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otez555 [7]
2 years ago
12

A student placed 11.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 40.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution

Chemistry
2 answers:
Alik [6]2 years ago
5 0

Answer:

0.459 gram

Explanation:

Find the attachment

Andreas93 [3]2 years ago
3 0

Answer:

There is 0.92 g of glucose in 100 mL of the final solution.

Explanation:

Initially, 11.5 g of glucose is added to the volumetric flask

Water is then added to 100 mL Mark,

The flask was then shaken until the solution was uniform.

The shaking of the mixture makes the concentration of glucose to become uniform all through the solution.

At this point, the concentration of this solution in g/mL is (11.5/100) = 0.115 g/mL

A 40.0 mL sample of this glucose solution was diluted to 0.500 L.

40.0 mL of the already mixed solution is then diluted to 0.500 L.

The mass of glucose in 40.0 mL of the mixed solution with concentration 0.115 g/mL is then given as

Mass = (conc in g/mL) × (volume) = 0.115 × 40 = 4.6 g

So, this mass is then diluted to 0.500 L mark.

New concentration = (mass)/(conc In mL) = (4.6/500) = 0.0092 g/mL

How many grams of glucose are in 100. mL of the final solution

Mass = (conc in g/mL) × (Volume in mL) = 0.0092 × 100 = 0.92 g

Hope this Helps!!!

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A certain electrochemical cell has for its cell reaction: zn + hgo → zno + hg which is the half-reaction occurring at the anode?
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When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
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Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

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Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

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