M₁=6.584 g
m₂=4,194 g
m(H₂O)=m₁-m₂
w(H₂O)=m(H₂O)/m₁
w(H₂O)=(m₁-m₂)/m₁
w(H₂O)=(6.584-4.194)/6.584=0.3630 (36.30%)
the percentage by mass of water in the hydrate 36.30
Answer: (22.98977 g Na/mol) + (1.007947 g H/mol) + (12.01078 g C/mol) + ((15.99943 g O/mol) x 3) = 84.0067 g NaHCO3/mol
9.
(1.20 g NaHCO3) / (84.0067 g NaHCO3/mol) = 0.0143 mol NaHCO3
10.
Supposing the question is asking about "how many moles" of CO2. And supposing the reaction to be something like:
NaHCO3 + H{+} = Na{+} + H2O + CO2
(0.0143 mol NaHCO3) x (1 mol CO2 / 1 mol NaHCO3) = 0.0143 mol CO2 in theory
11.
n = PV / RT = (1 atm) x (0.250 L) / ((0.0821 L atm/K mol) x (298 K)) = 0.0102 mol CO2
12.
(0.0143 mol - 0.0102 mol) / (0.0143 mol) = 0.287 = 28.7%
Explanation:
Answer:
572 g
Explanation:
Molar mass is the mass of 1 mol of an element or compound
molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound
molar masses of each element making up lithium sulphate
Li - 7 g/mol
S - 32 g/mol
O - 16 g/mol
molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )
molar mass = 110 g/mol
mass of 1 mol of Li₂SO₄ is 110 g
therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g
mass is 572 g
<span>2H2 + O2 → 2H2O</span>
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<span>okay???</span>
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