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3 years ago

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Which compound exhibits both cis trans and optical isomerism? A: CH3CH=CHCH2CH3B: CH3CHBrCH=CH2C: CH3CBr=CBrCH3D CH3CH2CHBrCH=CH

Br?

1 answer:

irinina [24]3 years ago

4 0

CH3CHBrCH=CBrCH3D
I think

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A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na

s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f: after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

$Normality=molarity \times n-factor$

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point: $N_1V_1=N_2V_2$

Before equivalence point : $N_1V_1>N_2V_2$

After the equivalence point: $N_1V_1$

$N_1V_1=100\times1=100$

case a: 5.00 mL of 1.00 M NaOH

$N_2V_2=5\times1=5$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

$N_2V_2=100\times1=100$

$N_1V_1=N_2V_2$ hence it is equivalence point

case c: 10.0 mL of 1.00 M NaOH

$N_2V_2=10\times1=10$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

$N_2V_2=150\times1=150$

$N_1V_1$ hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

$N_2V_2=50\times1=50$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

$N_2V_2=200\times1=200$

$N_1V_1$ hence it is after the eqivalence point

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3 years ago

A student is interested in resource recovery through composting. Which field

sdas [7]

I think it’s microbiology, since it would help the student understand the process of composting the best. Botany is plants- not for composting. Geophysics- he’s not researching tectonic plates. Oceanography- he isn’t interested in the marina trench

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2 years ago

Convert 0.02 g/mL to the unit g/L.

Vilka [71]

0,02 g/mL = 20 g/L

:)

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3 years ago

Shifts in the rock layer locations cannot account for gaps in the rock record<br><br> true or false

vivado [14]

Answer: That would be false because it is the contact between two layers representing a gap in the geologic record, usually from the erosion of the layers which would normally be expected to appear.

Explanation:

Have a good day

I hope this helps if not sorry :(

Stay motivated

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3 years ago

Which organism would have the most variation in the DNA of its offspring?

Arada [10]

The organism that would have the most variation in the DNA of its offspring is the cat (Option C). Meiosis is a type of cell division that generates more genetic variability than asexual types of reproduction.

Meiosis is a type of reductional cell division by which a parental cell produces 4 daughter cells (gametes), each containing half of the genetic material.

Animals (e.g., cats) generate gametes by meiosis which fuse during fertilization to produce new offspring.

Both amoeba and bacteria reproduce by a type of asexual reproduction called binary fission. Moreover, yeasts also reproduce asexually by a process called budding and fission.

Both asexual and sexual types of reproduction generate genetic variability by the emergence of new mutations in daughter cells.

Meiosis generates much more genetic variability than asexual types of reproduction due to two different processes:

Random assortment of chromosomes, which produces new allele combinations.

Recombination, i.e., by the exchange of genetic material (DNA) between non-sister chromatids during Prophase I.

Learn more in:

brainly.com/question/7002092

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3 years ago