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german
3 years ago
10

What is the product of the reaction of 3-isopropylcyclodecanol with hydrobromic acid? a. 1-bromo-3-isopropylcyclodecane b. 1-bro

mo-2-isopropylcyclodecane c. 1-bromo-3-isopropylcyclodecene d. none of these choices

Chemistry
1 answer:
zhenek [66]3 years ago
7 0

Answer: Correct answer is a - bromo-3-isopropylcyclodecane

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A diploid somatic ("body") cell has 2n = 20 chromosomes. At the end of mitosis, each daughter cell would have ______ chromosomes
kozerog [31]

Answer:

At the end of mitosis, 2n = 20

At the end of meiosis I, n = 10

At the end of meiosis II, n = 10

Explanation:

Mitosis is a type of cell division in which daughter cell produced are genetically identical to their mother cell. So, no. of chromosome does not change after mitosis.

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

Meiosis is a type of cell division in which mother cell produces two haploid cells ones with a single set of chromosomes.

Meiosis is a two step cell division, Meiosis I and Meiosis II.

In meiosis I, homologous pair separates, so no. of chromosomes becomes half.

In meiosis II, sister chromatids separates. So, the number of chromosomes remains same (i.e. Have same no. of chromosome as present in cell produced after meiosis I).

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

At the end of meiosis I, each daughter cell would have n = 10 chromosomes. At the end of meiosis II, each daughter cell would have n = 10 chromosomes.

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3 years ago
The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of
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Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = 1.83g/cm^3=1.83g/ml

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml

Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

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