<span> UV radiation are high energy radiations and they are mutation causing agents so
</span>Mutagen <span> best describes the relationship of solar UV radiation to the environment
so option A is correct
hope it helps</span>
Answer:
C. Lithium is most easily oxidized of the metals listed on the activity series and therefore it will most easily give electrons to metal cations
Explanation:
"Lithium" is a type of alkali metal that has a "single valence electron." Since it is a reactive element, it easily gives up an electron when it is combined with other elements. Such giving up of electron is meant to create compounds or bonds.
Among the common metals listed, "lithium" is the most easily oxidized. This means that it donates its electrons immediately. Such combination makes it exist as a<em> "cation"</em> or <em>"positively-charged."</em>
So, this explains the answer.
Answer:
Objects appear different colours because they absorb some colours (wavelengths) and reflected or transmit other colours. ... For example, a red shirt looks red because the dye molecules in the fabric have absorbed the wavelengths of light from the violet/blue end of the spectrum
Answer:
760 uM
Explanation:
<em>A biochemist carefully measures the molarity of magnesium ion in 47, mL of cell growth medium to be 97 uM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 6.0 mL. Calculate the new molarity of magnesium ion in the cell growth medium Be sure your answer has the correct number of significant digits.</em>
The problem here is that the amount of magnesium ion remains the same irrespective of the volume.
Amount of magnesium in the growth medium = <em>molarity x volume</em>
= 97 x
x 47 x
= 4.559 x 
Then, the volume reduced to 6.0 mL, the new molarity becomes;
<em>molarity = mole/volume </em>
= 4.559 x
/6 x
= 7.598333 x
M = 759.83333 uM
To the correct number of significant digits = 760 uM
Hello!
We have the following data:
ps: we apply Ka in benzoic acid to the solution.
[acid] = 0.235 M (mol/L)
[salt] = 0.130 M (mol/L)
pKa (acetic acid buffer) =?
pH of a buffer =?
Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = 
So:





Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:
![pH = pKa + log\:\dfrac{[salt]}{[acid]}](https://tex.z-dn.net/?f=%20%20pH%20%3D%20pKa%20%2B%20log%5C%3A%5Cdfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%20%20%20)





Note:. The pH <7, then we have an acidic solution.
I Hope this helps, greetings ... DexteR! =)