Answer:
Hg Pb are in one, Li and B and O are in one, and NaCl are in one, As is alone
Explanation:
Periods are horizontal
Hope this helps!
The crust
1. divergent (moves away from each other)
2. convergent (moves towards each other)
3. transform (slides past each other)
The answer is 6.25 x 10^4.
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
