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The reaction for the formation of the diamine-silver son is as follows: A® LÀ NHÀ ANH...
equilibrium help!! What is the equilibrium constant for the reaction CO(g) + 3 H2(g) - CH4(g)...
equilibrium help!!
What is the equilibrium constant for the reaction CO(g) + 3 H2(g) - CH4(g) + H2O(g) if at 20 °C the equilibrium molar concentrations are [CO] -0.613, [H2] = 1.839, (CH4) = 0.387, and [H20) - 0.387? Format Β Ι Ο The reaction for the formation of the diamine-silver ion is as follows: Ag (aq) + 2NH3(aq) + Ag(NH3)2(aq) a. Write the equilibrium constant expression for this reaction. for the above b. An experiment was carried out to...
Answer the following questions on this sheet or in your laboratory notebook. The reaction for the formation of diammin...
Answer the following questions on this sheet or in your laboratory notebook. The reaction for the formation of diamminesilver ion is as follows: Ag + (aq) + 2 NH3 (aq) Ag(NH3)2 + (2 points). 1. Write the equilibrium constant expression for the reaction 2. An experiment was carried out to determine the value of the equilibrium constant, Kc for the reaction. Total moles of Ag+ present= 3.6 x 10 -3 moles Total moles of NH3 present =...
For each equilibrium sample, calculate the initial concentrations (assuming no reaction occurred)...
For each equilibrium sample, calculate the initial
concentrations (assuming no reaction occurred) of SCN–
and Fe3+ based on the dilution factors used.
Enter the absorbance values for each sample.
For the standard solution, assume that, when equilibrium is
reached, the [FeSCN2+]eq is equal to
[SCN–]ini. Why is this a good
assumption?
For the equilibrium solutions, calculate the
[FeSCN2+]eq (Ceq) using Eq.
4.
Using Eq. 5 and 6, calculate
[Fe3+(aq)]eq and
[SCN–(aq)]eq for each equilibrium
solution.
Calculate values for the equilibrium...
4.81 mL of 0.00200 M Fe(NO3)3, 2.85 mL of 0.00200 M KSCN and 2.34 mL of...
4.81 mL of 0.00200 M Fe(NO3)3, 2.85 mL of 0.00200 M KSCN and 2.34 mL of distilled water were mixed. The resulting solution was allowed to attain equilibrium at 24 oC. The absorbance of the equilibrium solution was recorded, and the [FeSCN2+] was determined graphically to be 7.58 x 10-5 M. Part A: Calculate the number of moles of Fe3+ initially added to the solution. Part B: Calculate the number of moles of FeSCN2+ formed in the solution at equilibrium....
2. In Part A of this experiment, you prepare five FeSCN solutions (one that is just...
2. In Part A of this experiment, you prepare five FeSCN solutions (one that is just a blank) according to the reaction below Fe (aq) SCN (aq) > FeSCN (a) SCN formed We assume that the starting SCN determines the concentration of Fe (because Fe is in excess and SCN is limiting). Calculate the concentration ot FeSCN2 that forms for each of the solutions (Beakers 1-4) and fill out the table below. Show your calculations beneath the table. Concentration of...
A 25.0 mL volume of 0.0200 M Fe(NO3)3 is mixed with 50.0 mL of 0.00200 M...
A 25.0 mL volume of 0.0200 M Fe(NO3)3 is mixed with 50.0 mL of 0.00200 M NaSCN and 25.0 mL of 0.100 HNO3. The blood-red FeSCN2+ ion forms and the equilibrium is established: Fe3+(aq) + SCN-(aq) <---> FeSCN2+(aq) The equilibrium concentration of FeSCN2+ ([FeSCN2+]) was measured spectrophotometrically and found to be 7.0 x 10-4 mol/L. To calculate the equilibrium constant (Kc) for thr equilibrium system, proceed through the following steps: A. Moles of Fe3+, initial B. Moles of SCN-, initial...
Section Name Experiment 23 Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction...
Section Name Experiment 23 Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction 1. A student mixes 5.00 mL 2.00 X 10M Fe(NO), with 5.00 ml 2.00 x 10-M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN is 1.40 x 10M. Find K for the reaction Fe(aq) + SCN (aq) FeSCN2(aq). Step 1 Find the number of moles Fe and SCN initially present. (Use Eq. 3.) (5.00 x103 LX (300X163) = (x 103...
Name:. Date: Lab Section: 4. The blank solution used to calibrate the spectrophotometer is 5.0 mL...
