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:
The products are: KCl03 and H20.
Explanation:
The reaction between HC03 (chloric acid) and KOH (potassium hydroxide) is:
HC03 + KOH ----> KCl03 (KCl03 and H20) + H20 (water)
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<em>The reaction is of the double displacement type (in this case parts of the reagents are exchanged, producing two generating new compounds).</em>
River Banks? I'm not completely sure but hope this helps!
Answer : The concentration of A after 80 min is, 0.100 M
Explanation :
Half-life = 20 min
First we have to calculate the rate constant, we use the formula :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = 
t = time passed by the sample = 80 min
a = initial amount of the reactant = 1.6 M
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
Therefore, the concentration of A after 80 min is, 0.100 M
