B. 1, 1, 1, 2
Explanation:
You only need to balance the NaNO3 on the right. Since there is 2 NO3 on the left, you need to put a 2 in front of the NaNO3 on the right. Everything else is already balanced so the only coefficient needed is 2 in front of the NaNO3.
There's 18 atoms in ammonium phosphate
The question is incomplete, complete question is :
You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pka of acetic acid is 4.74.
Answer:
33.11 millimoles of acetate we will need to add to this solution.
Explanation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
Where :
tex]pK_a[/tex] = negative logarithm of acid dissociation constant of acid
[salt] = Concentration of salt
[Acid] = Concentration of salt
We have:
pH = 5.26
[salt] =![[CH_3COO^-]](https://tex.z-dn.net/?f=%5BCH_3COO%5E-%5D)
= ?
[acid] = ![[CH_3COOH]=10.0 mmol](https://tex.z-dn.net/?f=%5BCH_3COOH%5D%3D10.0%20mmol)
![5.26=4.74+\log(\frac{[CH_3COO^-]}{[10.0 mmol]})](https://tex.z-dn.net/?f=5.26%3D4.74%2B%5Clog%28%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5B10.0%20mmol%5D%7D%29)
![[CH_3COO^-]=33.11 mmol](https://tex.z-dn.net/?f=%5BCH_3COO%5E-%5D%3D33.11%20mmol)
33.11 millimoles of acetate we will need to add to this solution.
Answer:
2.53 L is the volume of H₂ needed
Explanation:
The reaction is: C₁₈H₃₀O₂ + 3H₂ → C₁₈H₃₆O₂
By the way we can say, that 1 mol of linolenic acid reacts with 3 moles of oxygen in order to produce, 1 mol of stearic acid.
By stoichiometry, ratio is 1:3
Let's convert the mass of the linolenic acid to moles:
10.5 g . 1 mol / 278.42 g = 0.0377 moles
We apply a rule of three:
1 mol of linolenic acid needs 3 moles of H₂ to react
Then, 0.0377 moles will react with (0.0377 . 3 )/1 = 0.113 moles of hydrogen
We apply the Ideal Gases Law to find out the volume (condition of measure are STP) → P . V = n . R . T → V = ( n . R .T ) / P
V = (0.113 mol . 0.082 L.atm/mol.K . 273.15K) 1 atm = 2.53 L
To answer this item, we must take note that the ligand that binds the tightest is the one with the lowest dissociation constant, Kd. Kd's for both A and B are already given so, we only need to solve Kds for C and D.
Kd of C
0.3 = (1x10⁻⁶)/(1x10⁻⁶ + Kd) ; Kd = 2.3x10⁻⁶
Kd of D
0.8 = (1x10⁻⁹)/(1x10⁻⁹ + Kd) ; Kd = 2.5x10⁻10
Since Ligand D has the least value of dissociation constant then, it can be concluded that it binds the tightest.
