NaCl would form because it’s a single replacement reaction
<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:

......(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: beta, gamma, alpha
Explanation: Beta is weakest gamma is middle alpha is strongest
The overall balanced
chemical reaction for this is:
Detonation of
Nirtoglycerin <span>
4 C3H5N3O9 --> 12 CO2 + 6 N2 + O2 + 10 H2O </span>
Therefore:
2.00 mL x 1.592 g/mL =
3.184 g <span>
3.184 g / 227.1 /mol = 0.0140 mol nitroglycerin
4 moles --> 12 + 6 + 1 + 10 = 29 moles of gas
<span>0.0140 mol x (29/4) = 0.1017 moles of gases or (0.102 mol) </span></span>
From the calculations, the heat of fusion of the substance is 0.73 kJ
<h3>What is is the heat of freezing?</h3>
The heat of freezing is the energy released when the substance is converted from liquid to solid.
Now we know that the molar mass of the substance is 82.9 g/mol hence the number of moles of the substance is; 13.3 g /82.9 g/mol = 0.16 moles
Now the heat of fusion shall be;
H = 4.60 kj/mol * 0.16 moles
H = 0.73 kJ
Learn more about freezing:brainly.com/question/3121416
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