Answer:
245 mL
Explanation:
Please see the step-by-step solution in the picture attached below.
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The Mass of oxygen in isolated sample is 8.6 g
<h3>What is the
Law of Constant composition?</h3>
The law of constant composition states that pure samples of the same compound contain the same element in the same ratio by mass irrespective of the source from which the compound is obtained.
Considering the given ascorbic acid samples:
Laboratory sample contains 1.50 gg of carbon and 2.00 gg of oxygen
mass ratio of oxygen to carbon is 2 : 1.5
Isolated sample will contain 2/1.5 * 6.45 g of oxygen.
Mass of oxygen in isolated sample = 8.6 g
In conclusion, the mass of oxygen is determined from the mass ratio of oxygen and carbon in the compound.
Learn more about the Law of Constant composition at: brainly.com/question/1557481
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Note that the complete question is given below:
A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains 1.50 g of carbon and 2.00 g of oxygen. Another sample of ascorbic acid isolated from citrus fruits contains 6.45 gg of carbon. According to the law of constant composition, how many grams of oxygen does this isolated sample contain?
Express the answer in grams to three significant figures.
8.47 g
Answer:
[OH-] = 1.0 x 10-10 M
Explanation:
The acidity of a solution can be determined directly from the concentration of the hydrogen ions and indirectly from the concentrations of the hydroxide ions.
Generally, for a neutral solution we have;
[H3O+] = [OH-] = 1.0 x 10-7 M
For an acidic solution;
[H3O+] > 1.0 x 10-7 M
[OH-] < 1.0 x 10-7 M
Comparing the options the correct option is;
[OH-] = 1.0 x 10-10 M
Answer:
we need 6.0 moles of zinc (Zn)
Explanation:
Step 1: Data given
Number of moles ZnO produced = 6.0 moles
Step 2: The balanced equation
2 Zn + O2 → 2 ZnO
For 2 moles Zinc we need 1 mol Oxygen to produce 2 moles Zinc oxide
Step 3: Calculate moles zinc
For 2 moles Zn we need 1 mol O2 to produce 2 moles ZnO
For 6.0 moles 2nO produced, we need 6.0 moles of zinc (Zn) and 3.0 moles of O2 to react.