Each enzyme's active site is suitable for one specific type of substrate – just like a lock that has the right shape for only one specific key. Changing the shape of the active site of an enzyme will cause its reaction to slow down until the shape has changed so much that the substrate no longer fits.
The 2nd ionization energy is removing a 2nd electron from that resulting cation:
<span>Li+ --> Li2+ + 1e- </span>
<span>Soda ash is sodium carbonate, Na2CO3. One chemical property of this compound is its basicity, which is measured by the pKb. The pKb for sodium carbonate is 3.67. It is the result of the dissociation of Na2CO3 in water: Na2CO3 + H2O = Na HCO3 + Na (+) + OH(-). This pKb means that it is a highly basic compound. pKb = log { 1 / [OH-] }, so pKb is a measure of the concentrations of OH- ions, which is the basiciity of the compound. </span>
Answer:
4.23.
Explanation:
<em>∵ pH = - log[H⁺].</em>
<em>For weak acids:</em>
∵ [H⁺] = √(ka)(c).
∴ [H⁺] = √(3.5 × 10⁻⁸)(0.10 M) = 5.92 x 10⁻⁵.
∴ pH = - log[H⁺] = - log(5.92 x 10⁻⁵) = 4.2279 ≅ 4.23.
Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5