Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
Answer:
2Ca₃(PO₄)₂ + 10C + 6SiO₂ → 6CaSiO₃ + P₄ + 10CO.
Explanation:
- To balance a chemical reaction, we should apply the law of conservation of mass.
- Law of conversation of mass states that the no. of atoms is equal in both sides of the chemical reaction.
- So, the balanced chemical reaction is:
<em>2Ca₃(PO₄)₂ + 10C + 6SiO₂ → 6CaSiO₃ + P₄ + 10CO.</em>
that 2 mol of Ca₃(PO₄)₂ react with 10 mol of C and 6 mol of SiO₂ to produce 6 mol of CaSiO₃, 1 mol of P₄ and 10 mol of CO.
We are provided with the amount of energy released when one mole of carbon reacts. We mus first convert the given mass of carbon to moles and then compute the energy released for the given amount.
Moles = mass / atomic mass
Moles = 23.5 / 12
Moles = 1.96 moles
One mole releases 394 kJ/mol
1.96 moles will release:
394*1.96
= 772.24
The enthalpy change of the reaction will be -772.24 kJ
