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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Answer:
Answer: A) .346 M
Explanation:
Given:
- 450 mL
- .5 M soln
-200 mL water
1) Convert mL to L
450 mL = .45 L
200 mL = .2 L
2) Find mols in solution
.5 M = x/.45 L
x = .225 mol
3) Find total volume of solution
.45 L + .2 L =.65 L
4) Find new molarity
molarity (M) = mols solute/ L solution
y = .225 mol (from step 2)/ .65 L (from step 3)
y = .346 M
Answer: A) .346 M
It is protected by a mucus layer formed on the surface regularly that ressist ph values.If it isn't formed our body will eat itself.
It is either a proton or a neutron
Answer:
We need 4.28 grams of sodium formate
Explanation:
<u>Step 1:</u> Data given
MW of sodium formate = 68.01 g/mol
Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L
pH = 3.74
Ka = 0.00018
<u>Step 2:</u> Calculate [base)
3.74 = -log(0.00018) + log [base]/[acid]
0 = log [base]/[acid]
0 = log [base] / 0.42
10^0 = 1 = [base]/0.42 M
[base] = 0.42 M
<u>Step 3:</u> Calculate moles of sodium formate:
Moles sodium formate = molarity * volume
Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles
<u>Step 4:</u> Calculate mass of sodium formate:
Mass sodium formate = moles sodium formate * Molar mass sodium formate
Mass sodium formate = 0.063 mol * 68.01 g/mol
Mass sodium formate = 4.28 grams
We need 4.28 grams of sodium formate
