Explanation:
Depression in Freezing point
= Kf × i × m
where m is molality , i is Van't Hoff factor, m = molality
Since molality and Kf remain the same
depression in freezing point is proportional to i
i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2
i=1 for C2h6O
i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)
So the freezing point depression is highest for MgCl2 and lowest for C2H6O
so freezing point of the solution = freezing point of pure solvent- freezing point depression
since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point
The cellular component responsible for energy production and metabolic processes is mitochondria.
What is Mitochondria?
A double-membrane-bound organelle known as a mitochondrion is found in the majority of eukaryotic organisms. The majority of the cell's adenosine triphosphate, which is then used as a source of chemical energy throughout the cell, is produced by mitochondria using aerobic respiration.
Oxidative phosphorylation, which produces ATP using the energy released during the oxidation of the food we eat, is the traditional function of mitochondria. For the majority of biochemical and physiological processes, including growth, mobility, and equilibrium, ATP is used as the main energy source in turn.
To learn more about mitochondria click the given link
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pH of solution = 13.033
<h3>Further explanation</h3>
Given
2.31 g Ba(OH)₂
250 ml water
Required
pH of solution
Solution
Barium hydroxide is fully ionized, means that Ba(OH)₂ is a strong base
So we use a strong base formula to find the pH
[OH ⁻] = b. Mb where
b = number of OH⁻
/base valence
Mb = strong base concentration
Molarity of Ba(OH)₂(MW=171.34 g/mol) :
Ba(OH)₂ ⇒ Ba²⁺ + 2OH⁻(b=valence=2)
[OH⁻]= 2 . 0.054
[OH⁻] = 0.108
pOH= - log 0.108
pOH=0.967
pOH+pH=14
pH=14-0.967
pH=13.033
Answer:
An increase in soil temperature increases microorganism growth because the rate of decomposition increases.
Explanation:
An increase in soil temperature increases microbial activities and the activity of extracellular enzymes that degrade polymeric organic matter within the soil. <u>An increase in the soil microbial activities means that the decomposition rate and microorganism growth would increase</u>. Increase in the activity of extracellular enzymes that degrade polymeric organic matter within the soil also means that more carbon dioxide would be produced and released into the atmosphere.
The conversion factor for volume at STP is
or \frac {22.4L}{1mol}. Since we want volume, we would use \frac {22.4L}{1mol}. We conclude with the following calculations:
The answer is 44.8L Cl2
