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emmasim [6.3K]
3 years ago
11

Which one of the following exhibits dipole-dipole attraction between molecules?

Chemistry
1 answer:
Maslowich3 years ago
8 0

Answer: No one. All these molecules are nonpolar.

Explanation: Dipole-dipole attractions occur only between polar molecules.  To know if a molecule is polar, you have to observe if their dipole moment (μ) is different from 0. You can do this by observing the vectors of charge in each molecule.

In all the molecules of the alternatives, the overall molecule is nonpolar (the sum of the vectors is equal to 0). So, these molecules are attracted only by induced dipole forces (weak forces that exist between nonpolar molecules).

For F2, this molecule is nonpolar because is composed of equal atoms. So, there isn't an atom which pulls the charge for itself.

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The chemical formula for magnesium oxide is 2MgO. The number 2 represents the number of ___.
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Answer:

magnesium oxide molecules

Explanation:

this is because the two is before the formula of MgO

7 0
3 years ago
How many molecules are in 145 g of aluminum sulfate?
Rina8888 [55]

Answer:2.55 x 10^23 molecules

Explanation:

i had the same question for my class lol, do u go to forest park?

5 0
3 years ago
The half-life of radon gas is approximately four days. Four weeks after the introduction of radon into a sealed room, the fracti
maksim [4K]

The fraction of the original amount remaining is closest to 1/128

<h3>Determination of the number of half-lives</h3>
  • Half-life (t½) = 4 days
  • Time (t) = 4 weeks = 4 × 7 = 28 days
  • Number of half-lives (n) =?

n = t / t½

n = 28 / 4

n = 7

<h3>How to determine the amount remaining </h3>
  • Original amount (N₀) = 100 g
  • Number of half-lives (n) = 7
  • Amount remaining (N)=?

N = N₀ / 2ⁿ

N = 100 / 2⁷

N = 0.78125 g

<h3>How to determine the fraction remaining </h3>
  • Original amount (N₀) = 100 g
  • Amount remaining (N)= 0.78125 g
  • Fraction remaining =?

Fraction remaining = N / N₀

Fraction remaining = 0.78125 / 100

Fraction remaining = 1/128

Learn more about half life:

brainly.com/question/26374513

7 0
2 years ago
Consider the following reaction at 298.15 K: Co(s)+Fe2+(aq,1.47 M)⟶Co2+(aq,0.33 M)+Fe(s) If the standard reduction potential for
fgiga [73]

Answer:

The correct answer is 0.186 V

Explanation:

The two hemirreactions are:

Reduction: Fe²⁺ + 2 e- → Fe(s)  

Oxidation : Co(s)  → Co²⁺ + 2 e-

Thus, we calculate the standard cell potential (Eº) from the difference between the reduction potentials of cobalt and iron, respectively,  as follows:

Eº = Eº(Fe²⁺/Fe(s)) - Eº(Co²⁺/Co(s)) = -0.28 V - (-0.447 V) = 0.167 V

Then, we use the Nernst equation to calculate the cell potential (E) at 298.15 K:

E= Eº - (0.0592 V/n) x log Q

Where:

n: number of electrons that are transferred in the reaction. In this case, n= 2.

Q: ratio between the concentrations of products over reactants, calculated as follows:

Q = \frac{ [Co^{2+} ]}{[Fe^{2+} ]} = \frac{0.33 M}{1.47 M} = 0.2244

Finally, we introduce Eº= 0.167 V, n= 2, Q=0.2244, to obtain E:

E= 0.167 V - (0.0592 V/2) x log (0.2244) = 0.186 V

4 0
4 years ago
Thisss for you blake :)))
Lady_Fox [76]
I’m not Blake but ok
4 0
3 years ago
Read 2 more answers
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