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Rudik [331]
3 years ago
7

What do carbon 12 carbon 13 and carbon 14 have in common?

Chemistry
1 answer:
tiny-mole [99]3 years ago
7 0

Answer:

They have all the same number of protons and electrons ( 6)

Explanation:

Carbon 12, Carbon 13 and Carbon 14 are all isotopes of Carbon.

The definition of an isotope is: They have the same number of protons (and electrons), but a different number of neutrons. Different isotopes of the same element have different masses.

Carbon 12 means an isotope with mass 12 u

it has 6 protons,(and 6 electrons) and 6 neutrons : 6 neutrons + 6 protons = 12

Carbon 13 is an isotope with mass 13 u

It has the same amount  of protons  ( and electrons) as Carbon 12, so 6 protons and 6 electrons

13 - 6 = 7 ⇒ Carbon 13 has 7 neutrons

Carbon 14 is an isotope with mass 14

it has the same of protons ( and electrons) as Carbon 12 and Carbon 13, so 6 protons and 6 electrons

14 - 6 = 8 ⇒ Carbon 14 has 8 neutrons

We can conclude that Carbon 12, Carbon 13 and Carbon 14 are 3 isotopes of Carbon. They all have the same number of protons (and electrons) = 6.

This means the isotopes will also have the same atomic number, because they are all isotopes of the same element.

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The possible products of a double displacement reaction in aqueous solution are SrSO4 and NaCl. Which of these possible products
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SrSO4

Explanation:

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3 years ago
An empty beaker has a mass of 75.0. Salt is added to the beaker until it has a total mass of 108.06 g. Using the
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1 year ago
A 2.00L flask was filled with 4.00 mol of HI at a certain temperature and given sufficient time to react. At equilibrium the con
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Answer:

The equilibrium concentration of I₂ is 0.400 M  and HI is 1.20 M, the Keq will be 0.112.

Explanation:

Based on the given information, the equilibrium reaction will be,  

2HI (g) ⇔ H₂ (g) + I₂ (g)

It is given that 4.00 mol of HI was filled in a flask of 2.00 L, thus, the concentration of HI will be,  

= 4.00 mol/2.00 L

= 2.00 mol/L

Based on the reaction, the initial concentration of 2HI is 2.00, H₂ is 0 and I₂ is O. The change in the concentration of 2HI is -x, H₂ is x and I₂ is x. The equilibrium concentration of 2HI will be 0.200-x, H₂ is x and I₂ is x.  

It is given that at equilibrium, the concentration of H₂ or x is 0.400 M.  

Now the equilibrium concentration of HI will be,  

= 2.00 -2x  

= 2.00 - 2 × 0.400

= 1.20 M

The equilibrium concentration of I₂ will be,  

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= 0.400 M

The equilibrium constant (Keq) will be,  

Keq = [H₂] [I₂] / [HI]²

= (0.400) (0.400) / (1.20)²

= 0.112

Thus, the Keq of the reaction will be 0.112.  

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3 years ago
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