Answer:
The red blood cells will burst
Explanation:
When the red blood cells are placed in pure water, they will gain water by osmosis, swell and finally burst due to their weak cell membranes. This process is referred to as hemolysis.
Mitosis is a stage of the cell cycle
The proper name for the following alkyl side group where the main carbon chain is denoted with a squiggly line is isopropyl.
In natural chemistry, an alkyl substituent is an alkane missing one hydrogen. The term alkyl is intentionally unspecific to include many viable substitutions. An acyclic alkyl has the overall formulation of CₙH₂ₙ₊₁.
An alkyl is a purposeful institution of an organic chemical that includes only carbon and hydrogen atoms, that are organized in a chain. Examples include methyl CH3 (derived from methane) and butyl C2H5 (derived from butane). they may be now not located on their own however are discovered attached to different hydrocarbons.
what is an alkyl group? Alkyl group is shaped through putting off a hydrogen atom from the molecule of alkane. Alkanes are quite regularly represented as R-H and here R stands for alkyl group. the overall method of the alkyl organization is CₙH₂ₙ₊₁. The smallest alkyl organization is CH3 referred to as methyl.
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The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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