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2 years ago

Which represents the correct equilibrium constant expression for the reaction below?

Chemistry

2 answers:

zubka84 [21]2 years ago

6 0

K(eq) = concentration of products/concentration of reactant
= [Cu+2] / [Ag+]^2

Activity of pure solid and liquid is taken as 1.

Hence last option is correct.

Hope this helps, have a great day ahead!

ryzh [129]2 years ago

3 0

Answer:

D. $Keq = \frac{[Cu^{2+}]}{[Ag^{+}]^{2} }$

Explanation:

this is the correct answer on ed-genuity, hope this helps! :)

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A patient needs to take 625 mg of ibuprofen twice daily. The pills in the bottle are each 250. mg. How many pills does the patie

WINSTONCH [101]

2 and a half
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250/2=125
and 125 is a half of 250
so your answer is
2 and a half

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3 years ago

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Aluminum chloride can be formed from its elements:

saul85 [17]

<u>Answer:</u> The $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of $AlCl_3$ is:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)$ $\Delta H^o_{formation}=?$

The intermediate balanced chemical reaction are:

(1) $HCl(g)\rightarrow HCl(aq.)$ $\Delta H_1=-74.8kJ$ ( × 6)

(2) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$ $\Delta H_2=-185kJ$ ( × 3)

(3) $AlCl_3(aq.)\rightarrow AlCl_3(s)$ $\Delta H_3=+323kJ$ ( × 2)

(4) $2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)$ $\Delta H_4=-1049kJ$

The expression for enthalpy of formation of $AlCl_3$ is,

$\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]$

Putting values in above equation, we get:

$\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ$

Hence, the $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

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2 years ago

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leonid [27]

Answer:

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Explanation:

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Heat never travels from a cooler object to a warmer object.

prohojiy [21]

Answer:

I would say False.

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3 years ago

Which of the following is not true of acids?

NeTakaya

Answer is: 2. dillute acids feel slipper.

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2) Dillute acids feel slippery is not correct. Bases, for example solution of sodium hydroxide feels slipery.

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3 years ago