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DiKsa [7]
4 years ago
8

Classify each of these compounds as a Brønsted-Lowry acid, a Brønsted-Lowry base, or neither.

Chemistry
1 answer:
chubhunter [2.5K]4 years ago
8 0

Answer:

Brønsted-Lowry acid : H2SO4, HF, HNO2

Brønsted-Lowry Base : NH3, C3H7NH2, CH3)3N

Neither : NaBr, CCl4

Explanation:

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When a solution containing 1.4000 g of Ba(NO3)2 and 2.4000 g of HSO3NH2 is boiled, a precipitate forms. One possible identity fo
Marta_Voda [28]

Answer:

See explanation for detailed solution

Explanation:

The balanced reaction equation is Ba(NO3)2 + 2HSO3NH2 → Ba(SO3NH2)2 + 2HNO3

Number of moles of Ba(NO3)2 = 1.4 g/ 261.337 g/mol = 5.36 × 10^-3 moles

From the reaction equation;

1 mole of Ba(NO3)2 yields 1 mole of Ba(SO3NH2)2

5.36 × 10^-3 moles of Ba(NO3)2 yields 5.36 × 10^-3 moles of Ba(SO3NH2)2

For HSO3NH2

Number of moles = 2.4g/97.10 g/mol =0.0247 moles

2 moles of HSO3NH2 yields 1 mole of Ba(SO3NH2)2

0.0247 moles of HSO3NH2 yields 0.0247 ×1/2 = 0.0137 moles

Hence, Ba(NO3)2 is the limiting reactant

The theoretical yield of Ba(SO3NH2)2 is 5.36 × 10^-3 moles × 329.4986 g/mol = 1.766 g

b)

Number of moles = mass/ molar mass

Molar mass = mass/ number of moles

Molar mass = 1.6925 g/5.36 × 10^-3 moles = 315.76 g

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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

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