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).
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C. 18.0g. all you really have to do is add both of them together and you get your answer.
Answer:
Mass of original sample = 100 g
Explanation:
Half life of cesium-137 = 30.17 years
Where, k is rate constant
So,
The rate constant, k = 0.02297 year⁻¹
Time = 90.6 years
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Initial concentration
= ?
Final concentration
= 12.5 grams
Applying in the above equation, we get that:-
![[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g](https://tex.z-dn.net/?f=%5BA_0%5D%3D%5Cfrac%7B12.5%7D%7Be%5E%7B-0.02297%5Ctimes%2090.6%7D%7D%5C%20g%3D100%5C%20g)
<u>Mass of original sample = 100 g</u>
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Answer:
Explanation:
You would have to add up the atomic masses of all the compounds in the compound, making sure you include how many molecules of each are in the compound
For example, in CuSOA we have 1 molecule of Cu and S, as 4 molecules of O
The atomic masses are as follows:
Cu = 63.55 u
S = 32.065 u
O = 15.99 units
This is how we would add it up:
(Atomic mass of Cu) + (Atomic mass of S) + 4(Atomic Mass of O)
(63.55) + (32.065) + 4(15.99)
(63.55) + (32.065) + 63.96
= 159.575 u