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Normality can be used for acid-base titrations. For example, sulfuric acid (H 2 SO 4) is a diprotic acid. Since only 0.5 mol of H 2 SO 4 are needed to neutralize 1 mol of OH −, the equivalence factor is: f eq (H 2 SO 4) = 0.5. If the concentration of a sulfuric acid solution is c(H 2 SO 4) = 1 mol/L, then its normality is 2 N. It can also be ...
Lilliefors test is a normality test based on the Kolmogorov–Smirnov test.It is used to test the null hypothesis that data come from a normally distributed population, when the null hypothesis does not specify which normal distribution; i.e., it does not specify the expected value and variance of the distribution. [1]
Standard solutions are generally prepared by dissolving a solute of known mass into a solvent to a precise volume, or by diluting a solution of known concentration with more solvent. [1] A standard solution ideally has a high degree of purity and is stable enough that the concentration can be accurately measured after a long shelf time. [2]
The null hypothesis of this chi-squared test is homoscedasticity, and the alternative hypothesis would indicate heteroscedasticity. Since the Breusch–Pagan test is sensitive to departures from normality or small sample sizes, the Koenker–Bassett or 'generalized Breusch–Pagan' test is commonly used instead.
Simple back-of-the-envelope test takes the sample maximum and minimum and computes their z-score, or more properly t-statistic (number of sample standard deviations that a sample is above or below the sample mean), and compares it to the 68–95–99.7 rule: if one has a 3σ event (properly, a 3s event) and substantially fewer than 300 samples, or a 4s event and substantially fewer than 15,000 ...
The second part of the Winkler test reduces (acidifies) the solution. The precipitate will dissolve back into solution as the H + reacts with the O 2− and OH − to form water. MnO(OH) 2 (s) + 4 H + (aq) → Mn 4+ (aq) + 3 H 2 O(l) The acid facilitates the conversion by the brown, Manganese-containing precipitate of the Iodide ion into ...
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Newton's method is ideal to solve this problem because the first derivative of (), which is an integral of the normal standard distribution, is the normal standard distribution, and is readily available to use in the Newton's method solution.