11.19%. This should be right. I have no doubt that you will get it right.
To calculate atomic mass, you have to take to weighted average of the isotopes' masses. What that means is M = RA*106 + (1 – RA)*104, where RA is relative abundance expressed in decimal form. If you simplify the right side of that equation, you get M = 2*RA + 104. Doing a little more algebra yields RA = (M –104)/2 = (104.4 – 104)/2 = 0.4 / 2 = 0.2, which is 20%. So the answer is B.
<span>c. Passing electric charge through the reactants Is the answer to you're question.
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Additional information
Relative atomic mass(Ar) : A=7, G=16
The empirical formula : A₂G
<h3>Further explanation</h3>
Given
3.5g of element A
4.0g of element G
Required
the empirical formula for this compound
Solution
The empirical formula is the smallest comparison of atoms of compound forming elements.
The empirical formula also shows the simplest mole ratio of the constituent elements of the compound
mol of element A :
mol of element G :
mol ratio A : G = 0.5 : 0.25 = 2 : 1
Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
