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wariber [46]
3 years ago
8

A sample of element X contains 90% X-35 atoms, 8.0% X-37 atoms, and 2.0% X-38 atoms. The average atomic mass will be closest to

which value?
Chemistry
2 answers:
Ne4ueva [31]3 years ago
6 0

To find average atomic mass you multiply the mass of each isotope by its percentage, and then add the values up.

35 * 0.90 + 37 * 0.08 + 38 * 0.02 = 35.22

Average atomic mass closest to 35.22 amu.

brilliants [131]3 years ago
4 0

Answer : The average atomic mass will be closest to 35 amu.

Explanation :

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

As we are given that,

Mass of isotope X-35 = 35 amu

Percentage abundance of isotope X-35 = 90 %

Fractional abundance of isotope X-35 = 0.90

Mass of isotope X-37 = 37 amu

Percentage abundance of isotope X-37 = 8.0  %

Fractional abundance of isotope X-37 = 0.08

Mass of isotope X-38 = 38 amu

Percentage abundance of isotope X-38 = 2.0  %

Fractional abundance of isotope X-38 = 0.02

Now put all the given values in above formula, we get:

\text{Average atomic mass of element}=\sum[(35\times 0.90)+(37\times 0.08)+(38\times 0.02)]

\text{Average atomic mass of element X}=35.22amu\approx 35amu

Therefore, the average atomic mass will be closest to 35 amu.

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5 0
3 years ago
Lt takes 4 hr 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. what is the half-life of radium-230?
Rufina [12.5K]
Radioactive decay => C = Co { e ^ (- kt) |

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Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min

Time conversion: 4 hr 39 min = 4.65 hr

1) Replace the data in the equation to find k

C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}

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2) Use C / Co  = 1/2 to find the hallf-life

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4 years ago
The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g
ra1l [238]

Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g)  ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ

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