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Anna71 [15]
3 years ago
7

The electrolysis of molten NaCl is an industrial process. What does the electrolysis produce?

Chemistry
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:

it is D

Explanation:

NaOH and chlorine gas

Electrolysis of Aqueous NaCl Since water can be both oxidized and reduced, it competes with the dissolved Na+ and Cl– ions. Rather than producing sodium, hydrogen is produced. ... The net process is the electrolysis of an aqueous solution of NaCl into industrially useful products sodium hydroxide (NaOH) and chlorine gas.

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The normal freezing point of a certain liquid
stepladder [879]

Answer : The molal freezing point depression constant of X is 4.12^oC/m

Explanation :  Given,

Mass of urea (solute) = 5.90 g

Mass of X liquid (solvent) = 450.0 g

Molar mass of urea = 60 g/mole

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = -0.5^oC

\Delta T^o = freezing point of liquid X= 0.4^oC

i = Van't Hoff factor = 1  (for non-electrolyte)

K_f = molal freezing point depression constant of X = ?

m = molality

Now put all the given values in this formula, we get

[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}

k_f=4.12^oC/m

Therefore, the molal freezing point depression constant of X is 4.12^oC/m

4 0
4 years ago
___________ a mixture of iron filings and sulfur will make the iron and sulfur atoms combine chemically
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5 0
3 years ago
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How much heat is required to raise the temperature of 250.0g of mercury 52.0°c if the specific heat of mercury is 0.140 j/(g x °
sasho [114]

Q = m c T

c= 0.140 j/(g x °c)

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Which type of wave has the lowest energy of the following:
Tcecarenko [31]

Answer:

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3 0
3 years ago
Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
3 years ago
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