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Anni [7]
3 years ago
12

2zns(s)+3o2(g)—>2zno(s)+2so2(g)

Chemistry
1 answer:
Paha777 [63]3 years ago
8 0

This is an example of displacement reaction

<u>Explanation:</u>

  • The chemical reaction in which the one element replaces the other element in a compound is called a displacement reaction. This reaction is also called a replacement reaction.

                                AB + C -----> AC +B

                        2ZnS + 3O2 -----> 2ZnO + 2SO2

  • This happens when A is more reactive than B and gives a stable product. Here the zinc sulfide compound reacts with the oxygen element to the stable product of zinc oxide and sulfur dioxide.
  • In short, the more reactive element displaces the less reactive element is called a displacement reaction.
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egoroff_w [7]

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3 years ago
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Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

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