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
#SPJ1
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:
B
Explanation:
i think leter B it because
Answer:
it could be 25degrees c... not sure
The intermolecular bonding for HF is van der Waals, whereas for HCL, the intermolecular bonding is hydrogen. Since the van der Waals bond is stronger than hydrogen, HF will have a higher boiling temperature. Since the covalent bond is stronger than van der Waals, HF will have a higher boiling temperature.
Answer : The
pH of a solution is, 8.56
Explanation : Given,

Concentration of ammonia (base) = 0.10 M
Concentration of ammonium nitrate (salt) = 0.55 M
First we have to calculate the value of
.
The expression used for the calculation of
is,

Now put the value of
in this expression, we get:



Now we have to calculate the pOH of buffer.
Using Henderson Hesselbach equation :
![pOH=pK_b+\log \frac{[Salt]}{[Base]}](https://tex.z-dn.net/?f=pOH%3DpK_b%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BBase%5D%7D)
Now put all the given values in this expression, we get:


The pOH of buffer is 5.44
Now we have to calculate the pH of a solution.

Thus, the pH of a solution is, 8.56