The term isoelectronic atoms means that if the given atoms are neutral, they would have the same number of electrons, which is relative to their sizes. Ions with positive charges are called cations, losing electrons. On the other hand, ions with negative charges are called anions, gaining electrons. The more electrons that the atoms have, the bigger it is in size. Hence, the arrangement of the radii of the atoms would have to be,
e²⁺, d⁺, c , b⁻, a⁻
The arrangement is from smallest to largest.
Answer:
9.36
Explanation:
Sodium formate is the conjugate base of formic acid.
Also,
for sodium formate is 
Given that:
of formic acid = 
And, 
So,
Concentration = 0.35 M
HCOONa ⇒ Na⁺ + HCOO⁻
Consider the ICE take for the formate ion as:
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻
At t=0 0.35 - -
At t =equilibrium (0.35-x) x x
The expression for dissociation constant of sodium formate is:
![K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5BOH%5E-%5D%5BHCOOH%5D%7D%7B%5BHCOO%5E-%5D%7D)
Solving for x, we get:
x = 0.44×10⁻⁵ M
pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64
pH + pOH = 14
So,
<u>pH = 14 - 4.64 = 9.36</u>
Answer:
D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.
Explanation:
Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.
<u>(1) Preparatory phase</u>
During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.
When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.
<u>(2) Pay-off phase</u>
During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.
It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.
If a metal is less reactive than carbon, it can be extracted from its oxide by heating with carbon. The carbon displaces the metal from the compound, and removes the oxygen from the oxide. This leaves the metal.
<span>The choices are as follows:
h2o + 2o2 = h2o2
fe2o3 + 3h2 = 2fe + 3h2o
al + 3br2 = albr3
caco3 = </span><span>cao + co2
The correct answers would be the second and the last option. The equations that are correctly balanced are:
</span> fe2o3 + 3h2 = 2fe + 3h2o
caco3 = cao + co2
To balance, it should be that the number of atoms of each element in the reactant and the product side is equal.
