Answer:
It should be acetic acid.
Explanation:
When you have ionic bonds, the ionic bonds will always be water soluble; the polarity doesn't matter for this case.
Explanation:
so moles = mass ÷ mr (1+ 79.9)
so 10.00g ÷ 80.9
which is 0.1236093943
so to 3 S.F is 0.124 moles
also there is 1 to 1 ratio for LiOH to HBr
hope this helps :)
Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%
Explanation: Given that the average atomic mass(M) of magnesium
= 24.3050amu
Mass of first isotope (M1) = 23.9850amu
Mass of middle isotope (M2)=24.9858amu
Mass of last isotope(M3)= 25.9826amu
Total abundance = 1
Abundance of middle isotope = 0.10
Let abundance of first and last isotope be x and y respectively.
x+0.10+y =1
x = 0.90-y
M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope
24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y
Substitute x= 0.90-y
Then
y = 0.11
Since y=0.11, then
x= 0.90-0.11
x=0.79
Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%
Concentration is the number of moles of solute in a fixed volume of solution
Concentration(c) = number of moles of solute(n) / volume of solution (v)
25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.
original solution molarity - 0.150 M
number of moles of LiOH in 1 L - 0.150 mol
number of LiOH moles in 0.125 L - 0.150 mol/ L x 0.125 L = 0.01875 mol
when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases
new volume - 125 mL + 25 mL = 150 mL
therefore new molarity is
c = 0.01875 mol / 0.150 L = 0.125 M
answer is 0.125 M
