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3 years ago

What happens when gasoline is used to power a vehicle?

1 answer:

3 0

The answer should be A. Because the energy in gasoline is called chemical. When burned it is heat, Then to power a vehicle, it is mechanical energy. But I don't know whether the question wants to mean that the energy in the gasoline will not convert totally to the heat, so it will lose. But if think like this, when heat energy transform to mechanical, it will lose again. So I think the answer is A.

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What are the Sun's layers, beginning with the innermost and moving out.

Lemur [1.5K]

Core, radiative zone, convective zone, photosphere, chromosphere, corona

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3 years ago

If 45 g of O2 gas in a 500 mL container is exerting a pressure of 5.2 atm, what temperature is the gas?

iogann1982 [59]

If 45 g of O2 gas in a 500 mL container is exerting a pressure of 5.2 atm, 22.53K is the temperature in the gas.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 5.2 atm

V= 500 mL =0.5 L

n=?

R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$

T=?

$Moles = \frac{mass}{molar \;mass}$

$Moles = \frac{45 g}{32}$

Moles = 1.40625

Putting value in the given equation:

$\frac{PV}{RT}=n$

$1.40625= \frac{5.2 \;atm\; X \;0.5 \;L}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X T}$

T= 22.53167034 K= 22.53K

Hence, If 45 g of O2 gas in a 500 mL container is exerting a pressure of 5.2 atm, 22.53K is the temperature in the gas.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

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2 years ago

Lewis structure for CO2

liq [111]

O = C = O Straight because there is no solitary electrons on C

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3 years ago

Read 2 more answers

Calculate the number of grams of deuterium in an 80,000-L swimming pool, given deuterium is 0.0150% of natural hydrogen

fiasKO [112]

Answer:

$m_{Deu}=2666.7g$

Explanation:

Hello,

In this case, we can apply the following mole-mass-density relationship in order to compute the required grams of deuterium, considering that it is the 0.0150% (molar basis) of natural hydrogen (H₂):

$m_{Deu}=80000LH_2O*\frac{1000gH_2O}{1LH_2O}*\frac{1molH_2O}{18gH_2O} *\frac{1molH_2}{1molH_2O} *\frac{2gH_2}{1molH_2} *\frac{2*0.0150g\ Deu}{100gH_2} \\\\m_{Deu}=2666.7g$