Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
----------------------------------
Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %
lights bullet electric heater radio Google
Answer:
The energy of the particles increase and the molecules move more quickly.
Explanation:
The molecules are moving from a solid (barely moves, molecules close together) to a liquid (molecules slide past each other and take any shape), so molecules are moving more and have more energy
To
determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample
of the compound with the given composition. Then, we calculate for the number
of moles of each element. We do as follows:<span>
mass moles
C 56.79 4.73
H 6.56 6.50
O 28.37 1.77
N 8.28 0.59
Dividing the number of moles of each element with
the smallest value, we will have the empirical formula:
</span> moles ratio
C 4.73 / 0.59 8
H 6.50 / 0.59 11
O 1.77 / 0.59 3
N 0.59 / 0.59 1<span>
</span><span>
The empirical formula would be C8H11O3N.</span>
Answer: 10.62g
Explanation:
First let us generate a balanced equation for the reaction.
HCl + NaOH —> NaCl + H2O
Molar Mass of HCl= 1 + 35.5 = 36.5g/mol
Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol
From the question,
Mass of HCl = 17g
Mass of NaOH = 6.99g
Converting these Masses to mole, we obtain:
n = Mass / Molar Mass
n of HCl = 17/36.5 = 0.4658mol
n of NaOH = 6.99/40 = 0.1748mol
From the question,
1 mole of NaOH requires 1mole of HCl.
Therefore, 0.1748mol of NaOH will also require 0.1748mol of HCl.
But we were told that 17g( i.e 0.4658mol) of HCl were mixed.
Therefore, the unreacted amount of HCl = 0.4658 — 0.1748 = 0.291mol
Converting this to mass, we have:
Mass of HCl = n x molar Mass
Mass of HCl = 0.291 x 36.5
Mass of HCl = 10.62g
Therefore the left over Mass of HCl is 10.62g
