That is correct hope that helps
Answer is Sodium Hydroxide.
Answer:
a) heat it from 23.0 to 78.3
q = (50.0 g) (55.3 °C) (2.46 J/g·°C) =
b) boil it at 78.3
(39.3 kJ/mol) (50.0 g / 46.0684 g/mol) =
c) sum up the answers from the two calculations above. Be sure to change the J from the first calc into kJ
Explanation:
Explanation:
First Reaction;
Ca + ZnCl2 --> CaCl2 + Zn
Oxidized Reactant: Ca. There is increase in oxidation number from 0 to +2
Reduced Reactant: Zn. There is decrease in oxidation number form +2 to 0
Second Reaction:
FeI2 + Mg --> Fe + MgI2
Oxidized Reactant: Mg. There is increase in oxidation number from 0 to +2
Reduced Reactant: Fe. There is decrease in oxidation number form +2 to 0
Third Reaction;
Mg + 2AgNO3 --> Mg(NO3)2 + Ag
Oxidized Reactant: Mg. There is increase in oxidation number from 0 to +2
Reduced Reactant: Ag. There is decrease in oxidation number form +1 to 0
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:
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