The molarity of the NaOH solution is 0.03 M
We'll begin by calculating the mole of the KHP
- Mass = 0.212 g
- Molar mass = 204.22 g/mol
- Mole of KHP =?
Mole = mass /molar mass
Mole of KHP = 0.212 / 204.22
Mole of KHP = 0.001 mole
Next, we shall determine the molarity of the KHP solution
- Mole of KHP = 0.001 mole
- Volume = 50 mL = 50/1000 = 0.05 L
- Molarity of KHP =?
Molarity = mole / Volume
Molarity of KHP = 0.001 / 0.05
Molarity of KHP = 0.02 M
Finally , we shall determine the molarity of the NaOH solution
KHP + NaOH —> NaPK + H₂O
From the balanced equation above,
- The mole ratio of the acid, KHP (nA) = 1
- The mole ratio of base, NaOH (nB) = 1
From the question given above, the following data were obtained:
- Volume of acid, KHP (Va) = 50 mL
- Molarity of acid, KHP (Ma) = 0.02 M.
- Volume of base, NaOH (Vb) = 35 mL
- Molarity of base, NaOH (Mb) =?
MaVa / MbVb = nA / nB
(0.02 × 50) / (Mb × 35) = 1
1 / (Mb × 35) = 1
Cross multiply
Mb × 35 = 1
Divide both side by 35
Mb = 1 / 35
Mb = 0.03 M
Thus, the molarity of the NaOH solution is 0.03 M
Complete question:
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The Total Amount Of Atoms In C6H6 Is 12 Atoms
That when water is boiled in a open beaker and it disappears that it evaporates into the air
Answer:
Mole fraction H₂O → 0.72
Mole fraction C₂H₅OH → 0.28
Explanation:
By the mass of the two elements in the solution, we determine the moles of each:
25 g . 1 mol/ 18g = 1.39 moles of water (solute)
25 g . 1 mol / 46 g = 0.543 moles of ethanol (solvent)
Mole fraction solute = Moles of solute / Total moles
Mole fraction solvent = Moles of solvent / Total moles
Total moles = Moles of solute + Moles of solvent
1.39 moles of solute + 0.543 moles of solvent = 1.933 moles → Total moles
Mole fraction H₂O = 1.39 / 1.933 → 0.72
Mole fraction C₂H₅OH= 0.543 / 1.933 → 0.28
Remember that sum of mole fractions = 1
Answer:
-30.7 kj/mol
Explanation:
The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula: ∆Go ’= -RTln K’eq
where,
R = -8.315 J / mo
T = 298 K
For reaction,
1. K′eq1=270,
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 270
= - 8.315 x 298 x 5.59
= - 13,851.293 J / mo
= - 13.85 kj/mol
2. K′eq2=890
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 890
= - 8.315 x 298 x 6.79
= - 16.82 kj/mol
therefore, total standard free energy
= - 13.85 + (-16.82)
= -30.7 kj/mol
Thus, -30.7 kj/mol is the correct answer.
