Chemistry
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New answer posted
4 months agoContributor-Level 10
Kf = 1.86
Using, density of water = 1g / mL
i = ?
i = 1.344
Now, using
n for ClCH2COOH = 2
a =
Using
So; x = 36
New answer posted
4 months agoContributor-Level 10
Anode :
Cathode : 2Ag+(aq) + 2e- ® 2Ag(s)
Zn(s) + 2Ag+(aq) ® Zn2+ (aq) + 2Ag(s)
x = 147
New answer posted
4 months agoContributor-Level 10
Initially -> 1 mol -
At eq. 1-x mol 2x mol
Here; molecules of Cl2 = atoms of Cl
i.e moles of Cl2 = moles of Cl
So : 1 – x = 2x x = 1/3
Moles of Cl2 at equibrium =
Moles of Cl at equilibrium =
Total moles =
No
New answer posted
4 months agoContributor-Level 10
Fraction of molecules having enough energy to form product =
Fraction of molecules having enough energy to form product =
So, x = 14
New answer posted
4 months agoContributor-Level 10
A (g) -> B (g) KP = 100
at 300 K and 1 atm
Using
=-R * 300 * 2 * 2.3
So; x = 1380.
New answer posted
4 months agoContributor-Level 10
Stability constant are :
K1 = 104
K2 = 1.58 * 103
K3 = 5 * 102
K4 = 102
Overall stability constant K will be
K = K1 * K2 * K3 * K4
= 7.9 * 1011
Now, overall equilibrium constant for dissociation of [Cu (NH3)4]2+ is
= 1.26 * 10-12
So; x = 1 (Rounded off to the nearest integer)
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