Please help me and explain. Show work. 1. An order is given to administer 53 g o

Question

Please help me and explain. Show work.

1. An order is given to administer 53 g of lactulose syrup, often used for treating complications of liver disease. The suspension contains 10. g/15 mL. What dose in milliliters should be given to the patient?
__ mL

2. Is ethanol, a polar organic molecule that can hydrogen bond, miscible with water?

3. The following equation shows the combustion of propane in a forklift engine. If five moles of propane undergo combustion in an excess of oxygen, how many moles of carbon dioxide can be formed?

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)

4. Balance the following equation. (Include states-of-matter under the given conditions in your answer. Use the lowest possible whole number coefficients.)

C6H14(l) + O2(g) CO2(g) + H2O(l)

Determine how many grams of water are produced when burning 1.32 g of hexane, C6H14(l), as a component of gasoline in automobile engines. Assume excess oxygen is present. __ g

Explanation / Answer

1 The suspension contains 10. g/15 mL

That means !5 mL of the suspension contains 10 g of lactulose

10 g of lactulose syrup = 15 mL of the suspension

Therefore 53 g of lactulose syrup = 15 mL x 53 g / 10 g of the suspension = 79.5 mL of the suspension

A dose of 79.5 mL of the suspension should be given to the patient

2.) YES. Ethanol, a polar organic molecule that can hydrogen bond is highly miscible with water.

The condition for a solute to be soluble/miscible in a solvent is that the solute should be able to form some kind of bond with the solvent .

Ethanol is a polar organic molecule that can form hydrogen bond with water.. Hence it is soluble in water.

3.)                         C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)

According to the stoichiometric equation 1 mole of propane [C3H8(g)] gives 3 moles of carbon dioxide [CO2(g)]

Therefore 5 moles of propane [C3H8(g)} will give 5x 3 moles of carbon dioxide [CO2(g)] = 15 mole of carbon dioxide [CO2(g)]

5 moles of propane [C3H8(g)} will give 15 moles of carbon dioxide [CO2(g)] of carbon dioxide [CO2(g)]

4.)              C6H14(l) + 19/2 O2(g) 6 CO2(g) + 7 H2O(l)

               2 [C6H14(l) + 19/2 O2(g) 6 CO2(g) + 7 H2O(l) ]

             2 C6H14(l) + 19 O2(g) 12 CO2(g) + 14 H2O(l)    Balanced equation

Molecular mass of hexane, C6H14 = 6 x12 + 14 x 1 = 86

1 mole of hexane, C6H14 = 86 g of hexane,

Molecular mass of water H2O= 2 x 1 + 1 x 16 = 18

1 mole of water H2O= 18 g of water

According to the stoichiometric equation 2 moles of hexane will give 14 moles of water

That is 2 x 86 g of hexane will give 14 x 18 g of water

Therefore 1.32 g of hexane, C6H14 will give 14 x 18 g x 1.32g / 2 x 86 g = 1.933 g of water

1.32 g of hexane, C6H14(l) on burning will produce 1.933 g of water

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