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Whitepunk [10]
3 years ago
7

Compare natural selection to evolution. Be sure to justify your response in two or more complete sentences in the essay

Chemistry
1 answer:
Wittaler [7]3 years ago
5 0

Answer:

Natural selection is a process in which organisms are fit to adapt and live in the environment. Natural selection was an idea developed by Charles Darwin, Darwin contributed this idea when he was studying birds and tortoises at the Galapagos Islands. He then wrote his famous book, <em><u>O</u></em><em><u>r</u></em><em><u>i</u></em><em><u>g</u></em><em><u>i</u></em><em><u>n</u></em><em><u> </u></em><em><u>o</u></em><em><u>f</u></em><em><u> </u></em><em><u>S</u></em><em><u>p</u></em><em><u>e</u></em><em><u>c</u></em><em><u>i</u></em><em><u>e</u></em><em><u>s</u></em><em><u>.</u></em> Evolution is also very similar, evolution is a process in which organisms reproduce and mutate in order to survive and adapt. As you can see, natural selection and evolution are alike in many ways.

☆anvipatel77☆

•Expert•

Brainly Community Contributor

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The answer is C. One item replacing another is a single replacement, but when two items trade places in two separate molecules it’s double replacement.
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3 years ago
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1. If you have a sample of gas at a pressure of 16 atm, what will the pressure be if the volume is halved?
vagabundo [1.1K]

Answer:

1. The pressure will be 32 atm, twice the initial pressure.

2. The pressure will be 1.83 atm, one third of the initial pressure.

Explanation:

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.

This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1=P2*V2

1. In this case:

  • P1= 16 atm
  • V1
  • P2= ?
  • V2= V1÷2= \frac{V1}{2} because the volume is halved.

So:

16 atm*V1= P2* \frac{V1}{2}

Solving:

\frac{16 atm*V1*2}{V1}=P2

16 atm*2= P2

32 atm= P2

<u><em>The pressure will be 32 atm, twice the initial pressure.</em></u>

2. Now

  • P1= 5.5 atm
  • V1
  • P2= ?
  • V2= V1*3 because the volume is tripled.

So:

5.5 atm*V1= P2* V1*3

Solving:

\frac{5.5 atm*V1}{3*V1}=P2

\frac{5.5 atm}{3}= P2

1.83 atm= P2

<u><em>The pressure will be 1.83 atm, one third of the initial pressure.</em></u>

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3 years ago
Breaking larger molecules into smaller molecules and carbon
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Answer: catabolism

Explanation:

Catabolism refers to the set of metabolic pathways which is necessary for the breaking down of molecules into smaller units. This is then oxidized for the release of energy or can be used to perform other anabolic reactions.

Catabolism is regarded as the opposite direction of anabolism which is simply the building-up of molecules. It should be noted that anabolism and catabolism work together in every living organisms and perform functions such as the production of energy and the repair of cells.

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3 years ago
Sodium hydroxide reacts with aluminum and water to produce hydrogen gas:2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2
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Answer: 0.147 grams

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{1.33g}{27g/mol}=0.049moles

\text{Moles of} NaOH=\frac{4.25g}{40g/mol}=0.106moles

2Al(s)+2NaOH(aq)+6H_2O(l)\rightarrow Na(Al(OH)_4(aq)+3H_2(g)

According to stoichiometry :

2 moles of Al require = 2 moles of NaOH/tex]Thus 0.049 moles of [tex]Al will require=\frac{2}{2}\times 0.049=0.049moles  of NaOH

Thus Al is the limiting reagent as it limits the formation of product and NaOH is the excess reagent.

As 2 moles of Al give = 3 moles of H_2

Thus 0.049 moles of Al give =\frac{3}{2}\times 0.049=0.0735moles  of H_2

Mass of H_2=moles\times {\text {Molar mass}}=0.0735moles\times 2g/mol=0.147g

Thus 0.147 g of H_2 will be produced from the given masses of both reactants.

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Answer:

31 moles

Explanation:

The balanced combustion reaction of the wax, C_{31}H_{64} is shown below as:

C_{31}H_{64}+47O_2\rightarrow 31CO_2+32H_2O

As seen from the reaction,

1 mole of wax, C_{31}H_{64} on combustion produces 31 moles of carbon dioxide, CO_2

<u>Hence, moles of CO_2 when 1 mole of wax, C_{31}H_{64} is burnt = 31 moles</u>

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