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3 years ago

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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If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M TRIS base, what will be the resulting pH?

Katyanochek1 [597]

Answer: The pH of resulting solution is 8.7

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For TRIS acid:

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol$

For TRIS base:

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

$0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol$

Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of TRIS acid = 8.3

$[\text{TRIS acid}]=\frac{0.005}{0.11}$

$[\text{TRIS base}]=\frac{0.012}{0.11}$

pH = ?

Putting values in above equation, we get:

$pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7$

Hence, the pH of resulting solution is 8.7

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3 years ago

I need help with problem 103 all of the questions.

AfilCa [17]

A. is decomposition so HCL = H2 + Cl2

not balanced cause hcl needs 2

2HCL = H2 + Cl2

balanced

b. Br2 + Al-i = AlBr3 + I2 single rep.

not balanced since br need 3 so watch carefully cause many changes needed

3Br2 + Al-i = AlBr3 + I2 not right is unbalanced so make it 2

3Br2 + Al-i = 2AlBr3 + I2 now left Al is unbal. so make 2 there

3Br2 + 2Ali = 2AlBr3 + I2

Balanced

C. Na + S = Na2S synthesis reaction is not bal. left Na needs 2

2Na + S = Na2S balanced.

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pav-90 [236]

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