Vapour pressure of system depends upon intermolecular forces of interaction. Greater the interaction, larger will the vapour pressure, more will be the boiling point.
Answer 1:
CH4 stands for methane and CH3Cl named as chloromethane. In methane, all the valances of C atom is satisfied by hydrogen. Due to this, it has zero dipole moment. While, in case of CH3Cl, one of the valance is satisfied by an electronegative element i.e. Cl. Due to this, it acquires a polar character. Also, it has a net dipole moment. Due to this, CH3Cl exhibits dipole-dipole intermolecular force of attraction, which is absent in CH4. Hence, CH3Cl has lower vapor pressure as compared to CH4.
Answer 2:
H2CO is named as formaldehyde, while CH3OH is named as methyl alcohol. In case of methyl alcohol, hydrogen atom (an electropositive atom) is bonded to oxygen (a highly electronegative element). This is absent in case of formaldehyde. Due the this, methyl alcohol as greater polarity as compared to formaldehyde. Due the greater polarity, vapour pressure of CH3OH is less as compared to H2CO.
Answer 3:
<span>CH3CH2CH2OH is named as propanol or propyl alcohol. Propyl alcohol, has longer chain length as compared to methyl alcohol (CH3OH). Both of this compounds has a polar character due to presence of -OH functional group. However, due to long chain of propyl alcohol, polar character increases. This can be attributed the +I effect of CH3 group. Due to this, intermolecular forces of interaction are higher in propanol, thereby decreasing its vapor pressure as compared to methanol. </span>
Answer:
The student should weigh out 61.2g of ethanolamine [6.12 * 10]
Explanation:
In this question, we are expected to calculate the mass of ethanolamine needed to make 60.0ml of it given that the density of the ethanolamine in question is 1.02g/cm^3
Mathematically, it has been shown that mass = density * volume
Hence, by multiplying the density by the volume, we get the mass.
Now, from the question we can see that we have the values for the density and the volume. We now need to get the mass.
Since cm^3 is same as ml, we need not perform any conversion.
Hence, the needed mass is:
60 * 1.02 = 61.2g
<span>Problem 1) From the given options the one that is proper of a chemist is the option B. testing a sample of water form a well. This activity belongs to the field called analytical chemistry which uses special techniques (instruments and processes) to study (separate, identify and quantify matter).
Problem 2) The only of the given procedures that involves a physical change in one substance is the option D. grinding chalk into a fine powder. In this process the molecules remain being the same (same atoms bonded in the number and way).</span>
Try your luck kid hope it helps
