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2 answers:
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Answer:- 3.84 grams
Solution:- Volume of the sample is 44.8 mL and the density is 1.03 gram per mL.
From the density and volume we calculate the mass as:
mass = volume*density
= 46.1 g
From given info, potassium bromide solution is 8.34% potassium bromide by mass. It means if we have 100 grams of the solution then 8.34 grams of potassium bromide is present in. We need to calculate how many grams of potassium bromide are present in 46.1 grams of the solution.
The calculations could easily be done using dimensional analysis as:
= 3.84 g KBr
Hence, 3.84 grams of KBr are present in 44.8 mL of the solution.
Answer:
See explanation
Explanation:
In this question, we have to follow the IUPAC rules. Lets analyze each compound:
a. 1-methylbutane
In this compound we have a chain of 5 carbons, so the correct name is <u>Pentane.</u>
b. 1,1,3-trimethylhexane
In this compound, we longest chain is made of 7 carbons, so, we have to use the name "heptane". Carbon one would be the closet one to the methyl group, so the correct name is <u>2,4-dimethylheptane.</u>
c. 5-octyne
In this case, carbon 1 would be the closet one to the triplet bond. With this in mind, the correct name is <u>oct-3-yne.</u>
d. 2-ethyl-1-propanol
In this compound, we longest chain is made of 4 carbons, so, we have to use the name "butane". Carbon one would be the carbon with the "OH" group, so the correct name is <u>2-methylbutan-1-ol.</u>
<u>e. 2.2-dimethyl-3-butanol</u>
In this case, carbon 1 would be the closet one to the "OH". With this in mind, the correct name is <u>3,3-dimethylbutan-2-ol.</u>
See figure 1
I hope it helps!
