Answer:
The correct answer is option B.
Explanation:
Michaelis–Menten 's equation:
![v=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}](https://tex.z-dn.net/?f=v%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D%3Dk_%7Bcat%7D%5BE_o%5D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D)
![V_{max}=k_{cat}[E_o]](https://tex.z-dn.net/?f=V_%7Bmax%7D%3Dk_%7Bcat%7D%5BE_o%5D)
v = rate of formation of products
[S] = Concatenation of substrate = ?
= Michaelis constant
= Maximum rate achieved
= Catalytic rate of the system
= initial concentration of enzyme
We have :

[S] =?

![v=V_{max}\times \frac{[S]}{(K_m+[S])}](https://tex.z-dn.net/?f=v%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D)
![\frac{V_{max}}{4}=V_{max}\times \frac{[S]}{(0.0050 M+[S])}](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7Bmax%7D%7D%7B4%7D%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%280.0050%20M%2B%5BS%5D%29%7D)
![[S]=\frac{0.005 M}{3}=1.7\times 10^{-3} M](https://tex.z-dn.net/?f=%5BS%5D%3D%5Cfrac%7B0.005%20M%7D%7B3%7D%3D1.7%5Ctimes%2010%5E%7B-3%7D%20M)
So, the correct answer is option B.
The given question is incomplete. the complete question is : If 5.15 g
is dissolved in enough water to make exactly 323 ml of solution, what is the molar concentartion of nitrate ion.
Answer: The molar concentartion of nitrate ion is 0.195.
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
Given : 5.15 g of
is dissolved
Volume of solution = 323 ml
where,
n= moles of solute
= volume of solution = 323 ml

As 1 M of
gives 3 M of
ions.
Thus 0.065 M of
gives =
of
ions.
The molar concentartion of nitrate ion is 0.195.
States that when a gas is held at a constant temperature and mass in a closed container, the volume and pressure vary inversely. The equation to use is P1V1=P2V2.
Given
V1=200mL×1L1000mL=0.2 L
P1=700 mmHg
V2=100mL×1L1000mL=0.1 L
Unknown
P2
Equation
P1V1=P2V2
Solution
Rearrange the equation to isolate P2 and solve.
P2=P1V1V2
P2=(700mmHg×0.2L)0.1L=1400 L, which must be rounded to 1000 L because all of the measurements have only one significant figure.
Explanation:
substance Q could be <em><u>oxygen (O2)</u></em>
substance R could be <em><u>carbon</u></em><em><u> </u></em><em><u>d</u></em><em><u>i</u></em><em><u>o</u></em><em><u>x</u></em><em><u>i</u></em><em><u>d</u></em><em><u>e</u></em><em><u> </u></em><em><u>(</u></em><em><u>C</u></em><em><u>O</u></em><em><u>2</u></em><em><u>)</u></em>