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FrozenT [24]
2 years ago
8

Which best explains why the trend in noble gas boiling points increases down the group?A) increasing dispersion interactions B)

increasing dipole-dipole interactions C) increasing ion-dipole interactions D) increasing hydrogen bonding interactions E) increasing ion-ion interactions

Chemistry
1 answer:
zhenek [66]2 years ago
8 0

Answer:

A) increasing dispersion interactions

Explanation:

Polarizability allows gases containing atoms or nonpolar molecules (for example,  to condense. In these gases, the most important kind of interaction produces <em>dispersion forces</em>, <em>attractive forces that arise as a result of temporary dipoles induced in atoms or molecules.</em>

<em>Dispersion forces</em>, which are also called <em>London forces</em>, usually <u>increase with molar mass because molecules with larger molar mass tend to have more electrons</u>, and <u>dispersion forces increase in strength with the number of electrons</u>. Furthermore, larger molar mass often means a bigger atom whose electron distribution is more easily disturbed because the outer electrons are less tightly held by the nuclei.

Because the noble gases are all nonpolar molecules, <u>the only attractive intermolecular  forces present are the dispersion forces</u>.

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2 years ago
When adding the measurements 42.1014 g + 190.5 g, the answer has ___ significant figures.
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Answer: The answer has 7 significant figures

Explanation:

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4 0
3 years ago
Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L​
Oduvanchick [21]

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___<u>Cu</u><u>+</u><u>²</u><u> </u>__|_<u>2</u><u>OH</u><u>-</u>____|

|<u>Initial concentration(M</u>)|___<u>0</u>__|_<u>0</u>______|

<u>|Change in concentration(M)</u>|_<u>+S</u><u> </u>|__<u>+2S</u>__|

|<u>Equilibrium concentration(M)|</u><u>_S</u><u> </u><u>_</u><u>|</u><u>2S___</u><u>|</u>

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

s =  \sqrt[3]{ \frac{2.2 \times  {10}^{ - 20} }{4} }  = 1.8 \times  {10}^{ - 7}

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

Cu (OH)2 =  \frac{1.8 \times  {10}^{ - 7} mol \:Cu (OH)2 }{1L}  \times  \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2}  \\  = 1.75428 \times 10 ^{ - 5}

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

I hope I helped you^_^

8 0
2 years ago
Give an example of chemical equilibrium you have come across into your day-to-day life. Explain how it meets the definition of e
Roman55 [17]
There are many examples of chemical equilibrium all around you. One example is a bottle of fizzy cooldrink. In the bottle there is carbon dioxide (CO2) dissolved in the liquid. There is also CO2 gas in the space between the liquid and the cap
7 0
2 years ago
A mixture of two compounds, A and B, was separated by extraction. After the compounds were dried, their masses were found to be:
Arlecino [84]

Answer:

The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.

Explanation:

Mass of compound A in a mixture  = 119 mg

Mass of compound A after re-crystallization = 83 mg

Percent recovery from re-crystallization :

\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100

Percent recovery of compound A:

\frac{83 mg}{119 mg}\times 100=69.74\%

Mass of compound B in a mixture  = 97 mg

Mass of compound B after re-crystallization = 79 mg

Percent recovery of compound B:

\frac{79 mg}{97 mg}\times 100=81.44\%

3 0
2 years ago
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