List the layers of the Earth and their characteristics (physical state, composition, position). How do scientists believe the Earth formed?
There are four layers in the Earth: the crust, the mantle, the inner core and the outer core. The crust is the outermost layer that makes up the ocean floor and the continents. It is solid, and is a cool, livable temperature for organisms. The mantle is below the crust and is composed of magma, which is liquid (molten) rock. In some parts of the crust, there are pockets of magma that are being fed from the mantle and when these pockets break through to the surface, volcanoes are formed. The mantle is very hot (1600 - 4000 degrees F), but very viscous. The movement of the crust floating on top of the mantle’s convection currents creates continental drift. The outer core is the next layer inside the Earth. It is composed of liquid nickel and iron. It is even hotter than the mantle, 4000-9000 degrees F. Lastly, in the very center is the inner core. Here it is so dense that even though it is around 9000 degrees F, the iron and nickel are turned back into their solid states. Scientists believe that the Earth formed from a hot ball of metals and rocks and when it started to cool off, the dense parts (like iron and nickel) sank to the center and the less dense parts (like other elements found in dirt and clay and rocks) formed the crust.
Ca(OH)2(s) + 2 HCl(aq) ---> CaCl2(aq) + 2 H2O(l) 1) What type of chemical reaction is taking place? 2) How many liters of 0.100 M HCl would be required to react completely with 5.00 grams of calcium hydroxide? 3) If I combined 15.0 grams of calcium hydroxide with 75.0 mL of 0.500 M HCl, how many grams of calcium chloride would be formed? 4) What is the limiting reagent from the reaction in problem #3? 5) How many grams of the excess reagent will be left over after the reaction in problem 3 is complete?
1. This is an acid - base reaction. Ca(OH) is basic because of the hydroxide ion and HCl is acidic because of the hydrogen ion. During the reaction HCl will donate the H+ to the OH- forming H20 and Ca will pair with Cl forming CaCl2. 2. This is a stoichiometry problem. Our initial value is 5.00 g Ca(OH)2 and we are looking for L (liters) of HCl. To find out what else we need to know (like molar masses) we can do a general conversion before plugging in any numbers. g Ca(OH)2 * (mol Ca(OH)2 / molar mass g Ca(OH)2) * (mol HCl / mol Ca(OH)2) = mol HCl Once we know the moles of HCl needed, we can use M = mol/L to find how many liters we need. L = mol HCl / M HCl In order to do this calculation, we need to find the molar mass of Ca(OH)2 . Everything else in the conversion is given in the problem. The molar mass of Ca is 40.078 g/mol, of O is 15.9994 g/mol, and of H is 1.0079 g/mol 1 x Ca + 2 x (O + H) = 40.078 + 2( 15.9994 + 1.0079) = 74.0926 g/mol 5.00 g Ca(OH)2 * (1 mol Ca(OH)2 / 74.093 g Ca(OH)2) * (2 mol HCl / 1 mol Ca(OH)2) = 0.135 mol HCl L HCl = 0.135 mol HCl / 0.100 M HCl = 1.35 L HCl 3. Since we do not know which reagent is limiting, we must do the stoichiometry for both initial values to find out how many grams of CaCl2 is formed. In both scenarios, we need the molar mass of CaCl2 so that is the first thing that we need to calculate. Molar mass of Ca is 40.078 g/mol, of Cl is 35.453 g/mol Molar mass of CaCl2 is 1 x Ca + 2 x Cl = 40.078 + 2(35.453) = 110.984 g/mol Starting with 15.0 g Ca(OH)2: 15.0 g Ca(OH)2 * (1 mol Ca(OH)2 / 74.093 g Ca(OH)2) * (1 mol CaCl2 / 1 mol Ca(OH)2) * (110.984 g CaCl2 / 1 mol CaCl2) = 22.5 g CaCl2 Starting with 75.0 mL of 0.500 M HCl: We must first convert this into moles of HCl and we know that M = mol/L 75.0 mL = 0.075 L Mol HCl = ML = (0.500)(0.075) = 0.0375 mol HCl Now we can do the stoichiometry to get grams of CaCl2: 0.0375 mol HCl * (1 mol CaCl2 / 2 mol HCl) * (110.984 g CaCl2 / 1 mol CaCl2) = 2.08 g CaCl2 2.08 g CaCl2 is less than 22.468 g CaCl2, so we know that the real result would be the formation of 2.081 grams of CaCl2. There is not enough HCl to make 22.5 g of CaCl2. 