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astraxan [27]
3 years ago
7

Calculate the mass of H2OH2O produced by metabolism of 2.4 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11

0O6C57H110O6), a typical animal fat, and assuming that during metabolism, tristearin reacts with O2O2 to form only CO2CO2 and H2OH2O.
Chemistry
1 answer:
otez555 [7]3 years ago
8 0

Answer:

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

Explanation:

2C_{57}H_{110}O_6+163O_2\rightarrow 114CO_2+110H_2O

Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g

1 kg = 1000 g

Molar mass of fat = M

M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]

Moles of fat = \frac{2400 g}{890 g/mol}=2.6966 mol

According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give  ;

\frac{110}{2}\times 2.6966 mol=148.31 mol of water

Mass of 148.31 moles of water ;

148.31 mol × 18 g/mol = 2,669.58 g

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

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2 years ago
Sodium carbonate reacts with silver nitrate according to the following balanced equation: Na2CO3 (s) + 2 AgNO3 (aq) → Ag2CO3 (s)
klemol [59]

Answer:

a) 2.01 g

Explanation:

  • Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃

First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:

  • 0.0302 mol AgNO₃ * \frac{1molNa_2CO_3}{2molAgNO_3}  = 0.0151 mol Na₂CO₃

So the remaining Na₂CO₃ moles are:

  • 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃

Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:

  • 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃

The closest answer is option a).

8 0
3 years ago
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2.Robert Hooke was the first to use the word “cell.”

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5.Theodor Schwann theorized that animals are made of cells.

6.Rudolf Virchow theorized that cells come from other pre-existing cells.

3 0
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Methanol can be produced by the following reaction: CO(g) 2 H2(g) CH3OH(g). How is the rate of disappearance of hydrogen gas rel
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Answer:

r_{H_2}=-2r_{CH_3OH}

Explanation:

Hello!

In this case, for the reaction:

CO(g)+ 2 H_2(g) \rightarrow CH_3OH(g)

In such a way, via the rate proportions, that is written considering the stoichiometric coefficients, we obtain:

\frac{1}{-1} r_{CO}=\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}

Whereas the reactants, CO and H2 have negative stoichiometric coefficients; therefore the rate of disappearance of hydrogen gas is related to the rate of appearance of methanol as shown below:

\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}\\\\r_{H_2}=\frac{-2}{1} r_{CH_3OH}\\\\r_{H_2}=-2r_{CH_3OH}

Which means that the rate of disappearance of hydrogen gas is negative and the rate of appearance of methanol is positive.

Regards!

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