Answer:
Enthalpy change for the reaction is -67716 J/mol.
Explanation:
Number of moles of 
in 50.0 mL of 0.100 M of
= Number of moles of HCl in 50.0 mL of 0.100 M of HCl
= 
moles
= 0.00500 moles
According to balanced equation, 1 mol of reacts with 1 mol of HCl to form 1 mol of AgCl.
So, 0.00500 moles of react with 0.00500 moles of HCl to form 0.00500 moles of AgCl
Total volume of solution = (50.0+50.0) mL = 100.0 mL
So, mass of solution = (
) g = 100 g
Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)
= 
= ![\frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}](https://tex.z-dn.net/?f=%5Cfrac%7B-100g%5Ctimes%204.18%5Cfrac%7BJ%7D%7Bg.%5E%7B0%7D%5Ctextrm%7BC%7D%7D%5Ctimes%20%5B24.21-23.40%5D%5E%7B0%7D%5Ctextrm%7BC%7D%7D%7B0.00500mol%7D)
= -67716 J/mol
[m = mass, c = specific heat capacity, 
= change in temperature and negative sign is included as it is an exothermic reaction]
Answer:
Explanation:
Given parameters:
Mass of CuCl₂ = 2.50g
Mass of Al = 0.50g
Unknown:
Number of moles of CuCl₂ and Al = ?
Solution:
To solve this problem, we must understand that the number of moles is a fundamental property used in stoichiometry calculations.
Number of moles = 
Molar mass of CuCl₂ = 63.6 + 2(35.5) = 134.5g/mole
Molar mass of Al = 26.98g/mole
Number of moles of CuCl₂ = 
= 0.019moles
Number of moles of Al = 
= 0.019moles
Answer:
Your answer would be B. Increasing the distance between the water molecules and increasing their phase energy.
Explanation:
Hope this helps you answer this question!
Omg i lost everything ugh
To do it again
1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol
2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol
3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol
4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol
5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g
6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g
7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g
8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g
I cant believe i had to do this all over
