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2 years ago

I really need the answers please!

Chemistry

1 answer:

krek1111 [17]2 years ago

4 0

1) D = 13.6 g / mL

2)ethyl alcohol weighs 158g

3)ρ _copper = 8.9 g $cm^{3}$

Explanation:

D = m / V

=306.0 g / 22.5 mL

D= 13.6 g / mL

density = mass / volume

mass = density × volume

=0.789g /ml × 200.0 ml

M=158g

Ethyl alcohol weighs 158g

ρ (density) = Mass / Volume

ρ _copper = 1896 g / 8.4cm × 5.5cm × 4.6cm

= 1896g / 212.5 $cm^{3}$

ρ _copper=8.9 g $cm^{3}$

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Part 1 :

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 $C_2H_6$ = $\frac{1000}{52}=19.23g$

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

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Part 2 :

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

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

Part 3 :

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:

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

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

Part 4 :

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

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

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