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Sever21 [200]
3 years ago
9

Copper metal reacts with hot concentrated sulfuric acid solution to form aqueous copper(ii) sulfate, sulfur dioxide gas, and wat

er. express your answer as a balanced chemical equation. identify all of the phases in your answer.
Chemistry
1 answer:
Alex777 [14]3 years ago
3 0
The unbalanced chemical equation with the corresponding phase for each substance is shown below: 

Cu₍s₎ + H₂SO₄₍aq₎ → CuSO₄₍aq₎ + SO₂₍g₎ + H₂O₍l₎

The copper metal is solid, the sulfuric acid and copper sulfate are in aqueous solution. Water is in the liquid phase while sulfur dioxide is in the gas phase. 

In order to obtain the balanced equation, we must ensure that the number of atoms present in the reactant side is equal to that in the product side. First, we list the elements involved and count their number on each side.

Reactant           Product
   1             Cu      1
   2              H       2
   1              S       2
   4              O       5

To balance the equation, we need to have 2 moles of H₂SO₄ and 2 moles of H₂O. The balanced chemical equation is then:

Cu₍s₎ + 2H₂SO₄₍aq₎ → CuSO₄₍aq₎ + SO₂₍g₎ + 2H₂O₍l₎

Reactant           Product
   1             Cu      1
   4              H       4
   2              S       2
   8              O       8
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Earth's oceans have an average depth of 3800 m, a total surface area of 3.63 x 108 km2, and an average concentration of dissolve
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a) Mass of gold in ocean = 8.1*10¹² g

b) Volume of gold = 4.2*10¹¹ cm³

c) Value of gold = $ 4.1*10¹³

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a) Depth of ocean = 3800 m

Surface area of ocean = 3.63*10⁸ km²

Now, 10⁶ m² = 1 km²

Therefore, the area in units of m² = 3.63*10¹⁴ m²

Volume\ of\ ocean = Depth*Area = 3800m*3.63*10^{14} m2=1.4*10^{18} m3

Average concentration of gold in ocean = 5.8*10⁻⁹ g/L

Now, 1 g/L = 1000 g/m³

Therefore, concentration of gold in g/m³ = 5.8*10⁻⁶ g/m³

Mass\ of\ gold =1.4*10^{18} m^{3} *5.8*10^{-6} g/m3 = 8.1*10^{12} g

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Density of gold = 19.3 g/cm³

Volume\ of\ gold = \frac{Mass}{density} =\frac{8.1*10^{12} }{19.3} =4.2*10^{11} cm^{3}

c) Price of gold = $1595/troy oz

Unit conversion

1 troy oz = 311 g

Therefore, 8.1*10¹² g of gold is equivalent to:

\frac{8.1*10^{12} g* 1\  troy\ oz}{311g} =2.6*10^{10} \ troy\ oz

The value of gold in the ocean is:

\frac{2.6*10^{10} troy\ oz*1595\ dollars}{1\ troy\  oz} =4.1*10^{13} \ dollars

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