We know that the element Z = 119 would be placed right below the Fr, in the column of the alcaline metals.
We also know that the trend in the electronegativity is to decrease when you go up-down ia group.
The known electronegativities of the elements of this group are:
Li: 0.98
Na: 0.93
K: 0.82
Rb: 0.82
Cs: 0.79
Fr: 0.70
Then the hypotetical element Z = 119 would probably have an electronegativity slightly below 0.70, for sure in the range 0.60 - 0.70.
Answer:
3 Pb(NO3)2 + Al2(SO4)3 = 2 Al(NO3)3 + 3 PbSO4
Explanation:
I think your equation is incorrect? This is balanced and the sum of coefficients is 9.
Answer:
The work done on the system is 17.75_KJ
Explanation:
To solve this question we need to know the required equations from the given variables, thus
Initial volume = 85L
Final volume = 12L
External pressure = 2.4 atm.
work done = - PΔV
2.4×(85-12) = 175.2L×atm
Converting from L•atm to KJ is given by
1 L•atm = 0.1013 kJ
(175.2 L•atm) * (0.1013 kJ / 1 L•atm)
= 17.75 kJ
The work done on the system is 17.75_KJ
