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IgorC [24]
4 years ago
12

How many sigma bonds does So2 have?

Chemistry
1 answer:
hichkok12 [17]4 years ago
5 0

Answer:

one sigma

As for the bonding, there is one sigma and one pi bond formed between sulphur and the two oxygen atoms. The atom will also accommodate one lone pair

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Lubov Fominskaja [6]

<u>Answer:</u>

<u>read below</u>

<u>Explanation:</u>

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Can someone please help me:( please dont scroll <br><br> create 2 quantitative observations
shusha [124]

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a quantitative observation implies that the subject can be measured by quantity, aka amount or in numbers.

Ex 1: adding one gram of salt to one gram of sugar makes two grams of seasoning. in this example, there are individual quantities (1 gram of each) and total quantity (2 grams). this only changes if the substances have a chemical reaction, such as one of them destroying the other, then the weight would change.

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2 years ago
A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

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