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vagabundo [1.1K]
3 years ago
9

Some molecules in air contain . Explain how atoms from the air can become part of human cells

Chemistry
1 answer:
Tcecarenko [31]3 years ago
5 0

The atoms in air are combined with oxygen present in the air and when we inhale the air, oxygen is moved into the body then atoms in oxygen are carried by red blood cells the blood is pumped to the lungs and when oxygen is transported to the body organ, the atoms in lungs become the part of human cells.

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Most chemical changes are _________ and cannot be reversed .
Pachacha [2.7K]

Answer:

irreversible changes

Explanation:

Most chemical changes are permanent and cannot be reversed. That is because a chemical change involves changing the molecular composition which can't be reversed. A common example of a chemical change would be burning wood. The wood burning into ash is changing the molecular composition of the wood transforming it to ash. Now can ash be changed back into wood? No, it is a "irreversible change."

Hope this helps.

5 0
3 years ago
What substances were found in the innermost regions (within about the inner 0.3 au) of the solar system before planets began to
coldgirl [10]

The answer would be Rocks, metals, hydrogen compounds, hydrogen and helium, all in gaseous form.

5 0
3 years ago
Which of the following is used to calculate the distance to a star based on subtle shifts in the star's position relative to an
Nana76 [90]
Parrallax is used to calculate the distance.
8 0
4 years ago
Calculate the mass (in g) of 2.8 X 10^22 H2O molecules.
siniylev [52]

Answer: wait im gonna search some info.

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6 0
3 years ago
Calculate the total energy, in kilojoules, that is needed to turn a 46 g block
gogolik [260]

<u>Answer:</u> The amount of heat absorbed is 141.004 kJ.

<u>Explanation:</u>

In order to calculate the amount of heat released while converting given amount of steam (gaseous state) to ice (solid state), few processes are involved:

(1): H_2O (s) (-25^oC, 248K) \rightleftharpoons H_2O(s) (0^oC,273K)

(2): H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)  

(3): H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)

(4): H_2O (l) (100^oC, 373K) \rightleftharpoons H_2O(g) (100^oC,373K)

Calculating the heat absorbed for the process having the same temperature:

q=m\times \Delta H_{(f , v)}       ......(i)

where,

q is the amount of heat absorbed, m is the mass of sample and \Delta H_{(f , v)} is the enthalpy of fusion or vaporization

Calculating the heat released for the process having different temperature:

q=m\times C_{s,l}\times (T_2-T_1)      ......(ii)

where,

C_{s,l} = specific heat of solid or liquid

T_2\text{ and }T_1 are final and initial temperatures respectively

  • <u>For process 1:</u>

We are given:

m=46g\\C=2.108J/g^oC\\T_2=0^oC\\T_1=-25^oC

Putting values in equation (i), we get:

q_1=46g\times 2.108J/g^oC\times (0-(-25))\\\\q_1=2424.2J

  • <u>For process 2:</u>

We are given:

m=46g\\\Delta H_{fusion}=334J/g

Putting values in equation (i), we get:

q_2=46g\times 334J/g\\\\q_2=15364J

  • <u>For process 3:</u>

We are given:

m=46g\\C=4.186J/g^oC\\T_2=100^oC\\T_1=0^oC

Putting values in equation (i), we get:

q_3=46g\times 4.186J/g^oC\times (100-0)\\\\q_3=19255.6J

  • <u>For process 4:</u>

We are given:

m=46g\\\Delta H_{vap}=2260J/g

Putting values in equation (i), we get:

q_4=46g\times 2260J/g\\\\q_4=103960J

Calculating the total amount of heat released:

Q=q_1+q_2+q_3+q_4

Q=[(2424.2)+(15364)+(19255.6)+(103960)]

Q=141003.8J=141.004kJ                  (Conversion factor: 1 kJ = 1000J)

Hence, the amount of heat absorbed is 141.004 kJ.

7 0
3 years ago
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