Answer:
The first thing that you need to do here is to figure out the mass of the sample.
To do that, you can use its volume and the fact that aluminium is said to have a density of
2.702 g cm
−
3
, which implies that every
1 cm
3
of aluminium has a mass of
2.702 g
.
Explanation:
Answer:
The new pressure is 53.3 kPa
Explanation:
This problem can be solved by this law. when the volume remains constant, pressure changes directly proportional as the Aboslute T° is modified.
T° increase → Pressure increase
T° decrease → Pressure decrease
In this case, temperature was really decreased. So the pressure must be lower.
P₁ / T₁ = P₂ / T₂
80 kPa / 300K = P₂/200K
(80 kPa / 300K) . 200 K = P₂ → 53.3 kPa
The answer is A. Mixtures can be separated by physical means
A pure substance cannot be separated.
J. J. Thomson is the corect awncer
<span>11.3 kPa
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant (8.3144598 L*kPa/(K*mol) )
T = Absolute temperature
We have everything except moles and volume. But we can calculate moles by starting with the atomic weight of argon and neon.
Atomic weight argon = 39.948
Atomic weight neon = 20.1797
Moles Ar = 1.00 g / 39.948 g/mol = 0.025032542 mol
Moles Ne = 0.500 g / 20.1797 g/mol = 0.024777375 mol
Total moles gas particles = 0.025032542 mol + 0.024777375 mol = 0.049809918 mol
Now take the ideal gas equation and solve for P, then substitute known values and solve.
PV = nRT
P = nRT/V
P = 0.049809918 mol * 8.3144598 L*kPa/(K*mol) * 275 K/5.00 L
P = 113.8892033 L*kPa / 5.00 L
P = 22.77784066 kPa
Now let's determine the percent of pressure provided by neon by calculating the percentage of neon atoms. Divide the number of moles of neon by the total number of moles.
0.024777375 mol / 0.049809918 mol = 0.497438592
Now multiply by the pressure
0.497438592 * 22.77784066 kPa = 11.33057699 kPa
Round the result to 3 significant figures, giving 11.3 kPa</span>
