Answer:
a. Kw = 1.0 × 10⁻¹⁴
Explanation:
a. Let's consider the self-ionization of water.
2 H₂O(l) ⇄ H₃O⁺(aq) + OH⁻
The ion-product of water (Kw) at 25 °C is:
Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴
c. Considering [H⁺] = [H₃O⁺] = 4.0 × 10⁻⁹ M, the concentration of OH⁻ is:
[OH⁻] = 1.0 × 10⁻¹⁴/[H₃O⁺] = 2.5 × 10⁻⁶ M
b. We can calculate the pOH using the following expression.
pOH = -log [OH⁻] = -log 2.5 × 10⁻⁶ = 5.6
d. We can calculate the pH using the following expression.
pH = -log [H⁺] = -log 4.0 × 10⁻⁹ = 8.4
The number of grams of NaOH that are needed to make 500 ml of 2.5 M NaOH solution
calculate the number of moles =molarity x volume/1000
= 2.5 x 500/1000 = 1.25 moles
mass = moles x molar mass of NaOH
= 1.25 x40= 50 grams of NaOH
We will use this formula for first order:
㏑[A] = - Kt +Ao
when we have t (given)= 30 min = 30 x 60 = 1800 s (we here convert time from min to second.
then we assume that the initial concentration Ao = 1
and the concentration of A (final concentration = 0.25
So by substitution:
㏑(0.25) = - K * 1800 + ㏑(1)
1.39 = K * 1800
∴ K = 0.00077 s^-1 or 7.7 x 10^-4
Hydrogen. Covalent bonds occur within each linear strand and strongly bond the bases, sugars, and phosphate groups (both within each component and between components). Hydrogen bonds occur between the two strands and involve a base from one strand with a base from the second in complementary pairing.
