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nikdorinn [45]
3 years ago
8

What functional group would be expected to be found in the major, and final, organic product upon completion of the reaction bet

ween 1-hexyne and a mixture of mercuric sulfate and aqueous sulfuric acid?
Chemistry
1 answer:
taurus [48]3 years ago
4 0

Answer:

Ketone

Explanation:

The reaction between 1-hexyne and mercuric sulphate in the presences of aqueous sulphuric acid is a reaction that leads to the formation of a product containing a carbonyl group(a ketone in this case).

This reaction occurs in alkynes but the product formed depends on the structure of the alkyne; an aldehyde or ketone may be formed depending on the structure of the alkyne.

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In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a
igor_vitrenko [27]

<u>Answer:</u> The equilibrium constant for this reaction is 1.068\times 10^{6}

<u>Explanation:</u>

The equation used to calculate standard Gibbs free change is of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]

For the given chemical reaction:

Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)

The equation for the standard Gibbs free change of the above reaction is:

\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]

We are given:

\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol

To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol  (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 58^oC=[273+58]K=331K

K_{eq} = equilibrium constant at 58°C = ?

Putting values in above equation, we get:

-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}

Hence, the equilibrium constant for this reaction is 1.068\times 10^{6}

4 0
3 years ago
What is the full number for 5.7x10^4
dsp73

Answer:

5.7*10^4 is equal to 57,000.

Explanation:

First, we must multiply 10 by its power, 4. That would be 10 4 times.

10*10*10*10 = 10,000.

Then, we multiply it by 5.7.

5.7*10,000 = 57,000.

Regards!

3 0
3 years ago
When the nuclide uranium-234 undergoes alpha decay: the name of the product nuclide is . the symbol for the product nuclide is?
Schach [20]
Answer is: <span>name of the product nuclide is thorium-230.
</span>

<span>Alpha particle is nucleus of a helium-4 atom, which is made of two protons and two neutrons.
²³</span>⁴U → ²³⁰Th + α (alpha particle).

Alpha decay is radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus) and transforms into an atom with an atomic number that is reduced by two and mass number that is reduced by four.

5 0
3 years ago
Read 2 more answers
Calculate the pH of a solution with a [H 3 O + ]=5.6x10 -9 M.
Gre4nikov [31]

Answer:

8.3

Explanation:

pH is the measure of the H+ or H30 (they r the same thing) ions in a solution. it is equal to -log[H+]. [H+]=  Molar concentration of H+ ions.

6 0
3 years ago
20.1 g of aluminum and 219 g of chlorine gas react until all of the aluminum metal has been converted to AlCl3. The balanced equ
Romashka-Z-Leto [24]

Answer:

<em>The amount of Cl2 gas left , after the reaction goes to completion is : </em><u><em>139.655 grams</em></u>

Explanation:

Molar mass : It is the mass in grams present in one mole of the substance.

Moles of the substance is calculated by:

Moles=\frac{Mass}{Molar\ mass}

2Al(s)+3Cl_{2}(g)\leftarrow 2AlCl_{3}(g)

According  to this equation:

2 mole of Al = 3 mole of Cl2 = 2 mole of AlCl3

Molar mass of Al = 27.0 g/mol

Mass of Al = 20.1 gram

Moles of Al present in the reaction :

Moles=\frac{Mass}{Molar\ mass}

Moles=\frac{20.1}{26.98}

Moles of Al = 0.744

Similarly calculate the moles of Cl2

Molar mass of Cl2 = 71.0 g/mol

Mass = 219 gram

Moles=\frac{Mass}{Molar\ mass}

Moles=\frac{219}{70.98}

Moles of Cl2 = 3.08 moles

According to equation,

2 mole of Al reacts with = 3 mole of Cl2

1 moles of Al reacts with = 3/2  mole of Cl2

0.744 moles of Al reacts with = 3/2(0.744) moles of Cl2

= 1.116 moles of Cl2

But actually present Cl2 = 3.08 moles

Hence Al is the limiting reagent , and Cl2 is the excess reagent.

The whole Aluminium Al get consumed during the reaction.

The amount of Cl2 in excess = Total Cl2 - Cl2 consumed

Cl2 in excess = 3.08 - 1.116 = 1.964 moles

<u>Cl2 in grams</u><u> </u>= 1.964 x 70.90 <u>= 139.655 grams</u>

6 0
4 years ago
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