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Ket [755]
4 years ago
7

A 4.215 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of oxygen gas, producing 9.58

2 g CO2 and 3.922 g H2O. What percent by mass of oxygen is contained in the original sample?
Chemistry
1 answer:
Scilla [17]4 years ago
5 0

Answer:

\% O=27.6\%

Explanation:

Hello,

In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:

- Moles of carbon are contained in the 9.582 grams of carbon dioxide:

n_C=9.582gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}  =0.218molC

- Moles of hydrogen are contained in the 3.922 grams of water:

n_H=3.922gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.436molH

- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:

m_O=4.215g-0.218molC*\frac{12gC}{1molC} -0.436molH*\frac{1gH}{1molH} =1.163gO

Finally, we compute the percent by mass of oxygen:

\% O=\frac{1.163g}{4.215g}*100\% \\\\\% O=27.6\%

Regards.

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3 years ago
Give the nuclear symbol (isotope symbol) for the isotope of bromine that contains 44 neutrons per atom. nuclear symbol:_________
Savatey [412]

Answer:79

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Explanation:

In the nuclear symbol of an element, the superscript represents the mass number. The mass number is the sum of the number of protons and the number of neutrons present in the atom. The subscript is the atomic number which is the number of protons present in the atom or the number of electrons in the neutral atom.

4 0
3 years ago
How many moles of iron (Fe) will be produced from 6.20 moles of carbon monoxide (CO) reacting with excess iron (III) oxide (Fe2O
bonufazy [111]

Answer:

4.13 moles of Fe

Explanation:

Step 1: Write the balanced equation

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8 0
3 years ago
1. The sodium-iodide symporter plays a role in the accumulation of iodide in the thyroid gland. Here, one iodide gets converted
Serggg [28]

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Answer is: a) 1s²2s²2p⁶3s¹ (sodium).

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