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WINSTONCH [101]
3 years ago
15

Classify each of the following as a strong acid or a weak acid and indicate how each should be written in aqueous solution. Clas

sify ... In solution this acid should be written as: weak 1. hydrocyanic acid H3O CN- _______ 2. hydrobromic acid
Chemistry
1 answer:
JulijaS [17]3 years ago
7 0

Answer:

HCN, weak acid

H⁺, Br⁻, strong acid

Explanation:

Hydrocyanic acid is a weak acid, according to the following equation.

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Thus, it should be written in the undissociated form (HCN).

Hydrobromic acid is a strong acid, according to the following equation.

HBr(aq) ⇒ H⁺(aq) + Br⁻(aq)

Thus, it should be written in the ionic form (H⁺, Br⁻).

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7 0
3 years ago
Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an
ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

TiCl_{4(l)} ⇒ TiCl_{4(g)}

ΔH°_{f} (TiCl_{4(l)}) = -804.2 kJ/mol

ΔH°_{f} (TiCl_{4(g)}) = -763.2 kJ/mol

Therefore,

ΔH°_{f} = ΔH°_{f} (TiCl_{4(g)}) - ΔH°_{f} (TiCl_{4(l)}) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(TiCl_{4(l)}) = 221.9 J/(mol*K)

s°(TiCl_{4(g)}) = 354.9 J/(mol*K)

Therefore,

s° = s° (TiCl_{4(g)}) - s°(TiCl_{4(l)}) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH°_{f} /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

5 0
3 years ago
Plz can anyone help me I'm a bit stuck in this one ​
trasher [3.6K]

Answer:

just see if i am not wrong

learning balancing in chemistry it take time

hope i am correct

8 0
3 years ago
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