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Ludmilka [50]
3 years ago
11

Antarctica, almost completely covered in ice, has an area

Chemistry
1 answer:
Marysya12 [62]3 years ago
5 0

<u>Answer:</u> The mass of ice is 2.39\times 10^{22}g

<u>Explanation:</u>

We are given:

Area of Antarctica = 5,500,000mi^2=5,500,000\times 2.59\times 10^{10}=142.45\times 10^{15}cm^2      (Conversion factor:  1mi^2=2.59\times 10^{10}cm^2  )

Height of Antarctica with ice = 7500 ft.

Height of Antarctica without ice = 1500 ft.

Height of ice = 7500 - 1500 = 6000 ft = 182.88\times 10^3cm     (Conversion factor:  1 ft = 30.48 cm)

To calculate mass of ice, we use the equation:

\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}

We are given:

Density of ice = 0.917g/cm^3

Volume of ice = Area × Height of ice = 142.45\times 10^{15}cm^2\times 182.88\times 10^3cm=26051.26\times 10^{18}cm^3

Putting values in above equation, we get:

0.917g/cm^3=\frac{\text{Mass of ice}}{26051.26\times 10^{18}cm^3}\\\\\text{Mass of ice}=(0.917g/cm^3\times 26051.26\times 10^{18}cm^3=2.39\times 10^{22}g

Hence, the mass of ice is 2.39\times 10^{22}g

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Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO
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Moles of O₂ = 4.93

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Mass of  produced H₂O = ?

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NH₃  :   NO                                   NH₃  :   H₂O

4     :    4                                          4    :      6

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Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:

O₂  :   NO                                               O₂ :   H₂O

5     :    4                                                  5     :    6

4.93   :   4/5×4.93 = 3.944 mol               4.93  : 6/5 × 4.93 = 5.916 mol

we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.

Mass of water = number of  moles × molar mass

Mass of water = 3.285 mol × 18 g/mol

Mass of water = 59.13 g

Mass of nitrogen monoxide  = number of  moles × molar mass

Mass of nitrogen monoxide = 2.19 mol × 30 g/mol

Mass of nitrogen monoxide = 65.7 g

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