Answer:
0.895 g/mL
Explanation:
Step 1: Given data
Mass of liquid Z (m): 2.763 lb
Volume of liquid Z (V): 5.93 cups
Step 2: Convert "m" to grams
We will use the relationship 1 lb = 453.59 g.
Step 3: Convert "V" to milliliters
We will use the relationship 1 cup = 236.59 mL.
Step 4: Calculate the density of the liquid Z
The density (ρ) of the liquid Z is equal to its mass divided by its volume.
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.
