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soldi70 [24.7K]
3 years ago
10

HELP ASAP 20 PTS!!!

Chemistry
1 answer:
sashaice [31]3 years ago
7 0
Basically they are asking you to find the element that is most closely associated with these clues.

1. Fireworkium: 15 protons and 15 electrons? No problem. We can see that the atomic number is 15, in which in real life, the element is Phosphorous. (Phosphorous is a common element in fireworks for an explosive personality and colors)

2. Toothium: A nonmetal with average mass number close to 19? We check the periodic table once more and find that Fluorine is our answer. (Notice that fluorine is in toothpaste, which is why they decided the name Toothier)

3. Breathium: Now honestly, I didn't even have to check the periodic table but just realize that we breath Oxygen. But if you want the straight answer, Oxygen is a nonmetal which has an atomic mass close to 16, and surprise, it has one less proton than Fluorine.

4.Lightium: Immediately we see that since this gas does not react with other elements, it is a noble gas. If it glows in a lamp or vacuum tube, we can also see it is of a light origin. If it has one more proton than Toothium (Fluorine) than we can immediately see that in the periodic table, our answer is Neon.

5. Rottoneggium: We can immediately guess sulfur. The straight way shows that it has 16 electrons, and the non-ion atom should also have 16 protons. Therefore, our answer is Sulfur.

6. I just realized I forgot Floatium: If it has 2 protons, we look in the periodic table and see that our answer is Helium.
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Hello,

In this case, given the reaction, we can directly compute the theoretically yielded grams of carbon dioxide, considering the 2:2 molar ratio between carbon monoxide (molar mass = 28 g/mol) and carbon dioxide (molar mass = 44 g/mol) and the initial reacting grams of carbon monoxide in excess oxygen:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}*\frac{44gCO_2}{1molCO_2}   =15.7gCO_2

Thus, as only 8 g were actually yielded, we compute the percent yield:

Y=\frac{8g}{15.7g}*100\% \\\\Y=50.9\%

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