They can’t have more than 8 valence electrons. Since fluorine is in group 7 it can only bound to one other molecule
Answer:
Explanation: the module will end up working out when you mix NO2CI and CI
Answer:
Explanation:
Utilizing Rydber's equation:
ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have
n=1 to n= infinity
ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)
n= 3 to n= infinity
ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J
b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .
1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹
1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m
λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm
The answer is A because it cannot be B because that is ethylene and not D because it needs a subscript of 2 since it is a diatomic and it is no C because it is not carbon dioxide its oxygen
