Answer:
67.1%
Explanation:
Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:
<em>Moles Na₂CO₃ - 105.99g/mol-:</em>
6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.
As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:
0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃
And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):
0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.
And percent of NaHCO₃ in the sample is:
10.06g NaHCO₃ / 15g Sample * 100 =
<h3>67.1%</h3>
The only one I know for sure is Mass is always conserved In a Chemical reaction.
Answer:
b) 49.48% C, 5.19% H, 28.85% N, and 16.48% O
Explanation:
we find the mass for each element in one mole by multiplying the number of atoms in one molecule with the atomic mass
mC=8Ac=8*12=96g
mH=10AH=10*1=10g
mN=4AN=4*14=56g
mO=2AO=2*16=32g
by adding the masses together we find the molar mass of the molecule
M=mC+nH+mN+mO=96+10+56+32=194g/mole
we apply the rule of threes to find the percentage of each element
194g..96gC..10gH...56gN....32gO
100g....a...........b...........c.............d
a=(100*96)/194=49.48%C
b=(100*10)/194=5.19%H
c=(100*56)/194=28.85%N
d=(100*32)/194=16.48%O
Answer:
Likely 
(indium.)
Explanation:
Number of atoms: 
.
Dividing, 
, the number of atoms by the Avogadro constant, 
, would give the number of moles of atoms in this sample:
.
The mass of that many atom is 
. Estimate the average mass of one mole of atoms in this sample:
.
The average mass of one mole of atoms of an element (
in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass 
. Indium, , is the closest match.
Answer:
The objects mass is 84, 105 g. See the explanation below, please.
Explanation:
We use the formula:
Density= mass/volume
Mass= Density x volume =0.801 g/mL x 105 mL = 84, 105 g
