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2 years ago

Radiation can cause long-term problems once it enters the environment.

Chemistry

1 answer:

tatuchka [14]2 years ago

7 0

Yes it can cause plants and animals to become un eatable leaving us hungry and the air toxicated

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The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

2 years ago

A cylinder was charged with 1.25 atm of oxygen gas, 6.73 atm of argon, and 0.895 atm of xenon. What is the mole fraction of each

katrin2010 [14]

Considering the Dalton's partial pressure, the mole fraction of each gas is:

$x_{oxygen}$ = 0.14
$x_{argon}$ = 0.76
$x_{xenon}$ = 0.10

<h3>Dalton's partial pressure</h3>

The pressure exerted by a particular gas in a mixture is known as its partial pressure.

So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

$P_{T} =P_{1} +P_{2} +...+P_{n}$

where n is the amount of gases in the gas mixture.

This relationship is due to the assumption that there are no attractive forces between the gases.

Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of a component to the number of moles of all the components present.

So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:

$P_{A} =x_{A} P_{T}$

In summary, the total pressure in a mixture of gases is equal to the sum of partial pressures of each gas.

Mole fraction of each gas

In this case, you know that:

$P_{oxygen }$ = 1.25 atm
$P_{argon}$ = 6.73 atm
$P_{xenon}$ = 0.895 atm
$P_{T} =P_{oxygen} +P_{argon}+P_{xenon}$ = 1.25 atm + 6.73 atm + 0.895 atm= 8.875 atm

Then:

$P_{oxygen} =x_{oxygen} P_{T}$
$P_{argon} =x_{argon} P_{T}$
$P_{xenon} =x_{xenon} P_{T}$

Substituting the corresponding values:

1.25 atm= $x_{oxygen}$ 8.875 atm
6.73 atm= $x_{argon}$ 8.875 atm
0.895 atm= $x_{xenon}$ 8.875 atm

Solving:

$x_{oxygen}$ = 1.25 atm÷ 8.875 atm= 0.14
$x_{argon}$ = 6.73 atm÷ 8.875 atm= 0.76
$x_{xenon}$ = 0.895 atm÷ 8.875 atm=0.10

In summary, the mole fraction of each gas is:

$x_{oxygen}$ = 0.14
$x_{argon}$ = 0.76
$x_{xenon}$ = 0.10

Learn more about Dalton's partial pressure:

brainly.com/question/14239096

brainly.com/question/25181467

brainly.com/question/14119417

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1 year ago

Describe what happens in a condensation reactions

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2 years ago

Why do you think steeples and roofs might appear green, but a copper square does not?

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Because the copper roofs, have been outside for a while and oxidization has occurred.
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2 years ago

What is environment?write in a sentence

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The natural environment or natural world encompasses all living and non-living things occurring naturally, meaning in this case not artificial.

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2 years ago

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