Answer:
A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.
Step by Step Explanation?
Boyle's law states that in constant temperature the variation volume of gas is inversely proportional to the applied pressure.
The formula is,
P₁ x V₁ = P₂ × V₂
Where,
P₁ is initial pressure = 1 atm
P2 is final pressure = ? (Not Known)
V₁ is initial volume = 10 L
V₂ is final volume = 15 L
Now put the values in the formula,
\begin{gathered}\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67\end{gathered]
Therefore, the answer is 0.67 atm.
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
Answer: The correct option is C ( is very hard and burns cleanly).
Explanation:
COAL is a form of rock that is made up of mostly carbon amongst other elements which includes sulphur, nitrogen, hydrogen and oxygen. There are different types of coal which include:
--> anthracite ( 90% carbon)
--> bituminous coal ( 70-90% carbon)
--> lignite ( 60- 70% carbon) and
--> peat (60 % carbon).
Anthracite is the type of coal that contains the highest carbon content ( 90% carbon). This makes it very hard and is often a times referred to as HARD COAL. Anthracite is a higher quality coal for domestic and open fire heating. This is because it contains less impurities than other type of coal and thereby making it to BURN CLEANLY avoiding atmospheric pollution.
I would say d
Hope that helps :)
The answer is visible rays. For example, when a rain storm stops, the sun comes out and you are likely to see a rainbow. This is because the sun shines on water molecules in the air and the colors appear. These colors are called visible light. The only ray that a naked Human eye can see