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:
547.5g
Explanation:
To get the mass, you need moles.
moles = (molarity)(Liters)
moles = (1.230M)(4.200L) = 5.166 moles Na2CO3
Now, just use stoichiometry
molar mass of Na2CO3 = 2(mass of Na) + (mass of C) + 3(mass of O)
= 2(22.9) + 12.01 + 3(16) = 105.99g/mol
5.166moles(105.99g/mol)
= 547.544
But, the measurements given had 4 significant figures, so in chemistry we write:
547.5g
Hey :)
The answer is B) Nitrogen, which has an electronegativity of 3.04
Hope this helps!
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