Examine the photo of quartz below in which way does the quartz break

Name the following alkyne.<br> Please help me <3

Alexandra [31]

Answer:

D. 7-methyl-3-octyne

Explanation:

1) Identify the parent chain and name it like an alkane.

• The longest chain of carbons, which consists of the functional group (which is the alkyne group in this case: C≡C).

• There are 8 carbons in the longest chain, so it is called an octane.

2) Now, identify the location of the functional group.

• The location number of the functional group should be as low as possible. Thus although we could count from the right, we start counting from the left since the functional group is closer to the left. From the left, the functional group would be at carbon 8 while from the left, it is on carbon 3.

• Replace 'ane' with 'yne' in octane for the alkyne group.

• This would give us 3-octyne.

3) Lastly, add in the name of the branch.

• Here we have one branch, -CH₃. This is read as methyl.

• Identify the location number of the branch by counting the number of carbons in the same direction as when we counted the location number of the functional group. The methyl branch has a location number of 7.

• Adding the name of the branch before the parent chain, we would arrive at 7-methyl-3-octyne as the IUPAC name of the alkyne.

Further Explanation:

A) This option is incorrect as there are only 8 carbons in the parent chain. Although there are 9 carbons in total, the 9th carbon is taken care of in '7-methyl'.

B) Location number of the functional group should be as low as possible, so start counting the number of carbons from the left!

C) Since the functional group is an alkyne, the word 'octane' should be 'octyne' instead.

8 0

3 years ago

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$