Answer:
2
Explanation:i did the test
The formula is m = D x V
D = <span>13.69 g/cm^3.
</span>V = <span>15.0 cm^3
the mass of the liquid mercury is m= </span>13.69 g/cm^3 x 15.0 cm^3 = 195g
the molar mass of Hg is 200,
1 mole of Hg = 200g Hg, so #mole of Hg= 195 / 200 = 0.97 mol
but we know that
1 mole = 6.022 E23 atoms
0.97 mole=?
6.022 E23 atoms x 0.97 / 1 mole = 5.84 E23 atoms
B3+ is isoelectronic with helium.
Isoelectronicity is the phenomenon whereby two or more molecular entities have the same number of electrons or similar electronic configuration regardless of the nature of the elements that are involved.
In the question given above, helium and B3+ have the same number of electrons. Helium has two electrons. Boron has five electrons but it has given away three of the electrons [that is why it has a charge of +3] and it now has only two left.
The water is formed from oxygen gas and...hydrogen gas, I'm assuming? It would have been nice for the question to have been a bit more explicit (not blaming you, of course).
Assuming that's the case, our chemical reaction would be:
2H₂(g) + O₂(g) → 2H₂O(l).
We are told that 1 mol of a gas has a volume of 24.0 dm³ at RTP. We can use this relation to determine the number of moles of O₂ gas that reacts given its initial volume, 33.5 dm³.
33.5 dm³ O₂(g)/24.0 dm³/mol = 1.396 mol O₂(g).
Since we are not given any information about H₂(g), or any other reactant for that matter, I am assuming that the O₂(g) is the limiting reactant. According to the equation, the stoichiometric ratio between O₂ and H₂O is 1:2. That is, for every one mole of O₂ that is consumed, two moles of H₂O are formed (i.e., the number of moles of H₂O formed is double the number moles of O₂).
Since 1.396 moles of O₂ reacts, 2(1.396) = 2.792 moles of H₂O are produced. To convert moles of water to grams, we multiply the number of moles of H₂O by the molar mass of H₂O:
(2.792 moles H₂O)(18.015 g/mol) = 50.3 g H₂O.
So, approximately 50.3 grams of water are formed from 33.5 dm³ of oxygen gas at RTP.