Conv10 Q29. Practice. Using conversion factors from the previous following probl

Question

Conv10 Q29. Practice. Using conversion factors from the previous following problems. units and labels. Remember to give the road map first and then numbers, question, solve the a. A fertilizer contains 14 % phosphorus by mass. A farmer needs to put 130 lb of phosphorus on a field. How many lb of fertilizer should he buy? b. A medicated lotion contains 0.50 % (m/v) cortisone. How many mL of the lotion is needed to provide 2.4 g of cortisone? C. A cleaning solution contains 33 % acetone by volume. How many mL of acetone are in 6.0 L of the cleaning solution? d-The average concentration of NaCl in sea water is 0.035 % Naci by mass. How tons of sea water would you need to collect 1.0 lb NaCr 030. Practice. Using everything learned in this activity, solve the following problems. a. The owner of a motorcycle claims he can travel 100 km on 4.3 L of gas on the highway. What does this equate to in MPG (miles per gallon)? b. A certain energy drink contains 13% sugar (m/v) and no other calorie supplying nutrients. This particular sugar provides 4.1 Calories per gram of sugar. There are 29.6 mL in one fluid ounce. How many Calories are in one 16 oz can of the energy drink? (The unit for the food Calorie is always abbreviated Cal.)

Explanation / Answer

Q29 a. A fertilizer contains 14% phosphorous by mass.

That means 14 gm phosphorous in 100 gm of fertilizer.

14 gm phosphorous ----> 100 gm fertilizer.

A farmer needs to put 130 Ib of phosphorous on a field.

Conversion of gm into Ib. (1gm = 0.0022 Ib)

14 gm = 0.0308 Ib (14 X 0.0022 / 1 = 0.0308 Ib)

100 gm = 0.22 Ib (100 X 0.0022 / 1 = 0.00 Ib)

Now we have all units in Ib

From above conversion, 0.0308 Ib Phosphours in 0.22 Ib Fertilizer.

Therefore to put 130 Ib of phosphours on the field the farmer needs,

130 Ib X 0.22 Ib / 0.0308 Ib = 928.57 Ib of fertilizer.

Thus 928.57 Ib of fertilizer the farmer should buy.

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b. A medicated lotion contains 0.50% (m/V) cartisone .

This means 0.50 gm cartisone in 100 ml of lotion.

0.50 gm cartisone ----> 100 ml lotion.

Therfore to provide 2.4 gm of lotion we need,

2.4 gm X 100ml / 0.50 gm = 480 ml of lotion.

Thus to provide 2.4 gm of cartisone, 480 ml of medicated lotion is needed.

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c. A cleaning solution contains 33% acetone by volume.

This means 33 ml of acetone in 100 ml of solution.

33 ml acetone -----> 100 ml solution.

Thus in 6 L of the solution, the ml of acetone will be,

Convert Litre into ml (1 L = 1000 ml)

Thus 6 L = 6000 ml

6000 ml X 33 ml / 100 ml = 1980 ml of acetone.

1980 ml of acetone are in 6L of the solution.

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d. The average concentration of Nacl in sea water is 0.035% Nacl by mass.

This means 0.035 gm Nacl in 100 gm sea water.

0.035 gm Nacl -------> 100 gm Sea water,

Conversion of grams into Ib (1 Ib = 453.592 gm)

0.035 gm X 453.592 / 1 = 0.00007716 Ib

100 gm X 453.592 / 1 = 0.22 Ib

From above conversion, 0.00007716 Ib Nacl -----> 0.22 Ib sea water

To collect 1 Ib of Nacl, we need

1 Ib X 0.22 / 0.00007716 Ib = 2851.21 Ib of sea water,

Conversion of Ib into tons (1Ib = 0.000453 ton)

2851.21 X 0.000453 / 1 = 1.2932 Tons of sea water.

Thus, 1.2932 tons of sea water need to collect 1 Ib of Nacl.

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