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Akimi4 [234]
2 years ago
15

1. Complete the balanced dissociation equation for the compound below. If the compound does not dissociate, write NR after the r

eaction arrow.
(NH4)2S(s) -> ______


2. Write the balanced NET ionic equation for the reaction when Al(NO₃)₃ and Na₃PO₄ are mixed in aqueous solution. If no reaction occurs, write only NR.
Chemistry
1 answer:
puteri [66]2 years ago
7 0

Answer:

1. (NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)

2. Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)

Explanation:

The dissociation of ammonium sulphide, (NH₄)₂S when dissolved in water is given in the equation below:

(NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)

However very little S²- ions are present in solution due to the very basic nature of the S²- ion (Kb = 1 x 105).

The ammonium ion being a better proton donor than water, donates a proton to sulphide ion to form hydrosulphide ion which exists in equilibrium with aqueous ammonia.

S²- (aq) + NH₄+ (aq) ⇌ SH- (aq) + NH₃ (aq)

Aqueous solutions of ammonium sulfide are smelly due to the release of hydrogen sulfide and ammonia, hence, their use in making stink bombs.

2. The reaction between aluminium nitrate and sodium phosphatein aqueous solution is a double decomposition reaction whish results in the precipitation of insoluble aluminium phosphate. The equation of the reaction is given below :

Al(NO₃)₃ (aq) + Na₃PO₄ (aq) ----> AlPO₄ (s) + 3 NaNO₃ (aq)

The net ionic equation is given below:

Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)

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5. What is the final "Celsius" temperature if 2.40 L of gas at 30.5 C is cooled until the volume reaches 1.00 L at constant pres
slega [8]

Answer:

Final temperature of the gas = -146.63 °C

Explanation:

At constant pressure, volume and temperature of the gases are related as:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

Where,

V1 = Initial volume = 1.00 L

V2 = Final volume = 2.40 L

T1 = Initial temperature = 30.5 °C = 30.5 + 273.15 = 303.65 K

Now, substitute the values in the above equation,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

\frac{2.40\;L}{303.65}=\frac{1.00\;L}{T_2}

T_2=\frac{1.00\times 303.65}{2.40}

T2 = 126.52 K

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

T( °C) = T(K) - 273.15

          = 126.52 - 273.15 = -146.63 °C

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