I think the answer would be trenches but I’m sorry if I’m wrong
A cold front is the leading edge of a cooler mass of air, replacing at ground level a warmer mass of air, which lies within a fairly sharp surface trough of low pressure.
Answer:
The ionization of 0.250 moles of H₂SO₄ will produce 0.5 moles of H⁺ (hydrogen ion)
Explanation:
From the ionization of H₂SO₄, we have
H₂SO₄ → 2H⁺ + SO₄²⁻
Hence, at 100% yield, one mole of H₂SO₄ produces two moles of H⁺ (hydrogen ion) and one mole of SO₄²⁻ (sulphate ion), therefore, 0.250 moles of H₂SO₄ will produce 2×0.250 moles of H⁺ (hydrogen ion) or 0.5 moles of H⁺ (hydrogen ion) and 0.25 moles of SO₄²⁻ (sulphate ion).
That is; 0.250·H₂SO₄ → 0.5·H⁺ + 0.250·SO₄²⁻.
Explanation:
Depression in Freezing point
= Kf × i × m
where m is molality , i is Van't Hoff factor, m = molality
Since molality and Kf remain the same
depression in freezing point is proportional to i
i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2
i=1 for C2h6O
i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)
So the freezing point depression is highest for MgCl2 and lowest for C2H6O
so freezing point of the solution = freezing point of pure solvent- freezing point depression
since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point
Need the answer to this question
