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vlada-n [284]
2 years ago
13

What pressure is required to reduce 40 ml of a gas at standard conditions to 20 ml at a temperature of 21?c? answer in units of

atm?
Chemistry
1 answer:
ElenaW [278]2 years ago
4 0
If you need half the volume, you need double the pressure (V * p). Normal air pressure in the atmosphere is 101300 Pa so that means you need double that, thus 202600 Pa (or 2,026 bars)

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Which of the following elements is not likely to form bonds?
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I don’t understand the question
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2. Would you measure a pencil in meters? A hallway in millimeters? Discuss
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Answer:

Measuring a pencil in meters would be very difficult, as a single meter is much longer than one pencil. Also, measuring a hallway in millimeters would be very difficult considering how small millimeters are in comparison to a hallway. However, if you switch these two then they would work very well.

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For an experiment, a student needs a compound that has the following properties.
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Read 2 more answers
Products: 1-methylcyclohexene, 3-methylcyclohexene, methylenecyclohexane
Lostsunrise [7]

Answer:

See explanation

Explanation:

The reaction that we are considering here is quite a knotty reaction. It is difficult to decide if the mechanism is actually E1 or E2 since both are equally probable based on the mass of scientific evidence regarding this reaction. However, we can easily assume that the methylenecyclohexane was formed by an E1 mechanism.

Looking at the products, one could convincingly assert that the reaction leading to the formation of the two main products proceeds via an E1 mechanism with the formation of a carbocation intermediate as has been shown in mechanism attached to this answer. Possible rearrangement of the carbocation yields the 3-methylcyclohexene product.

8 0
3 years ago
stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,
Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

8 0
3 years ago
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