3.0e23 atoms Ne
"E" means 10^
Then we multiply it by a mole of Ne. By the definetion of a mole, it is always 6.022e23 atoms of an element.
So now, we do this:
3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne)
After that, we use molar mass. A mole of Neon is equal, in terms of grams, to its avg. atomic mass. This goes true for any element.
It ends up like this:
3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne) x (20.1797 g Ne / 1 mol Ne)
Now cancel out the "atoms Ne" and "1 mol Ne"
You end up with a grand total of...
*plugs everything into a calculator*
10.05298... g Ne.
We need to round to 2 sig. figs. (3.0) so now it's....
10 g Ne.
Note that this method can only be used for converting atoms of an element to mass in grams.
Source(s):
A periodic table for the atomic mass of neon.
A chemistry textboook
A chemistry class.
Answer:I believe it’s because of the volcanos making up the ring
Explanation:
1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm
Ideal gas equation: pV = nRT => n = pV / RT
R = 0.0821 atm*liter/K*mol
V = 300 cm^3 = 0.300 liter
T = 298 K
p = 1 atm
=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol
2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type
X (+) + O2 (g) ---> X2O or
2 X(2+) + O2(g) ----> X2O2 = 2XO or
4X(3+) + 3O2(g) ---> 2X2O3
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)
So, lets probe those 3 cases.
3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol
=> x = 0.01226 moles of metal X
Now you can calculate the atomic mass of the hypotethical metal:
1.15 grams / 0.01226 mol = 93.8 g / mol
That does not correspond to any of the metal with valence 1+
So, now probe the case 2.
4) Case 2:
2moles X metal / 1 mol O2(g) = x / 0.01226 mol
=> x = 2 * 0.01226 = 0.02452 mol
And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol
That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.
4) Case 3
4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635
atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol
That does not correspond to any metal.
Conclusion: the identity of the metallic element could be titanium.
1.hydrogen
2.carbon
3.argon
4.titanium
5.cesium
