Answer:
Mass, m = 1.51 grams
Explanation:
It is given that,
The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm
Length of the cylinder, h = 4.2 cm
We know that the density of the Aluminium is 2.7 g/cm³
Circumference, C = 2πr
Density is equal to mass per unit volume.
m is mass of the cylinder
V is the volume of the cylinder
So,
So, the mass of the cylinder is 1.51 grams.
Answer:
Explanation:
Not likely to form any bonds because in it's last she'll it has 8 electrons and is therefore stable
The answer is A it's a basic because once you add another substance to a neutral it either becomes acidic or basic. this one becomes basic because the hydroxide ion concentrate increased.
1. LDFs
2. Intermolecular Forces
3. Intramolecular Forces
4. Linear
5. Tetrahedral
Edit: I'm new to this site and idk how to use it properly. I'm not sure about 2 and 3 currently because these forces are between molecules as well so INTERmolecular would be used twice (?)
<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>