Construct and calculate a theoretical titration curve, showing and discussing al
ID: 1066021 • Letter: C
Question
Construct and calculate a theoretical titration curve, showing and discussing all key areas, for the titration of 0.100 M NaOH with 50.0 mL 0.100 M HNO_2 (k_a HNO_2 = 7.1 times 10^-4) A 30.0 L of air sample (Density of 1.20 g/L) was passed through an absorption lower containing a solution of Cd^2+, where H_2S was retained as CdS. The mixture was acidified and treated with 10.0 mL of 0.01070 M l_2. After the reaction S^2-_ + I_2 = S(s) + 2I^- was complete, the excess iodine was titrated with 12.85 mL of 0.01344 M thiosulfate, S_2O_3^2-. Calculate the H_2S in ppm. Mol. Mass H_2S = 34 g/mol A 6.881-g sample containing MgCl_2 and NaCl was dissolved in sufficient water to give a 500-mL volume. Analysis of the chloride content in a 50-0 mL aliquot resulted in the formation of 0.5923-g of AgCl. The magnesium in a second 50-0 mL aliquot was precipitated as MgNH_4PO_4; on ignition 0.1796-g of Mg_2P_2O_7 was found. Calculate the percentages of MgCl_2.H_2O and NaCl in the original sample. Mol. Mass: MgP_2O_7 = 222.55 g/mol. NaCl = 58.44 g/mol, MgCl_2.6H_2O = 203.30 g/mol The homogeneity of a standard chloride solution was tested by analyzing portions of the material at the top and bottom with the following found Is homogeneity indicated at the 95% confidence level, use both t-test and f-testExplanation / Answer
Sample 1
n = 4
m = 26.355
s = 0.035118846
Sample 2
n = 3
m = 26.29
s = 0.045825757
"Unequal sample sizes, equal variance."
effective S for the two samples. That value will be
S = 0.00158
The S value is than plugged into the first formula to find the test value called t*. That value is
t* = 2.141048061 with 5 degrees of freedom
. A t table shows that the critical value for 95% confidence and 5 degrees of freedom is 2.5706.
Calculated t* value is less than the critical value,
Hence
there is not a significant difference in the means .
it can be concluded that the sample is homogeneous.