Problem set No.8 Name, surname Date Problem 1 Solve the equation cos.x-0.4x+0.3

Question

Problem set No.8 Name, surname Date Problem 1 Solve the equation cos.x-0.4x+0.3 using the Newton's method. 1.1 Find the smallest root (accurate to within 10" . 1.2 Find the largest root (accurate to within 10) 1.3 How many real roots does the equation have? Problem 2 sin(51+2x,-2 cos(x2-1) + x, = 0.7 using the Newton's method. Solve the system! 1.4 Find x, accurate to within 10 1.5 Find x, accurate to within 10 1.6 How many real roots does the system have? Problem 3. Solve the Cauchy problem y'(t) = 2x2 (1)-y(r) , on the interval (0,3). 1.7 Find x(1.1) 1.8 Find y(1.1). 1.9 Using the graph of the function x) find the maximum value of x on the interval (0.3) (accurate to within -0.1 ). 1.10 Using the graph of the function y() find the maximum value of y on the interval (0.3) (accurate to within 0.1 ). Answers.

Explanation / Answer

Solution :-

A 115 g of 0.180 m KBr

converting gram to kg

115 g * 1 kg / 1000 g = 0.115 kg

moles of KBr = molality * kg

                    = 0.180 mol per kg * 0.115 kg

                    = 0.0207 mol KBr

mass of KBr = moles * molar mass

                  = 0.0207 mol * 119 g per mol

                  = 2.46 g KBr

Therefore to make the desired solution dissolve 2.46 g KBr in 112.54 g water

B) 1.75 L of 12.0 % KBr by mass (density of solution = 1.10 g/ml)

1.75 L * 1000 ml / 1 L = 1750 ml

1750 ml * 1.10 g per ml = 1925 g

mass of KBr = 1925 g * 12.0 % / 100 % = 231 g KBr

Therefore use 231 g KBr and dissolve it to total volume of 1.75 L using water.

C) 0.130 M Solution that contains just enough KBr to precipitate 18.0 g AgBr

lets first calculate moles of AgBr

reaction equation

AgNO3(aq) + KBr(aq) ---- > AgBr(s) + KNO3(aq)

18.0 g AgBr * 1 mol / 187.77 g per mol = 0.09586 mol AgBr

since mole ratio of the AgBr and KBr is 1 :1 therefore moles of KBr needed = 0.09586 mol

now lets calculate the mass of this moles of KBr

mass of KBr =0.09586 mol * 119 g per mol = 11.4 g KBr

now lets calculate the volume to get desired molarity

volume in liter = moles / molarity

                    = 0.09586 mol / 0.130 mol per L

                   = 0.737 L

0.737 L * 1000 ml / 1 L = 737 ml

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