Enzymes are characterized to have weak bonds because their tertiary structure could easily bend and break because it will have to adjust to the shape of the substrate. It could be done via induced fitting or lock-and-key theory. These weak bonds are intermolecular forces like the London forces, electrostatic interactions and hydrogen bonding.
Answer:
94.325 g
Explanation:
We'll begin by converting 350 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
350 mL = 350 mL × 1 L /1000 mL
350 mL = 0.35 L
Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:
Volume = 0.35 L
Molarity of KC₂H₃O₂ = 2.75 M
Mole of KC₂H₃O₂ =?
Molarity = mole /Volume
2.75 = Mole of KC₂H₃O₂ / 0.35
Cross multiply
Mole of KC₂H₃O₂ = 2.75 × 0.35
Mole of KC₂H₃O₂ = 0.9625 mole
Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:
Mole of KC₂H₃O₂ = 0.9625 mole
Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)
= 39 + 24 + 3 + 32
= 98 g/mol
Mass of KC₂H₃O₂ =?
Mass = mole × molar mass
Mass of KC₂H₃O₂ = 0.9625 × 98
Mass of KC₂H₃O₂ = 94.325 g
Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g
Answer:
1 . 
2. 
Explanation:
The more stable the ionic compound, the more is it lattice energy.
- The more the charge on the cation and the anion, the greater is the lattice energy.
- The less the size of the cation and the anion, the greater is the lattice energy.
Scandium oxide (
) is an oxide in which 
behaves as cation and 
behaves as anion.
The compounds which has higher lattice energy than scandium oxide are:
1 .
This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation 
is smaller than . Thus, this corresponds to higher lattice energy.
2.
This is because the charge on the cation 
is greater than that of and also the size of the cation is smaller than . Thus, this corresponds to higher lattice energy.
Li2S + 2 HNO3 --> 2 LiNO3 + H2S
Li2 S + H2 N2 O2 --> Li2 N2 O5 + H2 S
Li S + H2 N2 O5 -> Li N2 O5 + H2 S
Li2 S2 + H4 N4 O10 --> Li2 N4 O10 + H4 S2
Li^2 S^2 + H^4 N^4 O^10 --> Li^2 N^4 O^10 + H^4 S^2
Answer: It depends equilibrium constant K
Explanation: You need to to have reaction formula.
If K >> 1 then concentrations of products are much bigger than
concentrations of reactants. If K < < 1, concentration of products is small.
