Answer:You can set up stoichiemetry using the following equation:
(15.6 g MgF2) x (38g F / 62g MgF2) x (6.022x10^23 / 19gF)
= 3.03 x 10^23 molecules of F
or 1.52 x 10^23 molecules of F2
The number of molecules of magnesium fluoride in 15.6 g of MgF2 has to be found.
The molecular mass of MgF2 is 62.3018. 15.6 g of MgF2 is equivalent to 15.6/62.3018 mole of MgF2.
One mole of a gas has 6.02214179*10^23 particles.
15.6/62.3018 mole of MgF2 has (15.6/62.3018)*6.02214179*10^23 molecules of the compound.
(15.6/62.3018)*6.02214179*10^23
=> 1.5079*20^23
If this is rounded to one decimal figure the result is 1.51*10^23.
The number of molecules of MgF2 in 15.6 g of the gas is 1.51*10^23.
Answer:
an ion is an element that has different numbers of protons and electrons
Explanation:
An ion is positive when it has more protons than electrons and negative when it has more electrons than ions.
(Hope this was helpful!) :)
+2, because Beryllium is in the Group II of the periodic table.
Hope this helps!
First, we need the no.of moles of O2 = mass/molar mass of O2
= 55 g / 32 g/mol
= 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2
So we can get the no.of moles of H2O = 2 * moles of O2
= 2 * 1.72 mol
= 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT
when P = 12.4 atm & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K
12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
∴ V ≈ 8.2 L
