0.08 because 4mol NO divided by 50g gives you your answer
Answer:
<em>Argon</em><em> </em><em>can</em><em> </em><em>exi</em><em>st</em><em> </em><em>freely</em><em> </em><em>in</em><em> </em><em>nature</em><em> </em><em>because</em><em> </em><em>it</em><em> </em><em>has</em><em> </em><em>a</em><em> </em><em>full</em><em> </em><em>octet</em><em> </em><em>of</em><em> </em><em>electron</em><em>s</em><em> </em><em>the</em><em> </em><em>way</em><em> </em><em>its</em><em> </em><em>found</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>nature</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>way</em><em> </em><em>its</em><em> </em><em>found</em><em> </em><em>in</em><em> </em><em>periodic </em><em>table</em><em> </em><em>of</em><em> </em><em>element </em><em>in</em><em> </em><em>vast</em><em> </em><em>amouts</em><em> </em><em>of</em><em> </em><em>stabilization</em><em>.</em>
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
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Potassium hydroxide (KOH) is formed when Potassium forms ionic bonds with OH-ions while Potassium Oxide (K2O) is formed when potassium forms ionic bonds with the Oxide (O2-) ions. i.e. This reaction is a neutralization reaction and occurs when an acid (HCl) reacts with a base (KOH).