Qualitatively explain the effect of changing concentration of Cu2+ on voltage using your experimental observations and the Keq values below. (No calculations are required). (a) Cu^2+(aq) + 4NH3(aq) ⇄ [Cu(NH^3)4]^2+(aq) Keq = 1.2 x 10^12
(b) CuS(s) ⇄ Cu^2+(aq) + S^2-(aq) , Keq = 8.5 x 10^-45
Cu/Zn = 0.525 V
after 1 drop NH3: 0.505 V
after 2 drops NH3: 0.435 V
Cu/Zn = 0.510 V
after 1 drop Na2S: 0.452 V
after 2 drops Na2S: 0.393 V

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.