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3 years ago

13

Which group of elements are the most important for living organisms?\

2 answers:

Drupady [299]3 years ago

6 0

B- Carbon nitrogen and oxygen. All organisms are made up of carbon, oxygen is essential to breath, and nitrogen makes up most of earth's atmosphere.

Serhud [2]3 years ago

4 0

The answer is b carbon nitrogen and oxygen

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when you put a book on a table the table pushes on the book is that newton's 1st,2nd,or 3rd law of motion ?

ohaa [14]

The table pushes the book and the book pushes the table
It's 3rd law

8 0

3 years ago

What happens if you cross a heterozygous organism with a heterozygous organism?

maw [93]

If you cross a heterozygous organism with another heterozygous organism, pair two Tt and Tt. Below is a punnet square to show the result:

T t
T TT Tt
t Tt tt

This means that you have two heterozygous recessive and one homogenous dominant and one homogenous recessive.

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3 years ago

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago

Another example of a parasitic

sasho [114]

Answer:a leech

Explanation: a parasitic relationship is one organism benefit on the other while the other is harmed

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3 years ago

Determine the mass of the H2CO3 produced.

FrozenT [24]

Answer:

62.03 g/mol

Explanation:

7 0

3 years ago