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cestrela7 [59]
3 years ago
11

Lithium-6 has a mass of 6.0151 amu and lithium-7 has a mass of 7.0160 amu. The relative abundance of Li-6 is 7.42% and the relat

ive abundance of Li-7 is 92.58%. Based on this data alone, calculate the average atomic mass for lithium to the correct number of significant digits.
Chemistry
2 answers:
horsena [70]3 years ago
3 0
Average atomic mass is the weighted average atomic masses with regard to the relative abundance of the isotopes 
average atomic mass of Li = relative abundance of Li-6 x mass of Li-6 + relative abundance of Li-7 x mass of Li-7
average atomic mass of Li = (7.42% x 6.0151 a.m.u) + (92.58% x 7.0160 a.m.u)
                                           = 0.446 + 6.495
                                           = 6.941 amu
average atomic mass of Li is 6.941 amu
Bezzdna [24]3 years ago
3 0

Answer: The average atomic mass of Lithium is 6.94 amu.

Explanation:

Mass of isotope Li-6 = 6.0151 amu

% abundance of isotope  Li-6 = 7.42%= \frac{7.42}{100}=0.0742

Mass of isotope Li-7=  7.0160

% abundance of isotope Li-7 = 92.58% = \frac{92.58}{100}=0.9258

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(6.0151\times 0.0742)+(7.0160\times 0.9258)]

A=6.94

Therefore, the average atomic mass of Lithium is 6.94 amu.

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Trava [24]

Answer:

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Explanation:

4 0
3 years ago
Why is there not a corresponding drop in ionization energies between Bi and Po in the sixth period as N and O?
Stels [109]

Answer:

See explanation

Explanation:

According to the Journal of Chemical Education, Volume 80, No.8 (2003); "The first ionization energy of bismuth appears to be  anomalous......It has been claimed that spin– orbit coupling by the Russell–Saunders scheme would lower  the ground state of Bi+ ..."

However, the involvement of  d and f orbitals in Bi and Po implies that the outermost orbitals are poorly screened hence the drop between nitrogen and oxygen is not observed between Bi and Po.

The same argument could be extended to explain the reason why  there not a corresponding drop between Ba and Tl is the sixth period even though they are in the same group as Be and B.

3 0
3 years ago
How many moles are equal to 3.5 grams of beryllium
tamaranim1 [39]

Answer:

w

Explanation:

7 0
3 years ago
A student investigated how much solid was dissolved in sea water.
SashulF [63]

Answer:

weighing balance/analytical balance

Graduated cylinder/buret

Explanation:

The mass of the evaporating basin could be measured using a weighing balance or an analytical balance. Both are classified as weighing scales but the analytical balance can measure the mass of objects up to 4 decimal places, thus, providing better accuracy in measurement than ordinary weighing balance that can only measure up to 2 decimal places.

In order to measure 50 cm3 of the sea water, a graduated cylinder or a buret can be used. Both equipment can measure up to the same decimal places and, thus, have virtually the same accuracy.

3 0
2 years ago
Calculate the number of milliliters of 0.710 M Ba(OH)2 required to precipitate all of the Mn2+ ions in 161 mL of 0.796 M KMnO4 s
USPshnik [31]

<u>Answer:</u> The volume of barium hydroxide is 183 mL.

<u>Explanation:</u>

To calculate the moles of cadmium nitrate, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}       .....(1)

Molarity of MnSO_4 = 0.796 M

Volume of MnSO_4 = 161 mL = 0.161 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)

By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with = \frac{1}{1}\times 0.13=0.13mol of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So, 0.183L=0.183\times 1000=183mL

Hence, the volume of barium hydroxide is 183 mL.

7 0
3 years ago
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