If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).
<h3>What is base dissociation constant? </h3><h3 />
The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 6.2× 10^(-10)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{6.2×10^(-10) }
= 1.6× 10^(-5)
Thus, the value of base dissociation constant at 25°C is 1.6 × 10^(-5).
learn more about base dissociation constant :
brainly.com/question/9234362
#SPJ4
The formula for density is:
D = m/v
We can use the formula to figure out the mass because we already know two of the three values (we are given the density and volume), so we only have to solve for <em>m. </em>If we plug our given values into the formula, we get:
2.70 = m / 264
Now, all we need to do is solve for <em>m</em>. The goal is to get <em>m</em> on one side of the equation, and all we have to do is multiply each side of the equation by 264:
264 × 2.70 = (m÷264) × 264
264 × 2.70 = m
m = 712.8
The mass of the piece of aluminum is 712.8 grams.
The much of the sample that would remain unchanged after 140 seconds is 2.813 g
Explanation
Half life is time taken for the quantity to reduce to half its original value.
if the half life for Scandium is 35 sec, then the number of half life in 140 seconds
=140 sec/ 35 s = 4 half life
Therefore 45 g after first half life = 45 x1/2 =22.5 g
22.5 g after second half life = 22.5 x 1/2 =11.25 g
11.25 g after third half life = 11.25 x 1/2 = 5.625 g
5.625 after fourth half life = 5.625 x 1/2 = 2.813
therefore 2.813 g of Scandium 47 remains unchanged.
