Tutor profile: Trent R.
Subject: Basic Chemistry
Part 1: Balance the chemical equation. _____ P4 + _____ O2 -> ______ P2O3 Part 2: If 5.3 grams of P4 is used for the reaction with oxygen, how much P2O3 is produced in mols. Part 3: Given the previous answer, find out how many molecules of P2O3 has been produced.
Part 1: In order to balance the equation, a great place to start is to create a chart and follow that as our roadmap in order to correctly balance the equation. We know then, that we are given the following: _____ P4 + _____ O2 -> ______ P2O3 Now let's begin categorizing what we have so that we can start massaging the coefficients to get the required balanced equation. _____ P4 + _____ O2 -> ______ P2O3 4 P 2O 2 P 3 O What we just did was categorize the individual atoms in the reaction. That is we have 4 phosphorus and 2 oxygens on the reactants side and 2 phosphorus and 3 oxygens on the reactant side. From this, we can see that by adding a 4 in front of P2O3 we would get 8 phosphorus and 12 oxygen. The reason for this is that we have to look a head and think like a scientist - what can we use that will make it so that we have even multiples of 2 so that we can simply multiply by a whole number coefficient on the reactants side. _____ P4 + _____ O2 -> 3P2O3 4 P 2O 8 P 12 O This is a great start and a general role of thumb is to being with the products. Now in order to match the oxygen and phosphorus on the reactant sides, we have to use a coefficient that will give us 8 phosphorus and 12 oxygen. 2 and 6 will appropriate coefficients as we will get our desired values. The balanced equation then looks like the following: 2P4 + 6O2 -> 3P2O3 Part 2: If 5.3 grams of P4 is used for the reaction with oxygen, how much P2O3 is produced in mols. A good first step when working with stoichiometric conversions and dimensional analysis is to recognize that we will be dealing with molar masses. Whether we need them or not, it is good practice to calculate the molar masses of every term in the chemical reaction. Using the periodic table, we get: P4 : 4*(30.974) = 123.896 g/mol O2: 2*(15.999) = 31.998 g/mol P2O3: 2*(30.974) + 3*(15.999) = 92.946 g/mol Now that we have found the molar masses of P4, O2 and 3P2O3, we can set up the problem. Though not necessary, we can deduce by our desired answer how the problem will progress. 5.3 grams P4( / )( / ) = ____________ mols The first step, almost always is to convert to mols. We have to ask ourselves, do we have the necessary information to accomplish this? The answer is yes! We do. The molar masse of P4 we first found allows us to convert grams to mols. So let's do just that. 5.3 g P4(1 mol P4 / 123.896 g P4)( / ) = ____________ mols P2O3 From the first step to the above calculation, we see that g P4 cancels with g P4 leaving us moles! Now, let's ask ourselves, how do we get from mols of P4 to mols of P2O3. We use the balanced chemical reaction! We see from the reaction that the following relationship holds true: 2 mol P4 = 3 mol P2O3 In order to cancel mols of P4, we would write the following: 5.3 g P4(1 mol P4 / 123.896 g P4)(3 mol P2O3 / 2 mol P4) by plugging these values into our calculators we get = 0.64 mols P2O3 Taking significant figures into consideration we can conclude that we will need 2 significant figures. Note: Chemistry is a wonderful subject because it is incredibly linear. Almost any chemistry problem can be checked if it is right when doing it by looking at your units. Did your units work out? If they do not, you are guaranteed not to have the correct answer. This can be used to help ease any exam anxiety if you know there is a good chance you got the problem correct before you even turn it in! Part 3: Given the previous answer, find out how many molecules of P2O3 has been produced. In order to find the number of molecules, we have to pull from our knowledge about Avogadro's number. Some relationships we can write are: 6.022E23 atom 6.022E23 molecule = > 1 mol 6.022E23 formula unit Because we are working with molar relationships, this is the only information we need in order to finish this problem. Proceeding ahead, we get: 0.64 mols P2O3 (6.022E23 molecule P2O3/ 1 mol P2O3) = 3.9 E 23 molecules of P2O3 Note! While working through chemistry problems, it is important to constantly compile any significant core ideas. For instance, let us call: 6.022E23 atom 6.022E23 molecule = > 1 mol 6.022E23 formula unit Tool Box B and our knowledge about molar mass in the second part of this problem as Tool Box A. Problems dealing with stoichiometry and dimensional analysis can be reduced to a single fundamental idea that they will ALWAYS pull at least one tool from Tool Box A and Tool Box B but NO MORE than two (one from each box). This allows us to condense any problem of this nature into an easy to remember rule to use as the beginning of our road map. Instead of needing to learn every single variation of a problem of this nature, we can just internalize a small handful of fundamental ideas that allow us to simply solve any problem that is given to us. This makes studying for exams incredibly easy and efficient and most of all makes tests far less stressful if you have every tool you need in your toolbox of knowledge.
Preform the matrix operation 5A - 2B on the following matrices: A = [ 7 3 -4 -1] and B = [ 9 6 3 10]
It is first important to remember how we distribute a coefficient to a matrix. Fortunately for us, this task is rather simple and only involves distributing the coefficient to every element in the matrix. The first step, then, in our solution of the previous problem is to begin the problem by writing out a roadmap to help guide us piece by piece through the problem. Roadmap: 5A - 2B = 5 [ 7 3 - 2 [ 9 6 -4 -1 ] 3 10 ] Now, let us first distribute each coefficient to the respective matrix separately: 5 [ 7 3 = [ 5(7) 5(3) -4 -1 ] 5(-4) 5(-1) ] = [ 35 15 -20 -5 ] 2 [ 9 6 = [ 2(9) 2(6) 3 10] 2(3) 2(10) ] = [ 18 12 6 20 ] With the first step of distribution, we can begin subtracting these new matrices that we just found. Matrix addition and subtraction, is just as simple as distributing each matrices' respective coefficient. We simply subtract each respective element. This is done by the following: 5A - 2B = 5 [ 7 3 - 2 [ 9 6 -4 -1 ] 3 10 ] = [ 5(7) 5(3) - [ 2(9) 2(6) 5(-4) 5(-1) ] 2(3) 2(10)] = [ 35 15 - [ 18 12 -20 -5 ] 6 20 ] = [ 35 - 18 15 - 12 -20 - 6 -5 -20] Then by simplifying we get: = [ 17 3 -26 -25 ]
Find the area of a rectangular box with the dimensions of: Width: (x-5) Height: (x-7) Reminder: The formula to find the area of a rectangular box is Area = (Width)*(Height)
In order to find the area of a box with the given dimensions, let us first remind ourselves what foil stands for. First Outer Inner Last Now we can proceed to fill in what we can to the area of a box formula given above. Area = (Width)*(Height) = (x-5)*(x-7). By using the foil method, we can begin to expand these binomials in order to find the desired area of the box. Area = (x-5)*(x-7) F O I L = (x)x) + (x)(7) + (-5)(x) + (-5)(7) = x^2 + 7x - 5x -35 = x^2 - 2x -35 Lastly, because this is an area, we must attach a unit to our answer. Because a unit is not given, we can write our final answer as: Area of the box is found to be x^2 - 2x - 35 units^2.
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