Answer: The element shown in the image is Helium (He).
Explanation: We are given a image of an atom having protons, neutrons and electrons.
Number of protons as shown in image = 2
Number of neutrons as shown in image = 2
Number of electron as shown in image = 2
Atomic number = Number of protons = Number of electrons
Atomic number of the element = 2
Atomic Mass = Number of protons + Number of neutrons
Atomic mass = 2 + 2 = 4
The element having Atomic number = 2 and mass number = 4 is Helium.
Element = 
It affects our everday life because of new inventions and easier ways to make/produce different stuff!
Answer:
The equilibrium concentration of NO is 0.001335 M
Explanation:
Step 1: Data given
The equilibrium constant Kc is 0.0025 at 2127 °C
An equilibrium mixture contains 0.023M N2 and 0.031 M O2,
Step 2: The balanced equation
N2(g) + O2(g) ↔ 2NO(g)
Step 3: Concentration at the equilibrium
[N2] = 0.023 M
[O2] = 0.031 M
Kc = 0.0025 = [NO]² / [N2][O2]
Kc = 0.0025 = [NO]² / (0.023)(0.031)
[NO] = 0.001335 M
The equilibrium concentration of NO is 0.001335 M
Question:
How many neutrons are there in 186W
Answer:
112
hope it helps (^^)
# Cary on learning
Answer:
Percentage abundance of 121 Sb is = 57.2 %
Percentage abundance of 123 Sb is = 42.8 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope, 121 Sb :
% = x %
Mass = 120.9038 u
For second isotope, 123 Sb:
% = 100 - x
Mass = 122.9042 u
Given, Average Mass = 121.7601 u
Thus,
Solving for x, we get that:
x = 57.2 %
<u>Thus, percentage abundance of 121 Sb is = 57.2 %
</u>
<u>percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %</u>
