A2+. Group 2 elements form cations with 2+ charge.
Answer:
Explanation:
What we need to do here is to determine the ratios by using the Rydberg equation starting with the transition to n1 = 1, 2,3, etc and see which one fits the data. Remember the question states that they are series and the wavelengths will be for increasing energy levels.
1/λ = Rh x ( 1/n₁² - 1/n₂²)
Lyman series ( n₁=1 and n₂= 2,3 etc) for the first two lines, the ratios will be:
1/λ₁ /1/λ₂ =(1/1 -1/ 2²) / (1/1 -1/ 3²) ⇒ 0.84 ≠ 0.74 (the first ratio)
For Balmer series n₁ = 2 and n₂ = 3,4,5, etc
1/λ₁ /1/λ₂ =(1/4 -1/3²) / (1/4 -1/4²) ⇒ 0.741 = 0.741 (match!)
Lets use the third line to check our answer:
1/λ₁ /1/λ₂ =(1/4 -1/3²) / (1/4 -1/5²) = 0.66
1. L
Number one because the lines match up
Answer:
5.79 × 10^23 Oxygen atoms
Explanation:
Number of Oxygen atom in the compound = 4×3 = 12
Molar mass of Al2(SO4)3 = 342 g/mol.
No of mole = mass/molar mass = 2.74/342 = 8.01×10^-03 mole
2.74g of Al2(SO4)3 × 1 mole of Al2 (SO4)3 / 342g of Al2 (SO4)3 * 12 mole of Oxygen/ 1mole of Al(SO4)3 * 6.02×10^23 Oxygen atom/ 1 mole of Oxygen
= 5.79×10^23 Oxygen atoms
Answer:
The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol
Explanation:
Fractional distillation is a separation process based on difference in boiling point of two compounds.
1-chlorobutane is a polar aprotic molecule due to presence of polar C-Cl bond. Hence dipole-dipole intermolecular force exists in 1-chlorobutane as a major force.
1-butanol is a polar protic molecule. Hence dipole-dipole force along with hydrogen bonding exist in 1-butanol.
Therefore intermolecular force is stronger in 1-butanol as compared to 1-chlorobutane.
So, boiling point of 1-butanol is much higher than 1-chlorobutane.
Hence mixture of 1-chlorobutane and 1-butanol can be separated by fractional distillation based on difference in boiling point.
So, option (D) is correct.
