Tutor profile: Sara O.
A 15.25 gram sample of an organic compound was combusted in oxygen and produced 34.71 grams of carbon dioxide and 14.20 grams of water. The compound also contained 27.59% oxygen. What is the empirical formula of the compound?
In order to solve this problem you must first determine the number of grams of both carbon and hydrogen produced by the combustion of the unknown compound. To do this you convert the grams of carbon dioxide into moles of carbon dioxide and then into moles of carbon. 0.7889 moles of carbon is produced because there is one carbon atom in carbon dioxide. 1.5778 moles of hydrogen are produced because there are two hydrogen atoms per one water molecule. We know that the compound is 27.59% oxygen, therefore we can multiply the mass of the compound by .2759 to obtain the mass and then divide by the molar mass of oxygen to determine the moles of oxygen in the compound. This gives 0.263 moles of oxygen. We then divide each number by the smallest number of moles produced. This gives 3 moles of carbon, 6 moles of hydrogen and one mole of oxygen. The empirical formula is C3H6O.
f(x) = 4x^3 + 10x^2 + 2x Find the derivative and evaluate at x=2.
In order to take the derivative of the equation, the power rule must be used. This means that the exponent of each individual part of the equation is multiplied by the coefficient, and then the original numerical exponent is increased by one digit. f'(x) = 12x^4 + 20x^3 +2x^2 f'(2) = 12(2)^4 +20(2)^3 + 2(2)^2 f'(2) = 360
Use substitution to solve for each variable in the following system of equations: 2x + 17y = 104 14x + 2(6x + y) = 200
In order to solve this system, you first need to work with each equation separately. It is usually easiest to start with the equation that is less complicated to solve. In this example, the first equation is simple to solve for either x or y, therefore it is up to the student which variable to solve, as the question asks for a numerical value for both variables once completed with the problem. If the first equation is solved for x and simplified completely, the value of x would be 52-9y. In order to solve the system, the value of x that was just found must be replaced in the second equation for the variable "x". The variable "y" can then be determined as a numerical value. When replacing "x" with 52-9y in the second equation, it should now look like this: 14(52-9y) + 2[6(52-9y) + y] = 200. In order to obtain a numerical value for "y", this equation must be solved. After solving this equation, the value of y should be -96/236. This value can then be used in the first equation to replace "y" and solve for "x". The value of "x" is 3284/59.
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