Explanation:
Formula for work done is as follows.
W =
where, k = proportionality constant = 


d = separation distance = 0.45 nm = 
Now, we will put the given values into the above formula and calculate work done as follows.
W =
=
= 
Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is
.
Answer:
d. increases PFK activity, decreases FBPase activity
Explanation:
Fructose-2,6-bisphophate is formed by the phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2, PFK-2.
Fructose-2,6-bisphophate functions as an allosteric effector of the enzymes phosphofructokinase-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase.
Fructose-2,6-bisphophate has opposite effects on the enzymes, PFK-1 and FBPase. When it binds to the allosteric site of the enzyme, PFK-1, it increases the enzymes's activity by increasing its affinity for its substrate fructose-6-phosphate and reduces its affinity for its allosteric inhibitors ATP and citrate. However, when it binds to FBPase, it reduces its activity by reducing its affinity for glucose, its substrate
2-methyl-3-hexyn-2-ol can be prepared from
Acetylene by treating Acetylene with <span>NaNH</span>₂<span> followed by CH</span>₃<span>CH</span>₂<span>Br, then treating the intermediate with NaNH</span>₂<span> followed by acetone and then doing aqueous workup. NaNH</span>₂ acts as base and abstracts proton from acetylene as the terminal alkynes are acidic in nature. While acetone on reduction gives tertiary alcohol.
<u>Given:</u>
Mass of Ag = 1.67 g
Mass of Cl = 2.21 g
Heat evolved = 1.96 kJ
<u>To determine:</u>
The enthalpy of formation of AgCl(s)
<u>Explanation:</u>
The reaction is:
2Ag(s) + Cl2(g) → 2AgCl(s)
Calculate the moles of Ag and Cl from the given masses
Atomic mass of Ag = 108 g/mol
# moles of Ag = 1.67/108 = 0.0155 moles
Atomic mass of Cl = 35 g/mol
# moles of Cl = 2.21/35 = 0.0631 moles
Since moles of Ag << moles of Cl, silver is the limiting reagent.
Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles
Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol
Ans: Formation enthalpy = 126.5 kJ/mol
Answer:
The limiting reactant is CuSO₄.
Explanation:
The reaction is:
Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s) (1)
To find the limiting reactant we need to find the number of moles of the reactants.

Where:
m: is the mass of iron = 3.26 g
: is the standard atomic weight of iron = 55.845 g/mol

Where:
M: is the concentration of the CuSO₄ = 0.200 mol/dm⁻³
V: is the volume of the solution = 80.0 cm³
First, we need to convert the units of the volume to dm³ knowing that 1 dm = 10 cm and 1 L= 1 dm³.

Now, the number of CuSO₄ moles is:
So, to determine the limiting reactant we need to use the molar ratio from equation (1), Fe:CuSO₄ = 1:1

Since we need 0.016 moles of Fe to react with 0.016 moles of CuSO₄ and initially we have 0.058 moles of Fe, then the limiting reactant is CuSO₄.
Therefore, the limiting reactant is CuSO₄.
I hope it helps you!