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jolli1 [7]
2 years ago
12

Be sure to answer all parts. this is a two-part question. first, draw the minor alkene product that should be formed in the reac

tion. second, draw a stepwise mechanism that shows the formation of the major product: part 1: 2xsafari + view structure major product minor product part 2: view structure h5mech30504 view structure + br+ br− ch3obr+ ch3oh2+ part 3 out of 3 edit structure ... arr edit structure ... + br+ ch3oh2+ ch3obr+ br−

Chemistry
1 answer:
kotykmax [81]2 years ago
8 0

Answer:

See explanation below

Explanation:

The question is incomplete, however, I found a question very similar to this, and I'm assuming this is the question you are asking to answer. If it's not, please tell me which one it is. Here's a tip for you to get an idea of how to solve.

Picture 1, would be the original question. Picture 2 is the answer of it.

Now, This is a E1 reaction where this type of reactions are taking place in two steps. The first step is the formation of the carbon cation, this step is always slow. The secon step is the addition of a nucleophyle, or, in this case, formation of a pi bond, and we get a alkene.

Hope this can help you

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1. A gas having the following composition is burnt under a boiler with 50% excess air.
jeka94

The composition of the stack gas are :

CH_4= 0.8713

C_3H_8 = 0.0202

CO = 0.107

<h3 /><h3>What is a mole fraction?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.

Mass of CH_4 : 70% of 100 g = 70 g

Mass of C_3H_8 : 15% of 100 g = 15 g

Mass of CO : 15% of 100 g = 15 g

Now calculate the number of moles of each species:

Number of moles of CH_4 : \frac{70 g}{16.04 g/mol} = 4.3 mole

Number of moles of C_3H_8: \frac{15 g}{144.1 g/mol} = 0.10 mole

Mass of CO : \frac{15 g}{28.01 g/mol} = 0.53 mole

Now to calculate the mole fraction of each we use the formula:

Mole fraction of CH_4: \frac{4.3}{4.935} = 0.8713

Mole fraction of C_3H_8 : \frac{0.10}{4.935} = 0.0202

Mole fraction of CO : \frac{0.53}{4.935} = 0.107

Hence, composition of the stack gas are:

CH_4 = 0.8713

C_3H_8 = 0.0202

CO = 0.107

Learn more about mole fraction here:

brainly.com/question/13135950

#SPJ1

8 0
2 years ago
PCl3(g) + Cl2(g) ⇋ PCl5(g) Kc = 91.0 at 400 K. What is the [Cl2] at equilibrium if the initial concentrations were 0.24 M for PC
Dmitry_Shevchenko [17]

Answer:

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Kc = 91

So let's analyse, all the process:

                PCl₃(g)        +        Cl₂(g)     ⇋        PCl₅(g)

Initially     0.24 M                 1.50M                 0.12 M

React           x                           x                         x

Some amount of compound has reacted during the process.

In equilibrium we have

              0.24 - x                  1.50 - x                  0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂] . [PCl₃]

91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)          

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 +x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b= -159.34; c = 32.64

(-b +- √(b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M

5 0
3 years ago
Consider the reaction 2CO * O2 —&gt; 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c
Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

Best regards!

8 0
2 years ago
LINKING IN<br> TECHNICAL OBJECTS<br> 1 a) What is linking?
d1i1m1o1n [39]

Answer:

A link is a fastening unit that attaches two parts of an object together

Different types of links have different characteristics

4 0
2 years ago
Read 2 more answers
A. If the mass of Hydrogen is 1 amu, what is the mass of Hydrogen in the reactant side of the equation above?
dimulka [17.4K]

Answer:

there are 4 hydrogen so

A.the mass of Hydrogen in the reactant side of the equation above is 1×4=4 amu.

B.the mass of Hydrogen on the product side of the equation above =1×4=4 amu.

<u>Note</u><u>:</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>reactant</u><u> </u><u>=</u><u>mass</u><u> </u><u>of</u><u> </u><u>product</u><u>.</u>

7 0
2 years ago
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