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oee [108]
2 years ago
14

What atomic or hybrid orbitals make up the sigma bond between c2 and c3 in propyne, chcch3 ? (c2 is the second carbon in the str

ucture as written.) orbital on c2 + orbital on c3 what is the approximate c1-c2-c3 bond angle ? ... °?

Chemistry
1 answer:
Anvisha [2.4K]2 years ago
4 0

Answer:

\boxed{\text{$\sigma$(sp-sp$^{3}$); 180$^{\circ}$}}

Explanation:

(a) Hybridization of orbitals

See the Lewis structure of propyne in the first diagram below.

C1 is directly bonded to two other atoms (H and C2) so it is sp hybridized.

C2 is directly bonded to two other atoms (C1 and C3) so it is sp hybridized.

C3 is directly bonded to four other atoms (C2 and 3 H) so it is sp³ hybridized.

(b) C2-C3 Sigma Bond

See the atomic and molecular orbitals in the second picture below.

C3 is using its hybrid atomic orbitals to form sigma molecular bonds. The C2-C3 sigma bond is formed by the overlap of the C2 sp atomic orbital with the C3 sp³ atomic orbital to make a σ(sp-sp³) molecular orbital.

(c) Bond angles

C1 and C2 are sp hybridized. Since the angle between sp orbitals is 180°, all atoms directly attached the C1 and C2 must be in a straight line. The C-C-C bond angle is 180°.

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Answer:

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Kinetic energy is the energy in motion, as such, a device that transfers kinetic energy transfers the energy the power source has into other energy forms

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C) The first law of thermodynamics states that energy is conserved and it can neither be created nor destroyed, but can be changed from one form to another.

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Explanation:

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3 years ago
Write the equilibrium constant expression for this reaction: nh4
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3 years ago
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3 years ago
Does changing the number of neutrons change the identity of the element you have built ?
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Answer:

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3 years ago
Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2
baherus [9]

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) Cu_{(s)} →Cu^{2+}_{(aq)} +2e E°=0.337 v

(reduction) MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}→MnO_{2 (aq)}+2H_{2}O E°=1.679 v

(overall) 2MnO_{4 (aq)}+3Cu_{(s)}+8H^{+}_{(aq)}→3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346

E=1.346

5 0
2 years ago
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