The answer to your question is A <span>4.184 J</span>
½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS
<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 : The volume of 3.0 M spinach solution added should be, 50 mL
Explanation :
Formula used :

where,
are the initial molarity and volume of spinach solution.
are the final molarity and volume of diluted spinach solution.
We are given:

Now put all the given values in above equation, we get:

Hence, the volume of 3.0 M spinach solution added should be, 50 mL