BRAINLIEST ANSWER.... How many bonding and nonbonding electrons are on the nitrogen atom in NO2?

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

otez555 [7]

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.

8 0

2 years ago

Is the first part "kilo" from the word kilograms a suffix, prefix or unit???

Nikolay [14]

Answer:

Kilo is a prefix

Explanation:

Prefixes are the beginning of words.

Thus KILOgram is at the beginning.

Suffix is at the END of a word

Unit is the body of the word

7 0

3 years ago

Read 2 more answers

A sample of air is slowly passed through aqueous Sodium hydroxide and then over heater copper. Which gases are removed by this p

bagirrra123 [75]

Carbon dioxide and oxygen are removed from the air.

Explanation:

When air is passed through aqueous sodium hydroxide solution the carbon dioxide is removed from the air.

First the carbon dioxide will dissolve and react with water to form carbonic acid ( H₂CO₃) :

CO₂ + H₂O → H₂CO₃

The the carbonic acid will react with sodium hydroxide to form sodium carbonate (Na₂CO₃):

H₂CO₃ + 2 NaOH → Na₂CO₃ + 2 H₂O

After this by passing the air over heated cooper the oxygen is removed.

2 Cu + O₂ → 2 CuO

Learn more about:

neutralization reaction

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7 0

3 years ago

The standard Gibbs free energy of formation of ________ is zero. (

zvonat [6]

The standard Gibbs free energy of formation ΔGf° of Rb(s), H2(g) and Pb(s) are all zero. Similar to enthalpies of formation, the values of the standard Gibbs energies of formation are zero for the elements in their most stable forms at room conditions 298 Kelvin and one atmosphere pressure.

6 0

2 years ago

At standard temperature and pressure (0 ∘C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L . What volume wo

Nostrana [21]

Taking into account the Charles's law, the same amount of gas at the same pressure and 65 ∘C would occupy a volume of 27.73 L.

<h3>Charles's Law</h3>

Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, at a constant pressure.

Volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles's law is a law that says that the quotient that exists between the volume and the temperature will always have the same value:

V÷ T= k

Considering an initial state 1 and a final state 2, it is satisfied:

V1÷ T1= V2÷ T2

<h3>Volume at 65°C</h3>

In this case, you know:

V1= 22.4 L
T1= 0 C= 273 K
V2= ?
T2= 65 C= 338 K

Replacing in Charles's law:

22.4 L÷ 273 K= V2÷ 338 K

Solving:

(22.4 L÷ 273 K) ×338 K= V2

Finally, the same amount of gas at the same pressure and 65 ∘C would occupy a volume of 27.73 L.

Learn more about Charles's law:

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3 0

1 year ago