Answer:
CaCO3 (s) → CaO (s) + CO2 (g)
The mass of carbonate that must have reacted was 43.03 grams
Explanation:
CaCO3 → CaO + CO2
Relation between reactant and product is 1:1
Let's apply the Ideal Gas Law to find out the moles of CO2 which were produced.
P . V = n . R . T
1 atm . 23 L = n . 0.082 L.atm/mol.K . 653K
(1atm . 23L) / (0.082 mol.K/L.atm . 653K) = n
0.43 moles = n
0.43 moles of CO2, were produced from 0.43 moles of CaCO3.
Molar weight of CaCO3 = 100.08 g/m
Mass = Molar weight . moles
Mass = 100.08 g/m 0.43 m = 43.03 g
Did you mean milliliters? if so your answer would still be 876 because they are equal. I don't think it is possible to convert millimeters and grams :)
A:- sn(s) => Sn +2(0.24 M) + 2e-
B:- Sn +2 (0.87 M) +2e- => Sn(s)
solution will become more concentrated and solution B become less concentrated
Sn(s)+ Sn +2(0.87 ) ----> Sn(s) + Sn +2(0.24)
E = Eo - 0.0592 / 2 * log [ (0.24 / 0.87 ) ]
E = 0.0 - 0.0592 / 2 * log ( 0.275)
( n=2 two electrons are transferred)
E = -0.0296 * ( - 0.560)
E = 0.0165 volts
