble gas. The chemical symbol for neon is Ne.
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Each student in a class placed a 2.00 g sample of a mixture of Cu and Al in a beaker and placed the beaker in a fume hood. The s
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Answer:

Percentage mass of copper in the sample = 32%

Explanation:

Equation of the reaction producing Cu(NO₃) is given below:

Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)

From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.

Molar mass of copper = 64 g/mol

mass of copper = number of moles * molar mass

mass of copper = 0.01 mol * 64 g/mol = 0.64 g

Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%

Percentage mass of copper in the sample = 32%

3 0
2 years ago
one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40
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<u>Answer:</u> The volume of chlorine gas produced in the reaction is 2.06 L.

<u>Explanation:</u>

  • <u>For potassium permanganate:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol

  • <u>For hydrochloric acid:</u>

To calculate the moles of hydrochloric acid, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of HCl = 6.00 M

Volume of HCl = 45.0 mL = 0.045 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol

  • For the reaction of potassium permanganate and hydrochloric acid, the equation follows:

2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with = \frac{2}{16}\times 0.27=0.033moles of potassium permanganate.

As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with = \frac{5}{16}\times 0.27=0.0843moles of chlorine gas.

  • To calculate the volume of gas, we use the equation given by ideal gas equation:

PV=nRT

where,

P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant = 0.0820\text{ L atm }mol^{-1}K^{-1}

T = temperature of the gas = 40^oC=[40+273]K=313K

Putting values in above equation, we get:

1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.

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For the reaction: MgF2(s) ⇌ Mg2+(aq) + 2F- (aq), Ksp= 6.4 × 10-9, the addition of 0.10 M NaF to the solution cause what effect o
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Answer:

Shifts the equilibrium to the left. reduces solubility.

Explanation:

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          S                   S              2S

∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²

⇒ 4S² * S = 6.4 E-9

⇒ 4S³ = 6.4 E-9

⇒ S³ = 1.6 E-9

⇒ S = 1.1696 E-3 M

  • NaF(s) → Na+(aq)  +  F-(aq)

        0.10M     0.10M        0.10M

  • MgF2(s) ↔ Mg2+(aq)  + 2F-(aq)

          S'                 S'              2S' + 0.10

⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²

If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:

⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'

⇒ S' = 6.4 E-7 M

∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption

We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility  is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).

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2 years ago
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