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zalisa [80]
3 years ago
13

..

Chemistry
2 answers:
My name is Ann [436]3 years ago
7 0

Answer:

b is the right answer of the following terms

FrozenT [24]3 years ago
5 0

Answer:

The correct option is b

Explanation:

During chemical equilibrium, when ΔH is positive, this means the forward reaction is an endothermic reaction but when ΔH is negative, this means the forward reaction is an exothermic reaction. An increase in temperature have different effects on these types of reaction.

An increase in temperature when ΔH is negative causes the equilibrium to shift backward causing more reactants to be formed and when ΔH is positive, an increase in temperature causes more products to be formed.  

The question shows the ΔH of the reaction is -188 KJ hence the forward reaction is an exothermic reaction and an increase in temperature (High temperature) will favour the formation of more reactants and not favour the formation of the product SO₃.

NOTE: An increase in pressure (since the reactants are both gases), removal of product as it is formed (decreasing SO₃) and increasing the concentration of the reactant(s) (increasing SO₂) all favour the forward reaction hence will lead to the formation of more product (SO₃).

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Answer:

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Explanation:

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Consider the two electron arrangements for neutral atoms A and B. What is the atomic number of A?
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For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
Aneli [31]
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
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3 years ago
A chemist determines by measurements that 0.030 moles of nitrogen gas participate in a chemical reaction. calculate the mass of
zzz [600]
Remember that: 
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Second we calculate the mass of the precipitate:
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