Tutor profile: Mitchell K.
Subject: Physics (Newtonian Mechanics)
A 3.0 kg box is sliding down a hill at an angle of 30 degrees. What is the force of friction on the box? (g = 9.8 N/kg, coefficient of friction = 0.4)
1) First, draw a diagram of the hill and the box. 2) Find the force of gravity. F(G) = gravitational force g * mass = 9.8 N/kg * 3.0 kg = 29 N 3) Break up the force of gravity vector using a 30 degree triangle. F(N) = normal force = vector perpendicular to hill incline = F(G) * cos(30) F(N) = 29 N * √3/2 = 25 N F(A) = force of acceleration = vector parallel to hill incline = F(G) * sin(30) F(A) = 29 N * 1/2 = 14 N 4) Use the normal force and coefficient of friction to find the force of friction. F(F) = force of friction = F(N) * coefficient of friction F(F) = 25 N * 0.4 F(F) = 10 N
A geologist is studying some new rocks near the Galapagos islands. She finds a new element that has not been studied with a ionization energy that is very similar to Osmonium (Os). If the properties of the element are discovered to be consistent with other properties of Osmonium, how would this new element relate to Sulfur (S)? Please refer to the periodic table A. Increased electronegativity, decreased ionization energy, decreased atomic radius B. Decreased electronegativity, decreased ionization energy, increased atomic radius C. Increased electronegativity, Increased ionization energy, decreased atomic radius D. decreased electronegativity, decreased ionization energy, decreased atomic radius
B. Electronegativity and ionization energy increase as you go up and to the right of the periodic table while atomic radius decreases. Since Osmonium is lower to the left of Sulfur on the periodic table, the new elements electronegativity will be decreased, its ionization energy will be decreased, and its atomic radius will be increased in comparison to Sulfur.
A chemist is hosting a dinner party and needs more salt (NaCl). The only things he finds in his pantry are 142.8 g Magnesium Chloride (MgCl2), 80.00 g Sodium Hydroxide (NaOH), and the following formula in his notebook. NaOH + MgCl2 -> NaCl + Mg(OH)2 How many grams of NaCl can he make and what is the limiting reagent?
1) The first step is to balance the equation: 2 NaOH + 1 MgCl2 -> 2 NaCl + 1 MgOH 2) Next, we find the molecular weights of the reagents and NaCl. MW MgCl2 = (1 mol Mg * 24.30 g/mol) + (2 mol Cl * 35.45 g/mol) = 95.20 g/mol MgCl2 MW NaOH = (1 mol Na * 22.99 g/mol) + (1 mol O * 16.00 g/mol) + (1 mol H * 1.01 g/mol) = 40.00 g/mol NaOH MW NaCl = (1 mol Na * 22.99 g/mol) + (1 mol Cl * 35.45 g/mol) = 58.44 g/mol NaCl 3) We can the moles of NaCl we can produce with each reagent to determine the limiting reagent. 142.8 g MgCl2 * (1 mol/95.20 g) = 1.500 mol MgCl2 80.00 g NaOH * (1 mol/40.00 g) = 2.000 mol NaOH 1.500 mol MgCl2 * (2 mol NaCl/1 mol MgCl2) = 3.000 mol NaCl 2.000 mol NaOH * (2 mol NaCl/2 mol NaOH) = 2.000 mol NaCl Even though there are more moles of NaOH available at the start than MgCl2, NaOH is the limiting reagent. This is because the ratio of moles NaOH:NaCl in the balanced equation is 1:1 while the ratio of moles MgCl2:NaCl in the balanced equation is 2:1. 4) We now have all the information we need to solve for grams of NaCl. 2.000 mol NaCl * (58.44 g/mol) = 116.88 g NaCl
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