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masya89 [10]
3 years ago
8

3.

Chemistry
1 answer:
sineoko [7]3 years ago
3 0

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

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Calculate the solubility of carbon dioxide in water at an atmospheric pressure of 0.400 atm (a typical value at high altitude).A
Bond [772]

Answer:

The molar solubility of carbon dioxide gas is 4.662\times 10^{-6} M.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{CO_2}=K_H\times p_{liquid}

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K_H = Henry's constant = 3.50\times 10^{-2}mol/L.atm

p_{CO_2} = partial pressure of carbonated drink

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Putting values in above equation, we get:

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8 0
3 years ago
Ill give brainliest :D<br><br> how many grams of kcl are produced from 11.5 grams?
eduard

Answer:

approximately 15.1 grams.

Explanation:

The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want.

change 25 grams of potassium chlorate to moles.

calculate the gram molecular mass of potassium chlorate.

Chlorate is Cl with 3 oxygens. ate = saturated. Chlorine has seven valance electrons when it is saturated six of these electrons are used by oxygen ( 2 electrons per oxygen) leaving only 1 electron.

1 K x 39 grams/mole

+1 Cl x 35.4 grams/ mole

+3 O x 16 grams/ mole

= 122.4 grams / mole Potassium Chlorate

25

122.4

= moles.

2.05 moles of Potassium Chlorate.

There is a 1:1 mole ratio. 1 mole of Potassium Chlorate will produce 1 mole of Potassium Chloride.

2.05 moles of Potassium Chlorate will produce 2.05 moles of Potassium Chloride.

Find the gram molecular mass of Potassium Chloride.

1 K x 39 = 39

+1 Cl x 35.4 = 35.4

= 74.4 grams / mole.

2.05 moles x 74.4 grams/ mole = 15.2 grams

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