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:
The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
Learn more about pH of buffer:
brainly.com/question/21881762
Answer:
The volume is 583.02 mL
Explanation:
Given that the density is:-
The mass, 
The expression for the calculation of density is shown below as:-
Using the above expression to calculate the volume as:-
Applying the values to calculate the volume as:-
<u>The volume is 583.02 mL.</u>
