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amid [387]
3 years ago
10

Gasoline has a density of 0.749 g/ml. how many pounds does 19.2 gallons of gasoline weigh?

Chemistry
2 answers:
Svetradugi [14.3K]3 years ago
6 0
A good first step is writing the amount in terms of ml. 

19.2 gallons = 72.68 L = 72680 ml

that would mean it weighs 0.749*720680g = 54437.32ml = 54.437 L

hope that helps :)
7nadin3 [17]3 years ago
5 0

Answer;

mass will be; 119.76 lbs

solution and Explanation;

1,000 g = 2.2 lb and 1 gal = 3785.41 mL.

Density = 0.749 g/ml * 2.2 lb/1,000 g * 3785.41 mL/1 gal

Density = 6.2376 lb/gal

Mass = 6.2376 lb/gal*19.2 gal

Mass = 119.76 lbs

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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

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

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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 '.

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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 '

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