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almond37 [142]
3 years ago
9

13. What is the total pressure inside the container​

Chemistry
1 answer:
nasty-shy [4]3 years ago
7 0

<u>Answer:</u> The total pressure inside the container is 77.9 kPa

<u>Explanation:</u>

Dalton's law of partial pressure states that the total pressure of the system is equal to the sum of partial pressure of each component present in it.

To calculate the total pressure inside the container, we use the law given by Dalton, which is:

P_T=p_{N_2}+p_{O_2}+p_{Ar}

We are given:

Vapor pressure of oxygen gas, p_{O_2} = 40.9 kPa

Vapor pressure of nitrogen gas, p_{N_2} = 23.3 kPa

Vapor pressure of argon, p_{Ar} = 13.7 kPa

Putting values in above equation, we get:

p_T=23.3+40.9+13.7\\\\p_{T}=77.9kPa

Hence, the total pressure inside the container is 77.9 kPa

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A major textile dye manufacturer developed new yellow dye. The dye has a percent composition of 75.9 5% C, 17.72% N
AnnZ [28]

Answer:

Such molecule must have molecular formula of C15N3H15

Explanation:

Mass of carbon in such molecule

0.7595*240_{g/mol} =182.28_{g C/mol}

The atomic mass of carbon is 12.01 g/mol, so in 182.28 g of carbon there is 15.18 mols of carbon.

Mass of Nitrogen in such molecule

0.1772*240_{g/mol} =37.73_{g C/mol}

The atomic mass of nitrogen is 14.01 g/mol, so in 42.53g of nitrogen there is 3.04 mols of nitrogen.

Mass of Hydrogen in such molecule

0.0633*240_{g/mol} =15.19 {g C/mol}

The atomic mass of Hydrogen is 1.00 g/mol, so in 15.19 g of Hydrogen there is 15.19 mols of Hydrogen.

Such molecule must have molecular formula of C15N3H15

5 0
3 years ago
3.
sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5.  V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a)  ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0
3 years ago
Jkgergzruge˜©∫fghxtgrhxghzghgzhgzhmhrghgmhzh
DedPeter [7]
Sissy imposter among us haha
6 0
3 years ago
Calculate e°cell for a silver-aluminum cell in which the cell reaction is al(s) + 3ag+(aq) → al3+(aq) + 3ag(s) –2.46 v 0.86 v –0
tekilochka [14]
When E° cell is an electrochemical cell which comprises of two half cells.
 
So,

when we have the balanced equation of this half cell :

Al3+(aq) + 3e- → Al(s)   and E°1 = -1.66 V 

and we have  also this balanced equation of this half cell :

Ag+(aq)  + e- → Ag(s)  and E°2 = 0.8 V 

so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)

when E° = E°2 - E°1

∴E° =0.8 - (-1.66)

      = 2.46 V

∴ the correct answer is 2.46 V




6 0
3 years ago
In 1953, who devopled the model that is shown below
PilotLPTM [1.2K]
What model? can you screenshot it or send a link?
7 0
3 years ago
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