Omitted options and they are
a) pure compound.
b. pure element.
c. pure substance.
d. homogeneous solution.
e. heterogeneous solution
Answer:d. homogeneous solution.
Explanation:
Pure substances or elements or compounds have a definite and sharp melting or boiling point, Any substance that is not pure is impure and will have different temperature of melting or boiling points.
To this effect, the clear colorless liquid cannot be a Pure substance, element or compound.
We can therefore say that the clear colorless liquid would be a homogeneous solution because a homogeneous solution is a mixture of constituents which completely mixes together such that each constituents cannot be seen with naked eye, When heated to boiling, each constituent in the mixture will give different boiling points.
A heterogeneous Solution, too is a mixture but contains constituents that can be seen and not a clear colourless solution.
Therefore On the basis of this information, we can say that the material in the beaker was a Homogeneous solution
Answer:
C. transition elements
Explanation:
Transition elements are groups of metallic elements that have partially filled d orbitals. They occupy the d-block of the periodic table (Group 3-12). Examples of elements that fall under this category are Nickel (Ni), Cobalt (Co), Copper (Cu), Zinc (Zn) etc. Transition metals have outstanding properties that distinguishes them from other elements.
One of these properties is their ability to form colored compounds due to their unfilled d electron shells. They form ions that are usually colored in solid compounds and in solution.
They are products of the reaction.
Answer:
Isopropyl propionate
Explanation:
1. Information from formula
The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.
2. Information from the spectrum
(a) Triplet-quartet
A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group
(b) Septet-doublet
A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group
(c) The rest of the molecule
The ethyl and isopropyl groups together add up to C₇H₁₂.
The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.
The compound is either
CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.
(d) Well, which is it?
The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.
The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.
The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.
We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.
3. Summary
My peak assignments are shown in the diagram below.