CalculationsGeneral Formulass=√∑x−xn−1; CV (relative standard deviation)=sx ; %error=|o−e|e×100 %s= sample standard deviation; x= sample mean; o=observed value; e= expected value1.29.5740+29.5739+29.5737+29.5742 +29.57385=29.5739 g; s=√∑x−xn−1;2.v (mL)=m(g)d (gmL) T 1+T 2+T 3+T 44=xof volumetrial weight of 5mL water Volume delivered1 4.9162 4.9271703452 4.951 4.9620483 4.9326 4.9436069414 4.9427 4.953729479avg 4.946638691sd 0.015010375trial weight of 50mL beaker1 29.5742 29.57393 29.57374 29.57425 29.5738avg 29.57392range 0.0005sd 0.000192354rsd 6.50417E-06rsd 0.00303446Percent error 1.07%trial weight of 10mL water Volume delivered1 9.9564 9.9786173922 9.9606 9.9828267643 9.9457 9.9678935154 9.9557 9.97791583avg 9.976813375sd 0.006329682rsd 0.000634439Percent error 0.23%3. trial Mass of 500 microL water1 0.49982 0.50683 0.50374 0.49885 0.4894avg 0.4997sd 0.006582553rsd 0.013173009% error 0.06%trial Mass of 1 mL water1 0.99972 0.98233 0.9834 1.0025 0.9955avg 0.9925sd 0.009292201rsd 0.009362419% error 0.75%4. Almost all of the pipettes performed better than the values provided in the table, except for the 1 mL micropipette. This can be attributed to improper technique used in the lab, which would lead to a systematic error. Questions1. Systematic error comes because of a flaw from equipment or design of experiment, so if the experiment is replicated the error is reproducible. For example, a poorly calibrated instrument such as a thermometer that reads 102 oC when immersed in boiling water and 2 oC when immersed in ice water at atmospheric pressure. Such a thermometer would result in measured values that are consistently too high. Random error occurs because of uncontrollable variables in the measurements such as the task of reading a scale. For example, different people will report there subjective reading between the markingswhen reading the scale measurement. Improving technique can reduce these types of errors, but they will always be present.trial Mass of 50 microL water1 0.05052 0.05033 0.05014 0.05045 0.0508avg 0.05042sd 0.000258844rsd 0.005133748% error 0.84%trial Mass of 100 microL water1 0.0992 0.10063 0.0994 0.09965 0.1002avg 0.09968sd 0.000715542rsd 0.007178388% error 0.32%2. It is advisable to report the median instead of the mean when outliers are present in the sample that can skew the mean. The median is resistant to outliers, so it will not be affected in the manner the meanis.3. TC glassware refers to glassware used to Contain liquid like volumetric flasks and graduated cylinders. TD glassware refers to glassware that is used to deliver liquid, such as pipettes and burets.4. I would propose to calibrate the volumetric flask to contain. I would propose to calibrate by measuring the mass of water filled in a flask and then using the density of water at that particular temperature to calculate volume. Dry the flask (with acetone and vacuum; not in the oven) and weigh it with its top. Then fill the flask to the line with water at the desired temperature and reweigh. A large volumetric flask probably will not fit onto the pan of an analytical balance. In such a case, a good top-loading balance may be used, because a tenth gram uncertainty is acceptable when the weight of water is 100 g or more.5. A volumetric pipette is used to transfer a desired volume of solution from one container to another. AMohr pipette is graduated for delivery from the zero mark to the final graduation mark. 6-7. One of the most important rules of weighing is that the sample must never come in direct contact with the balance pan. The best way to keep the balance pan clean is to use the method of weighting by difference. The weighing bottle plus dried sample is weighed; then, the required weight of sample is removed to a suitable receiver, the weighing bottle and contents are then reweighed. The weight of the sample in the receiver is simply the difference in the two weights.The sample can be removed from the weighing bottle by either of two methods. If the solid is crystalline, gentle tapping of the weighing bottle with its cap (Figure 5-3) will remove a controlled number of crystals at a time. To prevent fingerprints from changing the weight of the weighing bottle, it is usually handled with a paper loop. With powdery solids it is usually easier to remove the sample with a small spatula. With this method, make sure that no sample spills from the spatula and that the spatula is rinsed with deionized waterbefore use.In an experimental procedure,a typical weighing instruction might be: “Accurately weigh about 0.4 grams of sample”, where the actual number of grams will vary according to the experiment. This statement means that a sample of about 0.4 grams (usually within10%) is to be weighed to the nearest tenth of a milligram. A notebook entry forthis weighing procedure might look as follows:Initial weight21.9683 gFinal weight21.5537 gWeight of sample0.4146 gNote that the actual weights are recorded, not just the difference. If, by accident, too much sample is removed, that sample must be discarded and not returned to the weighing bottle.(Of course, in the case of very expensive or otherwise irreplaceable substances, a discarded sample would be saved and redried.)The method of weighing by difference may at first seem cumbersome but with practice, samples can be weighed efficiently. This method is used because:1.The balance pan is kept clean, because only the weighing bottle touches it.2.The sample is kept clean, because it remains in a capped weighing bottle as much as possible.3.The exact zero of the balance is not critical, because only the difference in weights is used.4.The sample can be weighed into a wet receiver.5.Time is saved when weighing several samples, because the final weight for sample 1 becomesthe initial weight for sample 2, etc. If samples were weighed on paper, two weighings would be required for each sample. (Found in Canvas Files)8.Average = 0.1283 MStandard deviation = 0.0009(See equations above)G1 = 0.5556 Keep data pointG2 = 0.1111 Keep data pointG3 = 0.8889 Keep data pointG4 = 1.444 Keep data point9. La2O3 + 6HNO3 → 2La(NO3)3 + 3H2OMoles La2O3 -Molar mass: 325.81g/mol-0.1968g * (1mol / 325.81g) = 6.04×10⁻⁴ moles La2O3Moles La(NO3)3:6.04×10⁻⁴ moles La2O3 * (2mol La(NO3)3 / 1mol La2O3) = 1.208×10⁻³ moles