Answer:
Theoretical yield = 49.45g
Percentage yield= 93.6%
Answer:
4Cr + 3O2 —> 2Cr2O3
Explanation:
Information from the question include:
Chromium + oxygen -> chromium(III) oxide
From the word equation given above, the equation can be written as follow:
Cr + O2 —> Cr2O3
The equation can be balance by doing the following:
There are 2 atoms of O on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of Cr2O3 and 3 in front of O2 as shown below:
Cr + 3O2 —> 2Cr2O3
Now, we have 4 atoms of Cr on the right side and 1 atom on the left. It can be balance by putting 4 in front of Cr as shown below:
4Cr + 3O2 —> 2Cr2O3
Now the equation is balanced
Answer:
The wavelength for the transition from n = 4 to n = 2 is<u> 486nm</u> and the name name given to the spectroscopic series belongs to <u>The Balmer series.</u>
Explanation
lets calculate -
Rydberg equation- 
where ,
is wavelength , R is Rydberg constant ( 
), 
and 
are the quantum numbers of the energy levels. (where 
)
Now putting the given values in the equation,
Wavelength 
=
= 486nm
<u> Therefore , the wavelength is 486nm and it belongs to The Balmer series.</u>
Answer:
a, g, c
Explanation:
The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.
The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.
