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:
Molecular geometry Vsepr
According to VSEPR, the valence electron pairs surrounding an atom mutually repel each other; they adopt an arrangement that minimizes this repulsion, thus determining the molecular geometry. This means that the bonding (and non-bonding) electrons will repel each other as far away as geometrically possible.
Explanation:
Its chemical formula H2O, indicates that each of its molecules contains one oxygen and two hydrogen atoms, connected by covalent bonds. The hydrogen atoms are attached to the oxygen atom at an angle of 104.45°. "Water" is the name of the liquid state of H2O at standard conditions for temperature and pressure.
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
<span>Balancing is making sure there are the same number of atoms on either side of the reaction.
Pb(NO3)2 + Li2SO4--> PbSO4 + LiNO3
There are 2 NO3 groups and 2 Li on the right side, need 2 on the left side.
Need a coefficient of 2 for LiNO3
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