Answer:
a. 174 mL
Explanation:
Let's consider the following reaction.
2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)
We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:
0.1550 L × 0.112 mol/L = 0.0174 mol
The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:
2 × 0.0174 mol = 0.0348 mol
The volume of a 0.200 M KI solution that contains 0.0348 moles is:
0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL
Answer:
Option (c).
Explanation:
Cellulose is one of the most important carbohydrate of the plant that provide structural framework to the plant cells. Cellulose is made of the monomer of the glucose that are linked through the beta glycosidic linkages.
Although cellulose contains large number of glucose but it cannot be used as the nutrient source for humans. Humans and other vertebrates lacks the enzyme cellulase that is required for the digestion of cellulose. Cellulose cannot be broken down in the human body.
Thus, the correct answer is option (c).
To find - Identify what kind of ligand (weak or strong), what kind
of wavelength (long or short), what kind of spin (high spin or
low spin) and whether it is paramagnetic or diamagnetic for
the following complexes.
1. [Mn(CN)6]4-
2. [Fe(OH)(H2O)5]2
3. [CrCl4Br2]3-
Step - by - Step Explanation -
1.
[Mn(CN)⁶]⁴⁻ :
Ligand - Strong
Wavelength - Short
Spin - Low spin
Number of unpaired electrons = 1 ∴ paramagnetic.
2.
[Fe(OH)(H₂O)₅]²⁺ :
Ligand - Weak ( both OH⁻ and H₂O )
Wavelength - Long
Spin - High spin
Number of unpaired electrons = 5 ∴ paramagnetic.
3.
[CrCl₄Br₂]³⁻ :
Ligand - Weak ( both Br⁻ and Cl⁻ )
Wavelength - Long
Spin - High spin
Number of unpaired electrons = 3 ∴ paramagnetic.
“Bonding molecular orbitals are formed by... in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule.”
The answer is a cation, I may be wrong but cation is plus and anion is subtract
