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alisha [4.7K]
3 years ago
5

Explain why fluorine has a smaller atomic radius than both oxygen and chlorine

Chemistry
1 answer:
Alex777 [14]3 years ago
5 0

Answer:

1) Has a smaller radius than oxygen because of the increased electromagnetic attraction of the nuclei

2) Has a smaller radius than chlorine because all the electrons of F have lower energy levels and have less repulsion of other electrons and hence are more attracted to the nuclei .

Explanation:

Further the electrons are from the nuclei , the bigger the atomic radius is.

(+) attraction of electrons to the nuclei, (-) repulsion of the electrons away from the nuclei.

1) From O to F:

(+) there is one more proton --> Stronger positive charge of the nuclei means that the electrons are attracted more , then they come closer to it and therefore the radius decreases

(-) There is one more electron --> Every electron is repulsed by others away from the nuclei --> the radius increases. But this effect is not so strong because the new electron is added at the same energy level.

Overall the (+) effect is stronger than the (-) effect --> Radius decreases from O to F

2) From F to Cl

(+) there is one more protons --> Same effect as before

(-) There is one more electron --> Every electron is repulsed by others away from the nuclei. But this time the new electrons have a higher energy level --> Meaning that they are less attracted and hence the radius increases.

And also the other inner layers of electrons (electrons of lower energy levels) repulse this new external layer of electrons more effectively than the case of O  --> Strong repulsion effect (called electron shielding effect) --> Radius increases

Overall the (-) effect is stronger than the (+) effect --> Radius increases from F to Cl  (or decreases from Cl to F)

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An oxide of phosphorus contains 56.4% phosphorus and 43.6% oxygen. It's relative molecular mass is 220. Find both the empirical
sladkih [1.3K]
Moles of P = 56,4g/30,974g/mole = 1,82 moles P
moles of O = 43,6/15,999 = 2,73 moles of O

converting to the simplest ratio:
For P : 1,82/1,82 = 1
For O : 2,73/1,82 = 1,5

1 P and 2 oxygens.
PO2 -> the empirical formula

hope this help
7 0
3 years ago
Read 2 more answers
How many moles of hydrogen are present in 5.30 moles of C5H10O2, propyl acetate, the compound that provides the odor and taste o
elena-s [515]

Answer:

Option D. 53 moles.

Explanation:

The following data were obtained from the question:

Number of mole of C5H10O2 = 5.3 moles

Number of mole of Hydrogen in 5.3 moles of C5H10O2 =?

From the chemical formula of propyl acetate, C5H10O2,

1 mole of C5H10O2 contains 10 moles of H.

Therefore, 5.3 moles of C5H10O2 will contain = 5.3 × 10 = 53 moles of H.

Thus, 5.3 moles of C5H10O2 contains 53 moles of H.

5 0
2 years ago
Which of these groups includes organisms that are MOST closely related?
gregori [183]
The answer is D, species!

Hope this helps!
8 0
3 years ago
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Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3
krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction (\Delta S^o).

\Delta S^o=S_{product}-S_{reactant}

\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0{(NH_3)} = standard entropy of NH_3

\Delta S^0{(H_2)} = standard entropy of H_2

\Delta S^0{(N_2)} = standard entropy of N_2

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]

\Delta S^o=-198.3J/K

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0
3 years ago
What is the correct name for Au3N?
Reptile [31]

Answer:

option D= Gold (I) nitride

Explanation:

The name of the given compound is gold(I) nitride.

Molar mass can be determine by following way:

molar mass Au3N = (molar mass of gold × 3) + (molar mass of nitrogen)

molar mass Au3N = (196.97 × 3 ) + ( 14 )

molar mass of Au3N =  590.91 g/mol + 14 g/mol

molar mass of Au3N = 604.91 g/mol

The nitrogen has valency of -3 so three Au(+1) will require while the valency of Au is (1+) one nitrogen will require to make the compound overall neutral.

Au3N

3(1+) + (-3) = 0

+3 - 3 = 0

0 = 0

The overall charge is 0, the compound will be neutral.

8 0
3 years ago
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