No of moles of MgCl2 = weight of MgCl2 / Molecular weight of MgCl2
Weight of MgCl2 =moles of MgCl2 x molecular mass of MgCl2
= 8.90 x 95=845.5 gm
Explanation:
The net equation will be as follows.
So, we are required to find 
for this reaction.
Therefore, steps involved for the above process are as follows.
Step 1: Convert K from solid state to gaseous state
, 
= 89 kJ
Step 2: Ionization of gaseous K
, 
= 418 KJ
Step 3: Dissociation of 
gas into chlorine atom
.
, 
= 122 KJ
Step 4: Iozination of chlorine atom.
, 
= -349 KJ
Step 5: Add 
ion and 
ion formed above to get KCl
.
, 
= -717 KJ
Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.
= 89 + 418 + 122 - 349 - 717
= - 437 KJ/mol
Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.
Answer:
SO₃²⁻ is the reducing agent and Cr₂O₇²⁻ is the oxidizing agent.
Explanation:
Oxidation reaction:
3SO₃²⁻ (aq) + 3H₂O (l) → 3SO₄²⁻ (aq) + 6H⁺ (aq) + 6e⁻
Reduction reaction:
Cr₂O₇²⁻ (aq) + 14H⁺ (aq) + 6e⁻ → 2Cr ³⁺ (aq) + 7H₂O (l)
Now, adding the oxidation and the reduction reactions we get the full net reaction:
Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺ (aq) → 2Cr ³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)
Since, the S in SO₃²⁻, present in the +4 oxidation state is oxidized to +6 oxidation state in SO₄²⁻, by the loss of 2e⁻.
<u>Therefore, SO₃²⁻ is the reducing agent. </u>
And, the Cr in Cr₂O₇²⁻, present in the +6 oxidation state is getting reduced to +3 oxidation state, Cr ³⁺, by the gain of 6e⁻.
<u>Therefore, Cr₂O₇²⁻ is the oxidizing agent.</u>
The answer is a day, a year is the earth's revolution around the sun
Answer:
The ion will repel the substance because it has more protons than electrons.
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