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Answer:
Conditioning two or three times will insure that the concentration of titrant is not changed by a stray drop of water.
Explanation:
"Check the tip of the buret for an air bubble. To remove an air bubble, whack the side of the buret tip while solution is flowing".
Answer:
The awnser is D
Explanation:
Since the black pieces absorb light better than white pieces, the white pieces will reflect light better. But since color doesn't affect sound, the sound waves are reflected the same.
For the reactants,
- The oxidation number of hydrogen = +1
- The oxidation number of oxygen = -2
- The oxidation number of arsenic = +5
- The oxidation number of carbon = +3
For the products,
- The oxidation number of hydrogen = +1
- The oxidation number of oxygen = -2
- The oxidation number of arsenic = +3
- The oxidation number of carbon = +4
Here, arsenic (+5 to +3) and carbon (+3 to +4) are the only oxidation numbers changing.
Note that an increase in oxidation number means electrons are lost. Thus oxidation is occurring, and a decrease in oxidation number means electrons are being gained, and thus reduction is occurring.
Also, the compound that contains the element being oxidized is the reducing agent, and the compound that contains the element being reduced is the oxidizing agent.
So, the answers are:
name of the element oxidized: Carbon
name of the element reduced: Arsenic
formula of the oxidizing agent: 
formula of the reducing agent: 
Answer: 120g/mol
Explanation:
The first step we are to take is to calculate the freezing point depression of the solution.
ΔT(f) = freezing point of pure solvent - freezing point of solution
ΔT(f) = 5.48 - 3.77
ΔT(f) = 1.71°C
Next we are to calculate the molal concentration of the solution using freezing point depression
ΔT(f) = K(f) * m
m = ΔT(f)/K(f)
m = 1.71/5.12
m = 0.333 molal
Now, we calculate the molecular weight of the unknown...
m = 0.333 mol = 0.333 mol X per kg of benzene
moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene
moles of X = 0.1665
molecular weight of X = 20g of X/0.1665
molecular weight of X = 120/mol