Answer:
2.
Explanation:
This should be right hopefully it is!
The pH of a solution is 9.02.
c(HCN) = 1.25 M; concentration of the cyanide acid
n(NaCN) = 1.37 mol; amount of the salt
V = 1.699 l; volume of the solution
c(NaCN) = 1.37 mol ÷ 1.699 l
c(NaCN) = 0.806 M; concentration of the salt
Ka = 6.2 × 10⁻¹⁰; acid constant
pKa = -logKa
pKa = - log (6.2 × 10⁻¹⁰)
pKa = 9.21
Henderson–Hasselbalch equation for the buffer solution:
pH = pKa + log(cs/ck)
pH = pKa + log(cs/ck)
pH = 9.21 + log (0.806M/1.25M)
pH = 9.21 - 0.19
pH = 9.02; potential of hydrogen
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Answer:
Molecular weight of the compound = 372.13 g/mol
Explanation:
Depression in freezing point is related with molality of the solution as:
Where,
= Depression in freezing point
= Molal depression constant
m = Molality
m = 0.26
Molality = 
Mass of solvent (toluene) = 15.0 g = 0.015 kg
Moles of compound = 0.015 × 0.26 = 0.00389 mol
Mass of the compound = 1.450 g
Molecular weight = 
C. released or absorbed.
When the energy is released the reaction is called exotermic reaction
When the energy is absorbed, the reaction is called endothermic reaction
An example of exotermic reaction is between water and H2SO4
Answer:
52.2 g
Explanation:
Step 1: Write the balanced equation
3 KOH + H₃PO₄ ⟶ K₃PO₄ + 3 H₂O
Step 2: Calculate the moles corresponding to 89.7 g of KOH
The molar mass of KOH is 56.11 g/mol.
89.7 g × 1 mol/56.11 g = 1.60 mol
Step 3: Calculate the moles of H₃PO₄ needed to react with 1.60 moles of KOH
The molar ratio of KOH to H₃PO₄ is 3:1. The moles of H₃PO₄ needed are 1/3 × 1.60 mol = 0.533 mol.
Step 4: Calculate the mass corresponding to 0.533 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
0.533 mol × 97.99 g/mol = 52.2 g
