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3 years ago

1 point

Chemistry

1 answer:

Elina [12.6K]3 years ago

6 0

Answer:

84672 J

Explanation:

From the question given above, the following data were obtained:

Height (h) = 72 m

Combined mass (m) = 120 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

We can obtain the energy by using the following formula:

E = mgh

Where

E => is the energy.

g => is the acceleration due to gravity

m => is the mass.

h => is the height.

E = 120 × 9.8 × 72

E = 84672 J

Thus, the energy is 84672 J

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Explanation:

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2 years ago

How would you prepare 250 mL of a 0.100 M solution of fluoride ions<br> from solid CaF2?

andrey2020 [161]

In 250 mL of volumetric flask add 0.975875 grams of $CaF_2$ and dissolve it in the 250 mL of water.

Given:

The solid of calcium fluoride.

To prepare:

The 250 mL solution of 0.100 M of fluoride ions from solid calcium fluoride.

Method:

Molarity of the fluoride ion solution needed = M = 0.100 M

The volume of the fluoride ion solution needed = V = 250 mL

$1 mL = 0.001L\\V=250 mL=250\times 0.001 L=0.250 L$

The moles of fluoride ion needed = n

According to the definition of molarity:

$M=\frac{n}{V}\\0.100M=\frac{n}{0.250 L}\\n=0.100M\times 0.250 L=0.025 mol$

Moles of fluoride ion = 0.025 mol

We know that solid calcium fluoride dissolves in water to give calcium ions and fluoride ions.

$CaF_2(s)\rightarrow Ca^{2+}(aq)+2F^-(aq)$

According to reaction, 2 moles of fluoride ions are obtained from 1 mole of calcium fluoride, then 0.025 moles of fluoride ions will be obtained from:

$=\frac{1}{2}\times 0.025 mol=0.0125 \text{mol of } CaF_2$

Moles of calcium fluoride = 0.0125 mol

Mass of calcium fluoride needed to prepare the solution :

$=0.0125 mol\times 78.07 g/mol=0.975875 g$

Preparation:

Weight 0.975875 grams of calcium fluoride
Add weighed calcium fluoride to a volumetric flask of the labeled volume of 250 mL.
Now add a small amount of water to dissolve the calcium fluoride completely.
After this add more water up to the mark of the volumetric flask of volume 250 mL.