<h2>Anabolic Reactions and Catabolic Reactions </h2>
Explanation:
Anabolic Reactions -
- Anabolic reactions are reactions that build larger molecules from smaller ones while the catabolic reactions are the reactions that break down molecules into smaller ones.
- Anabolic reactions are also called biosynthetic reactions because they synthesize larger molecules from smaller molecules or constituent parts.
- The energy used by these reactions to proceed is ATP.
- Anabolic reactions occur in cells for the synthesis of new proteins and other reactions.
Catabolic Reactions -
- Catabolic reactions are the reaction that includes the whole pathway of reaction that breaks down larger molecules into smaller units.
- These molecules are oxidized to release energy or sometimes used in other anabolic reactions occurring in the cell.
Answer:
In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limits the amount of products formed.
Explanation:
Answer:
(a) 1.92 moles of Bi produced.
(b) 80.6 grams
Explanation:
Balanced equation: Bi2O3(s) + 3C(s) → 2Bi(s) + 3CO(g)
1st find moles of Bi2O3:
Bi2O3 has Mr of 466 and mass of 447 g
![\hookrightarrow moles = \frac{mass}{Mr}](https://tex.z-dn.net/?f=%5Chookrightarrow%20moles%20%3D%20%5Cfrac%7Bmass%7D%7BMr%7D)
![\hookrightarrow moles = \frac{447}{466}](https://tex.z-dn.net/?f=%5Chookrightarrow%20moles%20%3D%20%5Cfrac%7B447%7D%7B466%7D)
2nd find moles of Bi:
Bi2O3 : 2Bi
→ 1 : 2 ------ this is molar ratio.
→ 0.959227 : (0.959227)*2
→ 0.959227 : 1.91845
→ 0.959227 : 1.92
Therefore 1.92 moles of Bi was produced.
3rd Find moles of 3CO:
Bi2O3 : 3CO
1 : 3
0.959227 : (0.959227 )*3
0.959227 : 2.87768
3CO has 2.87768 moles and we know the Mr is 28.
![\hookrightarrow mass = moles * Mr](https://tex.z-dn.net/?f=%5Chookrightarrow%20mass%20%3D%20moles%20%2A%20Mr)
![\hookrightarrow mass = 2.87768 * 28](https://tex.z-dn.net/?f=%5Chookrightarrow%20mass%20%3D%202.87768%20%2A%2028)
g
Therefore 80.575 grams of CO was produced.
Hello!
The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.
We have the following data:
mo (initial mass) = 43 g
m (final mass after time T) = ? (in g)
x (number of periods elapsed) = ?
P (Half-life) = 20 minutes
T (Elapsed time for sample reduction) = 80 minutes
Let's find the number of periods elapsed (x), let us see:
![T = x*P](https://tex.z-dn.net/?f=%20T%20%3D%20x%2AP%20)
![80 = x*20](https://tex.z-dn.net/?f=%2080%20%3D%20x%2A20%20)
![80 = 20\:x](https://tex.z-dn.net/?f=%2080%20%3D%2020%5C%3Ax%20)
![20\:x = 80](https://tex.z-dn.net/?f=%2020%5C%3Ax%20%3D%2080%20)
![x = \dfrac{80}{20}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cdfrac%7B80%7D%7B20%7D%20%20)
![\boxed{x = 4}](https://tex.z-dn.net/?f=%20%5Cboxed%7Bx%20%3D%204%7D%20)
Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:
![m = \dfrac{m_o}{2^x}](https://tex.z-dn.net/?f=%20m%20%3D%20%20%5Cdfrac%7Bm_o%7D%7B2%5Ex%7D%20%20)
![m = \dfrac{43}{2^{4}}](https://tex.z-dn.net/?f=%20m%20%3D%20%20%5Cdfrac%7B43%7D%7B2%5E%7B4%7D%7D%20%20)
![m = \dfrac{43}{16}](https://tex.z-dn.net/?f=%20m%20%3D%20%5Cdfrac%7B43%7D%7B16%7D%20%20)
![\boxed{\boxed{m = 2.6875\:g}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Cboxed%7Bm%20%3D%202.6875%5C%3Ag%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark%20)
I Hope this helps, greetings ... DexteR! =)