Answer:
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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
You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
The percent A by mass for substance AB =<u> 75%</u>
<h3>Further explanation</h3>
Proust states the Comparative Law that compounds are formed from elements with the same Mass Comparison, so that compounds have a fixed composition of elements
Empirical formula is the mole ratio of compounds forming elements.
From Substance AB₂ is 60.0% A by mass.
Let's say that AB₂ mass = 100 gram, then
mass A = 60 gram
mass B = 40 gram : 2 (coefficient in compound AB₂ = 2) = 20 gram
In compound AB:
Total mass = mass A + mass B
Total mass = 60 + 20 grams = 80 grams
Then the percentage of compound A = (60: 80) = 75%
<h3>Learn more</h3>
Grams of KO₂ needed to form O₂
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Keywords : percent mass, substance
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Answer:
1. Saturated hydrocarbons may be cyclic or acyclic molecules.
2. An unsaturated hydrocarbon molecule contains at least one double bond.
Explanation:
Hello,
In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).
Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true
Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.
Therefore, first and second statements are correct.
Best regards.