Class 11th
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New answer posted
a year agoContributor-Level 10
pH = – log [H+] or log [H+] = – pH = – 3.76 = 4.24
Therefore, [H+] = Antilog 4.24 = 1.738 x 10-4 = 1.74 x 10-4 M
New answer posted
a year agoContributor-Level 10
pH = – log [H+] = – log (3.8 x 10-3) = – log 3.8 + 3 = 3 – 0.5798 = 2.4202 = 2.42
New answer posted
a year agoContributor-Level 10
(a) OH– ions can donate an electron pair and act as Lewis base.
(b) F– ions can donate an electron pair and actas Lewis base.
(c) H+ ions can accept an electron pair and act as Lewis acid.
(d) BCl3 can accept an electron pair since Boron atom is electron deficient. It is a Lewis acid.
New question posted
a year agoNew answer posted
a year agoContributor-Level 10
An acid-base pair which differs by a proton only is known as conjugate acid-base pair. For example, HCl, and Cl represents conjugate acid base pair.
The conjugate acid/base for the species HNO2, CN–, HClO4F–, OH–, CO32-, and S2- are NO2–, HCN, ClO4–, HF, H2O, HCO3– and HS–.respectively.
New answer posted
a year agoContributor-Level 10
Kc = [CH4] [H2O] / [CO] [H2]3
3.90 = [CH4]*0.02 / (0.30) (0.10)3
[CH4] = 3.90*0.30* (0.10)3 / 0.02
=5.85*102 M
Thus, the concentration of methane in the mixture is 5.85*102 M.
New answer posted
a year agoContributor-Level 10
The equilibrium constant expression is
Kc = [O3]2 / [O2]3
2.0*1050 = [O3]2 / (1.6*102)3
[O3]2=2.0*1050* (1.6*102)3
=8.192*1056
Hence, the equilibrium concentration of ozone is
[O3]= 2.86*1028M
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