Answer:
Explanation:
If you insist on filling in the first blank you can put a one.
Eh I can't comment
That's y
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Only a cousin knows
Cz I've got a protective family lol
☃️ Chemical formulae ➝ 
How to find?
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
Solution:
❍ Molecular weight of
= 2 × 126.90
= 253.80
= 254 (approx.)
❍ Given weight: 12.7
Then, no. of moles,
⇛ No. of moles = 12.7 / 254
⇛ No. of moles = 0.05 moles
⚘ No. of moles of Iodine molecule in the given weight = <u>0.05</u><u> </u><u>moles </u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
B. The inhibitor wouldn’t stop sunlight or water, but it stops glucose production (from photosynthesis) and u could call glucose the plant’s food.
Missing in your question :
Ksp of(CaCO3)= 4.5 x 10 -9
Ka1 for (H2CO3) = 4.7 x 10^-7
Ka2 for (H2CO3) = 5.6 x 10 ^-11
1) equation 1 for Ksp = 4.5 x 10^-9
CaCO3(s)→ Ca +2(aq) + CO3-2(aq)
2) equation 2 for Ka1 = 4.7 x 10^-7
H2CO3 + H2O → HCO3- + H3O+
3) equation 3 for Ka2 = 5.6 x 10^-11
HCO3-(aq) + H2O(l) → CO3-2 (aq) + H3O+(aq)
so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3-(aq)
note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :
CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l) Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq) + CO3-2(aq) Ksp = 4.5 x 10^-9
∴ the overall equation will be as we have mentioned before:
when H3O+ = H+
CaCO3(s) + H+(aq) ↔ Ca2+ (aq) + HCO3-(aq) K= 80.55
from the overall equation:
∴K = [Ca2+][HCO3-] / [H+]
when we have [Ca2+] = [HCO3-] so we can assume both = X
∴K = X^2 / [H+]
when we have the PH = 5.6 so we can get [H+]
PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6
so, by substitution on K expression:
∴ 80.55 = X^2 / (2.5 x10^-6)
∴X = 0.0142
∴[Ca2+] = X = 0.0142
