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ElenaW [278]
2 years ago
11

Trihydrogen monophosphide is a covalent molecule that can also act as an acid. What is the correct acid name for trihydrogen mon

ophosphide ?
Chemistry
1 answer:
kati45 [8]2 years ago
4 0

Answer:

H₃P phosphidic acid

Explanation:

The Trihydrogen monophosphide, as stated in the exercise, can act as an acid. This is pretty similar to the case of hydrogen chloride, which is a gas but it can also be an acid, in this case, chloridic acid.

In the case of trihydrogen phosphide, we can write it molecular formula which is:

H₃P

Now, this is a binary compound because its composed of only two elements, in this case, hydrogen and phosphide. To name binary acid, we need to name the non metal with the sufije idic, and then, the word acid.

Following this simple rule, the trihydrogen phosphide would be, as acid:

H₃P: phosphidic acid

Hope this helps

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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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