Answer:
14.336 g MnF₂
Explanation:
number of moles = mass / molecular weight
number of moles of MnI₂ = 55 / 309 = 0.178 moles
number of moles of F₂ = 55 / 38 = 1.447 moles
From the reaction and the number of moles calculated we deduce that the fluorine F₂ is a limiting reactant.
So:
if 13 moles of F₂ reacts to produce 2 moles of MnF₃
then 1.447 moles of F₂ reacts to produce X moles of MnF₃
X = (1.447 × 2) / 13 = 0.223 moles of MnF₃ (100% yield)
For 57.2% yield we have:
number of moles of MnF₃ = (57.2 / 100) × 0.223 = 0.128 moles
mass = number of moles × molecular weight
mass of MnF₃ = 0.128 × 112 = 14.336 g
The type of bonding present in water (H2O) is hydrogen bonding.
I would say all of the above
12.5% of strontium-90 would remain in a sample after three half-lives have passed. Half-life automatically means 50% of the original amount would remain.
Answer:
The abundance of first isotope is 69.15 %
The abundance of second isotope is 30.85 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope:
% = x %
Mass = 62.9296 u
For second isotope:
% = 100 - x
Mass = 64.9278 u
Given, Average Mass = 63.546 u
Thus,
Solving for x, we get that:
x = 69.15 %
<u>The abundance of first isotope is 69.15 %</u>
<u>The abundance of second isotope is 100 - 69.15 % = 30.85 %</u>
