can be oxidized to form carbon dioxide
Explanation:
Oxygen is important in the oxidation of glucose because it can be oxidized to form carbon dioxide. Oxidation of glucose involves the reaction of oxygen with glucose in a process called respiration. This gives a product of water, carbon dioxide and energy which is stored as ATP.
- Oxidation involves the addition of oxygen.
- Any specie that undergoes oxidation, is a reducing agent and it is said to be oxidized.
- Oxygen is oxidized to form carbon dioxide and water.
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Answer:
hexane C6H14
IS YOUR ANSWER
Explanation:
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Answer:
= 0.28M
Explanation:
data:
volume = 0.250 L
= 0.250dm^3 ( 1litre = 1dm^3)
moles = 0.70 moles
Solution:
molarity = 
= 0.70 / 0.250
molarity = 0.28 M
Answer:
Carbon has a total of four bonded pairs of electrons around it.
Explanation:
Since there are four "lines" around the C (which stands for Carbon), we can conclude that Carbon has a total of four bonded pairs of electrons around it.
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%