Name:. Date: Lab Section: 4. The blank solution used to calibrate the spectrophotometer is 5.0 mL of 0.2 M Fe(NO)) diluted to 50 mt with H:O Why is this solution preferred to simply using de-ionized water for the calibration? Te celibutoneans ser He astanat os errors-free Te deionized woser alane sho no abtorpton 5. A reaction mixture of 4.00 mL. of 0.00200 M SCN and 5.00 mL of 0.00200 M Fe is uoreangth diluted to 10.00 mL with 0.1 M...
5.0 mL of 0.0020 M Fe(NO3)3 was mixed with 3.0 mL of 0.0020 M KSCN and...
5.0 mL of 0.0020 M Fe(NO3)3 was mixed with 3.0 mL of 0.0020 M KSCN and 2.0 mL of. The absorbance of this solution at 447 nm was measured as 0.35. A calibration curve was created using four standard solutions of Fe(SCN)2+. The equation for the best-fit line of [FeSCN2+] vs Abs was y = 5025x + 0.004. Using the data provided, calculate an equilibrium constant (Kc) for the formation of [Fe(SCN)2+]. For the above reaction, how were we able...
Data and Calculations: Determination of the Equilibrium Constant for a Chemical Reaction Method II Volume in...
Data and Calculations: Determination of the Equilibrium Constant for a Chemical Reaction Method II Volume in mL 2.00 x 103 M Fe(NO) Volume in mL, Depth in mm Volume in ml. 2.00 x 103 M Method I Mixture Unknówn KSCN Water Absorbance Standard FESCNP 4mL 1 5.00 x 10 M 1,00 .227 3mL 2 5,00 202 x 10 M 2,00 90 x 10 M .304 3 5,00 3.00 2mL 955 x 104 M I ImL 4 5.00 4,00 19x 10...
equilibrium help!!
What is the equilibrium constant for the reaction CO(g) + 3 H2(g) - CH4(g) + H2O(g) if at 20 °C the equilibrium molar concentrations are [CO] -0.613, [H2] = 1.839, (CH4) = 0.387, and [H20) - 0.387? Format Β Ι Ο The reaction for the formation of the diamine-silver ion is as follows: Ag (aq) + 2NH3(aq) + Ag(NH3)2(aq) a. Write the equilibrium constant expression for this reaction. for the above b. An experiment was carried out to...
For each equilibrium sample, calculate the initial
concentrations (assuming no reaction occurred) of SCN–
and Fe3+ based on the dilution factors used.
Enter the absorbance values for each sample.
For the standard solution, assume that, when equilibrium is
reached, the [FeSCN2+]eq is equal to
[SCN–]ini. Why is this a good
assumption?
For the equilibrium solutions, calculate the
[FeSCN2+]eq (Ceq) using Eq.
4.
Using Eq. 5 and 6, calculate
[Fe3+(aq)]eq and
[SCN–(aq)]eq for each equilibrium
solution.
Calculate values for the equilibrium...
2. In Part A of this experiment, you prepare five FeSCN solutions (one that is just a blank) according to the reaction below Fe (aq) SCN (aq) > FeSCN (a) SCN formed We assume that the starting SCN determines the concentration of Fe (because Fe is in excess and SCN is limiting). Calculate the concentration ot FeSCN2 that forms for each of the solutions (Beakers 1-4) and fill out the table below. Show your calculations beneath the table. Concentration of...
Section Name Experiment 23 Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction 1. A student mixes 5.00 mL 2.00 X 10M Fe(NO), with 5.00 ml 2.00 x 10-M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN is 1.40 x 10M. Find K for the reaction Fe(aq) + SCN (aq) FeSCN2(aq). Step 1 Find the number of moles Fe and SCN initially present. (Use Eq. 3.) (5.00 x103 LX (300X163) = (x 103...
Name:. Date: Lab Section: 4. The blank solution used to calibrate the spectrophotometer is 5.0 mL of 0.2 M Fe(NO)) diluted to 50 mt with H:O Why is this solution preferred to simply using de-ionized water for the calibration? Te celibutoneans ser He astanat os errors-free Te deionized woser alane sho no abtorpton 5. A reaction mixture of 4.00 mL. of 0.00200 M SCN and 5.00 mL of 0.00200 M Fe is uoreangth diluted to 10.00 mL with 0.1 M...
Data and Calculations: Determination of the Equilibrium Constant for a Chemical Reaction Method II Volume in mL 2.00 x 103 M Fe(NO) Volume in mL, Depth in mm Volume in ml. 2.00 x 103 M Method I Mixture Unknówn KSCN Water Absorbance Standard FESCNP 4mL 1 5.00 x 10 M 1,00 .227 3mL 2 5,00 202 x 10 M 2,00 90 x 10 M .304 3 5,00 3.00 2mL 955 x 104 M I ImL 4 5.00 4,00 19x 10...
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