PH= log[H3O+]
10.25=log [H3O+]
[H3O+] = 10^10.25
[H3O+]= 1.778 ×10^10
Percent (%) Composition of CuO
Cu = 1 x 50g - Multiply by one as there is one Cu
O = 1 x 12.5g - Multiply by one as there is one O
CuO = 62.5g
% for Cu = 50g over 62.5 multiplied by 100 = 80%
% for O = 12.5g over 62.5 multiplied by 100 = 20%
Final Answer :
<em>Percent (%) Composition of CuO = </em>80% (Cu) & 20% (O)
Interphase is not a phase of mitosis.
The correct answer should be B.
The correct answer would be B.
Answer:
Explanation:
1 mol of sodium = 23 grams (use the number on your periodic table).
0.7350 mol sodium = x
Cross multiply
1*x = 0.7350 * 23
x = 16.905
You will get slightly less than this, depending on your periodic table. But the method will be the same.
