Question #55b54

Question

851 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-06T15:12:39

$2.1 d m^{3}$ $2.1 dm^3$

$12 c m^{3} = 0.012 d m^{3}$ $12 cm^3=0.012 dm^3$ of "0.98 mol dm^-3 H_2 SO _4#

1

Will use the useful formula:

$C = \frac{n}{V} \to 0.98 = \frac{n}{0.012} \to n = 0.1176 m o l$ $C=n/V ->0.98=n/0.012 -> n=0.1176 mol$

Then we will use formula again to know how much volume of the first solution will need to prepare the second:

$C = \frac{n}{V} \to 0.056 = \frac{0.1176}{V} \to V = 2.1 d m^{3}$ $C=n/V ->0.056=0.1176/V -> V=2.1 dm^3$

$2$ $2$

Ww will use the other expression:

$C 1 \cdot V 1 = C 2 \cdot V 2$ $C1*V1=C2*V2$

$0.98 \cdot 0.012 = 0.058 \cdot V 2 \to V 2 = 2.1 d m^{3}$ $0.98*0.012=0.058*V2 ->V2=2.1 dm^3$