298 g of calcium carbonate CaCO₃
Explanation:
We have the following chemical reaction:
CaCN₂ (s) + 3 H₂O (l) → CaCO₃ (s)+ 2 NH₃ (g)
number of moles = mass / molar weight
number of moles of H₂O = 161 / 18 = 8.94 moles
Knowing the chemical reaction we devise the following reasoning:
if 3 moles of H₂O produces 1 mole of CaCO₃
then 8.94 moles of H₂O produces X moles of CaCO₃
X = (8.94 × 1) / 3 = 2.98 moles of CaCO₃
mass = number of moles × molar weight
mass of CaCO₃ = 2.98 × 100 = 298 g
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number of moles
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The molar mass of the unknown compound is calculated as follows
let the unknown gas be represented by letter Y
Rate of C2F4/ rate of Y = sqrt of molar mass of gas Y/ molar mass of C2F4
= (4.6 x10^-6/ 5.8 x10^-6) = sqrt of Y/ 100
remove the square root sign by squaring in both side
(4.6 x 10^-6 / 5.8 x10^-6)^2 = Y/100
= 0.629 =Y/100
multiply both side by 100
Y= 62.9 is the molar mass of unknown gas
Answer:
773.51495 grams
Explanation:
1 moles KBr to grams = 119.0023 grams
6.5*119.0023 = 773.51495 grams
Answer:-
0.229 L
Explanation:-
Molar mass of AgBr = 107.87 x 1 + 79.9 x 1
=187.77 grams mol-1
Mass of AgBr = 150 grams
Number of moles of AgBr = 150 grams / 187.77 gram mol-1
= 0.8 mol
The balanced chemical equation is
NaBr (aq) + AgNO3 (aq)--> AgBr(s) + NaNO3(aq)
From the equation we can see that
1 mol of AgBr is produced from 1 mol of AgNO3.
∴ 0.8 mol of AgBr is produced from 1 x 0.8 / 1 = 0.8 mol of AgNO3.
Strength of AgNO3 = 3.5 M
Volume of AgNO3 required = Number of moles / strength
= 0.8 moles / 3.5
=0.229 L
