Equation of trendline 1 calculation: m= rise/run = 12/15 = 0.8 c= 5.9 y= 0.8x + 5.9
Equation of trendline 2 calculation: m= rise/run = 25/45 = 0.5 c= 17.6 y= 0.5x +17.6
Point of intersection:
Equate both trendlines to find point of intersection
0.8x + 5.9 = 0.5x + 17.6 x= 39
Since x=39, x2= 9 as x2 represents x in descending order.
R2:
R2 for trendline 1 = 0.98
Correlation coefficient for trendline 1= 0.96
R2 for trendline 2 = 0.98
Correlation coefficient for trendline 2= -0.99
Ratio calculation from graph:
39.0 mL (± 0.3 mL) / 9.0 mL (± 0.3 mL)
= 4.3 (± 4.1%)
= 4.3 (± 0.2)
Hence, the ratio is 39:9 or 4.3 (± 0.2)
Qualitative and processed data from graph:
The ratio calculated from the graph is the volume of bleach to the volume of unknown. There are two trend lines present due to the two quantities present and the varying amounts of each.
The point of intersection of the two trend lines determines the ratio of volumes (bleach to unknown).
From the graph, the ratio is 39:9 or 4.3 (± 0.2). This matches with the qualitative data, as the solution was warm when settled after two minutes, was bubbling, and released a strong odour after one minute.
The X (two X axes) and Y axis were chosen as Volume of bleach and volume of unknown vs Temperature of solution. There are two X axes as the volumes of bleach and the unknown solvent so that the ratio of the reactants from the actual reaction can be identified accurately. This distinguishes the varying quantities of Bleach and the unknown solution. Percent error calculations: Percent error is calculated to measure the precision of the data collected from the experiment. % error for Volume of bleach = {|Experimental yield - theoretical yield|/ theoretical yield} x 100 = { |39 mL (± 0.5 mL) - 40 mL (± 0.5 mL) | / 40 mL (± 0.5 mL) } x 100 = {1.0 mL (± 1.0 mL)/ 40 mL ( ± 0.5 mL)} x 100 = {1 ml (± 100%)/ 40 mL (± 1.25 %) } x 100 = 2.5% (± 2.5 %) % error for volume of Sodium Hydroxide with Sodium Thiosulfate = {|Experimental yield - theoretical yield|/ theoretical yield} x 100 = { |9 mL (± 0.5 mL) - 10 mL (± 0.3 mL) | / 10 mL (± 0.3 mL) } x 100 = {1 mL ( ± 0.8 mL) / 10 mL (± 0.3 mL) } x 100 = {1 ml (± 60 %) / 10 ml (± 3%)} x 100 = 10.0 % (± 8.0 %) % error for ratio = {|Experimental yield - theoretical yield|/ theoretical yield} x 100 = [{ 4.3 (± 0.2) - 4.0 }/4.0] x 100 = 7.5 % (± 5.0%) Reaction rates: Reaction rate of products (Bleach and unknown) = Concentration/time (in seconds) = - (0.50 M/120 s) = - 0.00416 M/s The value is negative as the amount of each solution is disappearing in the redox reaction. (Moderate rate of reaction). The final ratio is 4.3 (± 0.2), Since the ratio from the graph seems to be close to 4:1, the solution is Sodium Hydroxide with Sodium Thiosulfate. Evaluation: Molar Ratio Lab The purpose of the lab was to determine if mole ratios in a reaction can be calculated if one of the reactants is unknown. …show more content…
To identify the correct ratio of reaction, the unknown solution and bleach were mixed in varying amounts. Then, the temperatures were recorded for each trial done. Since the amount before the reaction (reactants) is more than the amount after the reaction, the reaction is exothermic as it releases heat. This explains the increase in temperature from 21.2 (°C ± 0.1 °C ) to other respective temperatures. The volume of the solutions are supposed to be constant for one to calculate molar ratios. Hence, a concentration of 0.50 M is kept the same for each reactant. The temperature change is therefore proportional to the amount of reactants that are consumed. The formula for bleach is NaClO. The unknown is likely to be a solution of Sodium Hydroxide with Sodium Thiosulfate, Potassium iodide or Sodium Sulfite. Bleach is an oxidising agent, and the hypochlorite ion will be reduced as the unknown while the unknown is oxidised. Since the ratio from the graph seems to be close to 4:1, the solution is
Equation of trendline 2 calculation: m= rise/run = 25/45 = 0.5 c= 17.6 y= 0.5x +17.6
Point of intersection:
Equate both trendlines to find point of intersection
0.8x + 5.9 = 0.5x + 17.6 x= 39
Since x=39, x2= 9 as x2 represents x in descending order.
