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1 year ago

A student measures the pressure and volume of an empty water bottle to be 1. 4 atm and 2. 3 l

Chemistry

1 answer:

rewona [7]1 year ago

6 0

The new volume of the bottle is 5.07 L.

<h3>What is Boyle's Law?</h3>

It is a gas law that states the pressure decrease with the increase in the pressure.

By the formula

$\rm P_1V_1= P_1V_2$

Given that, volume1 is 2.3l

Pressure1 is 4 atm.

Pressure2 to 0.65 atm.

V2 is to find?

Putting the values in the equation

$\rm 1.4\times 2.3 = 0.65 \times V_2\\\\V_2 = 5.07 L$

Thus, the new volume of the bottle is 5.07 L.

Learn more about Boyle's Law

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A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0

miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

$Q_{metal} - Q_{water} = 0$

$Q_{metal} = Q_{water}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{metal}$ - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

$m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})$

The specific heat of the metal is cleared within equation above:

$c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}$

If we know that $m_{water} = 0.237\,kg$ , $m_{metal} = 0.244\,kg$ , $c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{w,o} = 10\,^{\circ}C$ , $T_{m,o} = 100.10\,^{\circ}C$ and $T = 15.30\,^{\circ}C$ , the specific heat of molybdenum is:

$c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}$

$c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}$

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

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I hope my answer has come to your help. God bless and have a nice day ahead!

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Explanation:

I'm not sure but I think idk so sorry if it's so wrong

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