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maria [59]
2 years ago
6

A gram of gasoline produces 45.0kj of energy when burned. gasoline has a density of 0.77/gml . how would you calculate the amoun

t of energy produced by burning 48.l of gasoline?
Chemistry
1 answer:
babymother [125]2 years ago
5 0
Given that there is 48 liters of gasoline to be burned and that 45 kJ of energy is released per gram of gasoline burned, the amount of energy that the gasoline fuel produces can then be calculated, First, we convert 48 liters of gasoline to units of mass (grams) in order to use the given conversion of 45 kJ per gram of gasoline. To do this, we use the density of gasoline which is 0.77 g/mL. The following expression is then used:

48 L gasoline x 1000 mL/L x 0.77 g/mL x 45 kJ/g gasoline = 1663200 kJ

<span>The amount of energy produced by burning 48 L of gasoline was then determined to be 1663200 kJ. </span>    
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Dima020 [189]
Mole ratio for the reaction is 1:1
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2 years ago
Increased motion in the particles in a liquid allows the particles to
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You have a 16.0-oz. (473-mL) glass of lemonade with a concentration of 2.66 M. The lemonade sits out on your counter for a coupl
salantis [7]

<u>Answer:</u> The new concentration of lemonade is 3.90 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

Molarity of lemonade solution = 2.66 M

Volume of solution = 473 mL

Putting values in equation 1, we get:

2.66M=\frac{\text{Moles of lemonade}\times 1000}{473}\\\\\text{Moles of lemonade}=\frac{2.66\times 473}{1000}=1.26mol

Now, calculating the new concentration of lemonade by using equation 1:

Moles of lemonade = 1.26 moles

Volume of solution = (473 - 150) mL = 323 mL

Putting values in equation 1, we get:

\text{New concentration of lemonade}=\frac{1.26\times 1000}{323}\\\\\text{New concentration of lemonade}=3.90M

Hence, the new concentration of lemonade is 3.90 M

7 0
2 years ago
How much water(in grams) at its boiling point can be vaporized by adding 1.50 kJ of heat? The molar heat of vaporization for wat
LekaFEV [45]

Answer:

0.66g of water

Explanation:

Molar heat of vaporization of any substance is defined as the heat necessary to vaporize 1 mole of the substance.

If heat of vaporization of water is 40.79kJ/mol and you add 1.50kJ, the moles you vaporize are:

1.50kJ × (1mol / 40.79kJ) = 0.0368 moles of water.

As molar mass of water is 18.01g/mol, mass of water that can be vaporized are:

0.0368 moles × (18.01g / mol) = <em>0.66g of water</em>

6 0
2 years ago
A carpet sells for 27.99 a square yard. What is the price of the carpet square per square meter ?
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There are 9 square meters in one square yard so divide 27.99 by 9

27.99/9=3.11

one square meter is $3.11

I hope I've helped!
3 0
2 years ago
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