Question
3.15 a) When the sodium metal element is combined with the non-metal element bromine, Br2 (1), how can the chemical formula of the product be determined? How do you know if the product is a solid, a liquid or a gas at room temperature? Write the balanced chemical equation for the reaction. b) When a hydrocarbon burns in the air, which reagent, besides the hydrocarbon, participates in the reaction? What products are formed? Write a balanced chemical equation for the combustion of benzene, C, H. (), In air.
3.15 a) Cuando el elemento metálico sodio se combina con el ele- mento no metálico bromo, Br2(1), ¿cómo se puede determinar l
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Answer #1

When sodium metal is reacted with Bromine sodium bromideis formed.

  • Sodium metal ( electropositive element) and Br is non metal ( elecronegative).
  • It is an ionic compound.
  • Since it is an ionic compond it will have higher melting and boiling points.
  • So NaBr can be said to be solid at room temperature.
  • The melting point of NaBr = 747oC , boiling point = 1396oC very high.
  • The valency of Na and Br both are equal to 1 ( oxidation states of Na is +1 and Br is -1)
  • Sodium atom dnates an electron( to satsfy octet ) and bromine( to satisfy octet) accepts single electron to foem a stable NaBr , ionic compond. This is how chemcial formula can be detrmined.

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2Na + Br2 -----------> 2NaBr

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b)

The other reagent is Oxygen that participates in the reaction with hydrocarbon, when a hydrocarbon is burnt in air.

The products formed when a hydrocarbon burns in air are aways water and csrbon dioxide.

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Combustion of benzene

2C6H6(l) + 15O2(g)-------------------> 12CO2(g) + 6H2O(g)

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