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3 years ago

Normal melting point: 658 C Normal boiling point: 2467 C

Chemistry

2 answers:

olchik [2.2K]3 years ago

8 0

Answer:

$Q = 1450089.956\,J$

Explanation:

The amount of heat required to melt the aluminium is the sum of latent and sensible heats:

$Q = Q_{lat} + Q_{sen}$

$Q = \left(12\,mol\right)\cdot \left(26.982\,\frac{g}{mol} \right)\cdot \left[\left(0.902\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (658\,^{\textdegree}C - 72\,^{\textdegree}C) + 3950\,\frac{J}{g} \right]$

$Q = 1450089.956\,J$

taurus [48]3 years ago

5 0

{ 12 mol * 27 g / mol * 0.902 j/gdeg C* ( 658 -72 ) deg C + 12 * 27 * 3.95 }
quantity of energy required to heat 12 mol of aluminium for 72 deg C to its melting point 658 C

so energy applied is 1451 057 . 33 J

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A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,

Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

$PV=nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 0.4 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

0.4 liter = $0.4\times10^{-3}=4\times10^{-4} m^{3}$

And initial temperature of balloon, $T_{1}$ = 20°C = (273 + 20)K

= 293 K

Let the final volume of balloon is $V_{2}$

And a given, final temperature of balloon, $T_{2}$ is 250°C = (273 + 250)K

= 523 K

Now, $V_{1}$ = $KT_{1}$

$4\times10^{-4}=K\times293\ (equation\ 1 )$

$V_{2}$ = $KT_{2}$

$=K\times523\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}$

K cancelled by K.

By cross multiplication:

$293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\ = \frac{2092\times10^{-4}}{293} \\ =7.14\times10^{-4}m^{3}$

Now convert it into liter with the help of calculation done above.

$7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter$

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

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2 years ago

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Alja [10]

Answer:

See explanation

Explanation:

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The Lewis structure of the molecule is shown in the image attached.

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Mashcka [7]

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sp2606 [1]

Answer:

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