Answer:
Concentrations:
1. 12 g salt in 3 L water = 0,4 g salt / 100 ml water
2. 12 g salt in 6 L water) = 0,2 g salt / 100 ml water
3. 30 g sugar in 5 L water) = 0,6 g sudar / 100 ml water
Explanation:
The given data are expressions to deal with the concentration of solutions (a kind of homogeneous mixtures).
There are many forms or units to express the concentrations of the mixtures.
The three cases show mass of solute (salt or sugar) in volume of solvent (water).
One form to express the concentration of the solutions is as mass of solute per 100 mililiters of solvent, which is useful to compare with solutibility tables.
You can do that using the formula:
- (mass of solute in grams / volume of solvent in mililiters) × 100.
To find the concentrations of the given solutions, first you need to convert the volume of the solvent to mililiters (ml):
- 3 L water × 1,000 ml / L = 3,000 ml water
- 6 L water × 1,000 ml / L = 6,000 ml water
- 5 L water × 1,000 ml / L = 5,000 ml water
Concentrations:
1. (12 g salt / 3,000 ml water) × 100 = 0,4 g salt / 100 ml water
2. (12 g salt / 6,000 ml water) × 100 = 0,2 g salt / 100 ml water
3. (30 g sugar / 5,000 ml water) × 100 = 0,6 g sugar / 100 ml water
Answer:
<h2>
False</h2>
Explanation:
The human body mechanisms try to excrete out the ethanol from the body before it gets circulated in the body through blood. The body does so by converting the ethanol to certain metabolic products which get filtered and excreted out through urine.
The metabolism of alcohol takes place in the liver in two steps- Ethanol is converted to acetaldehyde by the alcohol dehydrogenase enzyme. The acetaldehyde is toxic to cells and is the compound responsible for the hangover after drinking.
The acetaldehyde is then converted to acetic acid by an enzyme called acetaldehyde dehydrogenase. The acetic acid through respiration gets converted to water and carbon dioxide and rest is excreted out through urine.
Thus, the liver is the organ responsible for the metabolism of ethanol and not pancreas.
Answer:
This is because must of the cell must be in INTERPHASE.
Explanation:
The onion root tip is the growing part of the onion and many cells in it undergo mitosis. In the lab, it is prepared in a way that the cells become flat on the microscope slide so that you can observed them well.
In the cell cycle, most of the onions root tip cells are in interphase, they spend most of their life in interphase. In interphase, there is no observable changes in the celks, the cells keep on growing and the nucleus is been duplicated. This is because most of the onion root cells spend most of their time reproducing and growing in the interphase stage. During the interphase, the cells can still be seen in dark spots because the nucleus is still present and prophase has not started yet.
The dominant trait is hornless.
If this is correct, then the original bull is a heterozygote:
H = hornless
h = horned
Bull is Hh
Cross 1: Hh x hh (horned female) - offspring is hh (horned)
Cross 2: Hh x Hh (hornless female) - offspring is hh (horned)
Cross 3: Hh x hh (horned female) - offspring is Hh (hornless)
Cross 3 is the key to confirming this because the only way to get a hornless offspring would be if hornless is recessive.