Answer:
determining how many ozone molecules are lost in the atmosphere
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Answer:
Explanation:
Hello,
In this case, by using the ideal gas equation, we first compute the moles of oxygen at the given volume, pressure and temperature:
Then, since molar mass of gaseous oxygen is 32 g/mol, we compute the contained mass in grams as shown below:
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Answer:
When a Magnesium Ribbon is burnt, a powdery substance called magnesium oxide is formed.
Explanation:
There has obviously been a chemical change because several chemical properties of the magnesium have been modified: the color, the texture and the mass.
The increase in mass is due to the fact that oxygen from the air has combined with the magnesium to make magnesium oxide, MgO.
The chemical equation, Mg + O2 MgO shows this reaction but it needs to be balanced to make 2Mg + O2 2MgO.
Using stoichiometry, we can convert this eqation into an equation with moles:
2 mol Mg + 1 mol O2 2 mol MgO.
Next, we convert to grams using atomic masses obtained from the periodic table:
48g Mg + 32g O2 80g MgO
Lastly, we determine the same thing in the proportions we used. In other words, we used only 0.15g of Mg (not 48g) so everything needs to be divided by 320. So 80 / 320 = 0.25 g. If we burn 0.15 g of Mg, we obtain 0.25 g of MgO.
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This is my first answer.
Assuming we have 100g, this means that
39.97g Carbon * 1 mol / 12 g = 3.33 mol Carbon
13.41g Hydrogen * 1 mol/1 g = 13.41 mol Hydrogen
46.62g Nitrogen * 1 mol / 14 g = 3.33 mol Nitrogen
Dividing everything by 3.33, we get
1 mol Carbon, 4.03 mol Hydrogen, 1 mol Nitrogen.
Empirical formula is CH4N
<span>The mass of the empirical formula is
12 + 4 + 14 = 30
Since the molar mass is double, we multiply all our subscripts
The molecular formula is C2H8N2
The answers to this question are </span><span>an empirical formula of CH4N</span> and a molecular formula of C2H8N2 .
