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Diano4ka-milaya [45]
3 years ago
5

In the Bronsted-Lowry model of acids and bases, a(n) _____ is a hydrogen donor and a(n) _____ is a hydrogen acceptor.

Chemistry
1 answer:
ExtremeBDS [4]3 years ago
5 0
<span>In the Bronsted-Lowry model of acids and bases, a(n) _acid____ is a hydrogen donor and a(n) _base____ is a hydrogen acceptor.</span>
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John Dalton

Explanation:

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20000 Mg

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The normal boiling point of a substance is defined to be the temperature at which the liquid phase of the substance is in equili
boyakko [2]

Answer:

ΔSv = 0.1075 KJ/mol.K

Explanation:

Binary solution:

∴ a: solvent

∴ b: solute

in equilibrium:

  • μ*(g) = μ(l) = μ* +RTLnXa....chemical potential (μ)

⇒ Ln (1 - Xb) = ΔG/RT

∴ ΔG = ΔHv - TΔSv

⇒ Ln(1 -Xb) = ΔHv/RT - ΔSv/R

∴ Xb → 0:

⇒ Ln(1) = ΔHv/RT - ΔSv/R

∴ T = T*b....normal boiling point

⇒ 0 = ΔHv/RT*b - ΔSv/R

⇒ ΔSv = (R)(ΔHv/RT*b)

⇒ ΔSv = ΔHv/T*b

∴ T*b = 80°C ≅ 353 K

⇒ ΔSv = (38 KJ/mol)/(353 K)

⇒ ΔSv = 0.1075 KJ/mol.K

5 0
3 years ago
The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
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