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seraphim [82]
2 years ago
5

Calculate the mass of water produced by metabolism 84.0 g of glucose

Chemistry
1 answer:
scoundrel [369]2 years ago
8 0

The amount of water that will be produced is 50.36 grams

<h3>Stoichiometric problems</h3>

The metabolism of glucose is represented by the following equation:

C_6H_1_2O_6(s)+6O_2(g)--- > 6CO_2(g)+6H_2O(g)

The mole ratio of glucose metabolized to the water produced is 1:6.

Mole of 84.0 g glucose = 84/180.156 = 0.4662 moles

Equivalent mole of water = 0.4662 x 6 = 2.7975 moles

Mass of 2.7975 moles water = 2.7975 x 18 = 50.36 grams

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

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There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
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Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is -470.4 kJ/mol.

Explanation:

Step 1 : Calcium carbide and water react to form acetylene and calcium hydroxide

CaC_2(s)+2H_2O(g)\rightarrow

             C_2H_2(g)+Ca(OH)_2(s),\Delta H_1=-414 kJ..[1]

Step 2 : Acetylene, carbon dioxide and water react to form acrylic acid

6C_2H_2(g)+3CO_2(g)+4H_2O(g)\rightarrow 5CH_2CHCOOH(g),\Delta H_2=132 kJ..[2]

Using Hess's law:

[1] × 6 + [2]

6CaC_2(s)+16H_2O(g)+3CO_2\rightarrow 5CH_2CHCOOH(g)+6Ca(OH)_2(s),\Delta H_3=?

\Delta H_3=6\times \Delta H_1+\Delta H_2

\Delta H_3=6\times (-414 kJ)+132 kJ=-2352 kJ

The energy released on formation of 5 moles of acrylic acid = -2352 kJ

The energy released on formation of 1 moles of acrylic acid :

=\frac{-2352 kJ}{5 mol}=-470.4 kJ/mol

Hence, the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is -470.4 kJ/mol.

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Answer:

Explanation:

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Atom Y,

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3) True. Atom X and atom Y belong to the same period two on the periodic table

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