Covalent and hydrogen bonds
Answer:
option B.
Explanation:
Given,
V₁ = 156 L
P₁ =2 atm
Now, in the cylinder
P₂ = ?
V₂ = 36
Using relation between pressure and volume
Hence, pressure is equal to 8.67 atm.
Hence, the correct answer is option B.
Answer:
0.665 moles of CO₂
Explanation:
The balance chemical equation for the combustion of Ethane is as follow:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
Step 1: <u>Calculate moles of C₂H₆ as;</u>
Moles = Mass / M.Mass
Putting values,
Moles = 10.0 g / 30.07 g/mol
Moles = 0.3325 moles
Step 2: <u>Calculate Moles of CO₂ as;</u>
According to balance chemical equation,
2 moles of C₂H₆ produced = 4 moles of CO₂
So,
0.3325 moles of C₂H₆ will produce = X moles of CO₂
Solving for X,
X = 0.3325 mol × 4 mol ÷ 2 mol
X = 0.665 moles of CO₂
Sc (neutral) [Ar] 3d1 4s2
Sc+ [Ar] 3d1 4s1
Sc2+ [Ar] 3d1
hope this helped!
Given the percentage composition of HC as C → 81.82 % and H → 18.18 %
So the ratio of number if atoms of C and H in its molecule can will be:
C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8
So the Empirical Formula of hydrocarbon is:
C 3 H 8
As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol
Now let Molecular formula of the HC be ( C 3 H 8 ) n
Using molar mass of C and H the molar mass of the HC from its molecular formula is:
( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1
Hence the molecular formula of HC is C 3 H 8
Does that help?
