3 years ago

Free points just replie and yea

Chemistry

2 answers:

amm18123 years ago

8 0

Answer:

hi how's your day human being

anzhelika [568]3 years ago

8 0

Yes Daddy xD

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At a certain temperature, Kc = 0.0500 and ∆H = +39.0 kJ for the reaction below, 2 MgCl2(s) + O2(g) → 2MgO(s) + Cl2(g) Calculate

Minchanka [31]

Explanation:

Since, it is shown that the reaction has been reversed. Therefore, value of $K_{c}$ will become $\frac{1}{K_{c}}$ .

Hence, new $K_{c'} = \frac{1}{K_{c}}$

= $\frac{1}{0.0500}$

= 20

Also, the number of moles of each reactant has been halved. So, $K_{c''}$ for the reaction $MgO(s) + \frac{1}{2}Cl2(g) → MgCl_{2}(s) + \frac{1}{2} O2(g)$ will also get halved.

Therefore, $K_{c''}$ = $K_{c'}$ = $(20)^{0.5}$

= 4.47

As the value of $\Delta H$ is given as +39.0 kJ. So, it means that the reaction is endothermic in nature. So, energy of reactants will be more than the products. Hence, according to Le Chatelier's principle reaction will move in the forward direction.