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jok3333 [9.3K]
2 years ago
15

Copper has two naturally occurring isotopes, ⁶³Cu (isotopic mass = 62.9296 amu) and ⁶⁵Cu (isotopic mass = 64.9278 amu). If coppe

r has an atomic mass of 63.546 amu, what is the percent abundance of each isotope?
Chemistry
1 answer:
Stells [14]2 years ago
3 0

Copper has two naturally occurring isotopes, ⁶³Cu (isotopic mass = 62.9296 amu) and ⁶⁵Cu (isotopic mass = 64.9278 amu). If copper has an atomic mass of 63.546 amu,

Abundance of ⁶³Cu and ⁶⁵Cu ; 70% , 30.848%

<h3>Calculation : </h3>

Let the natural abundance of  63,  Cu isotope be x%.

The natural abundance of  

65

Cu isotope will be 100−x.

The average atomic weight is 63.546 g.

Hence, 100×63.546=63x+65(100−x)

6354.6=63x+6500−65x=6500−2x

2x=145.4

x=72.7≃70

Hence, the natural abundance of the  63Cu isotope must be approximately 70%.

For the ⁶⁵Cu calculation is same and the answer is 30.848%

<h3>What is isotope ?</h3>

Isotopes are members of a family of elements that all have the same number of protons and different numbers of neutrons. The number of protons in the nucleus determines the atomic number of an element on the periodic table.

Example : Magnesium has three natural isotopes : ²⁴Mg, ²⁵Mg, and ²⁶Mg.

<h3>What is Atomic mass ?</h3>

Atomic mass is the mass of an atom. The SI unit of mass is the kilogram, while atomic mass is often expressed in the non-SI unit dalton (synonymous with uniform atomic mass unit). 1 Da is defined as 1/12 the mass of a free carbon-12 atom at rest in the ground state.

To know more about Atomic mass please click here : brainly.com/question/338808

#SPJ4

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2.0L of hydrogen gas is mixed with 3.0L of nitrogen gas at STP in a rigid 5.0L vessel. A reaction occurs, producing ammonia gas
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Answer:

Moles NH₃: 0.0593

0.104 moles of N₂ remain

Final pressure: 0.163atm

Explanation:

The reaction of nitrogen with hydrogen to produce ammonia is:

N₂ + 3 H₂ → 2 NH₃

Using PV = nRT, moles of N₂ and H₂ are:

N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂

H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂

The complete reaction of N₂ requires:

0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = <em>0.402 moles H₂</em>

That means limiting reactant is H₂. And moles of NH₃ produced are:

0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = <em>0.0593 moles NH₃</em>

Moles of N₂ remain are:

0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = <em>0.104 moles of N₂</em>

And final pressure is:

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P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L

<em>P = 0.163atm</em>

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<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

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

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol

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

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

pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})

We are given:

pK_a = negative logarithm of acid dissociation constant of acetic acid = 4.74

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Putting values in above equation, we get:

5=4.74+\log(\frac{[CH_3COONa]}{0.200})

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To calculate the mass of sodium acetate for given number of moles, we use the equation:

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Putting values in above equation, we get:

0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g

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Answer:

                      Empirical Formula  =  C₁H₄ or CH₄

Explanation:

Data Given:

                     Mass of C  =  12.0 g  

                     Mass of H  =  4 g  

Step 1: <u>Calculate Moles of each Element;</u>

                     Moles of C  =  Mass of C ÷ At.Mass of C

                     Moles of C  = 12.0 ÷ 12.0

                     Moles of C  =  1.0 mol

                     Moles of H  =  Mass of H ÷ At.Mass of H

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                     Moles of H  =  4.0 mol

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Step 2: <u>Find out mole ratio and simplify it;</u>

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                              1                                                      4                  

Hence,

              Empirical Formula  =  C₁H₄ or CH₄

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