The heat that creates this temperature change coming from change in the internal energy of the system as per as first law of thermodynamics.
<h3>What is Boyle's law ?</h3>
A law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.
As we know, Boyle's law only works when the gas is kept at a constant temperature
Here,
When volume of gases decreased, it means work done has occurred on the system, so the work done is used for raising internal energy of the gas and the other is released as the thermal energy.
So,
According to 1st law of thermodynamics,
we know Q = ΔU + W i.e, change in internal energy and work done. So this is a reason. Changing temperature occurs.
Learn more about Internal enrgy here ;
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Answer:
Q = 143,921 J = 143.9 kJ.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:

Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:

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Answer:
moles H₂O = 10
Explanation:
The mass of Na₂CO₃⋅xH₂O is 3.837 g and the mass of Na₂CO₃ is 1.42g
Therefore the mass of xH₂O is 3.837 - 1.42 = 2.417 g
The molar mass of Na₂CO₃ is 106 g/mol and for H₂O is 18 g/mol
The moles of Na₂CO₃ and H₂O in the sample are:
Na₂CO₃ = 1.42 / 106 = 0.01340 moles
H₂O = 2.417 / 18 = 0.1343
Now using rule of three :
1 mole of Na₂CO₃ has x moles of H₂O
0.01340 moles of Na₂CO₃ has 0.1343 moles of H₂O
x = 1 * 0.1343 / 0.01340 = 10
Answer:
~1.5 g/cm3 and it does NOT float in water.
Explanation:
If you look at the graph, Object A weighs ~6 grams and is ~4 cm3 in volume
Density = Mass/Volume
So 6 grams/4 cm3 = 1.5 g/cm3
Water has a density of 1 g/cm3 and because Object A density is higher than that of water, it sinks.
:)