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enot [183]
3 years ago
11

A solution of naoh(aq) contains 8.1 g of naoh(s) per 100.0 ml of solution. calculate the ph and the poh of the solution at 25 °c

.
Chemistry
1 answer:
olga2289 [7]3 years ago
3 0
My Brother your answers is: 
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QUICK PLEASE
Trava [24]

Answer: 2.52 M

Explanation:

The product of molarity (moles/litre) and volume in litres yields moles, and the numbers of moles in two solutions means dilute and concentrated are equal, which is expressed by the following equation: 

M_1V_1=M_2V_2

$$Given that\\M $1=12.0 \mathrm{M}$ or mole $/ \mathrm{L}$\\$\mathrm{V} 1=420 \mathrm{ml}$\\$\mathrm{M} 2=$ ?\\$\mathrm{V} 2=2.0 \mathrm{~L}$ or $2000 \mathrm{ml}$\\\\$\mathrm{M} 1 \mathrm{~V} 1=\mathrm{M} 2 \mathrm{~V} 2$\\$\mathrm{M} 2=\mathrm{M} 1 \mathrm{~V} 1 / \mathrm{V} 2$\\$=12.0 * 420 / 2000$\\$=2.52 \ \mathrm{M}$

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If 3.0 moles of x and 4.0 moles of y react according to the hypothetical reaction below, how many moles of the excess reactant w
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Firstly the limiting reactant should be identified. Limiting reactant is the reactant that is in limited supply, the amount of product formed depends on the moles present of the limiting reactant.

the stoichiometry of x to y = 1:2
1 mole of x reacts with 2 moles of y
if x is the limiting reactant, there are 3 moles of x, then 6 moles of y should react, however there are only 4 moles of y. Therefore y is the limiting reactant and x is in excess.
4 moles of y reacts with 2 moles of x 
since there are 3 moles of x initially and only 2 moles are used up, excess amount of x is 1 mol thats in excess.


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