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BlackZzzverrR [31]
3 years ago
11

The abundance of 12c is 98.9%. 13c is another naturally occurring isotope. what is the percent abundance of 13c?

Chemistry
1 answer:
zepelin [54]3 years ago
8 0

Actually the total abundance of the isotopes of any element in this world must sum up to 100%. So we initially know that 12 C is 98.9 percent abundant, therefore the remaining of the 100 percent must be of 13 C, that is:

 

13 C = 100% - 98.9%

<span>13 C = 1.1% </span>

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The fact that some of the solid was transferred would decrease the mass of the limiting reactant.

<h3>What is the limiting reactant?</h3>

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Answer : The correct option is, (D) 89.39 KJ/mole

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First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

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From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

q=274g\times 4.18J/g^oC\times (325.8-298)K

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Now we have to calculate the enthalpy of neutralization.

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where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

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