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olga_2 [115]
3 years ago
10

Balance the following redox equation using the smallest integers possible and select the correct coefficient for the hydrogen su

lfite ion. MnO41-(aq) + HSO31-( aq) + H+( aq) → Mn2 ( aq) + SO42–( aq) + H2O( l)
Chemistry
1 answer:
goldfiish [28.3K]3 years ago
8 0

Answer:

Coefficient of HSO_{3}^{-} is 5

Explanation:

Oxidation: MnO_{4}^{-}(aq)\rightarrow Mn^{2+}(aq)

  • Balance O and H in acidic medium: MnO_{4}^{-}(aq)+8H^{+}(aq)\rightarrow Mn^{2+}(aq)+4H_{2}O(l)
  • Balance charge : MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow Mn^{2+}(aq)+4H_{2}O(l) .............(1)

Reduction: HSO_{3}^{-}\rightarrow SO_{4}^{2-}

  • Balance O and H in acidic medium : HSO_{3}^{-}(aq)+H_{2}O(l)\rightarrow SO_{4}^{2-}(aq)+3H^{+}(aq)
  • Balance charge : HSO_{3}^{-}(aq)+H_{2}O(l)-2e^{-}\rightarrow SO_{4}^{2-}(aq)+3H^{+}(aq) ..............(2)

[2\times eq(1)]+[5\times eq(2)] -

Balanced equation: 2MnO_{4}^{-}(aq)+5HSO_{3}^{-}(aq)+H^{+}(aq)\rightarrow 2Mn^{2+}(aq)+5SO_{4}^{2-}(aq)+3H_{2}O(l)

Coefficient of HSO_{3}^{-} is 5

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A solution is made by dissolving 62.42 g of ammonium sulfate in enough
natulia [17]

Answer:

9.46 M

Explanation:

Molarity is a measure of the molar concentration of a solution. It can be calculated by the following formula

Molarity = number of moles/volume

Using moles = mass/molar mass

Molar Mass of Ammonium sulfate (NH₄)₂SO₄ =

{14 + 1(4)}2 + 32 + 16(4)

= {14 + 4}2 + 32 + 64

= 18(2) + 96

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= 132g/mol

mole = mass/molar mass

mole = 62.42 ÷ 132

mole = 0.473mol

- Molarity = mole ÷ volume

Volume = 50.0 mL = 50/1000 = 0.05 L

Molarity = 0.473 ÷ 0.05

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3 0
3 years ago
2NH3(g) CO2(g) In an experiment carried out at this temperature, a certain amount of NH4OCONH2 is placed in an evacuated rigid c
dalvyx [7]

The question is incomplete, here is the complete question:

At 25°C, Kp = 2.9 × 10⁻³ for the reaction:

NH_4OCONH_2(s)\rightleftharpoons 2NH_3(g)+CO_2(g)

In an experiment carried out at this temperature, a certain amount of NH₄OCONH₂ is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.

<u>Answer:</u> The total pressure in the container at equilibrium is 0.2694 atm

<u>Explanation:</u>

Let the initial concentration of NH_4OCONH_2 be 'x'

The given chemical equation follows:

                    NH_4OCONH_2(s)\rightleftharpoons 2NH_3(g)+CO_2(g)

<u>Initial:</u>                       x

<u>At eqllm:</u>              x-y                         2y              y

The expression of K_p for above equation follows:

K_p=(p_{NH_3})^2\times p_{CO_2}

The partial pressure of pure solids and liquids are taken as 1 in equilibrium constant expression. So, the partial pressure of NH_4OCONH_2 is not seen in the expression.

We are given:

K_p=2.9\times 10^{-3}

Putting values in above expression, we get:

2.9\times 10^{-3}=(2y)^2\times y\\\\y=0.0898

So, the equilibrium partial pressure of ammonia = 2y = (2 × 0.0898) = 0.1796 atm

The equilibrium partial pressure of carbon dioxide = y = 0.0898 atm

Total pressure inside the container at equilibrium = p_{NH_3}+p_{CO_2}=[0.1796+0.0898]=0.2694atm

Hence, the total pressure in the container at equilibrium is 0.2694 atm

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Answer:

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Explanation:

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