If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy 25.48 L at STP.
<h3>How to calculate volume?</h3>
The volume of a gas at STP can be calculated using the direct proportion method.
According to this question, 50.75 g of a gas occupies 10.0 L at STP, then 129.3g of the same gas will occupy the following:
= 129.3 × 10/50.75
= 25.48L
Therefore, if 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy 25.48 L at STP.
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First, we need to get moles of NaOH:
when moles NaOH = volume * molarity
= 0.02573L * 0.11 M
= 0.0028 moles
from the reaction equation:
H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)
we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH
∴ X mol H3PO4 reacts with → 0.0028 moles NaOH
∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol
now we can get the concentration of H3PO4:
∴[H3PO4] = moles H2PO4 / volume
= 9.4 x 10^-4 / 0.034 L
= 0.028 M
Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus. This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals. These relativistic electrons gain mass and as a consequence, their orbits contract. As these s and (to some degree) p orbits are contracted, the other electrons in d and f orbitals are better screened from the nucleus and their orbitals actually expand.
Since the 6s orbital with one electron is contracted, this electron is more tightly bound to the nucleus and less available for bonding with other atoms. The 4f and 5d orbitals expand, but can't be involved in bond formation since they are completely filled. This is why gold is relatively unreactive.
Hope it helps
<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
<span>The answer is 200 mol of water.
The balanced reaction is 2(H2) + (O2) = 2(H2O)
The limiting reactant is O2 as it will be completely consumed first, before hydrogen gas. Hydrogen gas would need at least 105 mol oxygen gas to be consumed; in excess of the 100 mol O2.
Looking at the stoichiometric coefficients, the ratio between water and oxygen is 2:1.
Therefore, the water produced would be 200 moles.</span>
