1: Given that Ka for HCOOH is 1.8×10^-4 at 25°C what is the value of b for COOH− at 25 °C? Kb=
2: Given that b for (CH3)2 NH is 5.4×10^-4 at 25°C, what is the value of Ka for (CH3)2NH+2 at 25 °C?
Ka=

Explanation:

At 25 °C Dissociation constant of water or Kw = 1× 10^-14 mol2dm-6

Ka × Kb = Kw for all diluted solutions

Ka×Kb = 1 × 10^-14 mol2dm-6

Therefore for question 1 -

COOH- is the conjugate base for HCOOH ( we find the conjugate base by removing an H atom and the Conjugate acid by adding an H atom)

COOH- (aq) + H2O ( l) <> HCOOH( aq) + OH-(aq)

So we can apply the above mentioned equation to find Kb value for COOH-

Kb = 5.56 × 10^-11 moldm-3

For question 2-

We can see that (CH3)2NH+2 is the conjugate acid of (CH3)2 NH. (A H atom has been removed so the one with a H atom less is the conjugate base.)

(CH3)2NH+2 (aq) + H2O (l) <> (CH3)2 NH (aq) + H3O+ (aq)

So applying the previously mentioned equation we get Ka = 1.85 × 10^-11 moldm-3

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

so since they all have the same shape? i mean maybe because m& m and reeses are both chocolates