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Sladkaya [172]
2 years ago
7

Which compound most likely contains polar covalent bonds? NaCl AlF3 Br2 NF3

Mathematics
2 answers:
belka [17]2 years ago
7 0

Answer: The correct answer is NF_3

Explanation:

In NF_3 molecule, there are three bond pairs and one lone pair present in the molecule. In a molecule, lone pair present on the nitrogen atom will have a dipole moment in upward direction where as total dipole moment formed by three fluorine atoms will be more in magnitude and it will be directed in downward direction which means that the net dipole moment will be in downward direction. As shown in image attached.

Hence, covalent bond present between nitrogen and fluorine will become polar in nature.

So, the correct answer is NF_3.

tia_tia [17]2 years ago
4 0

Answer:

The correct option is NF_3

Step-by-step explanation:

we have to select the compound which most likely contains polar covalent bonds.

In polar covalent bond, the electrons shared by the atoms spend a greater amount of time, on the average. This is due to electronegativity difference between the two atoms in the molecule.

NaCl is purely ionic bond strong electrostatic forces of attraction between positively charged Na and negatively charged Cl.

AlF_3 is ionic.

Br_2 is a non-polar molecule both atoms are equally electronegative, neither of them withdraws electron density from the single covalent bond between them.

In molecule,NF_3 there are three bond pairs and one lone pair present in the molecule. In a molecule, lone pair present on the nitrogen atom will have a dipole moment in upward direction whereas total dipole moment formed by three atoms of fluorine will be more in magnitude and it will be directed in downward direction which means that the net dipole moment will be in downward direction. As shown in image attached.

Hence, covalent bond present between the atoms  nitrogen and fluorine will become polar in nature.

The correct option is NF_3

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a) Binomial distribution B(n=12,p=0.01)

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P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

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