First, let's count mole of 10 g Calcium Carbonate
mole = Mass / Molecular Mass
Calcium Carbonate = CaCO₃
Molecular Mass = Ar Ca + Ar C + (3 x Ar O)
Molecular Mass = 40 + 12 + (3 x 16)
Molecular Mass = 100
next
Mole of CaCO₃ = 10 gram / 100
Mole of CaCO₃ = 0,1 mol
then equal the reaction equation first
CaCO₃ + 2 HCl ==> CaCl₂ + CO₂ + H₂O (Equal)
To count the mass of carbon dioxide that produced we must know the mole of CO₂ first
we can count by coefficient comparison
mole CO₂ =
x mole CaCO₃
mole CO₂ = (1/1) x 0,1 mole
mole CO₂ = 0,1 mole
so
Mass of CO₂ = mole CO₂ x Molecular Mass of CO₂
Mass of CO₂ = 0,1 mole x (12 + (2 x 16))
Mass of CO₂ = 0,1 mole x 44
Mass of CO₂ = 4,4 g
so, mass of carbon dioxide that's produced by 10 g of calcium carbonate on reaction with chloride acid is 4,4 g.
If you were allowed to taste it sugar will be more sweeter whereas salt wouldnt. But if you were to touch it then sugar is more tinnier than salt.
Ka for weak acids is ---> Ka= (X)²/ (initial- X)
initial= 0.0400 M
to find X, you can find the H concentration using pH
[H+]= 10^-pH
[H+]= 10^-1.80= 0.016 which is X
ka = (0.016)²/ (0.0400 - 0.016) = 0.011
Answer:
This is the answer
Explanation: It's right here