Tutor profile: Shannon H.
Consider the following equation 2Al + 6 HCl --> 2AlCl3 + 3H2. What is the mass of H2 formed when 16g of Al reacts with excess HCl?
In order to answer this question we are going to need to look at the moles of each substance. 1. Make sure the equation is balanced. (It is) 2. Convert the 16g of Al to moles using the molar mass of Al (26.982 g/mol) Multiply 16g Al * 1 mol / 26.982 g Al. The grams will cancel to give us .593 mol Al 3. Next we are going to use the mole to mole ratio in order to figure out the moles of H2 formed. Logically we can determine that since Al is the limiting reagent (meaning it will all be react out) that for every 2 moles of Al, 3 moles of H2 will be formed. This is based on the coefficients of the equation. Therefore we can convert to moles of H2 by doing... .593 mol Al * 3 mol H2 / 2 mol Al. The Al will cancel to give us .889 mole H2. 4. Convert the moles of H2 to grams using the molar mass (1.008 * 2). Remember that when talking abut mass we use grams. .889 mol H2 * 2.016 g / 1 mol H2 = 1.793 g H2
Based on the following function f(x) = x^4 - 3x^2 + 12 find the local extrema and where the function is concave down.
First we need to understand what this question is asking us to find. Local extrema is talking about the maximums (highest points) and minimums (lowest points). To find max and mins we must first determine where the slope is = 0 and if the slope changes directions. Concavity is discussing which way the end points of a function (or part of one) is pointing. Basically is the graph pointing up or down. Remember that when looking at slope we are talking about the first derivative and concavity (which is the slope of the slope) is the second derivative. 1. Take the first derivative. f'(x) = 4x^3 - 6x 2. Since the first derivative describe the slope at a point and we are looking for where the slope = 0, set the first derivative = 0. 4x^3 - 6x = 0 3. Solve for x by first factoring out a 2x and set each part of the equation equal to 0. 2x (2x^2 -3), 2x = 0 and 2x^2 - 3 = 0 4. Now we have 2 solutions, x = 0, +/-sqrt of 3/2 (=1.22) 5. Next we must check to see if the signs change, remember to have a max the signs must change from positive to negative and a minimum must have it change from a negative to positive. It is easiest to use a number line for this. <--(-)1.22--0--(+)1.22--> 6. Plug in numbers to see if the signs of the slope changes. We find that using and numbers <-1.22 our slope is -, any number between -1.22 and 0 and slope is +, any number between 0 and 1.22 the slope is -, and any numbers >1.22 the slope is +. Therefore we have a maximum when x = 0 and minimums when x = +/- 1.22 7. For the concavity take the second derivative. f''(x) = 12x^2 - 6 8. For concavity we must find the inflection points. Inflection points are where the concavity changes and the 2nd derivative is equal to 0. Therefore set the 2nd derivative = 0. 12x^2 - 6 = 0 9. Solve for x by first factoring out a 6. 6 (2x^2 - 1) ---> 2x^2 - 1 = 0 ---> x = +/-sqrt(1/2) ---> +/-.707 10. Again we must check to how the signs change before and after this point. If the signs do not change then it is not an inflection point. Remember that when f''(x) > 0 it is concave up and when f''(x) < 0 it is concave down. Again use a number line. <---(-).707---(+).707---> 11. When plugging in a number less than -.707 we get a +, when plugging in a number between -.707 and .707 we get a -, and greater than .707 is positive. Therefore f(x) is concave down on the interval (-.707, .707)
Line 1 has a slope of 2/3 and contains the point (3,1). Line 2 has a slope of 1/6 and contains the point (0,5). Find where these 2 lines intersect.
We know that when 2 lines intersect they have both the same X and Y coordinates, therefore we need to figure out where the 2 lines equal each other. 1. Write the equations for the lines: In order to write the equation for a line we must have a slope and a point which is given. Therefore using the y = mx + b form and solving for b (using the point given to plug in for x and y) we find the 2 equations are y = (2/3)x - 1 and y = (1/6)x + 5. 2. Next, since we know when these lines intersect they will have the same Y coordinate/value (along with the X coordinate/value) we can set these equations equal to each by writing the following (2/3)x - 1 = (1/6)x + 5 3. Next move like terms to opposite sides. Therefore all the terms with an X we can move to left and all the constants (without a variable) we can move to the right to have... (2/3)x - (1/6)x = 1 + 5 4. Now combine like terms. Remember in order to add/subtract fractions the denominator must be the same. Since 3 and 6 have a least common multiple of 6 multiple (2/3) by (2/2) to get (4/6). Now we can combine the terms to get (3/6)x = 6. We can then simplify this to (1/2)x = 6. 5. Solve for X buy multiplying both sides by 2 (the reciprocal of 1/2) to get x = 12. 6. Use the value of X and plug it back in to one of our original equations to find the value of Y. It does not matter which equation is used as they both share the same coordinate value at this point. y = (1/6)(12) + 5 7. Solve for Y. y = 2 + 7 = 11 So the lines will intersect at the point (12,11)
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