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Dimas [21]
3 years ago
9

In a certain medical test designed to measure carbohydrate tolerance, an adult drinks 6 ounces of a 40% glucose solution. When t

he test is administered to a child, the glucose concentration must be decreased to 30%. How much 40% glucose solution and how much water should be used to prepare 6 ounces of 30% glucose solution?
Chemistry
1 answer:
Ostrovityanka [42]3 years ago
8 0

Answer:

4.5 ounces of 40% glucose solution and 1.5 ounces of water.

Explanation:

The new solution will be prepared by adding more solvent, so the relation between concentration and volume is given by:

C1xV1 = C2xV2

Where C is the concentration, V is the volume, 1 represents the initial solution and 2 the final solution.

We want 6 ounces of a solution which 0.3 glucose, so :

0.4xV1 = 0.3x6

0.4xV1 = 1.8

V1 = 4.5 ounces

So, it's necessary 4.5 ounces of the 40% glucose solution. The volume of water (Vw) will be the total volume less the volume of the solution:

Vw = 6 - 4.5

Vw = 1.5 ounces

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How much heat is required to change 25.0 g of water from solid to liquid at 0 oC? Water: ΔHfus = 334 J/g; ΔHvap= 2260J/g
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Explanation:

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Calculate the percentage yield for the reaction represented by the equation CH4 + 2O2 ? 2H2O + CO2 when 1000 g of CH2 react with
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Answer:

83.64%.

Explanation:

∵ The percent yield = (actual yield/theoretical yield)*100.

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  • We need to find the theoretical yield of CO₂:

For the reaction:

<em>CH₄ + 2O₂ → 2H₂O + CO₂,</em>

1.0 mol of CH₄ react with 2 mol of O₂ to produce 2 mol of H₂O and 1.0 mol of CO₂.

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<em>no. of moles of CH₄ = mass/molar mass</em> = (1000 g)/(16.0 g/mol) = <em>62.5 mol.</em>

<u><em>Using cross-multiplication:</em></u>

1.0 mol of CH₄ produces → 1.0 mol of CO₂, from stichiometry.

∴ 62.5 mol of CH₄ produces → 62.5 mol of CO₂.

  • We can calculate the theoretical yield of carbon dioxide gas using the relation:

∴ The theoretical yield of CO₂ gas = n*molar mass = (62.5 mol)(44.0 g/mol) = 2750 g.

<em>∵ The percent yield = (actual yield/theoretical yield)*100.</em>

actual yield = 2300 g, theoretical yield = 2750 g.

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