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1 year ago

My science exam is due tomorrow and I HAVE LIMITED TIME. can someone help me how to absorb anything you read and memorize it...

Chemistry

1 answer:

igor_vitrenko [27]1 year ago

7 0

Answer:

It won't let me type this for some reason but here it is.

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Help please i’ll give extra points

joja [24]

Answer:

okay no problem

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3 years ago

Whitch is a type of mining where steam beds are minded

Varvara68 [4.7K]

Answer: placer mining

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3 years ago

A 25.00 g solid sample of Ca(OH)2 was added into 1250 mL of 0.400 M HCl aqueous solution. The temperature of the solution was de

daser333 [38]

Answer:

86.735 kJ

Explanation:

Simply multiply the change in temperature by the Ccal;

(36.6 - 20.0)×5.225 = 86.735

5 0

3 years ago

Isotopes of the same element vary from each other in their number of:

Lina20 [59]

Answer: Neutrons vary from each other in their element

MARK ME BRAINLIST

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3 years ago

Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of

lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

$U_{2}-U_{1}=Q-W$

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

$dw=Pdv$

The pressure is constant, so: $w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg}$ (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

$U_{2}-U_{1}=500-(3*75)=275kJ$

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

$P_{1}V_{1}=nRT_{1}$

$P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}$

Subtracting the first from the second:

$1.5P_{1}V_{1}=nR(T_{2}-T_{1})$

Isolating $T_{2}$ :

$T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}$

Assuming that it is water steam, n=0.1666 kmol

$V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}$

$T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76$ ºC

7 0

3 years ago