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Sholpan [36]
3 years ago
11

HELPPPP! can we manufacture water? why or why not

Chemistry
1 answer:
Furkat [3]3 years ago
3 0

Answer:

No because water needs to have no salt

Explanation:

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I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

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Explanation:

What is true about an atom in an excited state? Select all that apply:

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B. It is more likely to enter chemical reactions.

C. It has less energy than a ground state atom.

D. It has more energy than a ground state atom.

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Separate the reaction Cl2 + 2Cs = 2CsCl into its component half reactions.
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