Ms.markas takes a thermos bottle filled with hot coffee to work each day. The thermos bottle keeps the coffe hot by

Complete and balance the reaction in acidic solution. equation: ZnS + NO_{3}^{-} -> Zn^{2+} + S + NO ZnS+NO−3⟶Zn2++S+NO Which

kotykmax [81]

Answer:

Balanced reaction: $3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O$

S is oxidized and N is reduced.

$NO_{3}^{-}$ is the oxidizing agent and ZnS is the reducing agent.

Explanation:

Reaction: $ZnS+NO_{3}^{-}\rightarrow Zn^{2+}+S+NO$

$\Rightarrow$ Oxidation: $ZnS\rightarrow Zn^{2+}+S$

Balance charge: $ZnS-2e^{-}\rightarrow Zn^{2+}+S$ ...............(1)

$\Rightarrow$ Reduction: $NO_{3}^{-}\rightarrow NO$

Balance H and O in acidic medium : $NO_{3}^{-}+4H^{+}\rightarrow NO+2H_{2}O$

Balance charge: $NO_{3}^{-}+4H^{+}+3e^{-}\rightarrow NO+2H_{2}O$ ...............(2)

[ $3\times$ Equation-(1)] + [ $2\times$ Equation-(2)]:

$3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O$

Oxidation number of S increases from (-2) to (0) for the conversion of ZnS to S. Therefore S is oxidized.

Oxidation number of N decreases from (+5) to (+2) for the conversion of $NO_{3}^{-}$ to NO. Therefore N is reduced.

$NO_{3}^{-}$ consumes electron from ZnS. Therefore $NO_{3}^{-}$ is the oxidizing agent and ZnS is the reducing agent.

7 0

4 years ago

Elaborate on the role of activation energy in chemical reactions. A) Decreasing the activation energy leads to higher randomness

Zina [86]

B is right answer ithink bro/sis

4 0

4 years ago

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jeffrey is on his skateboard and starts skating down a ramp. ten seconds later he is traveling at 20.0m/s calculate the accelera

Grace [21]

Velocity/time= acceleration

7 0

3 years ago

How many protons,neutrons and electrons are present in an atom of hafnium,hf,with a mass number of 178?

AleksAgata [21]

178 of each species
178 protons
178 neutrons
And 178 electrons

7 0

4 years ago

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A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac

UkoKoshka [18]

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

$pH = pKa + log(\frac{[KF]}{[HF]})$

$pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04$

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq) + F⁻(aq) ⇄ HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

$\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol$

$\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles$

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

$\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles$

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq) (1)

The number of moles of HF after the reaction of KF with HCl is:

$\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles$

And the concentration of HF after the reaction of KF with HCl is is:

$C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L$

Now, from the equilibrium of equation (1) we have:

$Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}$

$Ka = \frac{x^{2}}{0.531 - x}$ (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

$pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73$

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

8 0

4 years ago