A clear colorless liquid in an open beaker was heated to boiling. The liquid began to boil at 110°C, and as vapors escaped, the temperature of boiling gradually increased to 115°C, at which point the heating was stopped. On the basis of this information, we can say that the material in the beaker was a

Omitted options and they are

a) pure compound.

b. pure element.

c. pure substance.

d. homogeneous solution.

e. heterogeneous solution

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

is has three carbons, so the answer is propylamine.

meth-

eth-

prop-

but-

pent-

hex-

hept-

oct-

non-

explanation: