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>
Pure magnesium's formula would just be Mg because all elements except for 7 nonmetals are just left alone when they are by themselves in a formula. The 7 diatomic elements( means they have to have two of them without another element attached to it aka. a subscript two after it when it's by itself) are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. An easy way to remember the diatomic seven is that when looking at a periodic table if you trace over them from nitrogen over to fluorine and down to iodine all of those elements are diatomic + hydrogen.
And your unbalanced and balanced equations are correct.
(sorry I went on a tangent with the diatomic rules hopefully it will help you in the future though)
Answer: 1090°C
Explanation: According to combined gas laws
(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2
where P1 = initial pressure of gas = 80.0 kPa
V1 = initial volume of gas = 10.0 L
T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K
P2 = final pressure of gas = 107 kPa
V2 = final volume of gas = 20.0 L
T2 = final temperature of gas
Substituting the values,
(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2
T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)
T2 = 513 K × (1.3375) × (2)
T2 = 1372.275 K
T2 = (1372.275 - 273) °C
T2 = 1099 °C
