Instability of an atoms nucleus can result from an excess of either neutrons or protons . So neutrons and protons .
Answer:
22.7%
Explanation:
Given that last year stock for company A was $7200
The stock for company B last year was worth $3510
Stock in company A decreased by 24%
This means the new value of stock for company A became;
(100-24)/100 *$7200
76/100*$7200
0.76*7200 =$5472
Stock in company B decreased by 20%
This means the new value of stock for company B became;
(100-20)/100 *$3510
80/100*$3510
0.8*$3510
$2808
Original investors stock value was = $7200+$3510 =$10710
New investors stock value is = $5472+$2808=$8280
Decrease in value of stock = $10710-$8280 =$2430
percentage decrease in stock value = decrease in stock/original value of stock *100%
=2430/10710 *100 =22.689
=22.7%
<span>Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38
Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238
The total charge of all the vanadium ions would be +238.
Let x = number of of V+2
Then 100 – x = number of V+3
X(+2) + 100-x(+3) = +238
2x + 300 – 3x = 238
-x = 238-300 = -62
x = 62
Thus 62/100 are V+2
62/100 * 100 = 62%
</span>62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
