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noname [10]
3 years ago
7

Open the Balancing Chemical Equations interactive and select Introduction mode. Then choose Separate Water. Adjust the coefficie

nts until the equation is balanced.
What is the balanced chemical equation for the decomposition of water?
Open the Balancing Chemical Equations interactive and select Introduction mode. Then choose Make Ammonia. Adjust the coefficients until the equation is balanced.
What is the balanced chemical equation for the formation of ammonia?
Chemistry
1 answer:
aleksklad [387]3 years ago
5 0
<h2>H_2O  + H_2 + O_2</h2>

Explanation:

1. Water decomposition

  • Decomposition reactions are represented by-

       The general equation: AB → A + B.

  • Various methods used in the decomposition of water are -
  1. Electrolysis
  2. Photoelectrochemical water splitting
  3. Thermal decomposition of water
  4. Photocatalytic water splitting
  • Water decomposition is the chemical reaction in which water is broken down giving oxygen and hydrogen.
  • The chemical equation will be -

        H_2O  + H_2 + O_2

Hence, balancing the equation we need to add a coefficient of 2 in front of H_2O on right-hand-side of the equation and  2 in front of H_2 on left-hand-side of the equation.

     ∴The balanced equation is -

       2 H_2O → 2 H_2 + O_2

2. Formation of ammonia

  • The formation of ammonia is by reacting nitrogen gas and hydrogen gas.

      N_2 + H → NH_3

Hence, for balancing equation we need to add a coefficient of 3 in front of hydrogen and 2 in front of ammonia.

   ∴The balanced chemical equation for the formation of ammonia gas is as  follows -

     N_2+3H→ 2NH_3.

  • When 6 moles of N_2 react with 6 moles ofH_2 4 moles of ammonia are produced.

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Trimethylamine, (CH3)3N is a weak base (Kb = 6.3 × 10–5). What volume of this gas, measured at STP, must be dissolved in 2.5 Lof
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Answer:

8.9L is the volume of the gas that must be dissolved.

Explanation:

For a weak base, we can find [(CH₃)₃N] using the equation:

Kb = [OH⁻] [[(CH₃)₃NH⁺] / [(CH₃)₃N]

As [OH⁻] = [[(CH₃)₃NH⁺] and [OH⁻] = 10^-pOH = 3.16x10⁻³M:

6.3x10⁻⁵ = [3.16x10⁻³M][3.16x10⁻³M] / [(CH₃)₃N]

[(CH₃)₃N] = 0.1587M

As the volume is 2.5L, moles are:

2.5L * (0.1587mol / L) = 0.3968moles

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PV = nRT

We can solve for volume of the gas as follows:

P = 1atm at STP; n = 0.3968moles; R = 0.082atmL/molK; T = 273.15K at STP

V = 0.3968mol*0.082atmL/molK*273.15K/1atm

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