The answer is a strike-slip. More specifically a right-lateral strike-slip.
The answer is 0.59 M.
Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l
So, 1 mol has 95.2 g/l.
Our solution contains 55.8g in 1 l of solution, which is 55.8 g/l
Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M
The symbol %v/v means percent by volume. Assuming there is no volume effects when these substances are mixed, we calculate as follows:
% v/v = (25 mL ethanol / 25 mL + 150 mL ) x 100
%v/v = 14.29 mL ethanol / mL solution
Hope this answers the question.
By Using relative and radiometric dating methods hope this helps!!
