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<span>Kr has 8 valence electrons plus 1 for each Kr-F single bond. Total = 10 elcetrons
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Given the temperature 746 K and activity of Pb equal to 0.055. The mole fraction of Pb is 0.1. So, the mole fraction of Sn = 0.9.Activity coefficient, γ = 0.055 / 0.1 = 0.55.The expression for w=ln〖γ_Pb x RT〗/(X_Sn^2 )=(-0.5978 x 8.314 J/(mol K ) x 746 K)/(0.9 x 0.9)= -4577.7 J= -4578 J
Now we use the computed value above and new temperature 773 K. The mole fraction of Sn and Pb are 0.5 and 0.5 respectively. Calculate the activity coefficient in the following manner.lnγ_Sn=w/RT X_Pb^2=(-4578 J)/(8.314 J/mol x 773 K) x 0.5 x 0.5= -0.718lnγ_Sn=exp(-0.178)=0.386The activity of Sn= γ_Sn x X_Sn=0.386 x 0.5=0.418
w of the system is -4578 J and the activity of Sn in the liquid solution of xsn at 500 degree Celsius is 0.418
Answer:- 7 protons and 10 electrons.
Solution:- Atomic number for nitrogen atom is 7.
For neutral atom, atomic number = number of protons = number of electrons
An ion is formed if electrons are lost or gained by the neutral atom. Here we have nitride ion where nitrogen has -3 charge. This negative 3 charge indicates the gain of 3 electrons by nitrogen atom. The number of protons remains unchanged, means there are still 7 protons.
electrons for the given ion = 7 + 3 = 10
So, 
ion has 7 protons and 10 electrons.
D . on orbits or energy levels around the nucleus
