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4 years ago

Given that the density of iron is 7.9 g/cm3, what would be the volume of a 15.8 g piece of iron

Chemistry

1 answer:

Anuta_ua [19.1K]4 years ago

7 0

Answer:

2cm^3

Explanation:

Use the density triangle: D=MxV

Switch for variables, V=M/D

Plug in numbers, 15.8g/7.9g/cm^3=2cm^3

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Which term means human-made liquid, solid, or sludge waste that may endanger human health or the environment? A. sewage B. hazar

zmey [24]

The answer is B Hazardous waste.

4 0

3 years ago

Read 2 more answers

Calculate the mass, in grams, of a single xenon atom (mxe = 131.29 amu ).

Mariana [72]

Answer is $2.18017\times10^{-22}gms$ .

Explanation: To calculate the mass of 1 Xe-atom , we will require atomic mass of Xe which is 131.29amu.

This atomic mass is the mass of 1 mole of 1 Xe-atom. One mole of Xe-atom is $6.022\times10^{23}$ atoms of Xe. We will use this relation to convert Mass of Xe-atom into grams.

$\frac{\text{mass of 1 atom}}{atom}=\frac{\text{mass of 1 mole of atom}}{6.022\times10^{23}atoms}$

$\text{mass of 1 atom}=\frac{\text{mass of 1 mole of atom}}{6.022\times10^{23}}$

$\text{mass of 1 Xe-atom}=\frac{131.29}{6.022\times10^{23}}$

$\text{mass of 1-Xe atom}=21.8017\times10^{-23}gms$

= $2.18017\times10^{-22}gms$

5 0

3 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.