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Andreyy89
3 years ago
13

If the world suffered a huge catastrophic event (an event causing great and often sudden damage or suffering; a disaster) that w

iped out almost all the organisms. Who would have the best chance of repopulation the world with living organisms, an asexual reproducing organism or a sexually reproducing organism? Explain your answer.
Chemistry
1 answer:
Tom [10]3 years ago
3 0

Answer:

An asexually reproducing organism would have the best chance of repopulating the world.

Explanation:

Sexual reproduction is a method of reproduction that involves two parents and the formation and subsequent fusion of male and female gametes during fertilization. On the other hand, In asexual reproduction, only one parent is involved and produces offspring by making copies of itself through various methods such as binary fusion, budding and vegetatively.

In a case of a catastrophic event that wiped out almost all organisms, an organism that reproduces asexually have a greater chance of repopulating the world because of the following advantages in reproducing asexually:

1. Only one parent is needed, therefore reproduction is easier.

2. It isaves time and energy. Since almost all organisms have been wiped out, there is a lesser likelihood of finding a compatible mate, so reproducing asexually will give the organism a better chance.

3. It occurs at a very fast rate; much faster than sexual reproduction

4. The population can increase very quickly and colonization of new habitats is much easier.

All these advantages will give the esexually reproducig organism a fairer chance to repopulate the world.

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Calculate the mass of sodium phosphate in aqueous solution to fully react with 37 g of chromium nitrate(III) an aqueous solution
Alla [95]

Answer:

41 g

Explanation:

The equation of the reaction is;

Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)

Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles

1 mole of sodium phosphate reacts with 1 mole of chromium nitrate

x moles of sodium phosphate react as with 0.25 moles of chromium nitrate

x= 1 × 0.25/1

x= 0.25 moles

Mass of sodium phosphate = 0.25 moles × 163.94 g/mol

Mass of sodium phosphate = 41 g

4 0
3 years ago
1) 120KM/H A M/S Y CM/S
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4) would be your correct answer
8 0
2 years ago
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Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-c
IceJOKER [234]

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene  

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

5 0
3 years ago
Find mass of 3 moles of water​
EleoNora [17]

Answer:

54 g

Explanation:

1 mole of water = H2O

mass of 1 mole of H2O= mass of h2 + mass of o

= 2× mass of h +mass of o

= 2×1+16 =18 g

1 mole of water = 18g

3moles of water = 18×3g= 54g

6 0
3 years ago
7. Iron combines with 4.00 g of Copper (11) nitrate to form 6.01 g of Iron (I) nitrate and 0.400 g copper
olga55 [171]

Answer:

222325332

Explanation:

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3 years ago
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