Tutor profile: Jacob M.
Subject: Basic Chemistry
You have 5 grams of H2SO4. How any moles of oxygen are present in this volume of H2SO4?
First, we know our givens are 5g of H2SO4. Then, we also know the molecular weight of the compound, H2SO4, itself via the periodic table where the molecular weight of H is 1.01 g/mole, S is 32.06 g/mole and O is 16 g/mole. Thus, the total molecular weight of H2SO4 is 1.01*2+32.06+16*4 = 98.08 g/mole. Thus, we know that in 1 mole of H2SO4 equals 98.08g. Thus, in 5g, we can calculate the fraction of a mole of H2SO4 that is present by dividing 5/98.08 = 0.051, thus there are 0.051 moles of H2SO4 in 5g of the substance. But, we are not finished there, as the question asks for how many moles of oxygen, specifically, are there. Since one molecule of H2SO4 consists of 4 oxygen atoms, for each mole of H2SO4 there is 2 moles of H2 atoms, 1 mole of S atoms and 4 moles of oxygen atoms. So, if we have 0.051 moles of H2SO4, we have 0.051*4=0.24 moles of oxygen.
What is the derivative of the quantity (x^3)(x^2-1) between the values x=0 and x=2?
First, it is important to multiply out the terms in the two parenthesis, giving you x^5-x^3. Then, we derive using the standard method of deriving exponential terms, where the exponent is brought down in front of the variable and the exponent is then kept as one less than the original. So, for x^5, the 5 comes down and the exponent decreases by 1, yielding 5x^4. Likewise, the second term becomes 3x^2. So, the final set of terms for the derivative is 5x^4-3x^2, and plugging in first x=2, we get a value of 5*2*2*2*2 - 3*2*2 which equals 80-12 = 68.
You have a box of mass=M resting on inclined plane that is set at a 32 degree angle from the ground. If the box is resting at a distance x from the end of the inclined plane that is touching the ground, what must be the coefficient of friction so that, when released from rest, the box will remain still and not slide down the plane?
First, begin the question by writing out the givens in the question. Mass of box = M, acceleration due to gravity = g, coefficient of friction on the box = u(f), distance along box = x, height of box above ground = y, angel of the inclined plane = 32. Next, draw a free-body diagram consisting of an inclined plane with the angle designated as 32 degrees. Next draw the box and label its height above ground as y, and its distance along the inclined plane as x. Next, draw a force vector (arrow) from the center of the box pointing straight down to the ground (this is the weight force, W, of the box, which is equal to Mg). Next, draw a vector arrow pointing from the center of the box parallel to the inclined plane, toward the ground. Next, draw a force vector pointing from the center of the box straight up from the inclined plane, perpendicular to the plane's surface. This represents the normal force (N) of the box on the surface. Finally, draw a force vector from the center of the box pointing up the inclined plane, parallel to the plane's surface (this represents the friction force Fc, which equals the normal force multiplied by the coefficient of friction, N*u(f)). Now, draw the x and y components of the W vector, where the vertical (relative to the plane surface) component is Wcos(32) and the horizontal (relative to the plane surface) component is Wsin(32). So, for the box, the horizontal forces (forces acting down and up the surface of the plane) are Wsin(32), pointing down the plane, and Fc, pointing up the surface of the plane. Since we are curious as to what coefficient of friction must be present to have the box remain still, these forces must be equal. So, Wsin(32)-Fc=0, and Wsin(32)=Fc, where we know W=Mg and Fc=N*u(f). Likewise, the normal Force, N, equals the W, but only the vertical component (component pointing perpendicular to plane surface, which in this case is Wcos(32). So, the equality becomes M*g*sin(32)=M*g*cos(32)*u(f), and rearranging and cancelling terms, u(f) must equal sin(32)/cos(32), or u(f)=tan(32).
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