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Molodets [167]
3 years ago
6

Hydration of alkynes (via oxymercuration) gives good yields of single compounds only with symmetrical or terminal alkynes. draw

the major organic product(s) formed when hex-2-yne undergoes hydration in the presence of hgso4 and h2so4

Chemistry
1 answer:
nikitadnepr [17]3 years ago
3 0
Hex-2-yne is not symmetrical alkyne, So, it will give two products namely Hexan-3-one and Hexan-2-one.
The reaction along with mechanism is shown below,

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A slow chemical reaction will have...
yawa3891 [41]

Answer:

Ay high concentration of reactants

Explanation:

To slow down a reaction, you need to do the opposite. Factors that can affect rates of reactions include surface area, temperature, concentration, and the presence of catalysts and inhibitors. ... Concentration - another way to increase the rate of a chemical reaction is to increase the concentration of the reactants.

5 0
3 years ago
239 Pu 4 He<br> 94 2<br><br><br><br><br> Fill in the missing particle
lubasha [3.4K]

Answer:

la particula que queda es h2o

Explanation:

7 0
3 years ago
Use the balanced equation from number four above to determine the limiting reactant if we had a mixture of 5.0 moles of Fe and 4
Tom [10]

Answer:

2Fe + 3O₂  →  2Fe₂O₃

Limiting reactant: O₂

2Fe + O₂ → 2FeO

Limiting reactant: Fe

Explanation:

2Fe + 3O₂  →  2Fe₂O₃

2Fe + O₂ → 2FeO

These are the possible reactions:

In first case, 2 moles of Fe need 3 mol of oyxgen to react

If I have 5 moles of Fe, I will need (5 .3)/2 = 7.5 moles of O₂

Then, the oxygen is my limiting ( I only have 4 moles)

3 moles of O₂ need 2 moles of Fe

If I have 4 moles of O₂, I will need (4 .2)/3 = 2.66 moles (I have 5)

Fe, is the reactant in excess.

For second case, 2 moles of Fe need 1 mol of O₂, to react.

If I have 5 moles of Fe, I will need ( 5 .1) / 2 =2.5 moles of O₂

I have 4 moles of oxygen, so now it is my excess.

1 mol of O₂ need 2 moles of Fe, to react

If I have 4 moles of O₂, I will need the double of amount, 8 moles of Fe.

I have 5 moles, then the Fe is my limtiing reactant.

8 0
3 years ago
Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ca. Δ H°(kJ) Ca(s)→Ca(g) 1
Drupady [299]

Answer :  The value of second ionization energy of Ca is 1010 kJ.

Explanation :  

The formation of calcium oxide is,

Ca(s)+\frac{1}{2}O_2(g)\overset{\Delta H_f}\rightarrow CaO(s)

\Delta H_f^o = enthalpy of formation of calcium oxide = -635 kJ

The steps involved in the born-Haber cycle for the formation of CaO:

(1) Conversion of solid calcium into gaseous calcium atoms.

Ca(s)\overset{\Delta H_s}\rightarrow Ca(g)

\Delta H_s = sublimation energy of calcium = 193 kJ

(2) Conversion of gaseous calcium atoms into gaseous calcium ions.

Ca(g)\overset{\Delta H_I_1}\rightarrow Ca^{+1}(g)

\Delta H_I_1 = first ionization energy of calcium = 590 kJ

(3) Conversion of gaseous calcium ion into gaseous calcium ions.

Ca^{+1}(g)\overset{\Delta H_I_2}\rightarrow Ca^{+2}(g)

\Delta H_I_2 = second ionization energy of calcium = ?

(4) Conversion of molecular gaseous oxygen into gaseous oxygen atoms.

O_2(g)\overset{\Delta H_D}\rightarrow OI(g)

\frac{1}{2}O_2(g)\overset{\Delta H_D}\rightarrow O(g)

\Delta H_D = dissociation energy of oxygen = \frac{498}{2}=249kJ

(5) Conversion of gaseous oxygen atoms into gaseous oxygen ions.

O(g)\overset{\Delta H_E_1}\rightarrow O^-(g)

\Delta H_E_1 = first electron affinity energy of oxygen = -141 kJ

(6) Conversion of gaseous oxygen ion into gaseous oxygen ions.

O^-(g)\overset{\Delta H_E_2}\rightarrow O^{2-}(g)

\Delta H_E_2 = second electron affinity energy of oxygen = 878 kJ

(7) Conversion of gaseous cations and gaseous anion into solid calcium oxide.

Ca^{2+}(g)+O^{2-}(g)\overset{\Delta H_L}\rightarrow CaO(s)

\Delta H_L = lattice energy of calcium oxide = -3414 kJ

To calculate the overall energy from the born-Haber cycle, the equation used will be:

\Delta H_f^o=\Delta H_s+\Delta H_I_1+\Delta H_I_2+\Delta H_D+\Delta H_E_1+\Delta H_E_2+\Delta H_L

Now put all the given values in this equation, we get:

-635kJ=193kJ+590kJ+\Delta H_I_2+249kJ+(-141kJ)+878kJ+(-3414kJ)

\Delta H_I_2=1010kJ

Therefore, the value of second ionization energy of Ca is 1010 kJ.

6 0
3 years ago
What mass of carbon is present in 0.500 mol of sucrose?
AlexFokin [52]
C12H22O11
12×12×0.5=72
3 0
3 years ago
Read 2 more answers
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