The percent yield is 71.3 %.
Explanation:
Percent yield is the measure to analyze the success percentage of any experiment .The percent yield of any experiment can be obtained by the ratio of actual or experimental value to expected or theoretical value multiplied with 100.
So, in the present problem, we have obtained 10.7 g of adamantium nitrate from Wolverine's 10 pound claws. So the actual value or the experimental value is the amount of adamantium nitrate obtained from Wolverine's claws.
Thus, the experimental outcome is 10.7 g. While we had expected to recover 15 g of adamantium nitrate. So the theoretical outcome is 15 g.
Thus, the percent yield is 71.3 %.
Among the given options, Option (D) that is O, 
will have the same number of protons. Hence option (D) is correct.
<h3>What are
protons ?</h3>
A stable subatomic particle occurring in all atomic nuclei, with a positive electric charge equal in magnitude to that of an electron.
In option (D), All the three species (O, ) bears 8 protons but the number of electrons varies.
In any Ion, there will be no change in number of protons among the same atoms but number of electrons varies.
Therefore, Option (D) has same atoms but different ions.
Hence, it has same number of protons. Thus the correct answer is option (D)
Learn more about Protons here ;
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Answer:
Option b is show the chemical property of sodium....
Answer:
3: Lactone
Explanation:
Lactones are defined as carboxylic esters that contain the structure (−C(=O)−O−) which is essentially showing that an ester has now become part of the chemical structure of the ring.
Thus, looking at the question, it has the structure as defined in Lactones.
Thus, we can say that the compound is a Lactone.
Answer:
Explanation:
I'll assume 1.49 is liters.
All gases occupy 22.4 L for each mole of gas at STP. This makles a very useful, and important, conversion factor:
(22.4L/mole gas) for all gases at STP.
Therefore 1.49L of H2 would be:
(1.49L H2)/(22.4L/mole gas) = 0.0665 moles
The molar mass of H2 is 2, so:
(0.0665 moles)*(2g/mole) = 0.133 grams to 3 sig figs
