Answer: 173 g ( 0.17 kg in right accuracy)
Explanation: Amount in moles is n = N/Na = 2.0·10^24 / 6.022·10^23 (1/mol).
n = 3.32116 mol. M(Cr) = 52.00 g/mol and mass m = nM = 172.7 g
<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
Answer:
1.5 × 10² mL
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 1.9 atm
- Initial volume of the gas (V₁): 80 mL
- Final pressure of the gas (P₂): 1.0 atm (standard pressure)
- Final volume of the gas (V₂): ?
Step 2: Calculate the final volume of the gas
For an ideal gas, we can calculate the final volume of the gas using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 1.9 atm × 80 mL/1.0 atm
V₂ = 1.5 × 10² mL
Since the pressure decreased, the volume of the gas increased.
Is it the dry lab/wet lab week 1 or ?
