The loss of electron from an results in the formation of cation represented by the positive charge on the element whereas gaining of electron results in the formation of anion represented by the negative charge on the element.
The alkali earth metal beryllium (
) belongs to the second group of the periodic table. The ground state electronic configuration of
is:
From the electronic configuration it is clear that it has 2 valence electrons in its valence shell (
).
After losing all valence electrons that is 2 electrons from
orbital. The electronic configuration will be:

Since, lose of electron is represented by positive charge on the element symbol. So, the beryllium will have +2 charge on its symbol as
.
Hence, beryllium will have 2+ charge on it after losing all its valence electrons in the chemical reaction.
Answer:
B) number of electrons
Explanation:
The two species:
K+ and Cl- have the same number of electrons.
A neutral atom of K has 19 electrons
A neutral atom of Cl has 17 electrons
Now,
To form K+, K will have to lose an electron and then the number of electrons becomes 18
To form Cl-, Cl will have to gain an electrons and then the number of electrons becomes 18
So, therefore, the number of electrons on both species is 18
Answer:
C a chemical reaction formed
Explanation:
if something changes smell and nothing else it is a chemical reaction, every other answer does not make sense
Answer:

Explanation:
The energy of the photon can be found by using the formula
E = hf
where
E is the energy
f is the frequency
h is the Planck's constant which is
6.626 × 10-³⁴ Js
From the question we have
E = 6.626 × 10-³⁴ × 9.5 × 10¹³
We have the final answer as

Hope this helps you
The question is incomplete, complete question is :
The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia:
Step 1 : 
Step 2 : 
The net reaction is:

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants
and
. If you need to include any physical constants, be sure you use their standard symbols
Answer:
Equation that gives the overall equilibrium constant K in terms of the equilibrium constants
:

Explanation:
Step 1 : 
Expression of an equilibrium constant can be written as:
![K_1=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BNO%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E5%7D)
Step 2 : 
Expression of an equilibrium constant can be written as:
![K_2=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
The net reaction is:

Expression of an equilibrium constant can be written as:
![K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D)
Multiply and divide
;
![K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO]^4}{[NO]^4}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D%5Ctimes%20%5Cfrac%7B%5BNO%5D%5E4%7D%7B%5BNO%5D%5E4%7D)
![K=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO_2]^4}{[NO]^4}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D%5Ctimes%20%5Cfrac%7B%5BNO_2%5D%5E4%7D%7B%5BNO%5D%5E4%7D)
![K=K_1\times \frac{[NO_2]^4}{[O_2]^2[NO]^4}](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20%5Cfrac%7B%5BNO_2%5D%5E4%7D%7B%5BO_2%5D%5E2%5BNO%5D%5E4%7D)
![K=K_1\times (\frac{[NO_2]^2}{[O_2]^1[NO]^2})^2](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20%28%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BO_2%5D%5E1%5BNO%5D%5E2%7D%29%5E2)

So , the equation that gives the overall equilibrium constant K in terms of the equilibrium constants
:
