The question is incomplete, complete question is :
In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?
Structure is given in an image?
Answer:
There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.
Explanation:
Total numbers of carbon = 10
Number of primary carbons that is carbon joined to just single carbon atom = 6
Number of secondary carbons that is carbon joined to two carbon atoms = 1
Number of tertiary carbons that is carbon joined to three carbon atoms = 2
Number of quartenary carbons that is carbon joined to four carbon atoms = 1
So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.
Mass C₂H₂ needed : 22.165 g
<h3>Further explanation</h3>
Reaction
2C₂H₂+
5O₂ ⇒ 4CO₂ + 2H₂O
75.0 grams of CO₂ , mol CO₂ (MW=44 g/mol) :

mol C₂H₂ :

mass C₂H₂ (MW=26 g/mol) :

Answer:
H2 + I2 --> 2HI
Explanation:
The two reactants are diatomic molecules because they contain two atoms of the same element. Therefore, they would need to have a subscript of "2" next to their symbols.
When balancing an equation, you want the same amount and type of atoms on both sides. By adding a coefficient of "2" in front of the product, two H's and two I's are now on both sides.