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3 years ago

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A fertilized egg receives 50% of its chromosomes from a sperm and 50% from the egg. Which process ensures that the sperm and egg

contain only 50% of the chromosomes of an organism?
A.fertilization
B.meiosis
C. mitosis
D.mutation

1 answer:

Helen [10]3 years ago

5 0

Answer:

B

Explanation:

Meiosis ensures that the sperm and egg contain only 50-50.10% of the chromosomes of an organism.

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When 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because:

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7.5 mol of hydrogen would be needed to consume the available nitrogen.

Explanation:

When hydrogen reacts with nitrogen, ammonia is formed as shown below;

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As seen from the equation, every 3 moles of H₂ react with a mole of N₂ to form 2 moles of NH₃.

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Because the molar mass of nitrogen gas is approximately 28g/mol, 70g of nitrogen gas would be 2.5 moles.

The reaction ratio of nitrogen to hydrogen in the reaction is 1 : 3. The reaction would require 2.5 * 3 (7.5) moles of hydrogen for a complete reaction.

However since there are only 7g on hydrogen, (Remember 1 mole of H₂ is approximately 2g), the available moles of H₂ is 7 / 2 = 3.5

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The element chlorine has two stable isotopes, chlorine-35 with a mass of 34.97 amu and chlorine-37 with a mass of 36.95 amu. Fro

sergejj [24]

<u>Answer: </u>One isotope has a percentage abundance of 75.75 % and the percentage abundance of another isotope is 24.24%.

<u>Explanation:</u>

We are given the two stable isotopes of chlorine with their respective masses. The average atomic mass of chlorine is also given.

Average atomic mass of chlorine = 35.45 amu.

Let us assume the fractional abundance of one isotope be 'x' and the fractional abundance for another isotope will be (1 - x) because the total fractional abundance is always equal to 1.

For Chlorine-35 isotope:

Fractional abundance = x

Mass = 34.97 amu

For Chlorine-37 isotope:

Fractional abundance = 1 - x

Mass = 36.95 amu

The formula for the calculation of average atomic mass is given by:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

Putting values in above equation, we get:

$35.45=34.97(x)+36.95(1-x)\\\\35.45=34.97x+36.95-36.95\\\\-1.5=-1.98x\\\\x=\frac{-1.5}{-1.98}=0.7575$

$1-x=1-0.7575=0.2424$

Converting these two fractional abundances into percentage abundances by multiplying it with 100.

$x=0.7575\times 100=75.75\%\\\\1-x=0.2424\times 100=24.24\%$

Hence, one isotope has a percentage abundance of 75.75 % and the percentage abundance of another isotope is 24.24%.

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