<u>Answer:</u> The final temperature of water is 32.3°C
<u>Explanation:</u>
When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)
The equation used to calculate heat released or absorbed follows:
![m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
......(1)
where,
q = heat absorbed or released
= mass of solution 1 (liquid water) = 50.0 g
= mass of solution 2 (liquid water) = 29.0 g
= final temperature = ?
= initial temperature of solution 1 = 25°C = [273 + 25] = 298 K
= initial temperature of solution 2 = 45°C = [273 + 45] = 318 K
c = specific heat of water= 4.18 J/g.K
Putting values in equation 1, we get:
![50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K](https://tex.z-dn.net/?f=50.0%5Ctimes%204.18%5Ctimes%20%28T_%7Bfinal%7D-298%29%3D-%5B29.0%5Ctimes%204.18%5Ctimes%20%28T_%7Bfinal%7D-318%29%5D%5C%5C%5C%5CT_%7Bfinal%7D%3D305.3K)
Converting this into degree Celsius, we use the conversion factor:
Hence, the final temperature of water is 32.3°C
Answer:
Explanation:
Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:
Where 
is Henry's law constant.
Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.
Given initially:
Also, at sea level, we have an atmospheric pressure of:
Given mole fraction:
According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:
Then the equation becomes:
Solve for :
Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:
Apply Henry's law using the constant we found:
Answer : The molecular weight of a substance is 157.3 g/mol
Explanation :
As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.
Mass of solute = 7 g
Mass of solution = 100 g
Mass of solvent = 100 - 7 = 93 g
Formula used :
where,
= change in freezing point
= temperature of pure water = 
= temperature of solution = 
= freezing point constant of water = 
m = molality
Now put all the given values in this formula, we get
Therefore, the molecular weight of a substance is 157.3 g/mol
The answer is
<span>2PbS(s) + 3O2(g) = 2PbO(s) + 2SO2(g)
Your answer is not yet balanced because you have 3 oxygen atoms. it should be balanced by multiplying both side by 2 such as the balanced equation I made. To check it, I will explain why your answer is not yet balanced.
check: (from your equation)
</span> 1-Pb-1
1-S-1
2 -O-3
the difference between the reactant and the product of Oxygen will prove that it is not yet balanced.
If you use 2PbS(s) + 3O2(g) = 2PbO(s) + 2SO2(g), to check it:
2-Pb-2
2-S-2
6 -O-6
then this is now balance
