Atomic weight can be calculated by using following formula
atomic weight= massa × fracta + massb × fracb+ massc × fracc
Where, a,b and c are used for first,second and third isotopes.
We need to convert percentage into fractions by dividing each percentage term by 100.
Now,
Atomoic weight = 27.977 × 0.9223 + 28.976 × 0.04679 + 29.974 × 0.031
= 25.8031871 + 1.35578704 + 0.929194
= 28.0881681
= 28.09 amu
Therefore, first option is the right option.
Naturally occurring element X exists in three isotopic forms X-28 (27.977 amu 02 23% abundance) X-29...
References Naturally occurring element X exists in three isotopic forms: X-28 (27.970 amu, 92.21% abundance), X-29 (28.976 amu, 4.70% abundance), and X-30 (29.974 amu, 3.09% abundance). Calculate the atomic weight of X A. 25.79 amu B. 28.08 amu C. 86.92 amu D. 35.28 amu E. 29.08 amu Submit Answer Try Another Version 1 item attempt remaining
QUESTION 1 The formula mass of ammonium phosphite, (NH 4) 3PO 3, is a. 153.11 amu O b. 125.01 amu c. 97.01 amu d. 133.09 amu QUESTION 2 Naturally occurring element X exists in three isotopic forms: X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abundance), and X-30 (29.974 amu, 3.10% abundance). Calculate the atomic weight of X. a. 27.16 amu b.28.97 amu O c. 86.93 amu d. 48.63 amu e. 28.09 amu QUESTION 3 Balance the following equation...
In nature, the element X consists of two naturally occurring isotopes. 107X with abundance 54.84% and isotopic mass 106.9051 amu and 109X with isotopic mass 108.9048 amu. Use the given information to calculate the atomic mass of the element X to an accuracy of .001% (Report your answer like this yyy.yyyy) Atomic Mass = amu.
the element X has naturally occurring isotope. The masses (amu) and % abundance of the isotopes are given below. the average atomic mass of the element is ? Isotope: 221X, 220X, 218 X Abundance: 74.22, 12.78, 13.00 Mass: 220.9, 220.0, 218.1
The element carbon has two naturally occurring isotopes. The isotopic masses and abundances of these isotopes are shown in the table below. Isotope 12c isotopic mass (amu) Abundance (%) 12.00 13.00 98.93 1.07 Calculate the average atomic mass of carbon to two digits after the decimal point. Number = _______ amu
The element X has three naturally occurring isotopes. The masses (amu) and % abundances of the isotopes are given in the table below. The average atomic mass of the element is _ _amu. Isotope Abundance (%) Mass (amu) 221X 74.22 220.90 12.78 220.00 2187 13.00 218.10 2207 219.70 220.34 220.43 219.00 33.333
An unknown element (Element X) has three naturally occurring isotopes. Complete the table by filling in the missing percent abundance (2 decimal places). Then calculate the atomic mass of element X (1 decimal place) and determine its identity by filling in its atomic symbol (case sensitive). Isotope Abundance (%) Atomic Mass (amu) 23.985042 1 78.99 2 24.985837 10.00 3 25.982593 amu Atomic mass of element X (1 decimal place): Atomic symbol of element X:
A hypothetical element X has 3 naturally occurring isotopes: 41.20% of 21X, with an atomic weight of 21.016 amu, 6.83% of 22X, with an atomic weight of 21.942 amu, and 51.97% of 24X, with an atomic weight of 23.974 amu. On the basis of these data, calculate the average atomic weight of element X. Report your answer to two decimal places.
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