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Crazy boy [7]
2 years ago
8

A chiral C5H10O ether reacts with excess hot HI to give a C5H10I2 product. When this product is treated with KOH, 1,3-pentadiene

is produced. What is the structure of the starting ether
Chemistry
1 answer:
alisha [4.7K]2 years ago
7 0

Answer:

See answer and picture below

Explanation:

In this case, the key to get an idea of how is the starting ether is with the final product and the empirical formula of the ether.

First, the empirical formula of the ether gives us information of the structure of the ether. This ether could be either an alkene, alkane or cycloalkane.

Alkane cannot be because alkane's formula is CnH(2n+2), so it's either an alkene or cycloalcane (CnH2n).

However, as we can see, it reacts with HI in excess, so, it cannot be an alkene because in this case the acid will react with the double bond of the ether too forming another product. This discart the fact that the ether has a double bond. So it has to be a cycloalkane.

As the innitial structure it's an ether, it should be either a cyclohexane or cyclopentane with a methyl group attached. So, how can we know which of these structures is? well, we can actually do the reaction with both reactants and see which match. In the attached picture you have the answer.

Hope this helps.

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If 4.35 g of phosphoric acid are added to 5.25g of KOH, what is the percent yield of the reaction if only 3.15g of potassium pho
irina1246 [14]
We are given with
4.35 g Phosphoric acid
5.25 g KOH
3.15 g K3PO4 produced

The reaction is
H3PO4 + 3KOH => K3PO4 + 3H2O

First, convert masses into moles.
Then, determine the limiting reactant.
Next, determine the maximum amount of K3PO4 that can be produced from the limiting reactant.
Lastly, calculate the percent yield by dividing the actual amount produced by the theoretical amount produced.
5 0
1 year ago
Your friend looks at a piece of ice and says “Solids, like ice, have a fixed shape because the particles are not moving.” Is you
Alex_Xolod [135]

Answer:

yes

Explanation:

6 0
1 year ago
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At 73.0 ∘c , what is the maximum value of the reaction quotient, q, needed to produce a non-negative e value for the reaction so
damaskus [11]
Here we will use the general formula of Nernst equation:

Ecell = E°Cell - [(RT/nF)] *㏑Q

when E cell is cell potential at non - standard state conditions

E°Cell is standard state cell potential = - 0.87 V

and R is a constant = 8.314 J/mol K

and T is the temperature in Kelvin = 73 + 273 = 346 K

and F is Faraday's constant = 96485 C/mole

and n is the number of moles of electron transferred in the reaction=2  

and Q is the reaction quotient for the reaction 
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔  Br2(aq) + SO2(g) +2H2O(l)

so by substitution :

0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q      → solve for Q 


∴ Q = 4.5 x 10^-26 
6 0
1 year ago
2CH4(g)⟶C2H4(g)+2H2(g)
Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of CH_4 will be,

2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)    \Delta H^o=?

The intermediate balanced chemical reaction will be,

(1) CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-890.3kJ

(2) C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)     \Delta H_2=-136.3kJ

(3) 2H_2(g)+O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=-571.6kJ

(4) 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)     \Delta H_4=-3120.8kJ

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) 2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)     \Delta H_1=2\times (-890.3kJ)=-1780.6kJ

(2) C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)    \Delta H_2=-(-136.3kJ)=136.3kJ

(3) H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)    \Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ

(4) 2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)     \Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ

The expression for enthalpy of the reaction will be,

\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)

\Delta H=201.9kJ

Therefore, the enthalpy change for the reaction is, 201.9 kJ

5 0
1 year ago
How many grams of KClO3 are needed to produce of 4.26 moles of O2? 2 KClO3 2 KCl + 3 O2 a. 348 g b. 136 g c. 174 g d. 522 g e. 7
JulsSmile [24]
Ok so this is what we know :

2KClO3 -> 2KCl + 3O2         (Always check if equation is balanced - in this                                               case it is)
                              4.26moles
So we know that we have 4.26 moles of oxygen (O2). Now lets look at the ratio between KClO3 and O2.
We see that the ratio is 2:3 meaning that we need 2KClO3 in order to produce 3O2.
Therefore divide 4.26 by 3 and then multiply by 2.
4.26/3 = 1.42
1.42 * 2 = 2.84
Now we know that the molarity of KClO3 is 2.84 moles.
Multiply by R.M.M to find how many grams of KClO3 we have.

R.M.M of KClO3
K- 39
Cl- 35.5
3O- 3 * 16 -> 48
---------------------------
                      <span>122.5
</span>2.84 * 122.5 = 347.9 grams therefore the answer is (a)
                       348 grams needed of KClO3 to produce 4.26 moles of O2.
Hope this helps :).

8 0
1 year ago
Read 2 more answers
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