Please read the questions carefully, use only Unit factor method and show all yo

Question

Please read the questions carefully, use only Unit factor method and show all your work. 1. Chemist often gives the composition of a mixture as mass per unit volume. For example, if 1.55 g/5.0 ml.- 0.33 g/ml. of sugar in the solution. TExpress this concentration in each of the following units. kg/m3 (b) pgul. (d mg/ml. ug/ml. 2. The sun is estimated to have a mass of 2 x 1036 kg. Assuming it to be a sphere of average radius 6.96 x 105 km, calculate the average density of the sun in units of (a) kg/km, and (b) g/cm rm. 4r 3. In SI units land area is measured by hectare, defined as 1 hm2, 1 hectometer (hm)-100 m. How many acres correspond to 1.5 hectare? (use l mi2. 640 acres, l mi-5280 ft, I ft-12 in, 1 n-254 cm.) 4The volume of seawater on earth is about 3.3 x 108 mi, If seawater is 3.4% sodium chloride by mass and has a density of 1.03 g/ml, what is the approximate mass of sodium chloride, in tons dissolved in the seawater on earth? (use 1 ton- 2000 lh, 1 lb-453.6g) 5. A typical rate of deposit of dust Cdustal") from unpolluted air is ten tons per square mile per month. What is this dustall, expressed in mg per square meter per hour?

Explanation / Answer

1. Concentration = 0.33g/ml

a) 330 kg / (m^3)

b) 330 mg/ml

c) 330 000 000 pg / microlitre

d) 330 000 microgram / ml

e) 0.033 centigram / microlitre

2. V = volume of earth = 4*3.14*r^3/3

= ((4*3.14)*((6.96*10^5)^3)/(3))

= 1.41*10^18 km^3

Density = mass/volume = (2*10^36)/(1.41*10^18)

a) = 1.42*10^18 Kg/Km^3

b)     = 1 420 000 g / cc

3.   1.5 hectare = ? acres.

1 mi = (5280*12*2.54)*(1/100) m

     = 1609.34 metre

1 mi^2 = (1609.34)^2 m^2 = 640 acres

1 m^2 = 640/((1609.34)^2) = 0.0002471 acres

1 hectare = 1 hm^2 = 100^2 m

1.5 hectare = 1.5*10^4 m^2

1.5 hectare = (1.5*10^4)*(0.0002471) = 3.7065 acres

4. Volume of sea water = 3.3 X 10^8 mi^3 = 1.4 x10^ 18 m3

1m3= 1000L

1.4 x10^ 18 m3 = 1.4 x10^ 18 m3 X ( 1000L/1m3) = 1.4 x10^ 21 L

1.4 x10^ 21 L = 1.4 x10^ 21 L X (1000 mL/1L)= 1.4 x10^ 24 mL

Mass of seawater = volume X density

= 1.4 x10^ 24 mL X 1.03g/mL

= 1.44 X10^ 24g

1 Kg = 1000g

1.44 X10^ 24g= 1.44 X10^ 24g X (1kg/1000g) =1.44 X10^ 21Kg

1 ton = 1000kg

1.44 X10^ 21Kg= 1.44 X10^ 21Kg X ( 1 ton/1000kg) = 1.44 X10^ 18 ton

Amount of sodium chloride = 1.44 X10^ 18 ton x (3.4/100) = 4.896*10^16 tons

5. 10 ton / (mile.molth) = (10*10^6)/ ((2.6*10^6)*(30*24))

      = 0.00534 mg/sq.m * hr

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