A 2.00 L container of gas has a pressure of 1.00 atm at 300 K. The temperature of the gas is halved to 150K, and the measured pr
1 answer:
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Answer:
E) rate of appearance of C = 0.45 M/s
rate of the reaction = 0.15 M/s
Explanation:
2A + B → 3C
Writing rate law for the reaction:
<u>Rate of reaction</u> = - = -
=
→ equation 1
Given that the rate of disappearance of A is 0.3 M/s
⇒ - = 0.3 M/s
⇒Rate of reaction = - =
×0.3 M/s
⇒<u>Rate of reaction = 0.15 M/s</u>
From equation 1, = -
=
×0.3 M/s
⇒ = 0.45 M/s
or <u>the rate of appearance of C = 0.45 M/s</u>
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.
