Answer:
12.6.
Explanation:
- We should calculate the no. of millimoles of KOH and HCl:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
- It is clear that the no. of millimoles of KOH is higher than that of HCl:
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
Sodium lends 1 electron.
Phosphorus borrows 3 electrons.
Potassium lends one electron.
Oxygen borrows 2 electrons.
Iodine borrows one electron.
Cesium lends 1 electron.
Bromine borrows 1 electron.
Sulfur borrows 2 electrons.
And magnesium lends 2 electrons.
Answer:
RbI<RbBr<RbCl<RbF
Explanation:
As stated in the question, the latice energy depends on the relative size of the ions. When the action size is constant as in the question, the lattice energy now depends on the relative of the anions. The order of increase in ionic sizes among the halide ions is fluoride<Chloride<Bromide<Iodide. This order of increasing size means that the lattice energy will decrease accordingly as shown in the answer.
