All categories

3 years ago

What is the hybridization of the central atom of each of the following molecules? a.) H2O b.) PBr5 c.) COCl2? Thank you so much!

1 answer:

7 0

<span>The notion of hybrid orbitals was "invented" by Linus Pauling in 1931 to explain the observed geometry of mostly organic molecules. Today, we find that hybrid orbitals are pretty much confined to the central atoms found in the second period of the periodic table. Periods 3 and up .... not so much.

(a) H2O ..... O is the central atom, tetrahedral, sp3 hybridization
(b) PBr5 .... P is the central atom, maybe sp2 for the three equatorial bonds *
(c) COCl2.. C is the central atom, trigonal planar, sp2 hybridization </span>

You might be interested in

If a rock is thrown 0.8 meters into the air, How fast was it thrown

sergiy2304 [10]

V=0 m/s, u=?, a = 9.8 m/s² and s = 0.8 m
u²=2×9.8×0.8=15.68
<span>u=3.959 ≈ 3.96 m/s

</span>

3 0

3 years ago

Read 2 more answers

Joaq...

MrRa [10]

Answer:

all the statements are true of chemical changes

3 0

2 years ago

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0

3 years ago

An elimination reaction can best be described as a reaction in which an elimination reaction can best be described as a reaction

tresset_1 [31]

So I’m not 100% sure what you’re asking but I’m going to give it a go. The elimination reaction is a term used in organic chemistry that describes a type of reactions. The name kinda tells you what’s going to happen. Something is going to be removed/eliminated from initial reactant/substrate and as a result, an alkene (double bond containing compound) will form.

In elimination reactions a hydrogen atom is first removed (as a H+) from the beta carbon. As a result, the left behind electrons create a pi bond between the beta carbon and the neighboring alpha carbon. This promotes the electronegative atom, on the alpha carbon, to leaves the substrate taking both electrons from the shared sigma bond with the alpha carbon.

4 0

3 years ago

10. Isolation of a pure sample of the third product, which has been determined to be an isomer of the major and minor products,

Ymorist [56]

Answer:

Four possible isomers (1–4) for the natural product essramycin. The structure of compound 1 was attributed to essramycin by 1H NMR, 13C NMR, HMBC, HRMS, and IR experiments.

Explanation:

Three synthetic routes were used to prepare all four compounds (Figure 2A). All three reactions utilize 2-(5-amino-4H-1,2,4-triazol-3-yl)-1-phenylethanone (5) as the precursor, whereas each uses different esters (6–8) to construct the pyrimidinone ring. Isomer 1 was prepared by reaction A, which used triazole 5 and ethyl acetoacetate (6) in acetic acid. This was the reaction used in syntheses of essramycin by the Cooper and Moody laboratories.3,4 Reaction B produced compound 2 (minor product) and compound 3 (major product), which were separated chromatographically. This reaction allowed reagent 5 to react with ethyl 3-ethoxy-2-butenoate (7) in the presence of sodium in methanol, under reflux for 24 h. Compound 4 was prepared by reaction C, which was obtained by reflux of 5 and methyl 2-butynoate (8) in n-butanol.

7 0

3 years ago