Answer:
The answer to your question is P = 1.357 atm
Explanation:
Data
Volume = 22.4 L
1 mol
temperature = 100°C
a = 0.211 L² atm
b = 0.0171 L/mol
R = 0.082 atmL/mol°K
Convert temperature to °K
Temperature = 100 + 273
= 373°K
Formula

Substitution

Simplify
(P + 0.0094)(22.3829) = 30.586
Solve for P
P + 0.0094 = 
P + 0.0094 = 1.366
P = 1.336 - 0.0094
P = 1.357 atm
Answer:
0.11 mol
Explanation:
<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of oxygen in a sample of acetic acid. How many moles of hydrogen are in the sample?</em>
Step 1: Given data
- Formula of acetic acid: CH₃CO₂H
- Moles of oxygen in the sample of acetic acid: 0.054 moles
Step 2: Establish the appropriate molar ratio
According to the chemical formula of acetic acid, the molar ratio of H to O is 4:2.
Step 3: Calculate the moles of atoms of hydrogen
We will use the theoretical molar ratio for acetic acid.
0.054 mol O × (4 mol H/2 mol O) = 0.11 mol H
To determine the pOH assuming water is the universal solvent take the value of 10 ^ -14 and then divide it by the hydronium concentration and then take the negative logarithm of the final answer that is the solution to the hydroxide ion concentration in the solution.
#1 (As trash, synthetic polymers are not biodegradable)
#2 (Landfills can easily fill up with synthetic polymers)
#4 (Recycling synthetic polymers is costly)
Answer:
• pH = 3.0
• pH = 2.70
• pH = 3.61
• pH = 8.28
• pH = 1.40
Explanation:
El pH es una medida en química usada para determinar el grado de acidez o basicidad en una solución.
Se define como:
pH = -log₁₀ [H⁺]
<em>El - logaritmo de la concentración molar de H⁺</em>
<em />
Para las concentraciones de H⁺ dadas:
• [H+] = 0.001 M
pH = -log (0.001M) = 3
pH = 3.0
• [H+] = 0.002 M
pH = -log (0.002M)
pH = 2.70
• [H+] = 2.45X10-4 M
pH = -log (2.45X10-4 M )
pH = 3.61
• [H+] = 5.2X10-9 M
pH = -log (5.2X10-9 M)
pH = 8.28
• [H+] = 0.04 M
pH = -log (0.04M)
pH = 1.40