Draw structure for each of the following a) m-Xylene b) o-Cresol c) 3-Bromo-4-nitrobenzenesulfonic acid
1 answer:
Answer:
Explanation:
a) m-Xylene
It is an aromatic hydrocarbon having molecular formula, C8H10. in this compound, two methyl groups are present at 1 and 3 positions of benzene ring.
b) o-Cresol
It is an organic compound having molecular formula, C7H8O. It is a substituted phenol in which a methyl group is present at ortho position.
c) 3-Bromo-4-nitrobenzenesulfonic acid
In this aromatic compound, three functional groups: -SO3H, -NO2 and -Br are present on the benzene ring at 1, 4 and 3 postion respectively.
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Answer: The percent yield is, 93.4%
Explanation:
First we have to calculate the moles of Na.
Now we have to calculate the moles of
The balanced chemical reaction is,
As, 1 mole of bromine react with = 2 moles of Sodium
So, 0.189 moles of bromine react with = moles of Sodium
Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.
As, 1 mole of bromine give = 2 moles of Sodium bromide
So, 0.189 moles of bromine give = moles of Sodium bromide
Now we have to calculate the percent yield of reaction
Therefore, the percent yield is, 93.4%
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose =
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution =
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution =
Volume of the solution taken =
Molarity of the solution after dilution =
Volume of the solution after dilution=
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose =
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.