Answer:
They expressed it as rate of change in concentration of reactants or products in a chemical reaction
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
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2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
2. <span>1/8 = (1/2)³ </span>
<span>so it's 3 half lives.</span>
Answer: Dissolve approximately 3 mg of Sodium Chloride in 2 mL of water.
Add 0.4 mL of silver nitrate TS and acidify with diluted nitric acid.
Answer:
The number of atoms contained by one molecule of Iron (II) Sulfate are 6.
Explanation:
Iron (II) Sulfate is mage up of two parts. One is the Positive part which constitutes of Fe⁺² and a negative part which constitutes of a polyatomic anion i.e. SO₄²⁻. As there are four Oxygen and one sulfur atom in sulfate Ion so sulfate ion contains 5 atoms in total. Therefore, five atoms from sulfate iona dn one atom of Iron ion makes a total of 6 atoms in one molecule of Iron (II) Sulfate.
