Answer: Be= 2, C =4, Li = 1 and B=3
Explanation:
The valence shell can be define as the outermost shell of an atom that contains the valence electrons.
Beryllium (Be), electronic configuration; 1s2 2s2, = 2 electrons in its valence shell.
Carbon (C), electronic configuration; 1s2 2s2 2p2, = 4 electrons in its valence shell.
Lithium (Li), electronic configuration; 1s2 2s1 = 1 electron in its valence shell.
Boron (B) , electronic configuration; 1s2 2s2 2p1 = 3 electron in its valence shell.
Answer:
20N
Explanation:
Given parameters:
Force(N) Acceleration(m/s²)
10 0.2
? 0.4
Unknown:
The force applied when the acceleration is 0.4m/s²
Solution:
From newton's second law of motion;
Force = mass x acceleration
Since we are using the same box, let us find the mass of the box;
Force = mass x acceleration
10 = mass x 0.2
mass =
= 50kg
Now,
The force in the second instance will be;
Force = 50 x 0.4 = 20N
Answer:

Explanation:
In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:

Now we can identify the variables:




If we plug all the values into the equation:

And we solve for
:


I hope it helps!
<h2>C</h2>
Explanation:
The atomic number of S is 
So,number of electrons in S is 
The electronic configuration of S is 
The orbital electronic configuration of S is 
So,the number of orbitals involved is 9.