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Vinvika [58]
4 years ago
6

Which of the reagents listed below would efficiently accomplish the transformation of 2-methyl-3-cyclopentenol into 2-methyl-3-c

yclopentenone?
Chemistry
1 answer:
Anni [7]4 years ago
5 0

The answer to this question would be CrO3 or H2SO4. These reagents would efficiently accomplish the transformation of 2-methyl-3-cyclopentenol into 2-methyl-3-cyclopentenone. The synthetic strategy is used for the bond forming and transformation of the molecule. The reagent required depends on the type of molecule. 

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8.03 Solutions Lab Report<br> Does anyone have a PDF or Document of FLVS 8.03 Solutions Lab
GarryVolchara [31]

8.03 solutions report is described below.

Explanation:

8.03 Solutions Lab Report

In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.  

Pre-lab Questions:

In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.  

Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).

Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.

7 0
4 years ago
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I'm lost....can anyone explain please?
pashok25 [27]
I'll do A for you. They are all double displacement reactions. So recall: AB+CD--> CB+AD . Also, knowing the charges will help and make this so easy.
3 0
4 years ago
Is 4H2O an organic molecule ?
Ludmilka [50]

Answer:

it is an organic molecule

Explanation:

it's a compound called chromium, phosphate tetrahydrate, where tetra stands for 4 , as there are four water molecules attached to the compound.

8 0
4 years ago
Given the following balanced equation at 120°C: A(g) + B(g) ⇋ 2 C(g) + D(s)(a) At equilibrium a 4.0 liter container was found to
BlackZzzverrR [31]

Answer:

a) kc = 0,25

b) [A] = 0,41 M

c) [A] = <em>0,8 M</em>

[B] =<em>0,2 M</em>

[C] = <em>0,2M</em>

Explanation:

The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.

Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium.

If global reaction is:

A(g) + B(g) ⇋ 2 C(g) + D(s)

The kc = \frac{[C]^2}{[A][B]}

a) The concentrations of each compound are:

[A] = \frac{1,60 moles}{4,0 L} = <em>0,4 M</em>

[B] = \frac{0,40 moles}{4,0 L} = <em>0,1 M</em>

[C] = \frac{0,40 moles}{4,0 L} = <em>0,1 M</em>

<em>kc = </em>\frac{[0,1]^2}{[0,4][0,1]} = 0,25

b) The addition of B and D in the same amount will, in equilibrium, produce these changes:

[A] = \frac{1,60-x moles}{4,0 L}

[B] = \frac{0,60-x moles}{4,0 L}

[C] = \frac{0,60+2x moles}{4,0 L}

0,25 = \frac{[0,60+2x]^2}{[1,60-x][0,60-x]}

You will obtain

3,75x² +2,95x +0,12 = 0

Solving

x =-0,74363479081119   → No physical sense

x =-0,043031875855476

Thus, concentration of A is:

\frac{1,60-(-0,04 moles)}{4,0 L} = <em>0,41 M</em>

c) When volume is suddenly halved concentrations will be the concentrations in equilibrium over 2L:

[A] = \frac{1,60 moles}{2,0 L} = <em>0,8 M</em>

[B] = \frac{0,40 moles}{2,0 L} = <em>0,2 M</em>

[C] = \frac{0,40 moles}{2,0 L} = <em>0,2M</em>

I hope it helps!

8 0
4 years ago
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