Answer:
E. CH₄ < CH₃Cl < CH₃OH < RbCl
Explanation:
The molecule with the stronger intermolecular forces will have the higher boiling point.
The order of strength of intermolecular forces (strongest first) is
- Ion-Ion
- Hydrogen bonding
- Dipole-dipole
- London dispersion
RbCl is a compound of a metal and a nonmetal. It is an ionic compound, so it has the highest boiling point.
CH₃Cl has a C-Cl polar covalent bond. It has dipole-dipole forces, so it has the second lowest boiling point.
CH₃OH has an O-H bond. It has hydrogen bonding, so it has the second highest boiling point.
CH₄ has nonpolar covalent C-H bonds. It has only nonpolar bonds, so the only attractive forces are London dispersion forces. It has the lowest boiling point.
Thus, the order of increasing boiling points is
CH₄ < CH₃Cl < CH₃OH < RbCl
Answer:
- See the image attached. It is taken from an online chemistry textbook.
- See the explanation below.
Explanation:
<em>Sodium chloride</em> consits of sodium cations (positive ions), Na⁺, and chloride anions (negative ions), CL⁻.
<u>Pure sodium chloride</u> is packed in crystals: sodium ions and chloride ions are packed together and the ions are in fixed positions. There are not free electrons that can move. Thus, sodium chloride doesn't conduct electricity, because there are no electrons or ions which are free to move.
In aqueous solution, sodium chloride units dissociates into their ions:

Those ions are freely to move in the solution, and such they are charge carriers, which conduct the electricity.
As explained above, in solid sodium chloride, the ions cannot move and there is not flow of current.
That is why solid pure salt of NaCl does not conduct electricity and the solutions of NaCl do conduct electricity.
The image attached show both diagrams. In the diagram A, the ions are packed together, showing that they cannot move. In the diagram B, the ions are dissolved in water, showing that they can move and carry the charge, allowing the flow of current.
Answer:

Explanation:
We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.
- ΔT= final temperature - initial temperature
The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.
- ΔT= 88.9 °C - 58.8 °C = 30.1 °C
Now we know three variables:
- Q= 4500.0 J
- c= 0.4494 J/g°C
- ΔT = 30.1 °C
Substitute these values into the formula.

Multiply on the right side of the equation. The units of degrees Celsius cancel.

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

The units of Joules cancel.


The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

The mass of the sample of metal is approximately <u>333 grams.</u>
Hydrocarbons may vary from state to state depending on the length of the carbon chain. For methane to butane, they are usually in gas form. Starting from pentane, they are in the liquid form. For very long carbon chains, that occur as solids. Now, it depends if the pentane is in a nonpolar liquid or polar liquid. Since pentane is nonpolar, it is miscible in the liquid solvent. The movements would most likely be free flowing. But if he solvent is polar, the molecules would repel with one another. In the end, it will form two liquid phases, on for the pentane and one for the polar solvent.
Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.
Explanation:
1) Molarity of 0.250 L HCl solution : 0.0328 M

Moles of HCl in 0.250 L solution = 0.0082 moles
2) Molarity of 0.100 L NaOH solution : 0.0245 M

Moles of NaOH in 0.100 L solution = 0.00245 moles
3) Concentration of hydrochloric acid in the resulting solution.
0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.
Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L
Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles = 0.00575 moles

Molarity of HCl left un-neutralized :
0.0164 molar concentration of hydrochloric acid in the resulting solution.