Answer:
it was Millikan. He conducted the oil drop experiment. Thomson determined the electon charge not the quantity. Rutherford used the gold foil experiment to find positive charge and that most of the atom is empty space. Dalton proposed that matter was made of small particles called atoms but that was a concept already proposed by ancient greeks. Dalton also proposed the atomic theory.
Answer:
0.054 mol O
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 carbon in a sample of acetic acid. How many moles of oxygen are in the sample?</em>
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Step 1: Given data
- Chemical formula of acetic acid: CH₃CO₂H
- Moles of carbon in the sample: 0.054 moles
Step 2: Establish the appropriate molar ratio
According to the chemical formula, the molar ratio of C to O is 2:2.
Step 3: Calculate the moles of oxygen in the sample
We will use the molar ratio to determine the moles of oxygen accompanying 0.054 moles of carbon.
0.054 mol C × (2 mol O/2 mol C) = 0.054 mol O
We will balance the equation in the following order: metals, amethals, carbon, hydrogen and oxygen (the most common order).
The metal present in the equation is Sr, which is already balanced (there are 1 on each side of the equation).
The amethal present in the equation is Cl. There is 2 Cl in the left side and only one in the right side. So, we will multiply the quantity of the molecule that contains Cl by 2. Doing this, we'll obtain:
Looking at the equation, we can see that it is now fully balanced. Hence, a balanced equation of the reaction is:
Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
