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Contact [7]
2 years ago
7

Your body contains vast numbers of carbon atoms. How is it possible that some of these carbons may have been part of the body of

a prehistoric creature?
Chemistry
1 answer:
charle [14.2K]2 years ago
6 0
  • Carbon is reused through the carbon cycle.
  • There is carbon in oil, which is made of ancient animals, and this carbon is discharged into the air during ignition.
  • Carbon is then taken in by plants as CO2, which we then eat.

<u>Explanation</u>:

  • Carbon is reused through the carbon cycle. There is carbon in oil, which is made of ancient animals, and this carbon is discharged into the air during ignition. Carbon is then taken in by plants as CO2, which we then eat.  
  • Matter carries on as indicated by the guideline of conservation of matter: Under conventional conditions, any matter is neither made nor destroyed but instead is reused again and again. It tends to be changed or recombined, however, it doesn't vanish; everything heads off to someplace.  
  • Carbon particles of my body presumably used to be a piece of the body of a prehistoric creature, as chemical components have been utilized and reused by living animals.

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24 how many moles are in 2.04 × 1024 molecules of h2o?
Sergeeva-Olga [200]
The answer is 3.39 mol.

<span>Avogadro's number is the number of molecules in 1 mol of substance.
</span><span>6.02 × 10²³ molecules per 1 mol.
</span>2.04 × 10²⁴<span> molecules per x.

</span>6.02 × 10²³ molecules : 1 mol = 2.04 × 10²⁴ molecules : x
x = 2.04 × 10²⁴ molecules * 1 mol : 6.02 × 10²³ molecules
x = 2.04/ 6.02 × 10²⁴⁻²³ mol
x = 0.339 × 10 mol
<span>x = 3.39 mol
</span>
8 0
2 years ago
Exactly 500 grams of ice are melted at a temperature of 32°f. (lice = 333 j/g.) calculate the change in entropy (in j/k). (give
denpristay [2]
Entropy Change is calculated  by (Energy transferred) / (Temperature in kelvin) 
deltaS = Q / T 

Q = (mass)(latent heat of fusion) 
Q = m(hfusion) 
Q = (500g)(333J/g) = 166,500J 

T(K) = 32 + 273.15 = 305.15K 
deltaS = 166,500J / 305.15K 
deltaS = 545.63 J/K
3 0
2 years ago
Which of the following contains maximum number of atoms? a) 2.0g hydrogen b) 2.0g oxygen c) 2.0g nitrogen d) 2.0g methane
Fynjy0 [20]
A) 2.0g hydrogen is your answer
7 0
2 years ago
When a pH probe is inserted into a solution containing the chloride ion it is neutral. What is the pH of a solution containing t
vivado [14]

Answer:

a. the solution will be weakly basic.

b. Greater than 7 because CN⁻ is a stronger base than NH₄⁺ is an acid.

Explanation:

a. The fluoride ion (F⁻) reacts with water thus:

F⁻ + H₂O → HF + OH⁻

That means that fluoride ions produce OH⁻ ions in solution doing <em>the solution will be weakly basic.</em>

b. The acidic equilibrium of NH₄⁺ is:

NH₄⁺ ⇄ NH₃ + H⁺ with a ka of 5,6x10⁻¹⁰.

The basic equilibrium of CN⁻ is:

CN⁻ + H₂O → HCN + OH⁻ with a kb of 2x10⁻⁵

That means that the production of OH⁻ from CN⁻ is higher than production of H⁺ from NH₄⁺. The CN⁻ is a stronger base than NH₄⁺ is an acid.

Thus, the pH of a salt solution of NH₄CN would be <em>Greater than 7 because CN⁻ is a stronger base than NH₄⁺ is an acid.</em>

<em></em>

I hope ot helps!

3 0
2 years ago
When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at
STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

1.22 * 10^{5} = - 8.314* 2400 * ln K

\\ 1.22 * 10^{5} = -19953.6 * ln K

ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022

So, the equilibrium constant is 0.0022.

4 0
2 years ago
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