The relative molecular mass of lead (ii) triozonitrate (v) is [pb = 108, N = 14, O = 16]

I assume you are already familiar with the IUPAC rules. I will be referring to the rule number where applicable. If you do not have a copy of the IUPAC rules, please let me know by a comment or PM.

Alkane (A)
The first one is tricky, because the longest straight chain is not as drawn, but along a "branch". See first image.
Since there are 10 carbon atoms in the longest chain, the parent chain will be named decane.
The first substituent occurs in C3, that explains why the parent chain is named from right to left.
Now the substituents:
C3 : Cl and CH3, so 3-chloro 3-methyl
C6 : The Y-shape is named isopropyl, so 6-isopropyl
C7 : Substituent is a straight chain with 2 carbon atoms, so 7-ethyl
The complete name is therefore
3-chloro-7-ethyl-3-methyl-6-isopropyl-decane
Note that order is according to alphabetic order of the substituent name without the prefixes (explains why isopropyl comes after methyl)

Alkane (B)
Again, first step is to look for the longest chain, which is 11 (undecane), as numbered in image 2.
Numbering starts from the 2,2-dichloro end because it gives the lowest number for the locant.
C2 2,2-dichloro (2 branches, each is Cl)
C4 4-methyl (1 carbon atom)C5 5-ethyl (2 carbon atoms)
C8 8-ethyl
C9 9-methyl
So the complete name is
2-2-dichloro-5,8-diethyl-4,9-dimethyl-undecane

Alkane (C)
Here there are 9 carbon atoms (nonane) in the longest chain.
The locants of substituents are identical from either end. Here I choose the one with the smaller number of the first substitutent 2-methyl instead of 2,2-dimethyl, resulting the numbering starting from the right. I hope I am right here.
C2 1-methyl
C3 1-methyl
C4 1-methyl
C5 1-methyl
C6 6-fluro-6-iodo
C7 1-methyl
C8 8,8-dimethyl
So the complete name becomes
6-fluoro-6-iodo-2,3,4,5,7,8,8-heptamethyl nonane

To @study77 and anyone reading this answer:
Please check and understand everything I did in case I slipped somewhere. Any suggestion for improvement or corrections will be most appreciated.

8 0

3 years ago

Punnett Squares represent the probability of traits in offspring as a result of what type of

Xelga [282]

Answer:

A

Explanation:

B is incorrect because asexual reproduction produces an offspring exactly the same as it's parent. Thus the probability does not need to be calculated.

C and D logically do not make sense since B is incorrect and A is correct. I provided the explanation why it is incorrect above.

Thus A is the correct answer.

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Hoped that helped! If it did, choose me as Brainliest!

3 0

4 years ago

Read 2 more answers

Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ...

KatRina [158]

Answer:

The oxidation state of silver in $\rm Ag_2O$ is $+1$ .

The oxidation state of sulfur in $\rm SO_2$ is $+4$ .

Explanation:

The oxidation states of atoms in a compound should add up to zero.

There are two silver $\rm Ag$ atoms and one oxygen $\rm O$ atom in one formula unit of $\rm Ag_2O$ . Therefore:

$\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}$ .

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is $-2$ . Silver oxide $\rm Ag_2O$ isn't an exception. Therefore:

$\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}$ .

Solve this equation for the (average) oxidation state of $\rm Ag$ :

$\text{Oxidation state of $\rm Ag$} = 1$ .

Similarly, because there are one sulfur $\rm S$ atom and two oxygen $\rm O$ atoms in each $\rm SO_2$ molecules:

$\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}$ .

The oxidation state of $\rm O$ in $\rm SO_2$ is also $-2$ , not an exception, either.

Therefore:

$\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}$ .

Solve this equation for the oxidation state of $\rm S$ here:

$\text{Oxidation state of $\rm S$} = 4$ .

8 0

4 years ago