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WITCHER [35]
2 years ago
15

Can someone please help????nThis is sooooo hard

Chemistry
1 answer:
Rashid [163]2 years ago
6 0

Answer:

See explanation.

Explanation:

I highly suggest you watch OChem Tutor's videos on IUPAC nomenclature because the actual naming would take a lot of time to teach in text-based format. But here is how to name them:

1) I think there are two seperate pictures for number 1. The molecule on the left is 1-pentene and the one on the right is 4-methyl-1-pentene. If the whole thing is one molecule but there is just a bond missing where the red marker numbers are, that molecule would be 9-methyl-1,6-decadiene.

2) 4-methyl-2-pentene

3) 2,4-octadiene

4) 1,5-nonadiene

5) 2,5-dimethyl-3-hexene

6) 3,6-dimethyl-2,4-heptadiene

7) 2,5,5-trimethyl-2-hexene

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The four major attractive forces between particles are ionic bonds, dipole-dipole attractions, hydrogen bonds, and dispersion fo
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NaCl: ionic, HF: hydrogen bond,  HCl: dipole dipole , F2: dispersion force

Explanation:

complete question is:

The four major attractive forces between particles are ionic bonds, dipole-dipole attractions, hydrogen bonds, and dispersion forces. Consider the compounds below, and classify each by its predominant attractive or intermolecular force among atoms or molecules of the same type.Identify each of the following ( NaCl, HF, HCl, F2) as Ionic, H Bonding, Dipole or Dispersion.

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When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
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<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

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Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
3 years ago
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A change in color, odor, taste, chemical compound, and temperature all represent a chemical reaction, because these are all things that are happening within the the item that is being given the product of a chemical change.

Think of it this way: <em>internal changes within the product: chemical. External changes within the product: physical.</em>

I hope this helps.

4 0
2 years ago
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