Fictitious element X has an a average atomic mass of 254.9 and only 2 isotopes, One of its isotopes has an abundance of 72.00 and a mass of 250.9 amu. What is the atomic mass of the second isotope?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

Solution:

The average atomic mass of an element is the sum of the masses of its isotopes, each multiplied by its natural abundance.

Average atomic mass = f1M1 + f2M2 +… + fnMn where f is the fraction representing the natural abundance of the isotope and M is the mass number (weight) of the isotope.

In our case: Average atomic mass = f1M1 + f2M2

Average atomic mass = 254.9 amu

f1 = 72.00% or 0.72

M1 = 250.9 amu

f2 = (1 - f1) = (1 - 0.72) = 0.28

M2 = unknowm

Hence,

Average atomic mass of X = f1M1 + f2M2

254.9 amu = (0.72 × 250.9 amu) + (0.28 × M2)

254.9 amu = 180.648 amu + 0.28 × M2

M2 = (254.9 amu - 180.648 amu) / 0.28 = 265.186 amu = 265.2 amu

M2 = 265.2 amu

Answer: The atomic mass of the second isotope is 265.2 amu.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
<pre class="editor-code"><code class="prism">What is the molar mass of the fictitious compound Hg4Cl2N5 Which is the correct name for the fictitious compound P7O10 </code></pre>

Molar mass of the fictious compound Hg4Cl2N5(200.59X4)+(35.5X2)+(14X5)=943.36 gramsThe name of the fictious compound P7O10=Heptaphosphorus decoxide

1 Answers 1 views
lead-204 203.973amu, 1.40% abundance. lead-206 205.974 amu, 24.10% abundance. lead-207???, abundance 22.10%. lead-208 207.977, abundance 52.40%. Calculate the isotope mass of lead-207

=207.2 amu (Pb)

1 Answers 1 views
Natural boron is composed of two isotopes: 19.78% boron-10 (atomic mass = 10.0129 amu) and 80.22% boron-11 (atomic mass = 11.0093 amu). What is the average atomic mass of naturally occurring boron?

Average atomic mass=(19.78%*10.0129amu + 80.22%*11.0093amu)/100%=10.8122amu

1 Answers 1 views
Suppose that an element has two isotopes. The average atomic mass of the element is 49.237 u. One isotope has a mass of 41.834 u and an abundance of 35.94%. Calculate the mass of the second isotope.

Solution:The average atomic mass of an element is the sum of the masses of its isotopes, each multiplied by its natural abundance.Average atomic mass = f1M1 + f2M2 +… + fnMn,where f...

1 Answers 1 views
silver has two naturally occurring isotopes, one of mass 106.91 amu and the other of mass 108.90amu. FInd the fractional abundances for these two isotopes. The atomic weight is 107.87 amu

Notice that the abundance of Ag-107 (106.91 amu) is assigned “x” and the Ag-109 (108.90 amu) is “1 – x”. 106.91x + 108.90(1 – x) = 107.87 106.91x + 108.90...

1 Answers 1 views
What is the mass in amu of 2 molecules of Triuranium Octaoxide (U3O8)? A. 254 amu B. 208 amu C. 842 amu D. 1684 amu E. 1952 amu

The mass in amu of 2 molecules of Triuranium Octaoxide (U3O8) is 1684 (D).

1 Answers 1 views
Gallium has two isotopes. One isotope is 69-Ga and has a relative abundance of 60.50%. The relative atomic mass of gallium is 69.70 amu. Find the relative isotopic mass of the other isotope.

Ar(Ga)=0.3950*x + 0.6050*69=69.70;x=(69.70-41.745)/0.3950=70.77Answer: 70.77 amu

1 Answers 1 views
Calculate the nuclear binding energy per nucleon for the He nucleus. (atomic masses: He = 4.0026 amu; H = 1.0078 amu; n = 1.0087 amu; 1 amu = 1.6605 x 10-27 kg)

E= mc^2.E=1.66x10-27 x (3 x 10^8)^2=1,494e-10 kJ.

1 Answers 1 views
Element X has two naturally occurring isotopes, 10X (18.98 % abundance) and 11X (81.02 % abundance). What is the atomic mass of Element X?

We know the atomic mass= (1/100)× Σ(% abundance ×mass no.) of isotopesGiven,abundance of 10X isotope = 18.98 % Mass no. = 10abundance of 11X isotope = 81.02 %Mass no =...

1 Answers 1 views
What percentage of the mass of the isotope boron-10 do its neutrons contribute? proton = 1.0073 amu neutron = 1.0087 amu electron = 5.4858×10<sup>-4</sup> amu ............%

5 neutron 5 protons and 5 electrons.neutron = 5×1.0087 = 5.0435total = 5×1.0073 + 5×1.0087 + 5×5.4858x10-4total = 5.0365 + 5.0435 + 0.0027429total = 10.0827429percentage of neutron = (neutron/total )x100=[...

1 Answers 1 views
X