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Marina86 [1]
1 year ago
6

How much money will it cost to drive a school bus 94.00 miles if it gets 6.500 miles per gallon and gas costs $2.459/gallon?​

Chemistry
1 answer:
bagirrra123 [75]1 year ago
6 0

Answer:

It will cost $35.5602

Explanation:

$2.459 /6.5 (mpg) = 0.3783 (cents per mile)

$0.3783 (cost per mile) * 94 (total miles) = $35.5602 total

You might be interested in
Choose the words to finish the sentence. After learning about the law of conservation of mass, Sammy became interested in balanc
cestrela7 [59]

The chemical equation is unbalanced and synthesized.

<h3></h3><h3>What is a chemical equation?</h3>

A chemical equation is described as the symbolic representation of a chemical reaction in the form of symbols and chemical formulas.

In a chemical equation, the reactant entities are given on the left-hand side and the product entities is shown on the right-hand side with a plus sign between the entities in both the reactants and the products, and an arrow that indicates towards the products to show the direction of the reaction.

We can conclude that in the chemical equation shown is unbalanced because both amounts of the individual elements and compounds do not reflect on the reactant and product side.

Learn more about chemical equations at: brainly.com/question/11231920

#SPJ1

The complete question is below:

After learning about the law of conservation of mass, Sammy became interested in balancing equations. He knew that the symbol for aluminum was Al and silver tarnish was Ag2S. He also knew that mixing the two chemicals yielded pure silver, or Ag, in an aluminum sulfide solution. Here is the equation showing this reaction:

3 Ag2S + 2 Al → 6 Ag + Al2S3

This equation is (synthesis / unbalanced / replacement / balanced), and it represents a(n) (unbalanced / balanced / synthesized / replaced) chemical reaction.

answer choices:

  • synthesis; balanced

  • balanced; replacement

  • unbalanced; synthesized

  • balanced; balanced

6 0
1 year ago
What are the benefits and drawbacks to having a pouring temperature that is much higher than the metal’s melting temperature?
Mashcka [7]

Answer:

The main advantage would be that with the pouring temperature being much higher, there is very little chance that the metal will solidify in the mould while busy pouring. This will allow for moulds that are quite intricate to still be fully filled. The drawbacks, though, include an increased chance defects forming which relates to shrinkage (cold shots, shrinkage pores, etc). Another drawback includes entrained air being present, due to the viscosity of the metal being low because of the high pouring temperature.

3 0
3 years ago
A gas occupies 14.3 liters at a pressure of 45.0 mm Hg. What is the volume when the pressure is increased to
Simora [160]

Answer:

P1V1= P2V2

Explanation:

Inverse relationship

V2 = V1 X P1/P2

V2= 14.3 L x 45.0 mm Hg/63.0 mmHg= 8.99

8 0
3 years ago
A sample with a molar mass of 34.00 g/mol is found to consist of 5.9783% Hydrogen and 94.0217% oxygen. Find it’s molecular formu
kozerog [31]

Answer:

The formula is H202 (hydrogen peroxide, known as hydrogen peroxide)

Explanation:

100%----34g

5, 9783%---X= (5, 9783%x 34g)/100% =2 g

1g---1 atom of H

2g----x= 2g x 1 atom of H/1g = 2 atom of H

100%----34g

94, 0217%---X= (94,0217%x 34g)/100% =32 g

16g---1 atom of 0

32g----x= 32g x 1 atom of 0/16g = 2 atom of 0

5 0
2 years ago
Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
2 years ago
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