Answer:
Lose two electrons.
Explanation:
Barium is present in group 2.
It is alkaline earth metal.
Its atomic number is 56.
Its electronic configuration is Ba₅₆ = [Xe] 6s².
In order to attain the noble gas electronic configuration it must loses its two valance electrons.
When barium loses it two electron its electronic configuration will equal to the Xenon.
The atomic number of xenon is 54 so barium must loses two electrons to becomes equal to the xenon.
A. physical change took place during the experiment. (plato users)
Answer:
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?
A) BrCH2CH2Br
B) CH3CH2CH2Br
C) CH3CHBr2
D) CH3CH2CH2CH3
E) BrCH2CH2CH2CH2Br
Explanation:
The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.
This is an example of free radical substitution.
The structure of ethane and its bromination is shown below:
Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).
Remaining all other products are possisble to form on free radical substitution of ethane.
Answer:
I am looking for this as well. Can someone please help!
Explanation:
Answer:
7.5 × 10¹⁵ Hz
Explanation:
Given data
- Wavelength of the radio waves (λ): 40 nm = 40 × 10⁻⁹ m = 4.0 × 10⁻⁸ m
- Frequency of the radio waves (ν): ?
- Speed of light (c): 3.00 × 10⁸ m/s
We can determine the frequency of the radio waves using the following expression.
c = λ × ν
ν = c/λ
ν = (3.00 × 10⁸ m/s)/4.0 × 10⁻⁸ m
ν = 7.5 × 10¹⁵ s⁻¹ = 7.5 × 10¹⁵ Hz