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%
The statement that identifies an oxidation-reduction reaction is a reaction in which oxidation numbers change (option C).
<h3>What is a redox reaction?</h3>
A redox or oxidation-reduction reaction is a chemical reaction in which some of the atoms have their oxidation number changed.
In a chemical reaction that involves oxidation and reduction, the oxidation number of the involved ions either decreases or increases.
Therefore, the statement that identifies an oxidation-reduction reaction is a reaction in which oxidation numbers change.
Learn more about redox reaction at: brainly.com/question/13293425
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Answer:
A
Explanation:
What the equation is tell you is that for every 3 mols of NO2 you get 2 mol of HNO3
3 mol NO2 / 2 mol HNO2 ===> 300.00 mol NO2 / x Cross multiply
3x = 2 * 300
3x = 600 Divide by 3
3x/3 = 600/3 Do the division
x = 200.00
Answer:
6 x 10⁶ g Fe
Explanation:
Step 1: Set up dimensional analysis
7 x 10²⁸ atoms Fe (1 mol Fe/6.02 x 10²³ atoms Fe)(55.85 g Fe/1 mol Fe)
Step 2: Multiply, divide, and cancel out units
atoms Fe and atoms Fe cancel out.
mol Fe and mol Fe cancel out.
We should be left with g Fe.
7 x 10²⁸/6.02 x 10²³ = 116279 mol Fe
116279(55.85) = 6.49 x 10⁶ g Fe
Step 3: Sig figs
There is only 1 sig fig in this problem.
6.49 x 10⁶ g Fe ≈ 6 x 10⁶ g Fe
