Extensive hydrogen bonding.
It takes a lot of energy to break hydrogen bonds and turn liquid water into a gas, therefore the boiling point for water is high.
Hey there!:
Molar mass H3PO4 = <span>97.9952 g/mol
Atomic Masses :
H = </span><span>1.00794 a.m.u
</span>P = <span>30.973762 a.m.u
</span>O = 15.9994 a.m.u<span>
H % = [ ( 1.00794 * 3 ) / </span> 97.9952 ] * 100
H% = <span>3.0857 %
P % = [ ( </span>30.973762 * 1 ) / 97.9952 ] * 100
P% = <span>31.6074 %
O % = [ ( </span>15.9994 * 4 ) / 97.9952 ] * 100
O% = <span>65.3069 %
Hope this helps!</span>
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 %
