Answer:
The number of neutron in the Aluminium Isotope is :
B. 14
Explanation:
Isotopes : These are the atoms which have same atomic number but have different mass number.
<u>This image shows the average atomic mass of Al element because it is in decimals</u>.
Atomic mass = 26.98154
(Note : mass number of single isotope can never be in decimals)
It is the average of mass of different isotopes of Al
Major Isotopes of 
are :
......atomic mass = 26
.......atomic mass = 27
mass of Al given in image(26.98) is nearly equal to mass of 2nd isotope(27)
mass of 
Now calculate the neutron in
Number of neutron = mass number - atomic number
= 27 - 13
Number of neutron = 14
(Atomic mass is same as mass number)
Answer:
Yes
Explanation:
There is a difference between the homogeneous mixture of the hydrogen and the oxygen in a 2:1 ratio and the sample of the water vapor.
In the homogeneous mixture of the hydrogen and the oxygen which are present in the ratio, 2:1 , the elements are not chemically combined. They are explosive also as both shows their specific properties. They can be separated by physical means (Condensation, diffusion).
On the other hand, in water vapor, the two elements are chemically bonded in a specific mixture which cannot be separated via physical means. Water has its unique properties and they can be separated by chemical means only.
Answer:
The molality of the solution is 0.3716 mol/kg
The number of moles of solute is 0.0157 mol
The molecular weight of the solute is 129.30 g/mol
The molar mass of the solute is 129.32 g/mol
Explanation:
m (molality of the solution) = ∆T/Kf = (43.17 - 40.32)/7.67 = 0.3716 mol/kg
Number of moles of solute = molality × mass of solvent in kilogram = 0.3716 × 0.04219 = 0.0157 mol
Molecular weight of solute = mass/number of moles = 2.03/0.0157 = 129.3 g/mol
When Kf = 7.66 °C.kg/mol
Molar mass = 2.03 ÷ (2.85/7.66 × 0.04219) = 129.32 g/mol
Answer:
This means that waves need a medium to propagate, but light waves propagate where no particles are present. ...
Answer:
Row 1
![[H^+]=1.8\times 10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.8%5Ctimes%2010%5E%7B-6%7DM)
![pH=-\log[H^+]=-\log[1.8\times 10^{-6}]=5.7](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B1.8%5Ctimes%2010%5E%7B-6%7D%5D%3D5.7)
pOh=14-pH=14-5.7=8.3
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=0.5\times 10^{-8}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.5%5Ctimes%2010%5E%7B-8%7DM)
Hence, acidic
Row 2
![[OH^-]=3.6\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3.6%5Ctimes%2010%5E%7B-10%7DM)
![pOH=-\log[OH^-]=-\log[3.6\times 10^{-10}]=9.4](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D%3D-%5Clog%5B3.6%5Ctimes%2010%5E%7B-10%7D%5D%3D9.4)
pH=14-pOH=14 - 9.4 = 4.6
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=2.6\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D2.6%5Ctimes%2010%5E%7B-5%7DM)
Hence, acidic
Row 3
pH = 8.15
![[H^+]=0.7\times 10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.7%5Ctimes%2010%5E%7B-8%7DM)
pOH=14-pH=14 - 8.15 = 5.8
![[OH^-]=1.5\times 10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.5%5Ctimes%2010%5E%7B-6%7DM)
Hence, basic
Row 4
pOH = 5.70
![[OH^-]=1.8\times 10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.8%5Ctimes%2010%5E%7B-6%7DM)
pH=14-pOH=14 - 5.70 = 8.3
![[H^+]=0.5\times 10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.5%5Ctimes%2010%5E%7B-8%7DM)
Hence, basic
