Answer:
Explanation:
When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,
Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C
Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H
Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O
Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.
Moles C = 3.41/3.41 = 1
Moles H = 4.58/3.41 = 1.34
Moles O = 3.41/3.41 = 1
Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3
Moles C = 1x3 = 3
Moles H = 1.34x3 = 4
Moles O = 1x3 = 3
Empirical Formula
what grade is this because apparently i like to know what grade it is before i solve it
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%
Sunlight i think thats the answer
Answer:
Pro exercise con suffication
Explanation:
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