Mass of 60.009 g and radius of 1.02 cm what is the density of the sphere in (g/ml)

1 answer:

4 0

Density is given as mass / volume.

Mass is the sphere is 100 g.

Volume of the sphere = (pi∗r3)∗4/3
(
p
i
∗
r
3
)
∗
4
/
3

=(4∗22∗3∗3∗3)/(7∗3)cm3
=
(
4
∗
22
∗
3
∗
3
∗
3
)
/
(
7
∗
3
)
c
m
3

=792/7
=
792
/
7
cm3
3

Therefore, Density is 100/(792/7)g/cm3
100
/
(
792
/
7
)
g
/
c
m
3

Which gives: density = 0.883838 g/cm3
g
/
c
m
3

If you want to change the units to kg per cubic metres, then we need to divide this value by 1000( for g to kg) and multiply by 100 * 100 * 100 (for cm to m).

This makes the density to be 883.83 kg/m3

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What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?

zubka84 [21]

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g
H: 16.3% = 16.3 g

Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

$C: \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol$

$H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol$

We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:

$C: \dfrac{6.975}{6.975} = 1$

$H: \dfrac{16.3}{6.975} \approx 2.3$

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, t<span>hus, the minimum or empirical formula found for the compound will be:
</span>
$\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark$

I hope this helps. =)

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The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab. How many moles of dextrose is this equivalent t

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<h3>What are moles?</h3>

A mole is described as 6.02214076 × 1023 of a few chemical unit, be it atoms, molecules, ions, or others. The mole is a handy unit to apply due to the tremendous variety of atoms, molecules, or others in any substance.

To calculate molar equivalents for every reagent, divide the moles of that reagent through the moles of the restricting reagent. The calculation is follows:

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