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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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The molar heat capacity of an unknown substance is 92.1 J/mol-K. If the unknown has a molar mass of 118 g/mol, what is the speci

dlinn [17]

Answer : The specific heat (J/g-K) of this substance is, 0.780 J/g.K

Explanation :

Molar heat capacity : It is defined as the amount of heat absorbed by one mole of a substance to raise its temperature by one degree Celsius.

1 mole of substance releases heat = 92.1 J/K

As we are given, molar mass of unknown substance is, 118 g/mol that means, the mass of 1 mole of substance is, 118 g.

As, 118 g of substance releases heat = 92.1 J/K

So, 1 g of substance releases heat = $\frac{92.1}{118}=0.780J/g.K$

Thus, the specific heat (J/g-K) of this substance is, 0.780 J/g.K

8 0

3 years ago

Thermal energy always moves from ___ to ___

Inessa05 [86]

Answer:

hot to cold

Explanation:

pls mark brainliest

5 0

2 years ago

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The gas cyclobutane, C4H8(g), can be used in welding. When cyclobutane is burned in oxygen, the reaction is: C4H8(g) + 6 O2(g)4

Snowcat [4.5K]

Answer:

$\Delta H^o _{rxn} = -2568.9 \ kJ$

$H = 350 JK^{-1}$

$T_{max} = 32.4 ^o C$

Explanation:

From the question we are told that

The reaction of cyclobutane and oxygen is

$C_4H_8_{(g)} + 6 O_2_{(g)} \to 4 CO_2_{(g)} + 4 H_2O_{(g)}$

ΔH°f (kJ mol-1) : C4H8(g) = 27.7 ; CO2(g) = -393.5 ; H2O(g) = -241.8 ΔH° = kJ

Generally ΔH° for this reaction is mathematically represented as

$\Delta H^o _{rxn} = [[4 * \Delta H^o_f (CO_2_{(g)} ) + 4 * \Delta H^o_f(H_2O_{(g)} ] -[\Delta H^o_f (C_2H_6_{(g)} + 6 * \Delta H^o_f (O_2_{(g)}) ] ]$

=> $\Delta H^o _{rxn} = [[4 * (-393.5) + 4 * (-241.8) ] -[ 27.7 + 6 * 0]$

=> $\Delta H^o _{rxn} = -2568.9 \ kJ$

Generally the total heat capacity of 4 mol of CO2(g) and 4 mol of H2O(g), using CCO2(g) = 37.1 J K-1 mol-1 and CH2O(g) = 33.6 J K-1 mol-1. C = J K-1 is mathematically represented as

$H = [ 4 * C_{CO_2_{(g)}} + 6* C_{CH_2O_{(g)}}]$

=> $H = [ 4 * 37.1 + 6* 33.6 ]$

=> $H = 350 JK^{-1}$

From the question the initial temperature of reactant is $T_i = 25^oC$

Generally the enthalpy change( $\Delta H^o _{rxn}$ ) of the reaction is mathematically represented as

$|\Delta H^o _{rxn} |= H * (T_{max} -T_i)$

$2568.9 = 350 * (T_{max} -25)$

=> $\frac{2568.9 }{350} = T_{max} - 25$

=> $T_{max} = 32.4 ^o C$

4 0

2 years ago

What is the relationship of rocks to minerals.

Dovator [93]

Rocks are made up of minerals

8 0

3 years ago

Read 2 more answers

Practice 14.2: Predict the equilibrium constant for the first reaction given the equilibrium constants for the second and third

Finger [1]

Reverse the 2nd reaction,
CO2 + H2 -------> CO + H2O---- i

Now new equilibrium constant will be,
K' = 1/K2 = 10^-5

Adding i and last equation you'll get,
<span>CO2 (g) + 3 H2 (g)----->CH3OH (g) + H2O (g)
K1 = K' x K3 = 1.4 x 10^2 </span>

5 0

3 years ago

Read 2 more answers