How can one control the enzyme experiment?

Biology
How can one control the enzyme experiment?

ScanSolve AI Answer

Answer

0.1085 mol/L

Step 1: Calculate the average volume of NaOH used in the two titrations.

Vmeˊdio NaOH=V1+V22V_médio\ NaOH = \frac{V_1 + V_2}{2} Vmeˊdio NaOH=24.78 mL+25.80 mL2=50.58 mL2=25.29 mLV_médio\ NaOH = \frac{24.78 \, mL + 25.80 \, mL}{2} = \frac{50.58 \, mL}{2} = 25.29 \, mL

Step 2: Convert volumes to liters.

VNaOH=25.29 mL=25.29×10−3 L=0.02529 LV_NaOH = 25.29 \, mL = 25.29 \times 10^{-3} \, L = 0.02529 \, L VHCl=25 mL=25×10−3 L=0.025 LV_HCl = 25 \, mL = 25 \times 10^{-3} \, L = 0.025 \, L

Step 3: Calculate the number of moles of NaOH used at the equivalence point (1:1 stoichiometry with HCl).

nNaOH=cNaOH×VNaOHn_NaOH = c_NaOH \times V_NaOH nNaOH=0.1073 mol/L×0.02529 Ln_NaOH = 0.1073 \, mol/L \times 0.02529 \, L

First,

0.1073×0.02529=0.1073×(0.025+0.00029)=(0.1073×0.025)+(0.1073×0.00029)0.1073 \times 0.02529 = 0.1073 \times (0.025 + 0.00029) = (0.1073 \times 0.025) + (0.1073 \times 0.00029) 0.1073×0.025=0.00268250.1073 \times 0.025 = 0.0026825 0.1073×0.00029=0.0000311170.1073 \times 0.00029 = 0.000031117 nNaOH=0.0026825+0.000031117=0.002713617 moln_NaOH = 0.0026825 + 0.000031117 = 0.002713617 \, mol

Step 4: Since nHCl=nNaOHn_HCl = n_\mathrm{NaOH},

cHCl=nHClVHCl=nNaOHVHClc_HCl = \frac{n_HCl}{V_HCl} = \frac{n_NaOH}{V_HCl} cHCl=0.002713617 mol0.025 Lc_HCl = \frac{0.002713617 \, mol}{0.025 \, L} cHCl=0.0027136170.025=0.10854468 mol/Lc_HCl = \frac{0.002713617}{0.025} = 0.10854468 \, mol/L

Considering significant figures (4 significant figures),

cHCl=0.1085 mol/Lc_HCl = 0.1085 \, mol/L

0.1085 mol/L

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Quick Answer

Calculate the average volume of NaOH used in the two titrations. V_médio\ NaOH = (V_1 + V_2)/(2) V_médio\ NaOH = 24.78 \, mL + 25.80 \, mL2 = 50.58 \, mL2 = 25.29 \, mL Step 2: Convert volumes to liters.

How can one control the enzyme experiment?
Biology

This biology question covers important biological concepts and processes. The step-by-step explanation below helps you understand the underlying mechanisms and reasoning.

ScanSolve AI Answer
Step 1: Calculate the average volume of NaOH used in the two titrations. V_médio\ NaOH = (V_1 + V_2)/(2) V_médio\ NaOH = 24.78 \, mL + 25.80 \, mL2 = 50.58 \, mL2 = 25.29 \, mL Step 2: Convert volumes to liters. V_NaOH = 25.29 \, mL = 25.29 × 10^-3 \, L = 0.02529 \, L V_HCl = 25 \, mL = 25 × 10^-3 \, L = 0.025 \, L Step 3: Calculate the number of moles of NaOH used at the equivalence point (1:1 stoichiometry with HCl). n_NaOH = c_NaOH × V_NaOH n_NaOH = 0.1073 \, mol/L × 0.02529 \, L First, 0.1073 × 0.02529 = 0.1073 × (0.025 + 0.00029) = (0.1073 × 0.025) + (0.1073 × 0.00029) 0.1073 × 0.025 = 0.0026825 0.1073 × 0.00029 = 0.000031117 n_NaOH = 0.0026825 + 0.000031117 = 0.002713617 \, mol Step 4: Since n_HCl = n_NaOH, c_HCl = n_HClV_HCl = n_NaOHV_HCl c_HCl = 0.002713617 \, mol0.025 \, L c_HCl = (0.002713617)/(0.025) = 0.10854468 \, mol/L Considering significant figures (4 significant figures), c_HCl = 0.1085 \, mol/L 0.1085 mol/L