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Nostrana [21]
3 years ago
6

Which group on the periodic table would have zero electronegativity because they have a full octet?

Chemistry
1 answer:
bija089 [108]3 years ago
3 0
The group on the periodic table that would have 0 electronegativity due to the fact that their valence shell is full, i.e, have a full octet would be the inert or noble gases. They have a total of 8 electrons in their valence shell and are thus inert and cannot strongly attract electrons toward itself, from neighbouring atom electrons as it does not need to.
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HELP ASAP!!!!!!
harina [27]

that seems very false but I believe its the second one

7 0
3 years ago
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Is a process that could be used to seperate dissolved particles from the liquid in a solutoin?
kaheart [24]
Salutations!

_____ Is a process that could be used to separate dissolved particles from the liquid in a solution?

Distillation is the process that could be used to separate dissolved particles from the liquid in a solution. Distillation best works when liquids have unalike boiling points. It is widely used to to isolate ethanol.

Hope I helped :D
5 0
3 years ago
Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a
Inessa [10]

Different isotopes of the same element emit light at slightly different wavelengths, the minimum number of slits  is mathematically given as

N=1820slits

<h3>What minimum number of slits is required to resolve these two wavelengths in second-order?</h3>

Generally, the equation for the wave is mathematically given as

d\ sin\ (\theta\ m) \ = \ m\  \lambda

Where the chromatic resolving power (R) is defined by

R\  =\ \lambda\ / \ d \  \lambda

R = nN,

Therefore

\lambda_1 \ = \  (656.45)(1 \ * \ 10^{-9})/1mm

\lambda_1= 656.45*10^{-9}

and

\lambda_2= (656.27)(1*10^{-9})/1mm

\\\\\lambda_2= 656.27*10^{-9}m

In conclusion, the minimum number of slits is required to resolve these two wavelengths in second-order

N\ =\ \dfrac{\lambda}{m\ d\ T\ }\\\\

Therefore

N\ =\ \dfrac{656.45 \ * \ 10^{-9}}{2\ * \ (0.18*10^{-9})}

N=1820slits

Read more about slits

brainly.com/question/24305019

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7 0
2 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
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Using the principles of VSEPR theory, you can predict the geometry around any atom in any molecule, no matter how complex. Enant
Kaylis [27]

Explanation:

The O atom is sp3 in a water molecule, with two sigma bonds and two lone pairs of electrons like that in water. The steric integer is thus 4, and its structure is tetrahedral.

The C atom is sp hybridised into two identical bonds and two identical bonds in acetylene.

The steric integer is therefore 2 because only sigma bonds are engaged in deciding hybridization, and its structure is linear.

The C atom is sp2 hybridised in ethene with single pi bond and three sigma identical bonds.

Thus the steric integer is 3, and its structure is planar trigonal.

The C atom is sp2 hybridized in ethene, with one pi bond and three sigma identical bonds.

The steric integer would therefore be 3 and its structure is planar trigonal.

The O atom is sp3 in a water molecule with two bond pairs and two lone pairs of electrons like that. The steric integer is thus 4, and its structure is tetrahedral.

The C atom is sp3 in a methane ring, with 4 bond pairs and no solitary pairs of electrons like that. The steric integer is thus 4, and its structure is tetrahedral.

6 0
3 years ago
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