Answer:
The concentration of KOH is 0.186 M
Explanation:
First things first, we need too write out the balanced equation between HBr and KOH.
This is given as;
KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)
From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.
We use the acid base formular in calculating unknown concentrations. This is given as;

where;
Ca = Concentration of acid
Va = Volume of acid
Cb = Concentration of base
Vb = Volume of base
na = Number of moles of acid
nb = Number of moles of base
KOH is the base and HBr is acid.
Hence;
Ca = 0.225
Va = 35
Cb = ?
Vb = 42.3
na = 1
nb = 1
Making Cb subject of formular we have;

Cb = (0.225 * 35 * 1) / (42.3 * 1)
Cb = 0.186 M
Answer:
48 volts
Explanation:
Voltage (E) = Current (I) x Resistance (R), or E = IR.
The much of the sample that would remain unchanged after 140 seconds is 2.813 g
Explanation
Half life is time taken for the quantity to reduce to half its original value.
if the half life for Scandium is 35 sec, then the number of half life in 140 seconds
=140 sec/ 35 s = 4 half life
Therefore 45 g after first half life = 45 x1/2 =22.5 g
22.5 g after second half life = 22.5 x 1/2 =11.25 g
11.25 g after third half life = 11.25 x 1/2 = 5.625 g
5.625 after fourth half life = 5.625 x 1/2 = 2.813
therefore 2.813 g of Scandium 47 remains unchanged.