R2:
R2 for trendline 1 = 0.98
Correlation coefficient for trendline 1= 0.96
R2 for trendline 2 = 0.98
Correlation coefficient for trendline 2= -0.99
Ratio calculation from graph:
39.0 mL (± 0.3 mL) / 9.0 mL (± 0.3 mL)
= 4.3 (± 4.1%)
= 4.3 (± 0.2)
Hence, the ratio is 39:9 or 4.3 (± 0.2)
Qualitative and processed data from graph:
The ratio calculated from the graph is the volume of bleach to the volume of unknown. There are two trend lines present due to the two quantities present and the varying amounts of each.
The point of intersection of the two trend lines determines the ratio of volumes (bleach to unknown).
From the graph, the ratio is 39:9 or 4.3 (± 0.2). This matches with the qualitative data, as the solution was warm when settled after two minutes, was bubbling, and released a strong odour after one minute.
The X (two X axes) and Y axis were chosen as Volume of bleach and volume of unknown vs Temperature of solution. There are two X axes as the volumes of bleach and the unknown solvent so that the ratio of the reactants from the actual reaction can be identified accurately. This distinguishes the varying quantities of Bleach and the unknown solution. Percent error calculations: Percent error is calculated to measure the precision of the data collected from the experiment. % error for Volume of bleach = {|Experimental yield - theoretical yield|/ theoretical yield} x 100 = { |39 mL (± 0.5 mL) - 40 mL (± 0.5 mL) | / 40 mL (± 0.5 mL) } x 100 = {1.0 mL (± 1.0 mL)/ 40 mL ( ± 0.5 mL)} x 100 = {1 ml (± 100%)/ 40 mL (± 1.25 %) } x 100 = 2.5% (± 2.5 %) % error for volume of Sodium Hydroxide with Sodium Thiosulfate = {|Experimental yield - theoretical yield|/ theoretical yield} x 100 = { |9 mL (± 0.5 mL) - 10 mL (± 0.3 mL) | / 10 mL (± 0.3 mL) } x 100 = {1 mL ( ± 0.8 mL) / 10 mL (± 0.3 mL) } x 100 = {1 ml (± 60 %) / 10 ml (± 3%)} x 100 = 10.0 % (± 8.0 %) % error for ratio = {|Experimental yield - theoretical yield|/ theoretical yield} x 100 = [{ 4.3 (± 0.2) - 4.0 }/4.0] x 100 = 7.5 % (± 5.0%) Reaction rates: Reaction rate of products (Bleach and unknown) = Concentration/time (in seconds) = - (0.50 M/120 s) = - 0.00416 M/s The value is negative as the amount of each solution is disappearing in the redox reaction. (Moderate rate of reaction). The final ratio is 4.3 (± 0.2), Since the ratio from the graph seems to be close to 4:1, the solution is Sodium Hydroxide with Sodium Thiosulfate. Evaluation: Molar Ratio Lab The purpose of the lab was to determine if mole ratios in a reaction can be calculated if one of the reactants is unknown. …show more content…
To identify the correct ratio of reaction, the unknown solution and bleach were mixed in varying amounts. Then, the temperatures were recorded for each trial done. Since the amount before the reaction (reactants) is more than the amount after the reaction, the reaction is exothermic as it releases heat. This explains the increase in temperature from 21.2 (°C ± 0.1 °C ) to other respective temperatures. The volume of the solutions are supposed to be constant for one to calculate molar ratios. Hence, a concentration of 0.50 M is kept the same for each reactant. The temperature change is therefore proportional to the amount of reactants that are consumed. The formula for bleach is NaClO. The unknown is likely to be a solution of Sodium Hydroxide with Sodium Thiosulfate, Potassium iodide or Sodium Sulfite. Bleach is an oxidising agent, and the hypochlorite ion will be reduced as the unknown while the unknown is oxidised. Since the ratio from the graph seems to be close to 4:1, the solution is