In any case it is always critical to recognize that unit factoring is nothing

In any case it is always critical to recognize that

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In any case it is always critical to recognize that unit factoring is nothing else than the application of the rule of proportionality and that it should never been used without the understanding of each step. If not convinced, try to calculate the number of molecules of insulin in 1 ml of a 2.5 ppb solution. Insulin is a protein consisting of 21 amino acids. Its molar mass is 5508. g/mol. There are 273 billions molecules in 1 ml of 2.5 ppb insulin. Rule of three in chemistry Amongst the many conversion done by chemists is the conversion between moles and grams. This is a cornerstone of chemistry because the mole links the macroscopic to the microscopic. Chemistry deals with atoms and molecules which are entities too small to be directly measured. Therefore an amount of entities, the mole, was defined to represent the standard amount. A mole contains 6.022… *1023 entities. As we will see later the mass of the mole depends on the entity. A mole of lead is heavier than a mole of copper because the atom of lead is heavier than the atom of copper. This is a bit like a carton of 12 large eggs is heavier than a carton of 12 small eggs. It is often necessary to figure out how many moles are in a given mass. This is easy as long as we know the mass of a mole. To answer such a question we can follow several approaches. The easiest and most versatile is the mathematical approach. The other approaches as some time used. We will discuss these approach with a simple example: how many moles of water are in 1.0 L of pure water. We know that 1.0 L of water has a mass of 1000.0 g because 1.0 L equals 1000.0 ml and the density of pure water is 1.0 g/1 ml. We also know that the molar mass of water, some time called molecular weight, but actually the average molar mass, is 18.0 g/mol. The mathematical approach If we have pure water the ratio mass/volume is the same regardless of the amount. We can write an equality of two proportions: 18.0 g/1 mol = 1000.0 g/x mol and find the value of x x mol = (1000.0 g/18.0g) *1 mol
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The logic approach 18.0 g are in 1 mole 1.0 g is in 1 mol/18 1000.0 g are in (1 mol/18.0 )1000.0 The technical approach Going from gram to mole cancelling units 1000.0 g | 1 mol = | 18.0g All three approaches are obviously equivalent and can be used. However if you use the technical approach you must understand that you are actually writing an equality of proportion. Such an understanding is indispensable for the solution of real life problems. We will mostly use the mathematical approach in this course. Going from gram to mole cancelling units
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7- How to handle problems: A common myth is to believe that Science is about finding (or memorizing) formulas in which data will be plugged in. Such a one step linear approach is totally inadequate for real life situations. We will need to adopt a methodology based on analysis and deduction. 7-a Students are advised to adopt a standard procedure based on four activities: Sort Indicate clearly what needs to be found versus what is being given. Make sure that you have the correct units Strategize The most important and most difficult activity. A strategy consists in the series of steps leading to the solution. Real life problem cannot be answered with a single formula in
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