Answer : The final temperature of the metal block is, 
Explanation :

As we know that,

.................(1)
where,
q = heat absorbed or released
= mass of aluminum = 55 g
= mass of water = 0.48 g
= final temperature = ?
= temperature of aluminum = 
= temperature of water = 
= specific heat of aluminum = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]](https://tex.z-dn.net/?f=55g%5Ctimes%200.900J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%29%5EoC%3D-%5B0.48g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%29%5EoC%5D)

Thus, the final temperature of the metal block is, 
Just use the Heisenberg Uncertainty principle:
<span>ΔpΔx = h/2*pi </span>
<span>Δp = the uncertainty in momentum </span>
<span>Δx = the uncertainty in position </span>
<span>h = 6.626e-34 J s (plank's constant) </span>
<span>Hint: </span>
<span>to calculate Δp use the fact that the uncertainty in the momentum is 1% (0.01) so that </span>
<span>Δp = mv*(0.01) </span>
<span>m = mass of electron </span>
<span>v = velocity of electron </span>
<span>Solve for Δx </span>
<span>Δx = h/(2*pi*Δp) </span>
<span>And that is the uncertainty in position. </span>
Reduction is the process by which eletrons are gained by an atom.
Answer:
- <u>Freezing point: - 1.83ºC</u>
- <u>Boiling point: 100.50ºC</u>
Explanation:
The <em>freezing point</em> and<em> boiling point</em> of solvents, when a solute is added, will change accordingly to the concentration of the solute particles.
The freezing point will decrease and the boiling point will increase. These are two colligative properties.
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<u>Find attached the file with the whole answer, as the site is not uploading the answer in here.</u>
<h3>
Answer:</h3>
812 kPa
<h3>
Explanation:</h3>
- According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
- Mathematically,

At varying pressure and volume;
P1V1=P2V2
In this case;
Initial volume, V1 = 2.0 L
Initial pressure, P1 = 101.5 kPa
Final volume, V1 = 0.25 L
We are required to determine the new pressure;

Replacing the known variables with the values;

= 812 kPa
Thus, the pressure of air inside the balloon after squeezing is 812 kPa