Volatile organic compounds can be detected by hydrogeologists in the field or labs because of the odor of the vapors emitted from the groundwater and/or soil samples.
<h3>What are volatile substances?</h3>
Volatile substances are substances which can easily vaporize or change to gaseous state.
Volatile substances can either be solids or liquids but are mostly liquids.
Example of volatile substances include ether, petrol, chocolate.
The presence of volatile substances can be detected by the gases they release which may have characteristic odors.
Therefore, volatile organic compounds can be detected by hydrogeologists in the field or labs because of the odor of the vapors emitted from the groundwater and/or soil samples.
Learn more about volatile compounds at: brainly.com/question/25403770
The way how <span>data is not actually obtained from the experiment represented in a line graph is defnitely that </span><span>a colored line with a broken line. It is a well known fact that to obtain the actual data from the experiment you there should be plotted points on the line. Hope it will help you! Regards.</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
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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 %
