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2 years ago

A lone pair must reside in which orbital type(s) to participate in resonance/conjugation/delocalization?

1 answer:

5 0

If a lone pair is participating in resonance, this capability the atom is both Sp2 or Sp hybridized. Looking at the Sp2 case, that ability there are 3 Sp2 orbitals and one P orbital. Assuming there is no formal cost on the atom, the three hybrid orbitals are used for bonding and the lone P-orbital holds the lone pair. Sp3 orbitals are no longer involved with resonance so the lone pair worried in resonance should never be in an Sp3 orbital.

A mathematically described area round a nucleus in an atom or molecule that may also include zero, one, or two electrons Electrons organize themselves in cloudlike areas round the nucleus referred to as orbitals.

Learn more about orbitals here:

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Calculate the density of a water bottle with a mass of 4 grams and a volume of 100ml.

lilavasa [31]

Answer:

density is 40

Explanation:

6 0

3 years ago

Which of the following represents a compound? C20 H2O O2 H2

Andre45 [30]

The answer is C2O.............

7 0

3 years ago

Read 2 more answers

Draw the alcohol that is the product of the reduction of 3-methylpentanal.

natali 33 [55]

Answer: The product from the reduction reaction is

CH3-CH2-CH(CH3)-CH2-CH2OH

IUPAC name; 3- Methylpentan-1-ol

Explanation:

Since oxidation is simply the addition of oxygen to a compound and reduction is likewise the addition of hydrogen to a compound.

Therefore, hydrogen is added onto the carbon atom adjacent to oxygen in 3- methyl pentanal

CH3 CH2 CHCH3 CH2 CHO thereby -CHO( aldehyde functional group) are reduced to CH2OH ( Primary alcohol) which gives;

3-methylpenta-1-ol .

The structure of the product is:

CH3-CH2-CH(CH3)-CH2-CH2OH

7 0

4 years ago

Evaporation of Cane Sugar Solutions. An evaporator is used to concentrate cane sugar solutions. A feed of 10 000 kg/d of a solut

Vera_Pavlovna [14]

Answer:

Weight of solution produced = 5135 kg

Amount of water removed = 4865 kg

Explanation:

For the balance of mass, the incoming mass of sugar must be equal to the outgoing mass. So, the incoming mass (mi) is 38% of 10000 kg

mi = 0.38x10000 = 3800 kg

The outgoing mass (mo) must be 3800 kg, and it is 74% of the total mass (mt)

mo = 0.74xmt

0.74xmt = 3800

mt = 3800/0.74

mt = 5135 kg

This is the mass of solution produced.

The amount of water removed (wr) is the amount of water incoming (wi) less the amount of water outgoing (wo). Both will be the total mass less the mass of sugar :

wi = 10000 - 3800 = 6200 kg

wo = 5135 - 3800 = 1335 kg

wr = wi - wo

wr = 6200 - 1335

wr = 4865 kg

8 0

4 years ago

Calculate the amount in grams of Na2CO3 needed to react with HCL to produce 120g NaCl

worty [1.4K]

Let's start off with the balanced chemical equation:

$Na_2CO_{3_{(s)}} + \textbf2HCl_{_{(aq)}}\rightarrow CO_{2_{(g)}} + H_2O_{_{(l)}} + \textbf2NaCl_{_{(aq)}}$

Assuming the sodium carbonate is the limiting reagent, look at the coefficients of sodium carbonate and sodium chloride and use those as a ratio of sodium carbonate to sodium chloride: 1:2.

Since you have the required mass of NaCl, convert this to moles.

Assuming you know how to find the molar mass of NaCl:

$M_{NaCl} = 58.44g/mol$
$n = \frac{m}{M}$
$n_{NaCl} = \frac{120g}{58.44g/mol}$
$n_{NaCl} = 2.053mol$

Using the ratio, since 1 mole of sodium carbonate is required to produce 2 moles of sodium chloride, cross-multiply the ratios:

$1:2 = x:2.053mol$
$2x = 2.053mol$
$x = 1.027mol$

Therefore 1.027 moles of sodium carbonate is required to produce the required amount of sodium chloride. Convert to mass for your final answer:

$n_{Na_2CO_3} = 1.027mol$
$M_{Na_2CO_3} = 105.99g/mol$
$m = nM$
$m = (1.027mol)(105.99g/mol)$
$m_{Na_2CO_3} = 108.85g$

Hope this helps :)

6 0

3 years ago