<span>Step 1 is to determine the mass of each part
Mass of Ca is 40.08 g
Mass of C is 12.01 g
Mass of O is 16.00 x 3 = 48.00 g
Step 2 is to determine the total mass of the compound
Total mass of CaCO3 is 40.08 + 12.01 + 48.00 = 100.09 g
Step 3 is to determine the % of each part using the following formula:
Mass of part / total mass x 100 =
40.08 / 100.09 x 100 = 40.04 % Ca
12.01 / 100.09 x 100 = 12.00 % C
48.00 / 100.09 x 100 = 47.96 % O
Step 4 is to double check by adding all percentages. If they equal 100, then I probably did it right. :)
40.04
+12.00
+47.96
=100.00</span><span>
</span>
Alpha particles bouncing off of gold foil.
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%
Answer:sugar
Explanation:it is a homogeneous mixture
