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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.

You might be interested in
What is the [OH-] of a substance that has a pH of 11?
Deffense [45]

Answer:

0.001 M OH-

Explanation:

[OH-] = 10^-pOH, so

pOH + pH = 14 and 14 - pH = pOH

14 - 11 = 3

[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-

6 0
3 years ago
suppose that two hydroxides, moh and m′ (oh)2, both have a ksp of 1.39 × 10−12 and that initially both cations are present in a
11Alexandr11 [23.1K]

When base is added in order to increase the pH of a solution, it results in MOH salt <u>precipitating first.</u>

Equation :

MOH (s)  ⇄ M⁺ (aq)  ₊ OH⁻(aq)

Ksp = [M⁺] [OH⁻]

Ksp = 1.39×10⁻¹² and [M⁺] = 0.001M

∴ 1.39×10⁻¹² = (0.001M) [OH⁻]

[OH⁻] =  1.004×10⁻¹⁹M

pOH = -log [OH⁻] =  11.9

pH = 14 - pOH

= 14 - 11.9

<u>pH = 2.1</u>

To learn more about solubility equilibrium ;

https://brainly.in/question/9213521

#SPJ4

8 0
2 years ago
2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme
mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O

Next, we assign the oxidation numbers as follows:

2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

4 0
3 years ago
Please help! Super easy, I just forgot!<br> Fill in the two blanks
Alborosie

Answer:

nucleuas

Explanation:

6 0
3 years ago
Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. the initial temperature of the gas is 27.0c
Alla [95]

We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.

The final temperature is: 75.11 °C.

The work done at constant pressure, W=nR(T₂-T₁)

n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).

W=2.4 × 10³ Joule (Given)

From the expression,

(T₂-T₁)=\frac{W}{nR}

(T₂-T₁)= \frac{2.40 X 10^{3} }{6 X 8.314}

(T₂-T₁)= 48.11

T₂=300+48.11=348.11 K= 75.11 °C

Final temperature is 75.11 °C.


6 0
3 years ago
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