Answer:
10.875L
Explanation:
The problem here is a simple conversion. The conversion is from gal to liters Liter is a SI unit for recording volume as stated in the problem.
Given that:
1 gallon of water = 3.75L
2.90 gallon will be 2.9 x 3.75; 10.875L
Methods Of Separating Mixtures
Handpicking.
Threshing.
Winnowing.
Sieving.
Evaporation.
Distillation.
Filtration or Sedimentation.
Separating Funnel.
228 grams
start with mass of Cr multiply by molar mass of Cr mole to mole ratio between Cr and Cr2O3 times molar mass of Cr2O3
<u>Answer:</u> The pH of the buffer is 5.25
<u>Explanation:</u>
Let the volume of buffer solution be V
We know that:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[\text{conjugate base}]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7Bconjugate%20base%7D%5D%7D%7B%5Bacid%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.90
![[\text{conjugate base}]=\frac{2.25}{V}](https://tex.z-dn.net/?f=%5B%5Ctext%7Bconjugate%20base%7D%5D%3D%5Cfrac%7B2.25%7D%7BV%7D)
![[acid]=\frac{1.00}{V}](https://tex.z-dn.net/?f=%5Bacid%5D%3D%5Cfrac%7B1.00%7D%7BV%7D)
pH = ?
Putting values in above equation, we get:
Hence, the pH of the buffer is 5.25
It’s DEFINITELY 2 like DEFINITELY
