Answer:
A. The energy stored in atmospheric carbon dioxide is conserved because it is used to create new forms of energy present in decomposed plants.
Explanation:
In the carbon cycle image, the result of an industry's work releases carbon dioxide into the atmosphere (this is represented by the letter G), this carbon dioxide is stored in the atmosphere (letter C) and then absorbed by plants during the process. of photosynthesis (letter A).
The carbon cycle is constituted by the absorption of carbon dioxide by plants in the photosynthesis process. Half of this absorbed carbon is released into the atmosphere and the other half the vegetable uses to produce sugars (glycoses). By ingesting the plants, the animals ingest together the carbon to their body, being released through respiration or decomposition. Because some fungi and bacteria are responsible for the decomposition of both animals and vegetables, they ingest part of this carbon, releasing it into the atmosphere and soil. In addition to bacteria, the burning process also releases carbon dioxide into the soil and atmosphere. Vegetables, through the breathing process, also absorb carbon dioxide and release oxygen unlike animals.
Oxygen is the only element in the list.
<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.
<u>Explanation:</u>
All radioactive decay processes undergoes first order reaction.
To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:
![k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%20%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = ?
t = time taken = 1.52 hrs
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= Initial concentration of reactant = 100 g
[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g
Putting values in above equation, we get:
To calculate the half life period of first order reaction, we use the equation:
where,
= half life period of first order reaction = ?
k = rate constant = 
Putting values in above equation, we get:
Hence, the half life of the sample of silver-112 is 3.303 hours.
Answer:
Explanation:
Hello there!
In this case, according to the rules for the oxidation states in chemical reactions, it is possible to realize that lone elements have 0 and since magnesium is in group 2A, it forms the cation Mg⁺² as it loses electrons and oxygen is in group 6A so it forms the anion O⁻²; therefore resulting oxidation numbers are:
Best regards!
