Answer:
3.50 molal
Explanation:
Molality → Moles of solute / kg of solvent.
Let's convert the solvent's mass from g to kg
16.2 g . 1kg / 1000 g = 0.0162 kg
Let's determine the moles from the solute
2.61 g . 1 mol / 46 g = 0.0567 moles
Molality → 0.0567 mol / 0.0162 kg = 3.50 m
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.
<span>B) The crystals did not phosphoresce within the drawer but did expose the film</span>
The charge on Pb in Pb(SO3)2 is Lead (IV) Sulfite.
(mass)(ΔT)(Cs) = (Cs)(ΔT)(mass)
(140.0g)(95.00−x)(4.184)=(4.184)(x−25.00)(400.0g)
Solve for x
55647.2 - 585.76x = 1673.6x – 41840
-585.76x – 1673.6x = -41840 – 55647.2
-2,259.36x = -97,487.2
-2,259.36x / -2,259.36 = -97,487.2 / -2,259.36
X = 43.15
