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The general calculation method is: C standard · V standard = C to be measured · v to be measured, of course, if the measurement coefficient is different, it should be proportional. So c to be tested = c standard · v standard v to be tested in this expression.
, the C standard must have specific data, and the V to be tested must be taken in advance and put into the Erlenmeyer flask.
There are also specific data, only the v standard is determined according to the titration process, there are accidents, and it will also be affected by many factors, so the analysis of the error of c to be measured, generally closely related to the v standard to analyze, especially pay attention to the fact that the two change trends are the same.
Then the indicator discoloration will stop the titration, it may be that the concentration of the local standard solution is large, so that the indicator will change color, once shaken, it will return to the original color, so it must be added in the process of continuous oscillation after the last drop of the standard solution, the indicator changes color and does not recover within half a minute, it is recorded as the end point of the titration. Therefore, if the titration is stopped as soon as the indicator changes color, the V standard will be smaller, and the C to be measured will also be smaller.
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The exact requirement is that the indicator changes color and does not return to its original color within half a minute before the titration endpoint is reached. If the titration is stopped when the color changes, it may be that the local neutralization is complete, but it changes back to the original color after shaking, and the neutralization is not completely completed, that is, the volume of the standard liquid is small, and the concentration of the liquid to be measured is calculated.
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Stopping immediately may cause temporary discoloration of the indicator due to excessive local concentrations. If this is the case, we calculate the volume and concentration without the original solution being neutralized, resulting in a lower calculated concentration than the theoretical concentration.
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The indicator changes color and does not return to its original color within half a minute, i.e., the titration endpoint is reached.
If the indicator is discolored, the titration will be stopped, which may be a local excess, and the volume of the standard solution added dropwise is small, according to the formula.
C (solution to be measured) ==C (standard solution) V (standard solution) V (solution to be tested) indicates that the measured concentration is low.
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Because different titration methods have different indicator discoloration principles for the end point of titration analysis, it will be indicator discoloration.
For example, the indicator of acid-base titration has two types (acid type and basic type), and each has different colors (acid type has acid color, basic type has basic color). When the indicator is added to the acid solution value, the acid color of the acid form of the indicator is presented, and since the solution becomes basic at the end of the titration, the indicator also changes to the basic color of the basic form.
Typically, the discoloration range of the indicator is selected, which is within the abrupt range of the titration curve. For example, the abrupt range of sodium hydroxide standard solution titration hydrochloric acid solution is pH = , then the phenolphthalein indicator can be selected by selecting the indicator, and its discoloration range is pH = , and the color of the indicator can be changed at the end of the titration (there is an acid type acid color "colorless", which changes to the basic color of the basic type, red. )
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Each has its own chemical principles, and it cannot be generalized.
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In order to measure the data more accurately.
As an example, the concentration of 100 ml of HCl solution is determined with standard NaOH in acid-base titration. Assuming that the initial pH of the solution to be tested is = 1, that is, the slowness of hydrochloric acid concentration is . 1 drop of water is approximately.
It is assumed that there are 3 concentrations of standard NaOH used, which are 10mol L and 1mol L. The final error in the titration is 1 drop of the standard solution (which will be reflected in the final calculation formula), then the error values of these three standard solutions are 5*10 (-3)mol l, 5*10 (-4)mol l, 5*10 (-5)mol l (error = volume of 1 drop of water * NaOH concentration volume of liquid to be measured). Obviously, the lower the concentration of the standard solution, the higher the accuracy.
However, due to the selection of a standard solution with too low concentration, the titration process will be very long and cumbersome, and the excessive amount of standard solution is not conducive to the selection of burettes, but also will bring about the slow accumulation error of Tiling (for example, in the above example, if the standard solution is used, it theoretically needs 100ml to be fully titrated, so that the volume of the original solution changes from 100ml to 200ml, and the dilution may bring about a decrease in pH (phenolphthalein requires discoloration under alkaline conditions), thus affecting the experimental results. Therefore, the standard solution of the appropriate concentration should be selected during the titration, which should not be too high or too low.
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(1) The reaction phenomenon is different: sulfur burns in the air and emits a light blue flame, while pure oxygen burns and emits a blue-purple flame.
2) The reaction rate is different: charcoal burns faster and more violently in pure oxygen than in air (3) The product is different: the product of charcoal combustion when oxygen is sufficient, and the product of combustion when oxygen is not sufficient is CO
Therefore, the answer is: the combustion phenomenon and rate of sulfur and charcoal in air and oxygen are different, and the combustion products of charcoal in sufficient oxygen and insufficient oxygen are different
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In this reaction, there is only the electron gain and loss of nitrogen element, the valency has risen and decreased, so C is wrong, the increase of nitrogen in metadimethylhydrazine is oxidized, it is a reducing agent, so a is wrong, the reaction is the recombination of atoms, so there is an endothermic and exothermic process, C is wrong, D is left, of course, there are eight electrons transferred from two N2O4 to N2 D is correct.