Question

calculate the activity cofficient of $NH_{4} OH$ at 0.01 M conc

Answer 1

Answer:

Explanation:

as mean activity coefficient $f$ ± =?

log $f$± = -0.5091 z +z=$\sqrt{u}$

$u=\frac{1}{2}$∑$C_{i} z_{\\i^{2}$

here

$u =\frac{1}{2} (CNH_{4^{+} } Z^{2} NH_{4} +COH Z^{2} O^{-} H)$

$u= \frac{1}{2} (0.01*1^{2} +0.01*1^{2} )$

$u=\frac{1}{2} * 0.02 = 0.01$

$logf$±$= -.5091*1*1*\sqrt{.01}$ = 0.05091

$f$±$Antilog(-0.050901) = 0.889$

Answer 2

Answer:

i dono ask someother bye

Explanation: