A certain element XX has four isotopes. 5.845% of XX has a mass of 53.93961 amuamu. 91.75% of XX has a mass of 55.93494 amuamu. 2.123% of XX has a mass of 56.93539 amuamu. 0.2820% of XX has a mass of 57.93328 amuamu. What is the average atomic mass of element XX? Express your answer numerically to four signif

55.85 amu

Explanation:

Let A, B, C and D represent the four isotopes of element X.

The following data were obtained from the question:

Isotope A:

Abundance (A%) = 5.845%

Mass of A = 53.93961 amu

Isotope B:

Abundance (B%) = 91.75%

Mass of B = 55.93494 amu

Isotope C:

Abundance (C%) = 2.123%

Mass of C = 56.93539 amu

Isotope D:

Abundance (D%) = 0.2820%

Mass of D = 57.93328 amu

Average atomic mass of element X =.?

The average atomic mass of X can be obtained as follow:

Average Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100] + [(Mass of D × D%)/100]

Average Atomic mass = [(53.93961 × 5.845)/100] + [(55.93494 × 91.75)/100] + [(56.93539 × 2.123)/100] + [(57.93328 × 0.2820)/100]

= 3.15277 + 51.32031 + 1.20874 + 0.16337

= 55.84519 ≈ 55.85

Therefore, the average atomic mass of X is 55.85 amu

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

l dont know l am s sorry