Answer:
Ka3 for the triprotic acid is 7.69*10^-11
Explanation:
Step 1: Data given
Ka1 = 0.0053
Ka2 = 1.5 * 10^-7
pH at the second equivalence point = 8.469
Step 2: Calculate Ka3
pKa = -log (Ka2) = 6.824
The pH at the second equivalence point (8.469) will be the average of pKa2 and pKa3. So,
8.469 = (6.824 + pKa3) / 2
pKa3 = 10.114
Ka3 = 10^-10.114 = 7.69*10^-11
Ka3 for the triprotic acid is 7.69*10^-11
7 × 10² pounds of Cl₂ would be produced in a typical 8-h operating day.
<h3>What is Stoichiometry ?</h3>
Stoichiometry helps us use the balanced chemical equation to measures quantitative relationships and it is to calculate the amount of products and reactants that are given in a reaction.
<h3>What is Balanced chemical equation ?</h3>
The balanced chemical equation is the equation in which the number of atoms on the reactants side is equal to the number of atoms on the product side in an equation.
2Cl⁻ (aq) → Cl (g) + 2e⁻
According to stoichiometry, moles of Cl₂

= 4477 moles
Pounds Cl₂

= 7 × 10² lb
Thus from the above conclusion we can say that 7 × 10² pounds of Cl₂ would be produced in a typical 8-h operating day.
Learn more about Stoichiometry here: brainly.com/question/14935523
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<span>Assume
p=735 Torr
V= 7.6L
R=62.4
T= 295
PV-nRT
(735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K)
0.30346 moles of NH3
Find moles
0.300L solution of 0.300 M HCL = 0.120 moles of HCL
0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind
Find molarity
0.120 moles of NH4+/0.300L = 0.400 M NH4+
0.18346 moles of NH3/0.300L = 0.6115 M NH3
NH4OH --> NH4 & OH-
Kb = [NH4+][OH]/[NH4OH]
1.8 e-5=[0.300][OH-]/[0.6115]
[OH-]=1.6e-5
pOH= 4.79
PH=9.21
.</span>
Answer:
0.50 moles of Cu
Explanation:
The balanced chemical equation for given synthetic reaction is,
2 Cu + S → Cu₂S
According to balance chemical equation,
1 mole of Cu₂S is produced by = 2 moles of Cu
So,
0.25 moles of Cu₂S will be produced by = X moles of Cu
Solving for X,
X = 0.25 mol × 2 mol / 1 mol
X = 0.50 moles of Cu
Hence, as the molar ratio of Cu to Cu₂S is 2:1 hence, to produce 0.25 moles of Cu₂S we will need 0.50 moles of Cu.