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Answer:
Boyle's Law

Explanation:
Given that:
<u><em>initially:</em></u>
pressure of gas, 
volume of gas, 
<em><u>finally:</u></em>
pressure of gas, 
volume of gas, 
<u>To solve for final volume</u>
<em>According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.</em>
<em>According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.</em>
But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.
Here nothing is said about the temperature, so we consider the Boyle's Law which states that <em>at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.</em>
Mathematically:



I would think that you would multiply then divide
1) 29.8 C
At the beginning, the metal is at higher temperature (70.4 C) while the water is at lower temperature (23.6 C). When they are put in contact, the metal transfers heat to the water, until they reach thermal equilibrium: at thermal equilibrium the two objects (the metal and the water have same temperature). Therefore, since the temperature of the water at thermal equilibrium is 29.8 C, the final temperature of the metal must be the same (29.8 C).
2) 6.2 C
The temperature change of the water is given by the difference between its final temperature and its initial temperature:

where

Substituting into the formula,

And the positive sign means that the temperature of the water has increased.
3) -40.6 C
The temperature change of the metal is given by the difference between its final temperature and its initial temperature:

where

Substituting into the formula,

And the negative sign means the temperature of the metal has decreased.
Part a)
At t = 0 the position of the object is given as

At t = 2

so displacement of the object is given as

so average speed is given as

Part b)
instantaneous speed is given by


now at t= 0

at t = 1


at t = 2

Part c)
Average acceleration is given as



Part d)
Now for instantaneous acceleration
As we know that

at t = 0

at t = 1

now we have

At t = 2 we have



<em>so above is the instantaneous accelerations</em>