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:
1: due to difference in reactivity and melting point
2 during the reaction both combine with oxygen in this case magnesium remain solid but some potassium vaporise
86 elections so 43 orbitals
The balanced chemical equation is:
2H2 + O2 ---> 2H2O
We are given the amount of the product produced from the reaction. This will be the starting point for the calculations.
355 g H2O ( 1 mol H2O/ 18.02 g H2O) ( 1 mol O2 / 2 mol H2O ) ( 32 g O2 / 1 mol O2 ) = 315.205 g O2