Answer: 4.21×10⁻⁸
Explanation:
1) Assume a general equation for the ionization of the weak acid:
Let HA be the weak acid, then the ionization equation is:
HA ⇄ H⁺ + A⁻
2) Then, the expression for the ionization constant is:
Ka = [H⁺][A⁻] / [HA]
There, [H⁺] = [A⁻], and [HA] = 0.150 M (data given)
3) So, you need to determine [H⁺] which you do from the pH.
By definition, pH = - log [H⁺]
And from the data given pH = 4.1
⇒ 4.10 = - log [H⁺] ⇒ [H⁺] = antilog (- 4.10) = 7.94×10⁻⁵
4) Now you have all the values to calculate the expression for Ka:
ka = 7.94×10⁻⁵ × 7.94×10⁻⁵ / 0.150 = 4.21×10⁻⁸
The group is might be labeled as VIIB or VIIA.
Answer:
See Explanation
Explanation:

Hence the mass defect is;
[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]
= 236.0526 - 235.85361
= 0.19899 amu
Since 1 amu = 1.66 * 10^-27 Kg
0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg
Binding energy = Δmc^2
Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J
ii) 
Hence the mass defect is;
[10.01294 + 1.00867] - [7.01600 + 4.00260]
= 11.02161 - 11.0186
= 0.00301 amu
Since 1 amu = 1.66 * 10^-27 Kg
0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg
Binding energy = Δmc^2
Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J
Answer:
la verdad no se no ablo ingles solo espanis