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liq [111]
1 year ago
12

A certain element consists of two stable isotopes. The first has atomic mass of 7.02 amu and a percent natural abundance of 92.6

%. The second has an atomic mass of 6.02 amu and a percent natural abundance of 7.42%. What is the atomic weight of the element? in amu
Chemistry
1 answer:
statuscvo [17]1 year ago
8 0

<u>Answer:</u> The average atomic mass of the element is 6.95 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

We are given:

Mass of isotope 1 = 7.02 amu

Percentage abundance of isotope 1 = 92.6 %

Fractional abundance of isotope 1 = 0.926

Mass of isotope 2 = 6.02 amu

Percentage abundance of isotope 2 = 7.42 %

Fractional abundance of isotope 2 = 0.0742

Putting values in equation 1, we get:

\text{Average atomic mass of element}=[(7.02\times 0.926)+(6.02\times 0.0742)]

\text{Average atomic mass of element}=6.95amu

Hence, the average atomic mass of the element is 6.95 amu

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A student has two samples of NaCl, each one from a different source. Assume that the only potential contaminant in each sample i
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Answer:

The correct option is;

A. Which sample has the higher purity

Explanation:

The information given relate to the presence of two samples of NaCl, from different sources

The only potential contaminant in each of the sources = KCl

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The molar mass of NaCl = 58.44 g/mol

The molar mass of KCl = 74.5513 g/mol

Let the number of moles of KCl in the sample = X

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The molar mass of potassium = 39.0983 g/mol

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Therefore;

Each mole of KCl, will yield 35.453 g/mol per 74.5513 g/mol of KCl

While each mole of NaCl will yield 35.453 g/mol per 58.44 g/mol of NaCl

Therefore, the pure sodium chloride sample will yield more chlorine per unit mass of sample.

As such if the two samples have the same mass, the sample with the contaminant of KCl will yield less mass of chlorine per unit mass of the sample, from which the student will be able to tell the purity of the solution.

The sample with the higher purity will yield  a higher mass chlorine per unit mass of the sample.

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2 years ago
Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t
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Answer:

Explanation:

N₂       + 3H₂     =     2 NH₃

1 vol                         2 vol

786 liters               1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }

V_2 =303.44 liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

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5 0
2 years ago
Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.
Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

For the given chemical reaction:

C(s)+2H_2(g)\rightarrow CH_4(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ

Hence, the enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

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