Question

Drag conversion units onto the boxes in the equation to make conversions. Some boxes can be left empty. Click on a
unit to remove it from its position.
1 cmCu
9 g Cu
9.5 x 1021 atoms Cu
1 g Cu
1 kg
1000 g
1 cm
1 ml
1
1000 cm3
70.0 kg x
X
X
II
70.0 kg•L

Answer 1

We have that 70kg of Copper Cu is having a volume of $V=7.76442liters$

1 cmCu

=9 g Cu

9.5 x 10^21 atoms Cu

=1 g Cu

1 kg  =1000 g

1 cm

=1 ml

1

=1000 cm3

70.0 kg x  X  _X  II

70.0 kg•L

Generally We First determine the No of Atoms in 70kg of cu

No of Atoms of c$u=\frac{weight}{atomic mass of cu}*n$

Where

n=Avogadro's Constant

$n= 6.02214076 x 10^{23}$

Therefore

No of Atoms of cu$=\frac{70*10^3}{63.5}*6.02214076 * 10^{23}$

No of Atoms of cu=6.63858037x10^{26} atoms of Cu

Hence

We determine the the grams of copper derived from $6.63858037x10^{26}$atoms of Cu

No of grams of cu$=\frac{6.63858037x10^{26}}{9.5 x 10^{21}}*1$

No of grams of cu=69879.79337 g of Cu

Generally

No of grams of Cu=liters

Where

9 g of Cu= 1cm^3 Cu

Therefore

Volume of cu

$V=(69879.79337)*\frac{1}{9}*\frac{1}{1000}$

$V=7.76442liters$

In conclusion

70kg of Copper Cu is having a volume of $V=7.76442liters$

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Answer 2

We are to drag the given conversion units onto the boxes in the equation in order to make conversions in the boxes that are left empty.

The dimensional analysis for the conversion ratio is given as follows

1cm^3 Cu = 9g Cu        9.5*10^21  atoms of Cu = 1 g Cu

1 kg     =  1000 g                                       1 cm^3 = 1 mL

1 L      =  1000 cm^3

 Here is the box equation

        box         box        box

70 g    ×-----  ×  ------   × -----  =  

        box         box        box

From the given information, the volume of the Cu sample in Liters is:

= 7.765  liters

 From the parameters given:

• the mass of the sample of Cu = 70 kg

The first thing we are to do is to calculate the number of atoms of Cu in 70 kg by using Avogadro's constant.

We know that the number of atoms of Cu is:

$\mathbf{= \dfrac{mass \ weight \ of \ Cu}{atomic \ mass \ of \ cu}\times avogadro's \ constant}$

where;

• weight in Cu = $\mathbf{70 * 10^{3}}$
• the atomic mass of Cu = 63.5
• Avogadro's constant = $\mathbf{6.023 \times 10^{23} \ Cu \ atoms}$

the number of Cu atoms is:

$\mathbf{= \dfrac{70*10^3}{63.5}\times 6.023 * 10^{23}}$

$\mathbf{= 6.6395 \times 10^{26} \ atoms \ of \ Cu}$

However, from the given conversion ratios:

If $\mathbf{9.5\times10^{2}}$  atoms of Cu = 1 gram of Cu  

∴

atoms of Cu will be:

$\mathbf{= \dfrac{6.6395 \times 10^{26} \ atoms \ of \ Cu \times 1 \ gram \ of \ Cu}{9.5*10^{21} \ atoms \ of \ Cu }}$

$\mathbf{= 0.69889 *10^{5} \ grams \ of \ Cu}$

Since:

• 1cm³ Cu = 9g Cu and;    
• 1 L      =  1000 cm³

∴

The correct setup of the sample in the box can be computed as follows:

$\mathbf{= 0.69889 *10^{5} \ gram \ of \ Cu \ \times \dfrac{1 cm^3 \ of \ Cu }{9 \ g \ of \ Cu }\times \dfrac{1 \ Liter}{1000 \ cm^3}}$

= 7.765  liters

Therefore, we can conclude that the Volume of Cu sample in liters is:

= 7.765 liters

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