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Hatshy [7]
2 years ago
9

When heating a flammable or volatile solvent for a recrystallization, which of these statements are correct? More than one answe

r may be correct.
a. You should not use an open flame to heat the solvent.
b. You should heat the solvent in a stoppered flask to keep vapor away from any open flames.
c. You should ensure that no one else is using an open flame near your experiment.
Chemistry
2 answers:
Citrus2011 [14]2 years ago
6 0

Answer:

You should not use an open flame to heat the solvent and the solvent should be heat in a stoppered flask to vapour away from the open flame

Explanation:

ruslelena [56]2 years ago
4 0

Answer:

A

Explanation:

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❃ The following points should be kept in mind to write and balance a chemical equation :

Step 1 : Write the molecular formula of all the reactants and products correctly.

Step 2 : Separate reactants and products by a sign of arrow. If reactants or products are more than one, connect them by a sign of a plus.

Step 3 : Balance the atoms of O and H at last [ The atoms used at many places in an equation should be balanced at last ]. For balancing , the number should be added as coefficient i.e in the front of the molecules.

[ Remember those substance that take part in a chemical reaction are called reactants. Likewise , those substances which are formed after a chemical reaction are called products ]

\large{ \tt{❁ \: LET'S \: GET \: STARTED}} :

1. Carbon disulfide + Oxygen gas gives carbon dioxide + Sulfur dioxide.

Step 1 : The molecular formula of carbon disulfide is CS₂ , molecular formula of Oxygen gas is 0₂ [ Since oxygen is a diatomic element ] molecular formula of carbon dioxide is CO₂ and molecular formula of sulfur dioxide is SO₂.

Step 2 : CS₂ + O₂ ⟶ CO₂ + SO₂

Step 3 : In the reactant side , there is two ' S ' but on the other side , there is one ' S '. So , add 2 as a coefficient before S on the product side. Now , There are two ' O ' in the reactant side but six ' O ' in the product side. So , add 3 as a coefficient before O on the reactant side. Now , there are equal atom of C , S and O on both sides

i.e CS₂ + 3O₂ ⟶ CO₂ + 2SO₂

Answer : \boxed{ \tt{CS₂ + 3O₂ ⟶ CO₂ + 2SO₂}}

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2. Silver + nitric acid gives silver nitrate + nitrogen dioxide + water

Step 1 : The molecular formula of Silver is Ag, molecular formula of nitric acid is HNO₃ , molecular formula of Silver nitrate is Ag ( No₃ ) , molecular formula of nitrogen dioxide is NO₂ and molecular formula of water is H₂O.

Step 2 : Ag + HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O

Step 3 : In the reactant side , There is one ' H ' but on the other side , there are two ' H '. Now add 2 before H on the reactant side. There are equal atom of ' Ag ' , ' H ' , ' N ' , and ' O '.

i.e Ag + 2HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O

Answer : \boxed{ \tt{Ag + 2HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O }}

  • Yay! We're done ! :)

- The last step is a bit more confusing I guess. So , which balancing , count the atoms in following ways :

  • The number written at the right lower corner of an atom is counted for that atom only. For example : In MgSO₄ , there are one ' Mg ' , one ' S ' and four ' O '

  • The number written at the right lower corner of a bracket is for all the atoms enclosed in the bracket. For example : In Al₂ ( SiO₃ ) has two Al , three ' S ' and nine ' O '.

  • The coefficient number is for all the atoms of the molecule. For example , in 2 Al ₂( SiO₃ )₃ , there are four ' Al ' , six ' Al ' and eighteen ' O '.

- Hope this helps! :)

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