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irina1246 [14]
3 years ago
6

Write the Lewis structure and molecular formula for each of the following hydrocarbons:

Chemistry
1 answer:
dusya [7]3 years ago
4 0

Answer:

molecular formula:

a) Hexane: C6H14

b) 3-methylpentane: C6H14

c) cis-3-hexene: C6H12

d)4-methyl-1-pentene: C6H12

e) 3-hexyne: C6H10

f) 4-methyl-2-pentyne: C6H10

Explanation:

structure:

a) hexane : CH3-CH2-CH2-CH2-CH2-CH3

b) 3-metilpentane: CH3-CH2-CH-CH3-CH3

                                                 CH3

c) cis-3-hexene:   CH3-CH2-CH=CH-CH2-CH3

d)4-methyl-1-pentene: CH2=CH-CH2-CH-CH3

                                                              CH3

e) 3-hexyne: CH3-CH2-C≡C-CH2-CH3

f)4-methyl-2-pentyne: CH3-C≡C-CH-CH3

                                                      CH3

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How many moles are in 187.54 grams of magnesium chlorate?
Svetlanka [38]

Hey there!

Magnesium chlorate: Mg(ClO₃)₂

Find molar mass.

Mg: 1 x 24.305 = 24.305

Cl: 2 x 35.453 = 70.906

O: 6 x 16 = 96

------------------------------------

                      191.211 g/mol

We have 187.54 grams.

187.54 ÷ 191.211 = 0.9808

There are 0.9808 moles in 187.54 grams of magnesium chlorate.

Hope this helps!

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3 years ago
There are two physical properties of minerals that both result in smooth, flat surfaces with specific angles between them. The f
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3. crystal habit and cleavage.

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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
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