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3 years ago

How many total carbon atoms are found in a molecule of 2-methyl-2-butene?

Chemistry

2 answers:

ExtremeBDS [4]3 years ago

7 0

Equation of 2 methyl 2 butene:
H3CC(CH3) = CHCH3

this formula show there are 5 carbon (C) atoms.

hope this help

IceJOKER [234]3 years ago

5 0

Answer: The total number of carbon atoms found in the given molecule is 5.

Explanation:

We are given a chemical compound having chemical name 2-methyl-2-butene.

This means that the given hydrocarbon has double bond between $C_2-C_3$ and a methyl substituent is attached to $C_2$ atom.

Structure of the given compound is attached below and the total number of carbon atoms that are found in the given molecule is 5.

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At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

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Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago

What is the standard cell notation for a galvanic cell made with silver and nickel?

MA_775_DIABLO [31]

Answer:

Ni(s)∣∣Ni2+(aq) ∣∣∣∣ Ag+(aq)∣∣Ag(s)

Explanation:

Start by finding the standard reduction potential for the Ag+ and Ni2+ ion. Normally, the values are listed at the back of most chemistry textbooks.

Ag+(aq)+1e−→Ag(s) Eo=0.80 V
Ni2+(aq)+2e−→Ni(s) Eo=−0.23 V

In the galvanic cell, the reaction is spontaneous and for a spontaneous reaction Eocellmust be a positive quantity.

Eocell=EoAnode+Eocathode

Manipulate the two equations so that Eocell is positive. Note that the anode is the site of oxidation (where electrons are lost) and the cathode (where electron are gained) is the site for reduction.

Ag+(aq)+1e−→Ag(s) Eo=0.80 V
Ni(s)→Ni2+(aq)+2e− Eo=0.23 V

2×{Ag+(aq)+1e−→Ag(s)} Eo=0.80 V (Cathode)
Ni(s)→Ni2+(aq)+2e− Eo=0.23 V (Anode)
 −−−
2Ag+(aq)+Ni(s)→Ni2+(aq)+2Ag(s) Eocell=1.03 V

Start with the anode components (site of oxidation) - the cathode components are listed to the right.

The single vertical lines indicate the boundary (phase difference) between solid Niand Ni2+ ions in the aqueous solution of the first compartment and between solid Ag and Ag+ ions present in the aqueous solution of the second compartments.

The double vertical lines refer to the salt bridge - note that the salt bridge must be an inert salt to both ions present in both compartments of the galvanic cell ...

5 0

3 years ago