The answer for the following question is explained.
<u><em>Therefore the number of electrons present with the values n = 5, l = 2, m = -2, s = +1/2 is</em></u><u> </u><u><em>one(1).</em></u>
Explanation:
Here;
n represents the principal quantum number
l represents the Azimuthal quantum number
m represents magnetic quantum number
s represents spin quantum number
n = 5,
l = 2,
m = -2,
s = +1/2
Here, it implies 5d orbital.
In the 5d orbital, 10 electrons.
As the magnetic quantum number is -2, and so it can have 1 electron.
<u><em>Therefore the number of electrons present with the values n = 5, l = 2, m = -2, s = +1/2 is</em></u><u> </u><u><em>one(1)</em></u>
Answer:
One of the main uses of the cathode ray tube is in the Cathode ray oscilloscope
Explanation:
Cathode rays are produced when a gas in an evacuated glass at very low pressure and high pressure breaks up into positive and negative ions. the negative ions move towards the anode(positive electrode) while the positive ions move towards the cathode(negative electrode), and there they knock off electrons (which are known as cathode rays) from the metal plate of the cathode.
Cathode ray tubes are mainly used in oscilloscopes, television picture tubes and in computer screens.
The cathode ray oscilloscope is used in a.c. and d.c. voltage measurements, observation of waveforms, time measurements, etc.
one mole of P weights about 31 grams
in one mole there are 6.022*10^23 atoms
we use the rule of threes
6.022*10^23atoms......weight..........31 grams
3.45*10^23 atoms.........weight...........x grams
x=(3.45*10^23*31)/6.022*10^23
x=106.95/6.022=<u><em>17.76 grams</em></u>
Answer:
Acid(BSA) = CH₃COOH
Base (BSB) = H₂O
Conjugate base (CB) = CH₃COO⁻
Conjugate acid (CA) = H₃O⁺
Explanation:
Equation of reaction;
CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺
Hello,
From my understanding of the question, we are required to identify the
1) Acid
2) Base
3) conjugate acid
4) conjugate base in the reaction
Acid (BSA) = CH₃COOH
Base (BSB) = H₂O
CA = conjugate acid = H₃O⁺
CB = conjugate base = CH₃COO⁻
