Three that is the answer to your question
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
Answer:
B
Explanation:
as temperature rises, the particles gain kinetic energy (they will move faster) so if temperatures stays constant, so will the movement or vibration of the particles
Answer:
107.5 amu
Explanation:
isotopes: fractional Wt Avg
isotopes: isotopic mass %Abundance abundance (amu)
X110 110 60 0.60 66.0
X105 105 30 0.30 31.5
<u> X100 100 10 0.10 10.0</u>
∑ atm mass contributions = 107.5 amu*
*amu = atomic mass units
Answer:
12.044 X 10^23 molecules of NaOH
Explanation:
because NaOH is an ionic bond, we should be asking how many <em>formula units </em> are in 2 moles of NaOH, there are 0 molecules since NaOH is measured in formula units.
but for the sake of the problem I'll assume NaOH is measured in molecules
for every mol of something there are 6.022 X 10^23 of something of that something.
so there are 6.022 X 10^23 molecules for every mol of NaOH
that means we have 2 X 6.022 X 10^23 molecules in 2 moles of NaOH = 12.044 X 10^23 molecules of NaOH
