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maria [59]
3 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]3 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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During which change of state would the volume of a substance increase the most?
Alika [10]

Answer:

The most state has the high volume is the gas state.

So here the answer is vaporization!!!!

Explanation:

The most state has the high volume is the gas state.

So here the answer is between vaporization and sublimation???

and when the sublimation is: the change of the material from the solid state to the gas state directly without pass to liquid state from the solid first then to the gas

and the vaporization is : The change of the material from the liquid state to the gas state by heating Ex: water vaporize at 100°

here, the answer is sublimation because the difference of volume which resulted from this process is more than the volume resulted from vaporization.

Plz mark brainlist!!

4 0
3 years ago
Read 2 more answers
When supply and demand for a product increase simultaneously, we can predict that both the market clearing price and the equilib
Whitepunk [10]

Answer: Explanation:

When supply and demand for a product increase simultaneously, we cannot predict the market clearing price, but know that the equilibrium quantity will increase.

NB: We cannot predict the market clearing price because no adequate information such as government policy is provided. such information would likely affect the market clearing price either positively or negatively.

P.S: This is a business question.

3 0
3 years ago
CH4 + 20 + CO2 + 2 H 0
yKpoI14uk [10]

Answer:

50g

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

6 0
3 years ago
How is heating of sugar and heating of ammonium chloride different from each other? Explain your answer.
algol13
When sugar is heated it melts and then caramelize giving of water. After this, it turns blank (carbon) and then coverts to co2 (carbon dioxide) .
So, heating of sugar is an endothermic , decomposition and oxidation reaction.
  When Ammonium Chloride is heated , it directly goes from solid state to vapour state without changing into liquid state.
So, heating of Ammonium Chloride is an example of sublimation reaction.
3 0
4 years ago
For the following reaction, 13.1 grams of glucose (C6H12O6) are allowed to react with 10.6 grams of oxygen gas. glucose (C6H12O6
mart [117]

Answer:

There will be formed 14.58 grams of CO2

O2 is the limiting reagent

There will remain 3.151 grams of glucose

Explanation:

Step 1: Data given

Mass of glucose = 13.1 grams

molar mass of glucose 180.156 g/mol

Mass of oxygen = 10.6grams

molar mass of oxygen = 32 g/mol

<u>Step 2</u>: The balanced equation

C6H12O6 + 6 O2 → 6H2O + 6CO2

<u>Step 3</u>: Calculate moles of glucose

Moles of glucose = mass glucose / molar mass glucose

Moles of glucose = 13.1 grams / 180.156 g/mol

Moles of glucose = 0.0727 moles

Step 4: Calculate moles of oxygen

Moles O2 = 10.6 grams / 32 g/mol

Moles O2 = 0.33125 moles

Step 5: Calculate the limiting reactant

For 1 mol of glucose , we need 6 moles of O2 to produce 6 moles of H2O and 6 moles of CO2

Oxygen is the limiting reactant. It will completely be consumed ( 0.33125 moles).

Glucose is in excess. There will be consumed 0.33125/6 = 0.05521 moles

There will remain 0.0727 - 0.05521 = 0.01749 moles

This is 0.01749 moles * 180.156 g/mol = 3.151 grams

Step 6: Calculate moles of CO2

For 1 mol of glucose , we need 6 moles of O2 to produce 6 moles of H2O and 6 moles of CO2

For 0.33125 moles of O2 we'll get 0.33125 moles of CO2 produced

Step 7: Calculate mass of CO2

Mass of CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.33125 moles * 44.01 g/mol

Mass CO2 = 14.58 grams

7 0
3 years ago
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