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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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Answer:

Hydrogen bonded materials

Explanation:

Metals have metallic bonds, polymers usually have covalent bonds and ceramic materials have covalent and ionic bonds; on the other hand, hydrogen bonds are a type of bonding characterized for have Van der Waals interactions, a type of intermolecular interactions, so the correct answer is Van der Waals interactions.

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3 years ago
A container of gas has a volume of 3.5 L and a pressure of 0.8 atm. Assuming the temperature remains constant, what volume of ga
strojnjashka [21]

Answer:

The correct answer is option A.

Explanation:

Initial volume of the gas =V_1=3.5 L

Final volume of the gas = V_2=?

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Final volume of the gas = P_2=0.5 atm

Using Boyle's law:

P_1V_1=P_2V_2

0.8 atm \times 3.5 L=0.5 atm\times V_2

V_2=5.6 L

Hence,the correct answer is option A.

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3 years ago
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A. By referring to forces at work in the atomic nucleus, how can you
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If a substance conducts heat easily, it is considered to be a
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A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet
stira [4]

Explanation:

The given data is as follows.

           Mass of apple sauce mixture = 454 kg

           Heat added (Q) = 121300 kJ

 Heat capacity (C_{p}) of apple sauce at 32.8^{o}C = 4.0177 kJ/kg^{o}C

So, Heat given by heat exchanger = heat taken by apple sauce

                            Q = mC_{p} \Delta T

or,                    Q = mC_{p} (T_{f} - T_{i})  

Putting the given values into the above formula as follows.

                     Q = mC_{p} (T_{f} - T_{i})  

              121300 kJ = 454 kg \times 4.0177 kJ/kg^{o}C \times (T_{f} - 10)

                      T_{f} = 76.5^{o}C

Thus, we can conclude that outlet temperature of the apple sauce is 76.5^{o}C.

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