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1 answer:
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Answer:
Explanation:
This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:
The volume of the gas starts at 250 milliliters and the temperature is 137 °C.
The volume of the gas is increased to 425 milliliters, but the temperature is unknown.
We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.
Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.
The units of milliliters (mL) cancel.
The temperature changes to <u>232.9 degrees Celsius.</u>
Answer : The volume of pure diamond is
Explanation : Given,
Density of pure carbon in diamond =
Moles of pure diamond = 23.7 moles
Molar mass of carbon = 12 g/mol
First we have to calculate the mass of carbon or pure diamond.
Molar mass of carbon = 12 g/mol
Now we have to calculate the volume of carbon or pure diamond.
Formula used:
Now putting all the given values in this formula, we get:
Volume =
As we know that:
So,
Volume =
Volume =
Therefore, the volume of pure diamond is
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.
