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DanielleElmas [232]
4 years ago
8

If 17.8 grams of KOH dissolve in enough water to make a 198-gram solution, what is the concentration in percent by mass?

Chemistry
1 answer:
sleet_krkn [62]4 years ago
7 0
Solute of solution = 17.8 g

Solvn = 198 g

% = 17.8 / 198

w% = 0.089 x 100 = 8.9%  by mass

hope this helps!
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What is the molarity of a solution in which 25.3 grams of potassium bromide is dissolved in 150. mL of solution?
irga5000 [103]

Answer:

~1.417M

Explanation:

Molarity=(number of moles of solute)/(litres of solution)

In this case, we need to find moles of potassium bromide.

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        =(25.3)/(119)

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Hope this helps:)

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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}}

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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}

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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.

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