• Aerobic respiration, photosynthesis
It is aerobic because it requires oxygen to be carried out. Anaerobic on the other hand does not require oxygen.
It’s two products would be the reactants for photosynthesis because in photosynthesis the reverse reaction takes place as glucose is produced along with water from carbon dioxide (CO2) and water (H2O) in the presence of chlorophyll and sunlight.
You can use a separating funnel. The person will go the top and the water to the bottom.
The molar mass of the gene fragment is 19182 g/mol.
What is osmotic pressure ?
Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. It is also defined as the measure of the tendency of a solution to take in a pure solvent by osmosis. Potential osmotic pressure is the maximum osmotic pressure that could develop in a solution if it were separated from its pure solvent by a semipermeable membrane.
We employ the osmotic pressure equation to determine the solute's concentration, which is:
π = iMRT
Using the values in the equation above, we obtain: 19182 g/mol.
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<h3><u>Answer;</u></h3>
Phloem
<h3><u>Explanation;</u></h3>
- <u>Club moss</u> plant belongs to the the family Lycopodiaceae, Lycophyte includes any spore-bearing vascular plant.
- <u>Liverworts</u> on the other hand are bryophytes which belongs to the division bryophyta. Bryophytes are small, non-vascular plants which includes mosses, hornworts and liverworts.
- <em><u>Vascular plants contain vascular tissues which play an important role of transportation in plants. </u></em>The major vascular tissues are phloem and xylem. <em><u>Non-vascular plants</u></em> on the other hand lacks the vascular tissues for transportation of substances.
Answer:
0.11%
Explanation:
Without mincing words, let us dive straight into the solution to the question/problem. The first step to solve this question is to write out the chemical reaction, that is the reaction showing the dissociation of acetic acid.
CH3COOH <=======================================> CH3COO⁻ + H⁺
Initially, the amount present in the acetic acid which is = 12M, the concentration for CH3COO⁻ and H⁺ is 0 respectively.
At equilibrium, the amount present in the acetic acid which is = 12 - x, the concentration for CH3COO⁻ = x and H⁺ = x respectively. Note that the ka for acetic acid = 1.8 × 10⁻⁵.
1.8 × 10⁻⁵ = x²/ 14 - x. Therefore, x = 0.0158 M.
The next thing to do is to calculate for the percentage of dissociation, this can be done as given below:
percentage of dissociation = x/14 × 100. Recall that the value that we got for x = 0.0158 M. Hence, the percentage of dissociation = 0.0158 M/ 14m × 100 = 0.11%
