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.
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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:
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Answer: HCl
Explanation:
calcium carbonate dissolves in HCl acid producing CO 2 gas. It will not dissolve in pure water. The Ksp for calcium carbonate in water is 3.4 x 10-9 moldm-3 which is very low. What takes place here is actually a chemical reaction:
CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + H 2CO 3(aq)
This reaction accounts for the solubility of the Calcium carbonate in HCl and not in pure water.
The volume of the dry gas at stp is calculated as follows
calculate the number on moles by use of PV =nRT where n is the number of moles
n is therefore = Pv/RT
P = 0.930 atm
R(gas contant= 0.0821 L.atm/k.mol
V= 93ml to liters = 93/1000= 0.093L
T= 10 + 273.15 = 283.15k
n= (0.930 x0.093) /(0.0821 x283.15) = 3. 72 x10^-3 moles
At STp 1 mole = 22.4L
what about 3.72 x10^-3 moles
by cross multiplication
volume = (3.72 x10^-3)mole x 22.4L/ 1 moles = 0.083 L or 83.3 Ml