Answer:
One: <u>Selenium</u> is Paramagnetic
Explanation:
Those compounds which have unpaired electrons are attracted towards magnet. This property is called as paramagnetism. Lets see why remaining are not paramagnetic.
Electronic configuration of Scandium;
Sc = 21 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹
Sc³⁺ = 1s², 2s², 2p⁶, 3s², 3p⁶
Hence in Sc³⁺ there is no unpaired electron.
Electronic configuration of Bromine;
Br = 35 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁵
Br⁻ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶
Hence in Br⁻ there is no unpaired electron.
Electronic configuration of Magnesium;
Mg = 12 = 1s², 2s², 2p⁶, 3s²
Mg²⁺ = 1s², 2s², 2p⁶
Hence in Mg²⁺ there is no unpaired electron.
Electronic configuration of selenium;
Se = 34 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴
Or,
Se = 34 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4px², 4py¹, 4pz¹
Hence in Se there are two unpaired electrons hence it is paramagnetic in nature.
Base pairs of DNA are A with T, and C with G
here's how to remember:
Apple - Tree
Car - Garage
I'm having trouble with this one. At the very least, they both purify water. If I'm not mistaken, distillation is part of the process of purifying water.
Really sorry, I hope I helped at least a little! If you have any other questions I might be able to answer better, let me know..
The empirical formula : C₁₂H₄F₇
The molecular formula : C₂₄H₈F₁₄
<h3>Further explanation</h3>
mol C (MW=12 g/mol)

mol H(MW=1 g/mol) :

mol F(MW=19 g/mol)

mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7
Empirical formula : C₁₂H₄F₇
(Empirical formula)n=molecular formula
( C₁₂H₄F₇)n=562 g/mol
(12.12+4.1+7.19)n=562
(281)n=562⇒ n =2
Molecular formula : C₂₄H₈F₁₄
Answer:
a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe
b. Glu-Ala-Phe + Gly-Ala-Tyr
Explanation:
In this case, we have to remember which peptidic bonds can break each protease:
-) <u>Trypsin</u>
It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.
-) <u>Chymotrypsin</u>
It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.
With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the <u>"Lis"</u> and <u>"Arg"</u> (See figure 1).
In "peptide b", the peptidic bond that would be broken is the one in the <u>"Phe"</u> (See figure 2). The second amino acid that can be broken is <u>tyrosine</u>, but this amino acid is placed in the <u>C terminal spot</u>, therefore will not be involved in the <u>hydrolysis</u>.