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
4.The other light bulb will stay on and glow brightly.
D all are 1 because the left side equals the right
Answer:
B
Explanation:
First of all it is important to know that a half filled orbital is particularly stable. In phosphorus all the electrons occur singly in the 3p sublevel minimizing inter electronic repulsion hence it is more difficult to remove an electron from this energetically stable arrangement. In sulphur, electrons are paired in one of the 3p orbitals thereby lowering the energy of that level due to instability caused by interelectronic repulsion between two electrons in the same orbital.