Various Math #41 - Q&A

Question 1 Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as Ft = PairgV balloon will push the balloon upward. If the balloon has a diameter of 12 m and carries two people, 85 kg each, determine the acceleration of the balloon when it is first released. Assume the density of air is ρ ropes and the cage. Answer: 22.4 m 42 1.16k, and neglect the weight of the m³' 3 Answer 1 Givens and Knowns: Pa = 1.16k g m³ D = 12m mp = 85kg Solution: Fft a= m→t Fft = FB - W Where FB is the bouyancy force and W is the total weight. FB = Pax Vol × g = 1.16 × X π× 4 × 6³ 3 x 9.81 10296.02 mt = 2 × mp + Phe × Vol = 170 + 1.16 × 904.78 = 319.934 7 W = g × mt = 9.81 × 319.934 = 3138.56 (Note: Phe = Pa) 7 Now, we can calculate the net force exerted as follows, Fft = 10296.02 – 3138.56 = 7157.46 7157.46 a = = 22.372 319.934 Answer 2 1 Helium weighs (weigth of air) 7 the radius of the balloon r = 6m Hence the volume of the balloon is V = 4 13 πr³ (3)π(6) 3 4 = 904.32m³ the mass of each person m = 85kg from Newton's law F = Fb - Fg ma = pair g Vballoon - 2.85 + (1/7) (pair) (9.8) = (1.16) (9.8) (904.32) = 10280.33134.6 = 7145.7 2.85+ (1/7) (904.32) (1.16) (9.8) 2.85+ (1/7) (Vballoon) (pair) a = 7145.6 the acceleration is as a result 7145.7 a = = 22.34 319.85 m 2 s Answer 3 D = 12m Phe = Pair / 7Pair = 1.16kg / m³ mpeople = 85 × 2 = 170kg FB = Pover ballon w = mtotal mhe Vballoon PheV ballon mhe = 1 x Rbaloon 4 ×1.16×904.78 = 149.93kg Mtotal = mhe + mpeople = 149.93 + 170 = 319.93kg Applying Newton's 2nd law. Ef = mtotala FB-W = mtotala PairgVballon-mtotalg = mtotala a = Pair ball - Ntotal g mtotal (1.16) 4.78) - 319.93 319.93 = 22.4m/s² Question 2 Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) limh→0 (2 + h)² - 4 h Answer 1 I believe you wanted to write following limit: lim h⇒0 (−2+h)² - 4 h If so, then by plugging h = 0 we obtain: (-2+0)²-4 4-4 0 = 0 0 0 Answer 2 Take the limit of each term. 2 limh⇒0h + limh⇒04 Evaluate the limits by plugging in 0 for all occurences of h. Evaluate the limit of h by plugging in 0 for h. 2.0 + lim 4 h→0 Evaluate the limit of 4 which is constant as h approaches 0. 2⋅ 0 + 4 Simplify the answer. Multiply 2 by 0. 0+4 Add 0 and 4. Answer: 4 Question 3 Determine the mass of water formed when 12.5 L NH3 (at 298 K and 1.50 atm) is reacted with 18.9 L of O2 (at 323 K and 1.1 atm). 4NH3(g)+5O2(g) → 4NO(g)6H2O(g) Answer 1 Step 1 The balanced reaction taking place is given as, ⇒ 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) According to ideal gas law, PV = nRT where P = pressure in atm V = volume in L n = moles R = gas constant = 0.0821 atm. L mol K T = temperature in K = T ∈ C° + 273 Step 2 a) For NH3, P = 1.50 atm, V = 12.5L and T = 298K Hence substituting the values we get, ⇒ 1.50 x 12.5 = n x 0.0821 × 298 ⇒ n = moles of NH3 taken = 0.7664 mol approx. b) For O2, P = 1.10 atm, V = 18.9L and T = 323K Hence substituting the values we get, ⇒1.10 x 18.9 = n x 0.0821 × 323 ⇒ n = moles of O2 taken = 0.784 mol approx. Step 3 ⇒4 NH3(g) + 5O2(g) → 4 NO(g) + 6H2O(g) From the above balanced reaction we can see that 5 moles of O2 is required to react with 4 moles of NH3 Hence moles of O2 required = moles of NH3 taken × 5 = 0.7664 x 5/4 = 0.958 mol 4 Since moles of O2 taken is less than required. Hence O2 is the limiting reagent and will react completely. From the above reaction we can see that 5 moles of O2 is reacting to produce 6 moles of H2O. Hence moles of H2O that can be produced 6 = moles of O2 taken x 5 = 0.784 x 6/5 = 0.9408 mol Molar mass of H2O = Atomic mass of H x 2 + Atomic mass of O = 1×2 + 16 = 18g/mol Hence mass of H2O that can be produced = moles x molar mass = 0.9804 × 18 = 16.9g approx. Hence the answer is 16.9 g. Answer 2 The mass of water formed is calculation Use the ideal gas equation to calculate the moles of NH3 and O2 that is Pv = nRT where; P = pressure, V = volume, n = number of moles, R = gas constant = 0.0821L. at m/ol. K make n the formula of the subject by diving both side by RT PV n= RT The moles of NH3 n = 1.50atm × 12.5L 0.0821L. at mol. k × 268K = 0.766 moles The moles of 02 n = 1.1atm x 18.9L 0.0821L. at mol. k × 323K = 0.784 moles write the reaction between NH3 and O2 4NH3 +502 ⇒ 4 No + 6H2O from equation above 0.766 moles of NH3 reacted to produce 0.766 x 6/4 = 1.149 moles of H2O 