Hydrocarbons
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Answer:
Mass = 182.4 g
Explanation:
Given data:
Number of moles of Al₂O₃ = 3.80 mol
Mass of oxygen required = ?
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Now we will compare the moles of aluminum oxide and oxygen.
Al₂O₃ : O₂
2 : 3
3.80 : 3/2×3.80 = 5.7
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 5.7 mol × 32 g/mol
Mass = 182.4 g
<h3>Answer:</h3>
There is One electrophilic center in acetyl chloride.
<h3>Explanation:</h3>
Electrophile is defined as any specie which is electron deficient and is in need of electrons to complete its electron density or octet. The main two types of electrophiles are those species which either contain positive charge (i.e. NO₂⁺, Cl⁺, Br⁺ e.t.c) or partial positive charge like that contained by the sp² hybridized carbon of acetyl chloride shown below in attached picture.
In acetyl chloride the partial positive charge on sp² hybridized carbon is generated due to its direct bonding to highly electronegative elements *with partial negative charge) like oxygen and chlorine, which tend to pull the electron density from carbon atom making it electron deficient and a good electrophile for incoming nucleophile as a center of attack.
<u><em>Answer:</em></u>
- The correct option is C.
- Formation of a precipitate
<u><em>Explanation:</em></u>
During a chemical reaction, new substances are formed known as a products, mostly reaction occur and their product is obtained as precipitates.
<u><em>Example</em></u>
Arylidene-2-thiobarbituric acid is obtained as precipitates when aldehyde and thiobarbituric acid react to each other.
melting of a substance
It is just indication of physical changes, like melting of ice, composition remained same as before.
boiling of a substance
It is just indication of physical changes, like boiling of water into vapors, composition remained same as before.
freezing of a substance
It is just indication of physical changes, like freezing of water into ice, composition remained same as before
Answer:
C2H5NO
Explanation:
constituent elements N O C H
Mass composition 0.420 0.480 0.540 0.135
mole ratio 0.42/14 0.48/16 0.54/12 0.135/1
= 0.03 0.03 0.045 0.135
dividing by the smallest 0.03/0.03 0.03/0.03 0.045/0.03 0.135/0.03
ratio = 1 1 1.5 4.5
= 1 1 2 5
EMPERICAL FORMULA = C2H5NO
