A student tritrates 20mL of 1.0 M NaOH with 2.0 M formic acid, HCO2H (Ka = 1.8*10^-4). Formic acid is a monoprotic acid. How much formic acid, in mL, is needed to reach the equivalence point?

This problem seems to be much more complicated than it really is, so let's not over analyze things here. First, you know that the equivalence point is defined as the point at which the moles of acid is equal to the moles of base. The amount of NaOH is equal to (1.0M)*(0.0200L) = 0.0200 mol. To calculate the volume of acid: 2.0M = 0.0200 mol/V (By definition of Molarity) V = 0.0100 L = 10.0 mL Unless the problem states otherwise, you don't need to be too picky about your significant figures. Remember, just focus on what you NEED to solve the problem, which is especially true of Acid-Base questions.

After eating a large plate of carbohydrate-filled pasta, your blood is drawn and analyzed for two hormones: glucagon and insulin. How would the levels of these two hormones change before and after eating?

Digestion breaks down carbohydrates into glucose quickly after eating. This would increase blood sugar levels, which would cause the pancreas to secrete insulin to lower blood sugar. Glucagon, which raises blood sugar, would decrease as glucose levels rise after eating. This is a tightly regulated example of how the body maintains homeostasis.

Find the angle that maximizes the area under a projectile path's curve with an initial velocity $$v_0$$, angle theta. Use the formula for projectile motion for the 2nd dimension $$y-y_0=v_{y0}t+(a_{y}/2)t^2$$, re-written by taking into account the angle produced in the y-direction and assuming $$a_y$$= -9.8 m/s^2, $$y_0$$=0, $$y=v_{o}\sin\theta t+(g/2)t^2$$ Let the final vertical position be $$y=0$$

The projectile formula must be used to consider the case of maximizing area. But we must create a parametric solution to include theta and t. So considering both formulas for horizontal and vertical motion, we integrate the following by substitution $$\int_{t_0}^{t_1} (v_o\sin\theta-0.5gt^2) t\cos\theta dt$$ $$Which$$ yields$$ $$ $$({2v_o^4}/{3g^2})\cos\theta (sin\theta)^3$$ $$$$Taking the derivative of the expression to find the maximum area results in$$$$ $$\frac{2v_o^4}{g^2}((4\cos\theta)^2-1)(\sin\theta)^2$$ $$$$Setting the latter expression to 0 while not considering sine of theta and considering the first quadrant yields$$$$ $$\cos\theta=1/2$$ $$$$Thus, the angle for maximizing the area under the curve is$$$$ $$\boldsymbol{\theta=\frac{\pi}{3}}$$ or $$\boldsymbol{60^{\circ}}$$