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zvonat [6]
3 years ago
14

If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct

ure The central X atom is single bonded to two Y atoms, which are 180 degrees apart. The X atom has two lone pairs. could be abbreviated as XY2Z2. Classify each molecule according to its shape.Molecular Geometries: linearbent 120bent 109trigonal pyramidalT-shapedsee sawsquare planarsquare pyramidalPossiblities:XY4Z2XY5ZXY2ZXY2Z3XY4ZXY2Z2XY3Z2XY3Z
Chemistry
1 answer:
Snowcat [4.5K]3 years ago
8 0

Answer:

XY4Z2 ----- square planar

XY5Z ------- square pyramidal

XY2Z----- bent < 120°

XY2Z3 ----- Linear

XY4Z ---- see saw

XY2Z2 ----- bent <109°

XY3Z2 ----- T shaped

XY3Z ----- Trigonal pyramidal

Explanation:

The valence shell electron pair repulsion theory ( VSEPR) gives the description of molecular geometry based on the relative number of electron pairs present in the molecule.

However, electron pairs repel each other, the repulsion between two lone pairs is greater than the repulsion between a lone pair and a bond pair which is also greater than the repulsion between two lone pairs.

The presence of lone pairs distort the bond angle and molecular geometry from the expected geometry based on VSEPR theory. Hence, in the presence of lone pairs of electron, the observed molecular geometry may be different from that predicted on the basis of the VSEPR theory, the bond angles also differ slightly or widely depending on the number of lone pairs present.

All the molecules in the question possess lone pairs, the number of electron pairs do not correspond to the observed molecular shape or geometry due to lone pair repulsion. Usually, the molecular geometry deals more with the arrangement of bonded atoms in the molecule.

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The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A
liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

p^{OH}=14-56.17

      =8.823

The volume of the NH_{3} = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated NH_{3} =0.10 M

The volume of the NH_{4}Cl= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated NH_{4}Cl= 0.10 M

The volume of the H_{2}So_{4}= 30 ml

convert into the liter= 0.030L  

The value of concentrated H_2So_4=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of NH_3 = \frac{0.10\times 0.040}{0.120}= 0.33 \ M

calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

5 0
3 years ago
Lindsey is trying to gain credibility for her studies. She completed her experiment and discussed her finding with colleagues. H
noname [10]

Answer:

Option B:Publishing scientific journals

Explanation:

We are told that Lindsey is trying to gain credibility for her studies.

Since she completed her experiment and discussed her finding with colleagues, the most logical next step would be to publish scientific journals. This is because the other options given are not steps that should be taken because she has completed the research and therefore has no need to speak at a conference next nor even create new charts which they must have done during the research. No need for her to make sure the topic is popular.

Option B is correct

8 0
3 years ago
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