Question

hyy solve this fastly.... Q no 17 and 18
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Answer 1

${ \bold{ \green{ \underline{ \underline{ANSWER}} : - }}}$

${ \bold{ \pink{ ADD \: \: BOTH \: \: EQUATION - } }}$

${ \bold{ \orange{ {Ag}^{ + } + 2 \: NH_{3}⇌ {[Ag (NH_{3})_{2} ]}^{ + } }}}$

${ \bold{ \pink{so \:, k_{eq} = k_{1} \times k_{2} }}}$

${ \bold{ \pink{ k_{eq} = (3.5 \times {10}^{ - 3}) \times (1.7 \times {10}^{ - 3} )}}} \\ \\ { \bold{ \pink{ \boxed{ \mathfrak{ k_{eq} = 6.08 \times {10}^{ - 6} }}}}}$

Answer 2

Explanation:

ADDBOTHEQUATION−

{ \bold{ \orange{ {Ag}^{ + } + 2 \: NH_{3}⇌ {[Ag (NH_{3})_{2} ]}^{ + } }}}Ag

+

+2NH

3

⇌[Ag(NH

3

)

2

]

+

{ \bold{ \pink{so \:, k_{eq} = k_{1} \times k_{2} }}}so,k

eq

=k

1

×k

2

\begin{lgathered}{ \bold{ \pink{ k_{eq} = (3.5 \times {10}^{ - 3}) \times (1.7 \times {10}^{ - 3} )}}} \\ \\ { \bold{ \pink{ \boxed{ \mathfrak{ k_{eq} = 6.08 \times {10}^{ - 6} }}}}}\end{lgathered}

k

eq

=(3.5×10

−3

)×(1.7×10

−3

)

k

eq

=6.08×10

−6