Answer:
carbon dioxide, water, sunlight
Explanation:
The given data is as follows.


Now, according to Michaelis-Menten kinetics,
![V_{o} = V_{max} \times [\frac{S}{(S + Km)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%20V_%7Bmax%7D%20%5Ctimes%20%5B%5Cfrac%7BS%7D%7B%28S%20%2B%20Km%29%7D%5D)
where, S = substrate concentration =
M
Now, putting the given values into the above formula as follows.
![V_{o} = V_{max} \times [\frac{S}{(S + Km)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%20V_%7Bmax%7D%20%5Ctimes%20%5B%5Cfrac%7BS%7D%7B%28S%20%2B%20Km%29%7D%5D)
![V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%206.8%20%5Ctimes%2010%5E%7B-10%7D%20%5Cmu%20mol%2Fmin%20%5Ctimes%20%5B%5Cfrac%7B10.4%20%5Ctimes%2010%5E%7B-6%7D%20M%7D%7B%2810.4%20%5Ctimes%2010%5E%7B-6%7DM%20%2B%205.2%20%5Ctimes%2010%5E%7B-6%7D%20M%29%7D%5D)

= 
This means that
would approache
.
Answer:
The number of copper atoms 12.405 ×10²³ atoms.
The number of silver atoms 13.13 ×10²³ atoms.
Beaker B have large number of atoms.
Explanation:
Given data:
In beaker A
Number of moles of copper = 2.06 mol
Number of atoms of copper = ?
In beaker B
Mass of silver = 222 g
Number of atoms of silver = ?
Solution:
For beaker A.
we will solve this problem by using Avogadro number.
The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.
While we have to find the copper atoms in 2.06 moles.
So,
63.546 g = 1 mole = 6.022×10²³ atoms
For 2.06 moles.
2.06 × 6.022×10²³ atoms
The number of copper atoms 12.405 ×10²³ atoms.
For beaker B:
107.87 g = 1 mole = 6.022×10²³ atoms
For 222 g
222 g / 101.87 g/mol = 2.18 moles
2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms