Answer:
714 nm
Explanation:
Using the equation: nλ=d<em>sin</em>θ
where
n= order of maximum
λ= wavelength
d= distance between lines on diffraction grating
θ= angle
n is 1 because the problem states the light forms 1st order bright band
λ is unknown
d= 
or 0.0000014 (meters)
sin(30)= 0.5
so
(1)λ=(0.0000014)(0.5)
=0.000000714m or 714 nm
Answer: Wavelength associated with the fifth line is 397 nm
Explanation:
= Wavelength of radiation
E= energy
For fifth line in the H atom spectrum in the balmer series will be from n= 2 to n=7.
Using Rydberg's Equation:
Where,
= Wavelength of radiation
= Rydberg's Constant =
= Higher energy level = 7
= Lower energy level = 2 (Balmer series)
Putting the values, in above equation, we get
Thus wavelength λ associated with the fifth line is 397 nm
Answer:
136 g Al₂O₃
Explanation:
Assuming you do not need to find the limiting reactant, to find the mass of Al₂O₃, you need to (1) convert grams O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles Al₂O₃ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles Al₂O₃ to grams Al₂O₃ (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value (64.0 g).
Molar Mass (O₂): 32 g/mol
Molar Mass (Al₂O₃): 102 g/mol
4 Al + 3 O₂ -----> 2 Al₂O₃
64.0 g O₂ 1 mole 2 moles Al₂O₃ 102 g
----------------- x -------------- x ------------------------ x ------------- = 136 g Al₂O₃
32 g 3 moles O₂ 1 mole
