Answer:
The volume of a sample of gas (2.49 g) was 752 mL at 1.98 atm and 62∘C 62 ∘ C .
Answer:
1) positive
2) carbocation
3) most stable
4) faster
Explanation:
A common test for the presence of alcohols can be achieved using the Lucas reagent. Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride.
The reaction of Lucas reagent reacts with alcohols leading to the formation of an alkyl chloride. Since the reaction proceeds via a carbocation mechanism, tertiary alcohols give an immediate reaction. Once a tertiary alcohol is mixed with Lucas reagent, the solution turns cloudy almost immediately indicating an instant positive reaction.
Secondary alcohols may turn cloudy within five minutes of mixing the solutions. Primary alcohols do not significantly react with Lucas reagent obviously because they do not form stable carbocations.
Therefore we can use the Lucas reagent to distinguish between primary, secondary and tertiary alcohols.
Answer:
Explanation:
In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:
Now we can substitute any known values into the formula.
Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.
The volume is decreased to 14.5 liters, but the pressure is unknown.
Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .
The units of liters cancel.
The original values of volume and pressure have 3 significant figures, so our answer must have the same.
For the number we found, that is the hundredth place.
The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.
The new pressure is approximately <u>3.53 atmospheres.</u>
B. I’m sorry if I’m wrong ! Wasn’t sure about this question .
<span>The </span>equilibrium<span> will </span>shift<span> to favor the side of the reaction that involves fewer moles of gas.
Its C
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