Answer:
ΔU = 25.8 J
Explanation:
The gas absorbs 33.3 J of heat, that is, Q = 33.3 J.
The work (W) of expansion can be calculated using the following expression:
W = -P. ΔV
where,
P is the external pressure
ΔV is the change in volume
W = -1.45 × 10⁴ N . m⁻² × (8.40 × 10⁻⁴ m³ - 3.24 × 10⁻⁴ m³) = -7.48 J
The change in the internal energy (ΔU) is:
ΔU = Q + W
ΔU = 33.3 J + (-7.48 J) = 25.8 J
F the solubility of a gas in water is 5.0g/L when the pressure of the gas above the water is 2.0 atm, what is the pressure of the gas above the water <span>when the solubility of the gas is 1.0 g/L</span>
Here's how to solve this one.
The formula for solubility is
<span>P1 / P2 = solubility1 / solubility2 </span>
P1=2*1/5= .4 atm
So the correct answer is 0.4 atm.
The answer is (2). If you recall Rutherford's gold foil experiment, remember that a stream of positively charged alpha particles were shot at a gold foil in the center of a detector ring. The important observation was that although most of the particles passed straight through the foil without being deflected, a tiny fraction of the alpha particles were deflected off the axis of the shot, and some were even deflected almost back to the point from which they were shot. The fact that some of the alpha particles were deflected indicated a positive charge (because same charges repel), and the fact that only a small fraction of the particles were deflected indicated that the positive charge was concentrated in a small area, probably residing at the center of the atom.
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.