4. Because of this, we know that HCl is what is limiting this reaction and that there will be excess Ca(OH)2. All of the HCl gets used up in the reaction to form 2.08 g of CaCl2, but there is enough Ca(OH)2 to makes 22.5 g which is more than 10 times as much. 5. To find out how much Ca(OH)2 there will be remaining, we have to work backwards. We will now start with the product of 2.08 g of CaCl2 and find out exactly how many grams of Ca(OH)2 were needed to make it. Then we can subtract from the total amount to find out how much was left over. 2.08 g CaCl2 * (1 mol CaCl2 / 110.984 g CaCl2) * (1 mol Ca(OH)2 / 1 mol CaCl2) * (74.093 g Ca(OH)2 / 1 mol Ca(OH)2 ) = 1.39 g Ca(OH)2 We started with 15.0 g Ca(OH)2 and used 1.39 g Ca(OH)2. 15.0 - 1.39 = 13.6 g Ca(OH)2 is left over after the reaction
For t ≥ 0, a particle is moving along a curve so that its position at time t is (x(t), y(t)). At time t = 2, the particle is at position (1, 5). It is known that (dx/dt) = ((t + 2)^0.5)/(e^t) and that (dy/dt) = sin^2(t). (a) Is the horizontal movement of the particle to the left or to the right at time t = 2 ? (b) Find the slope of the path of the particle at time t = 2. (c) Find the x-coordinate of the particle’s position at time t = 4. (d) Find the speed of the particle at time t = 4.
a. The horizontal movement of the particle is governed by the x coordinate's rate of change. To the right is defined to be the positive x direction and to the left is defined to be negative. If the derivative (rate of change) of the x coordinate at a certain time is positive, then the particle is moving to the right. If it is negative, it is moving to the right. We know that (dx/dt) = ((t + 2)^0.5)/(e^t), so at t = 2, (dx/dt) = ((2 + 2)^0.5)/(e^2) = 2/(e^2) Since 2 and e^2 are both positive, (dx/dt) is positive. This means that the particle is moving to the right at t = 2. b. The slope of a curve is the change in y for every change in x, (dy/dx). Since we have the change in y (dy/dt) and the change in x (dx/dt), we just divide (dy/dt)/(dx/dt) and get (dy/dx). (dy/dt) = sin^2(t) (dx/dt) = ((t + 2)^0.5)/(e^t) (dy/dx) = (dy/dt)/(dx/dt) = (sin^2(t)) / ((t + 2)^0.5)/(e^t) = sin^2(t)(e^t) / ((t + 2)^0.5) at t = 2, (dy/dx) = sin^2(2)(e^2) / ((2 + 2)^0.5) = 3.055 (dy/dx) = 3.055 c. To find the x coordinate of the particle at any given time, we must anti-differentiate the function given for the rate of change of the x coordinate, (dx/dt). In this case, it is nearly impossible to do this by hand so we can have a calculator do the anti-differentiation for us. We know that at t = 2, the x coordinate is 1. So, we only need to find out how far it has gone (in the x direction) from t = 2 to t = 4. We have the calculator compute the integral from t = 2 to t = 4 of ((t + 2)^0.5)/(e^t)dt The calculator tells us this is equal to 0.253. If it moved 0.253 units from t = 2 to t = 4, then its final position, x(4), is 1 + 0.253 = 1.253 x(4) = 1.253 d. We are given the x and y components of the speed of the particle, (dx/dt) and (dy/dt). To find the overall speed of the particle, we must use Pythagorean Theorem c^2 = a^2 + b^2, where c is the overall speed, and a and b are the x and y components. We do this because the components inherently make a right triangle which is then applicable to Pythagorean Theorem. We will designate the overall speed to be the variable v. v^2 = (dx/dt)^2 + (dy/dt)^2, v = ((dx/dt)^2 + (dy/dt)^2)^0.5 v = [(((t + 2)^0.5)/(e^t))^2 + (sin^2(t))^2]^0.5 at t = 4, v = [(((4 + 2)^0.5)/(e^4))^2 + (sin^2(4))^2]^0.5 = 0.575 v(4) = 0.575