GaBr3
Gallium=Ga
Bromine= Br
Bromide=Br3
Given :
An atom of a certain element has 12 protons, 14 neutrons, and 2 valence electrons.
To Find :
The name of this element.
Solution :
We know, element with atomic number ( number of protons ) 12 and valance electrons is Magnesium.
Now, isotope of magnesium of proton with 14 neutrons is Mg-16 .
Therefore, name of element is Mg-16 .
Hence, this is the required solution.
Answer:
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Explanation:
For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.
For a hypothetical reaction:
xA + yB ⇄ zC
The equilibrium constant is :
![Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D%7D%7B%5BC%5D%5E%7Bz%7D%20%7D)
The given reaction involves the decomposition of H2O into H2 and O2
The equilibrium constant is expressed as :
![Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Since Keq = 5.31*10^-10
It would be 35.8 Calories or calories. Not sure about that part. Hope this helps though.
Answer:
This element is Rubidium (Rb) and has an average atomic mass of 85.468 u
Explanation:
The average mass of an element is calculated by taking the average of the atomic masses of its stable isotopes.
The enitre atomic mass = 100 % or 1
⇒ this consists of X-85 with 72.17 % abundance with atomic massof 84.9118 g/mol
72.17 % = 0.7217
⇒ this consists of X-87 with 27.83 % abundance with atomic mass of 86.9092 g/mol
27.83 % = 0.2783
To calculate the mass of this isotope we use the following:
0.7271 * 84.9118 + 0.2783 * 86.9092 =85.468 g/mol
This element is Rubidium(Rb) and has an average atomic mass of 85.468 u
