Answer:
The given isotope is Br 80
If 35 protons present then its atomic number is 35
Mass number is 80 so its meutrons are 45 because proton+ neutron = mass number
Its electron are 36 because of negative one charge
Explanation:
Explanation:
Using Beer-Lambert's law :
Formula used :

where,
A = absorbance of solution
c = concentration of solution
l = length of the cell
= molar absorptivity of this solution
According to question:
A = (C) : absorbance measured by the spectrometer
c = (B) : concentration, in mol/L, of the stock solution from which the sample was made
l = (A): pathlength of light through the cell
ε = (D) : molar absorptivity, a constant unique to that substance at that wavelength
Answer: ![K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3Cl%5D%5Ctimes%20%5BOH%5E-%5D%7D%7B%5BCH_3OH%5D%5Ctimes%20%5BCl%5E-%5D%7D)
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients. Pure solids are assumed to have a concentration of 1.
The given balanced equilibrium reaction is:

The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3Cl%5D%5Ctimes%20%5BOH%5E-%5D%7D%7B%5BCH_3OH%5D%5Ctimes%20%5BCl%5E-%5D%7D)
Thus the equilibrium constant expression for this reaction is ![K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3Cl%5D%5Ctimes%20%5BOH%5E-%5D%7D%7B%5BCH_3OH%5D%5Ctimes%20%5BCl%5E-%5D%7D)
Answer:
An ionic bond
Explanation:
Metals and nonmetals form ionic bonds, in which the metal (low electronegativity) loses electrons to form a cation and the nonmetal (high electronegativity) gains electrons to form an anion. The cation and the anion attract each other through electrostatic forces.
'Heavy' hydrogen (the isotope deuterium) , 2/1H is involved ,
[2= mass number, 1= atomic (proton) number]
2/1H + 2/1H → 3/2He + 1/0n