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3 years ago

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

Chemistry

1 answer:

Nadusha1986 [10]3 years ago

5 0

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)

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A neutral atom posseses an atomic number of 15 and an atomic mass of 31 3 electrons are gained

kodGreya [7K]

Answer:I believe the correct answer from the choices listed above is option D. A

Explanation:neutral atom possesses an atomic number of 15 and an atomic mass of 31.Three electrons are gained. This conversion results to a negatively charged ion.

8 0

3 years ago

When converting from moles to molecules OR molecules to moles you use what quantitative value?

miv72 [106K]

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3 years ago

Calcium carbonate is often used as an antacid. Your stomach acid is composed of HCl at a pH of 1.5. If you ate toooo much Turkey

stiks02 [169]

<u>Answer:</u> 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

<u>Explanation:</u>

pH is defined as the negative logarithm of hydrogen ion concentration present in the solution

$pH=-\log [H^+]$ .....(1)

Given value of pH = 1.5

Putting values in equation 1:

$1.5=-\log[H^+]$

$[H^+]=10^{(-1.5)}=0.0316M$

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Number of moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

We are given:

Volume of solution = 15.0 mL

Molarity of HCl = 0.0316 M

Putting values in equation 2:

$0.0316=\frac{\text{Moles of HCl}\times 1000}{15.0}\\\\\text{Moles of HCl}=\frac{0.0316\times 15.0}{1000}=4.74\times 10^{-4}mol$

The chemical equation for the reaction of HCl and calcium carbonate follows:

$2HCl+CaCO_3\rightarrow H_2CO_3+CaCl_2$

By the stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of calcium carbonate

So, $4.74\times 10^{-4}mol$ of HCl will react with = $\frac{1}{2}\times 4.74\times 10^{-4}=2.37\times 10^{-4}mol$ of calcium carbonate

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of calcium carbonate = $2.37\times 10^{-4}mol$

Molar mass of calcium carbonate = 100.01 g/mol

Putting values in the above equation:

$\text{Mass of }CaCO_3=(2.37\times 10^{-4}mol)\times 100.01g/mol\\\\\text{Mass of }CaCO_3=0.0237g$

Hence, 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

7 0

2 years ago

Consider the following mechanism for the reaction between nitric oxide and ozone:

swat32

Answer:

A . 2 O₃(g) + 2 NO ⇒ 2 O₂ (g) + 2 NO₂(g)

B . Yes

C. O and NO₃

Explanation:

A. The overall reaction is obtained by adding the individual steps in the reaction mechanism where we will get the reactants and product and the intermediates will cancel.

Thus, adding 1+ 2 +3 we get

2 O₃(g) + 2 NO ⇒ 2 O₂ (g) + 2 NO₂(g)

B. The reaction intermediates are those that are produced from the initial and/or subsequent steps and are consumed later on in the reaction mechanism, but are neither reactants nor products, they just participate.

From this definition it follows that O(g) and NO₃ are reaction intermediates.

C. O and NO₃

6 0

3 years ago

What is required for storing chemicals

AnnZ [28]

A chemical storing thing

4 0

2 years ago

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