If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
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What is base dissociation constant?
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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 2.8× 10^(-9)
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) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
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(1) Ocean to Continent
(2)Continent to Continent
(3)Ocean to Ocean
are the three sub types of convergent plate boundaries.
Answer:
The standard cell potential is 2.00 V
Explanation:
<u>Step 1:</u> Data given
Cu is cathode because of
higher EP
Al3++3e−→Al E∘=−1.66 V anode
Cu2++2e−→Cu E∘=0.340 V cathode
<u>Step 2:</u> Balance both equations
2*(Al → Al3+-3e−) E∘=1.66 V
3*(Cu2++2e−→Cu) E∘=0.340 V
<u>Step 3:</u> The netto equation
2 Al + 3Cu2+ +6e- → 2Al3+ + 3Cu -6e-
2 Al + 3Cu2+ → 2Al3+ + 3Cu
<u>Step 4:</u> Calculate the standard cell potential
E∘cell = E∘cathode - E∘anode
E∘cell = E∘ Cu2+/Cu - E∘ Al3+/Al
E∘cell =0.340 V - (-1.66) = 2.00 V
The standard cell potential is 2.00 V
N<span>et ionic equation for ammonia and phosphoric acid</span> is
3 NH4OH + H3PO4 >> (NH4)3PO4 + 3 H2O
hope this helps
Hi there!
It can take years to remove all of the harmful substances from the water.
I hope my answer helped :)
