Is magnesium a pure substance????
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The Lewis structure of Carbonic Acid (H₂CO₃) is given below. In structure it is shown that carbon has a double bond with one oxygen atom and two single bonds with hydroxyl groups.
Formal Charge;
Formal charge is caculated as,
Formal charge = # of valence e⁻ - [# of lone pair of e⁻ + 1/2 # of bonded e⁻]
Formal charge on Carbon;
Formal charge = 4 - [ 0 + 8/2]
Formal charge = 4 - [4]
Formal charge = Zero
Formal Charge;
Formal charge is caculated as,
Formal charge = # of valence e⁻ - [# of lone pair of e⁻ + 1/2 # of bonded e⁻]
Formal charge on Carbon;
Formal charge = 4 - [ 0 + 8/2]
Formal charge = 4 - [4]
Formal charge = Zero
Answer:
41.54 grams of oxygen are required to burn 13.5 g of acetylene
Explanation:
The balanced reaction is:
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- C₂H₂: 2 moles
- O₂: 5 moles
- CO₂: 4 moles
- H₂O: 2 moles
Being the molar mass of the compounds:
- C₂H₂: 26 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- C₂H₂: 2 moles* 26 g/mole= 52 grams
- O₂: 5 moles* 32 g/mole= 160 grams
- CO₂: 4 moles* 44 g/mole= 176 grams
- H₂O: 2 moles* 18 g/mole= 36 grams
You can apply the following rule of three: if by stoichiometry 52 grams of acetylene react with 160 grams of oxygen, 13.5 grams of acetylene react with how much mass of oxygen?
mass of oxygen= 41.54 grams
<u><em>41.54 grams of oxygen are required to burn 13.5 g of acetylene</em></u>
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Answer:
The IUPAC name of compound is 6-Ethyl-2-Octene.
Explanation:
First of all draw a straight chain of 8 carbon atoms.
Make a double bond between carbon number 6 and 7 numbering from left.
Add ethyl group at position 3 starting from left.
The structure sketched is attached below,
According to rules the longest chain containing a double bond is selected. Numbering is started from the end to which double bond is nearer. So, in our case the double bond starts at carbon 2, hence parent name of compound is 2-Octene. Then the substituent at position 6 is named.
The IUPAC name of compound is 6-Ethyl-2-Octene.
Explanation:
First of all draw a straight chain of 8 carbon atoms.
Make a double bond between carbon number 6 and 7 numbering from left.
Add ethyl group at position 3 starting from left.
The structure sketched is attached below,
According to rules the longest chain containing a double bond is selected. Numbering is started from the end to which double bond is nearer. So, in our case the double bond starts at carbon 2, hence parent name of compound is 2-Octene. Then the substituent at position 6 is named.