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SashulF [63]
3 years ago
14

Which electron notation is incorrect?

Chemistry
2 answers:
Allisa [31]3 years ago
8 0
The first one is wrong
Ne4ueva [31]3 years ago
8 0
The first one my guy is wrong
You might be interested in
What is the correct iupac name of CH3-CH2-CH2CHO ?
kow [346]

the correct IUPAC name of the compound is 1-Butanal.

<h3>What are IUPAC names?</h3>

It is a system of naming organic compounds based on the longest carbon-to-carbon single bonds. It does not matter whether these longest chains are continuous or in a ring.

Thus, when the compound with the chemical formula, CH3-CH2-CH2CHO is considered. The longest carbon-to-carbon chain is 4. The 1st carbon carries a functional group known as an aldehyde.

Aldehydes are equipped with the carbonyl group and have the general formula R−CH=O. They are also sometimes referred to as formyl.

Aldehydes are named after their parent alkane chains with a slight modification. The 'e' is replaced with 'al'

The aldehyde in this case has four carbons. This means that the parent alkane is Butane. Therefore, the name of the compound will be 1-Butanal.

More on IUPAC names can be found here: brainly.com/question/16631447

#SPJ1

4 0
2 years ago
What is the ph of a 0.0055 m ha (weak acid) solution that is 8.2% ionized?
Trava [24]
Answer is: pH value of weak is 3.35.
Chemical reaction (dissociation): HA(aq) → H⁺(aq) + A⁻(aq).
c(HA) = 0.0055 M.
α = 8.2% ÷ 100% = 0.082.
[H⁺] = c(HA) · α.
[H⁺] = 0.0055 M · 0.082.
[H⁺] = 0.000451 M.
pH = -log[H⁺].
pH = -log(0.000451 M).
pH = 3.35.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity <span>an aqueous solution.</span>
3 0
3 years ago
You weigh out 0.1183 g of a complex salt to analyze for the percentage of cyanide ion in your complex salt. After dissolving the
Stels [109]

Answer:

25.35%

Explanation:

Again let me restate the the equation of the reaction;

H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)

Amount of potassium permanganate  reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles

If 2 moles of MnO4 - reacts with 3 moles of CN-

8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2

= 1.229 * 10^-3 moles of CN-

Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol

= 0.03 g

Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100

= 25.35%

8 0
3 years ago
Maleic acid is an organic compound composed of 41.39% , 3.47% , and the rest oxygen. If 0.378 mole of maleic acid has a mass of
storchak [24]

Answer:

Empirical CHO

molecular C4H4O4

Explanation:

From the question, we know that it contains 41.39% C , 3.47% H and the rest oxygen. To get the % composition of the oxygen, we simply add the carbon and hydrogen together and subtract from 100%.

This means : O = 100 - 41.39 - 3.47 = 55.14%

Next is to divide the percentage compositions by their atomic masses.

C = 41.39/12 = 3.45

O = 55.14/16 = 3.45

H = 3.47/1 = 3.47

Now we divide by the smallest value which is 3.45. We can deduce that this will definitely give us an answer of 1 all through as the values are very similar.

Hence the empirical formula of Maleic acid is CHO

Now we go on to deduce the molecular formula.

To do this we need the molar mass. I.e the amount in grammes per one mole of the compound.

Now we can see that 0.378mole = 43.8g

Then 1 mole = xg

x = (43.8*1)/0.378 = 115.87 = apprx 116

[CHO]n = 116

(12 + 1 + 16]n = 116

29n = 116

n = 116/29 = 4

The molecular formula is thus C4H4O4

5 0
3 years ago
When 150. g zinc sulfide are burned in excess oxygen, 68.5 g of zinc oxide are actually produced, along with sulfur dioxide. Det
Bumek [7]

%yield = 54.6%

<h3>Further explanation</h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

(theoretical)

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

<h3 />

Reaction

2ZnS+3O₂ ⇒ 2ZnO+2SO₂

MW ZnS = 97.474 g/mol

  • mol ZnS

\tt \dfrac{150}{97.474}=1.54

MW ZnO = 81.38 g/mol

  • mol ZnO (from mol ZnS as limiting reactant, O₂ excess)

\tt \dfrac{2}{2}\times 1.54=1.54

  • Actual ZnO produced

\tt 1.54\times 81.38=125.33~g

Theoretical production = 125.388

  • %yield

\tt \dfrac{68.5}{125.33}\times 100\%=\boxed{\bold{54.6\%}}

4 0
3 years ago
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