The Discussions in this course are set up deepen your understanding of the material as you make real world connections and emplo
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Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.
Answer 1) σ Bond a : C has O has
.
Explanation : The orbitals of oxygen and carbon which are involved in hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.
Answer 2) π Bond a: C has π orbitals and O also has π orbitals.
Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.
Answer 3) Bond b: O H has only S orbital involved in bonding.
Explanation : The bonding at 'b' position involves oxygen hydrogen atoms in it. It has
hybridized orbitals and S orbital of hydrogen involved in the bonding.
Answer 4) Bond c: C is O is also
Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of hybridized bonds.
Answer 5) Bond d: C atom has C atom has
Explanation : At the position of 'd' the bonding between two carbon atoms is found to be . Therefore, the orbitals that undergo
hybridization are
.
Answer 6) Bond e : C1 containing O
C2 is
Explanation : The carbon atom which contains oxygen along with a double bond has hybridized orbitals involved in the bonding process; whereas the carbon at C2 has
hybridized orbitals involved during the bonding. This is for the 'e' position.
Answer:
Explanation:
The frequency and the wavelength of a wave are related by the equation
where
v is the speed
f is the frequency
is the wavelength
For the sound wave in this problem, we have
f = 634 Hz (frequency)
(wavelength)
Therefore, the speed of the wave is
The speed of sound in air depends on the temperature according to the equation
where T is the temperature.
In this problem, we know the speed, so we can calculate the temperature: