Given that the volume of one drop is approximately 0.05 mL, compare the number of moles...
Given that the volume of one drop is approximately 0.05 mL, compare the number of moles for each ligand when one drop of 2.5 M KSCN is added to a test tube and 1.0 mL of 0.125 M EDTA added.
Enter the number of moles of each ligand. Support your answers with appropriate calculations.
Moles SCN− :
Moles EDTA:
Homework Answers
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Given that the volume of one drop is approximately 0.05 mL,
compare the number of moles...
One drop of 2.5 M KSCN solution is added to 0.5 mL of 0.10 M Fe(NO3)3. After the solution is mixed, 1.0 mL of 0.125 M ED...
One drop of 2.5 M KSCN solution is added to 0.5 mL of 0.10 M Fe(NO3)3. After the solution is mixed, 1.0 mL of 0.125 M EDTA solution is added. Enter the chemical formula of the major iron complex in solution after the addition of EDTA
A drop of water has a volume of approximately 7x10-2 mL Exercise 3.110 A drop of...
A drop of water has a volume of approximately 7x10-2 mL Exercise 3.110 A drop of water has a volume of approximately 7x10-2 mL Part A How many water molecules does it contain? The density of water is 1.0 g/cm Express your answer using one significant figure. AEDO ? molecules Submit Request Answer Provide Feedback
One drop has a volume of about 0.05 mL, thus the concentration of Cu2+ in the...
One drop has a volume of about 0.05 mL, thus the concentration of Cu2+ in the dilute half-cell is about (0.05 mL/25 mL)(0.1 M)=2x10-4 M. At 298K use this equation: E= -(0.0592/n)xlog(An+ Mdil/ An+ Mconc) to calculate it.
A student made solution #3 using the experimental method in this lab, and measured an absorbance of 0.559. The starting...
A student made solution #3 using the experimental method in this
lab, and measured an absorbance of 0.559. The starting reagents are
2.00 x 10-3 M Fe(NO3)3 and 2.00 x
10-3 M KSCN.
The amount of absorption is proportional to the concentration
of
FeSCN2+. This relationship – true for many solutions – is called
“Beer’s Law”, and has the simple equation:
A = bc
where “A” is the absorption, “b” is 5174.6 for FeSCN2+ and “c”
is molarity
Make Five...
Table A. Preparation of Standard solutions of FeSCN2+ 1.0 M HNO3 0.002 M 0.200 M Solution...
Table A. Preparation of Standard solutions of FeSCN2+ 1.0 M HNO3 0.002 M 0.200 M Solution KSCN (mL) Fe(NO3)3 (mL) 0.5 5 [FeSCN2+] (mol/L)* 1 4.0x10^-5 Add 1.0 M 2 1.0 5 8.0x10^-5 HNO3 3 1.5 5 1.2x10^-10 4 2.0 5 1.6x10-4 to each to adjust the volume to 25 mL. 5 2.5 5 2.0x10-4 * Calculate the concentrations of FeSCN2+ in each beaker, assuming that all SCN-ions exist as FeSCN2+. In other words, [FeSCN2+] (in Soln 1) = [SCN-]...
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...
Q&A CH2630 LB (5 points) Name: Given the information in Table 1 below, calculate the concentrations...
Q&A CH2630 LB (5 points) Name: Given the information in Table 1 below, calculate the concentrations of each species to complete Table 2 Iron(III) nitrate and KSCN are initial concentration values, while the concentration of FeSCN is after the reaction is complete. As the concentration of Felis significantly larger than that of SCN, the reaction conditions depicted are non-equilibrium. Since there is a large excess of one reactant, the concentration of FeSCN2+ is dependent upon the limiting reactant. This is...
Calculate the initial concentration of Fe3+ in test tubes 1-4. (you will need to take the...
Calculate the initial concentration of Fe3+ in test tubes 1-4. (you will need to take the dilution in to account) Calculate the initial concentration of SCN-in each of test tubes 1-4. (you will need to take the dilution in to account) 2. Label four 20 x 150 mm test tubes 1-4. Pour about 30 mL of 0.0020 M Fe(NO3)3 into a clean, dry 100-ml beaker. Pipet 5.0 mL of this solution into each of the four labeled test tubes. Use...
Number 3 contains 0.0050 moles of 2. You are given two solutions. Solution X has a...
Number 3
contains 0.0050 moles of 2. You are given two solutions. Solution X has a volume of 100. ml and imidazole and 0.0030 moles of imidazolium chloride. Solution Y has a volume of 100. ml and contains 0.050 moles of imidazole and 0.030 moles of imidazolium chloride. What are the pH values of solutions X and ? 3. Calculate the properties of solution X as you titrate it with a 0.100 M solution of HCI: moles Volume of added...
Data and Results Data Sheet 1-Titration Trial 1 Saturated potassium hydrogen tartrate solution volume 25.00 mL...
Data and Results Data Sheet 1-Titration Trial 1 Saturated potassium hydrogen tartrate solution volume 25.00 mL NaOH Solution Concentration 0.0167 mol L mL Initial Volume NaOH 0.10 mL Final Volume NaOH 29.69 mL Volume of NaOH added Volume of NaOH added Moles of NaOH added mol Moles of HC,H,O, mol Moles of K* Saturated potassium hydrogen tartrate solution volume Concentration of HC,H,O, mol mol Concentration of K* Kap Enter Data Sheet I Calculations Here 2/11 Calculations Perform the following calculations...
A drop of water has a volume of approximately 7x10-2 mL Exercise 3.110 A drop of water has a volume of approximately 7x10-2 mL Part A How many water molecules does it contain? The density of water is 1.0 g/cm Express your answer using one significant figure. AEDO ? molecules Submit Request Answer Provide Feedback
A student made solution #3 using the experimental method in this
lab, and measured an absorbance of 0.559. The starting reagents are
2.00 x 10-3 M Fe(NO3)3 and 2.00 x
10-3 M KSCN.
The amount of absorption is proportional to the concentration
of
FeSCN2+. This relationship – true for many solutions – is called
“Beer’s Law”, and has the simple equation:
A = bc
where “A” is the absorption, “b” is 5174.6 for FeSCN2+ and “c”
is molarity
Make Five...
Table A. Preparation of Standard solutions of FeSCN2+ 1.0 M HNO3 0.002 M 0.200 M Solution KSCN (mL) Fe(NO3)3 (mL) 0.5 5 [FeSCN2+] (mol/L)* 1 4.0x10^-5 Add 1.0 M 2 1.0 5 8.0x10^-5 HNO3 3 1.5 5 1.2x10^-10 4 2.0 5 1.6x10-4 to each to adjust the volume to 25 mL. 5 2.5 5 2.0x10-4 * Calculate the concentrations of FeSCN2+ in each beaker, assuming that all SCN-ions exist as FeSCN2+. In other words, [FeSCN2+] (in Soln 1) = [SCN-]...
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...
Q&A CH2630 LB (5 points) Name: Given the information in Table 1 below, calculate the concentrations of each species to complete Table 2 Iron(III) nitrate and KSCN are initial concentration values, while the concentration of FeSCN is after the reaction is complete. As the concentration of Felis significantly larger than that of SCN, the reaction conditions depicted are non-equilibrium. Since there is a large excess of one reactant, the concentration of FeSCN2+ is dependent upon the limiting reactant. This is...
Calculate the initial concentration of Fe3+ in test tubes 1-4. (you will need to take the dilution in to account) Calculate the initial concentration of SCN-in each of test tubes 1-4. (you will need to take the dilution in to account) 2. Label four 20 x 150 mm test tubes 1-4. Pour about 30 mL of 0.0020 M Fe(NO3)3 into a clean, dry 100-ml beaker. Pipet 5.0 mL of this solution into each of the four labeled test tubes. Use...
Number 3
contains 0.0050 moles of 2. You are given two solutions. Solution X has a volume of 100. ml and imidazole and 0.0030 moles of imidazolium chloride. Solution Y has a volume of 100. ml and contains 0.050 moles of imidazole and 0.030 moles of imidazolium chloride. What are the pH values of solutions X and ? 3. Calculate the properties of solution X as you titrate it with a 0.100 M solution of HCI: moles Volume of added...
Data and Results Data Sheet 1-Titration Trial 1 Saturated potassium hydrogen tartrate solution volume 25.00 mL NaOH Solution Concentration 0.0167 mol L mL Initial Volume NaOH 0.10 mL Final Volume NaOH 29.69 mL Volume of NaOH added Volume of NaOH added Moles of NaOH added mol Moles of HC,H,O, mol Moles of K* Saturated potassium hydrogen tartrate solution volume Concentration of HC,H,O, mol mol Concentration of K* Kap Enter Data Sheet I Calculations Here 2/11 Calculations Perform the following calculations...
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