Molarity is one of the method of expressing concentration of solution. Mathematically it is expressed as,
Molarity =
Given: Molarity of solution = 5.00 M
Volume of solution = 750 ml = 0.750 l
∴ 5 =
∴
number of moles = 3.75Answer: Number of moles of KOH present in solution is 3.75.
Answer:
Explanation:
A solution of a weak base and its conjugate acid is a buffer.
The equation for the equilibrium is
The Henderson-Hasselbalch equation for a basic buffer is
![\text{pOH} = \text{p}K_{\text{b}} + \log\dfrac{[\text{BH}^{+}]}{\text{[B]}}](https://tex.z-dn.net/?f=%5Ctext%7BpOH%7D%20%3D%20%5Ctext%7Bp%7DK_%7B%5Ctext%7Bb%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5B%5Ctext%7BBH%7D%5E%7B%2B%7D%5D%7D%7B%5Ctext%7B%5BB%5D%7D%7D)
Data:
[B] = 0.400 mol·L⁻¹
[BH⁺] = 0.250 mol·L⁻¹
Kb = 4.4 × 10⁻⁴
Calculations:
(a) Calculate pKb
pKb = -log(4.4× 10⁻⁴) = 3.36
(b) Calculate the pH
Answer:
53.11× 10²³ molecules
Explanation:
Given data:
Number of molecules of CO₂ = ?
Mass of CO₂ = 388.1 g
Solution:
Formula:
Number of moles = mass/ molar mass
Molar mass of CO₂ = 12× 1 + 16×2
Molar mass of CO₂ = 44 g/mol
Now we will put the values in formula.
Number of moles = 388.1 g/ 44 g/mol
Number of moles = 8.82 moles
Now we will calculate the number of molecules by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
8.82 mol × 6.022 × 10²³ molecules / 1 mol
53.11× 10²³ molecules
To determine the number of potassium laid side by side by a given distance, we simply divide the total distance to the diameter of each atom. The diameter is twice the radius of the atom. We calculate as follows:
number of atoms = 4770 / 231x10^-12 = 2.06x10^13 atoms
Answer:
Option 2, Half of the active sites are occupied by substrate
Explanation:
Michaelis-Menten expression for enzyme catalysed equation is as follows:
![V_0=\frac{V_{max\ [S]}}{k_M+[S]}](https://tex.z-dn.net/?f=V_0%3D%5Cfrac%7BV_%7Bmax%5C%20%5BS%5D%7D%7D%7Bk_M%2B%5BS%5D%7D)
Here, 
is Michaelis-Menten constant and [S] is substrate concentration.
When [S]=Km
Rearrange the above equation as follows:
![V_0=\frac{V_{max}[S]}{k_M+[S]}\\V_0=\frac{V_{max}[S]}{[S]+[S]}\\V_0=\frac{V_{max}[S]}{2[S\\]}\\V_0=\frac{V_{max}}{2}](https://tex.z-dn.net/?f=V_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7Bk_M%2B%5BS%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7B%5BS%5D%2B%5BS%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7B2%5BS%5C%5C%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%7D%7B2%7D)
when [S]=Km, the rate of enzyme catalysed reaction becomes half of the maximum rate, that means half of the active sites are occupied by substrate.
Therefore, the correct option is option 2.
