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NikAS [45]
2 years ago
8

An experimenter found that for the hydrolysis of lactose, ΔH° = +0.44 kJ mol–1 and ΔS° = +0.031 kJ mol–1 K–1. ΔG°for this reacti

on is:
Chemistry
1 answer:
Ksivusya [100]2 years ago
7 0

Answer:

ΔG° = -8.8 kJ/mol

Explanation:

The standard Gibbs free energy of reaction (ΔG°) can be calculated using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

ΔH°: standard enthalpy of reaction

T: absolute temperature

ΔS°: standard entropy of reaction

At 298 K (the temperature that is usually used), ΔG° for the hydrolysis of lactose is:

ΔG° = ΔH° - T.ΔS°

ΔG° = 0.44 kJ/mol - 298 K × 0.031 kJ/mol.K

ΔG° = -8.8 kJ/mol

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What is a group of electrons orbiting at roughly the same distance from the nucleus
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A shell of electrons.
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A sample of oxygen gas in one container has a volume of 20.0mililiter at 297 K and 101.3 kPa. The entire sample is transferred t
VashaNatasha [74]

Answer: V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=\frac{P_1V_1T_2}{T_1P_2}

where,

P_1 = initial pressure of gas = 101.3 kPa

P_2 = final pressure of gas = 94.6 kPa

V_1 = initial volume of gas = 20.0 ml

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 297K

T_2 = final temperature of gas = 283K

Now put all the given values in the above equation, we get the final volume of gas.

V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

V_2=20.4ml

Thus the correct numerical setup for calculating the new volume is \frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

3 0
3 years ago
How many neutrons does 3919K have?
Igoryamba
19 protons and 19 electrons but neutrons are vary
5 0
3 years ago
Enter ionic and net equations: feso4(aq)+ na3po4(aq) arrow fe3(po4)2(s)+na2so4(aq)
stepan [7]

Answer:

<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

<em> net ionic equation: </em>3Fe(2+)(aq)  + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

Explanation:

The balanced equation is

3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)

<em>Ionic equations: </em>Start with a balanced molecular equation.  Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions . Indicate the correct formula and charge of each ion. Indicate the correct number of each ion . Write (aq) after each ion .Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation

3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

<em>Net ionic equations: </em>Write the balanced molecular equation.  Write the balanced complete ionic equation.  Cross out the spectator ions, it means the repeated ions that are present.  Write the "leftovers" as the net ionic equation.

3Fe(2+)(aq)  + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

6 0
3 years ago
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