Answer:
The atomic number of silicon is 14 while atomic mass of carbon is 14.
Explanation:
An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.
In given atoms ¹⁴₆C and ²⁸₁₄Si the atomic mass of carbon is 14 while the atomic number of silicon is 14. It means silicon has 14 electrons or protons while carbon has 6 protons or electrons because its atomic number is 6. Carbon has 6 protons and 8 neutrons in its nucleus while silicon has 14 protons and 14 neutrons in its nucleus.
In C:
Number of neutrons + protons = 8 + 6 = 14 amu (mass number)
Number of electrons = 6
In Si:
Number of neutrons + protons = 14 + 14 = 28 amu (mass number)
Number of electrons = 14
Remembering that
d = m ÷ v
d = ?
m = 89 g
v = 10 cm³
Therefore:
d = 89 ÷ 10
d = 8,9 g÷cm³
The masses are always equal. Since matter can not be created nor destroyed, you will have the same amount of mass as you did before the reaction as you do after.
Answer: 
Explanation:
cM 0 0
So dissociation constant will be:
Given: c = 0.15 M
pH = 1.86
= ?
Putting in the values we get:
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![1.86=-log[H^+]](https://tex.z-dn.net/?f=1.86%3D-log%5BH%5E%2B%5D)
![[H^+]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01)
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
As ![[H^+]=[ClCH_2COO^-]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BClCH_2COO%5E-%5D%3D0.01)
![K_a=1.67\times 10^{-3]](https://tex.z-dn.net/?f=K_a%3D1.67%5Ctimes%2010%5E%7B-3%5D)
Thus the vale of for the acid is