0.784moles of O2 reacted to produce0.784 x 6/5 = 0.9408moles of H20 since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced = 0.9408 moles mass of H2O = moles x molar mass from periodic table the molar mass of H2O = (1 x 2) + 16 = 18 g/mol mass = 18 g/ol x 0.9408moles = 16.93grams m Answer 3 Solution: First, let's convert the given volumes of NH3 and O₂ to the number of moles using the ideal gas law: For NH3: PV = nRT where: P = pressure of NH3 = 1.50 atm v = volume of NH3 = 12.5 L n = number of moles of NH3 R = ideal gas constant = 0.0821 L atm/(mol K) T = temperature of NH3 = 298 K Rearranging the equation to solve for n: PV n= RT Substituting the given values: nNH3 = (1.50 atm) (12.5 L) (0.0821 L atm/(mol K))(298 K) Now, let's calculate the number of moles of O2 using the same approach: For O2: PV = nRT where: P = pressure of O2 = 1.1 atm V = volume of O2 = 18.9 L n = number of moles of O2 R = ideal gas constant = 0.0821 L atm/(mol K) T = temperature of O2 = 323 K Rearranging the equation to solve for n: n= PV RT Substituting the given values: nO2 = (1.1 atm) (18.9 L) (0.0821 L atm/(mol K)) (323 K) Next, let's determine the limiting reactant by comparing the number of moles of NH3 and O 2. The balanced equation shows that 4 moles of NH3 react with 5 moles of O2. Therefore, the ratio of moles is 4:5. We have: nNH3 nO2 (1.50 atm) (12.5 L) (0.0821 L atm/(mol K)) (298 K) (1.1 atm) (18.9 L) (0.0821 L atm/(mol K)) (323 K) Comparing the moles of NH3 and O2, we find that the moles of NH3 are multiplied by 5/4 to balance the equation. Now, let's determine the moles of water formed. From the balanced equation, we see that 4 moles of NH3 yield 6 moles of H2O. Therefore, the number of moles of H2O formed can be calculated as: nH2O = 6/4 ΧηΝΗ3 Now, we can calculate the mass of H2O formed using the molar mass of water, which is approximately 18.015 g/mol . mH2O = nH20Xmolar mass of H2O Finally, we substitute the calculated values and solve for the mass of water formed. Question 4 A -12 nC charrge is located at the origin. a) What is the strength of the electric field at the position (x, y) = (0ст, 5.0cm)? b)What is the strength of the electric field at the position (x, y) = (-5.0cm, -5.0cm)? c) What is the strength of the electric field at the position (x, y) = (-5.0cm, 5.0cm)? Answer 1 Step 1 The charge located at the origin is given as, q = -12nC = -12nC X 10⁻⁹C /1nC = -12 × 10 °C Step 2 (a) The position at which electric field required is given as, (x, y) = (0cm, 5cm) The distance of the required position from the origin can be determined as, r = √ x² + y² 2 = √(0cm)² + (5cm)² = 5cm × 10⁻²m/1cm = 0.05cm The electric field strength at the required location can be determined as, E = kq / r² = (9×10⁹N.m²/C²)|-12×10⁻⁹C| (0.05m)² N = 43200 C Answer 2 (b) The position at which electric field required is given as, (x, y) = (-5cm, -5cm) The distance of the required position from the origin cam be determined as, r = √ x² + y² = √(-5cm)² + (-5cm)² = 7.1cm x 10⁻²m/1cm = 0.071cm The electric field strength at the required location can be determined as, E = kq/r² (9×10⁹N.m²/C²)-12×10⁻⁹C (0.071m)² N = 21424.3 C Answer 3 (b) The position at which electric field required is given as, (x,y) = (-5ст, 5ст) The distance of the required position from the origin cam be determined as, r = √x² + y² = √(-5cm)² + (5cm)² = 7.1cm × 10⁻²m/1cm = 0.071cm The electric field strength at the required location can be determined as, E = kq/r² = (9×10⁹N.m²/C²)-12×10⁻⁹C (0.071m)² = 21424.3N/C Question 5 What are the ion concentrations in a 0.12 M solution of AlCl3? Answer 1 Explanation: To answer this question, we will need to write the dissociation equation of aluminum trichloride. AlCl3 → Al³+ + 3Cl⁻ It can be seen that when aluminum chloride dissociates, it gives one mole of aluminum ion and three moles of the chloride ion. From here we can see that the concentration of the aluminum chloride equals that of the aluminum ion while that of the chloride ion is thrice that of the aluminum chloride. This means we simply multiply 0.12M by 3 to get the molarity of the chloride ion while that of the aluminum ion remains the same Answer 2 AlCl3 has one Al+3 and 3Cl⁻ion Consentration of AlCl3 = 0.12M So; the lonic concentration of Al+3 = 0.12M and lonic concentration of Cl⁻ = 3 x 0.12M = 0.36M