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2 years ago

15

Human offspring begin with 1 cell that contains 46 chromosomes. This is because each sperm cell contains _____ chromosomes and e

ach egg cell contains _____ chromosomes.
A. 23; 23
B. 23; 46
C. 46; 46
D. 46; 69

1 answer:

Elza [17]2 years ago

3 0

Answer:

A

Explanation:

A man has to give 23 and women has to also give 23.

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If the average atomic mass of hydrogen in nature is 1.0079, what does that tell you about the percent composition of H-1 and H-2

vovangra [49]

Answer:

That the isotope H-1 is the most abundant in nature.

Explanation:

Hello!

In this case, since the average atomic mass of an element is computed considering the mass of each isotope and the percent abundance each, for hydrogen we would set up something like this:

$m_H=m_{H_1}*\%abund_{H_1}+m_{H_2}*\%abund_{H_2}$

Moreover, since the isotope notation H-1 and H-2 means that the atomic mass of H-1 is 1 amu, that of H-2 is 2 amu and the average one is 1.0079 amu, we can infer that the most of the hydrogen in nature is H-1 as the most of it composes the average hydrogen atom.

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2 years ago

Under the same conditions of temperature and pressure, which of the following gases would behave most like an ideal gas? A)He(g)

umka21 [38]

Answer: He(g)

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2 years ago

When 0.100 mol of carbon is burned in a closed vessel with8.00

antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of $O_2$ = 8.00 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of CO = 28 g/mole

First we have to calculate the moles of $O_2$ .

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C+O_2\rightarrow 2CO$

From the balanced reaction we conclude that

As, 2 mole of $C$ react with 1 mole of $O_2$

So, 0.1 moles of $C$ react with $\frac{0.1}{2}=0.05$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO$

From the reaction, we conclude that

As, 2 mole of $C$ react to give 2 mole of $CO$

So, 0.1 moles of $C$ react to give 0.1 moles of $CO$

Now we have to calculate the mass of $CO$

$\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO$

$\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g$

Therefore, the mass of carbon monoxide form can be 2.8 grams.

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