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3 years ago

In the first step of the reaction, which compound is the limiting reagent: sodium hydroxide or the nucleophile

1 answer:

5 0

Since the nucleophile is the actual attacking molecule or molecule that starts the reaction and allows for further steps in the mechanism to occur, it is the limiting reagent, as based on the amount of the nucleophile you have, the reaction will tend to proceed until you run out. The excess would be the sodium hydroxide, it is union part of the solution.

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Name the following compound: 2-ethyl-4-methylheptene 3,5-dimethyl-2-octene 2-ethyl-4-methylheptane 3-methyl-5-propyl-2-hexene

Nadusha1986 [10]

Answer:

3,5-dimethyl-2-octene

Explanation:

The parent chain will be choosen based on the highest value. In this case, if we count from top to bottom, we'll get seven carbon, however if we count from the second carbon, going left and then down, we'll get eight carbon. So the parent chain is octene

The double bond is located at the second carbon and the methyl groups are located on carbon 3 & 5. Since there are two methyl groups, we add di- in front of methyl to indicate two methyl groups present.

Note: The functional group has to be prioritise and it needed to be a part of the parent chain. In this case, the functional group is the double bond. (alkene)

5 0

3 years ago

A cylinder with a piston is full of gas which of the following actions would result in an increase in the pressure of the gas be

SpyIntel [72]

The answer is B. Decreasing the cylinder volume decreases the amount of space the gas has to occupy there for increasing pressure

7 0

3 years ago

Where are Uranium and plutonium found? (IN NATURE)

Basile [38]

Wanted to write more both there is a problem with the editor. Uranium is formed naturally in the crust of rocks and seawater. Plutonium does not occur in nature. It is found in the biosphere.

8 0

2 years ago

Read 2 more answers

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

2 years ago

Draw the lewis structure for so42-. How many equivalent resonance structures can be drawn?

Elena-2011 [213]

Answer : The Lewis-dot structure and resonating structure of $SO_4^{2-}$ is shown below.

Explanation :

Resonance structure : Resonance structure is an alternating method or way of drawing a Lewis-dot structure for a compound.

Resonance structure is defined as any of two or more possible structures of the compound. These structures have the identical geometry but have different arrangements of the paired electrons. Thus, we can say that the resonating structure are just the way of representing the same molecule.

First we have to determine the Lewis-dot structure of $SO_4^{2-}$ .

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $SO_4^{2-}$

As we know that sulfur and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $SO_4^{2-}$ = 6 + 4(6) + 2 = 32

According to Lewis-dot structure, there are 12 number of bonding electrons and 20 number of non-bonding electrons.

Hence, the Lewis-dot structure and resonating structure of $SO_4^{2-}$ is shown below.

8 0

3 years ago