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mihalych1998 [28]
3 years ago
15

Three isotopes of argon occur in nature – 36 18Ar, 38 18Ar, 40 18Ar. Calculate the average atomic mass of argon to two decimal p

laces, given the following relative atomic masses and the abundances of each of the isotopes: argon36 (35.97 amu; 0.337%), argon-38 (37.96 amu; 0.063%), argon-40 (39.96 amu; 99.600%). 1. 119.89 amu 2. 39.95 amu 3. 39.96 amu 4. 35.97 amu 5. None of these 6. 37.96 amu 7. 37.95 amu 8. 35.96 amu
Chemistry
1 answer:
likoan [24]3 years ago
4 0

Answer: 3) 39.96 amu

Explanation:

Mass of isotope Ar- 36 = 35.97 amu

% abundance of isotope Ar- 36= 0.337% = \frac{0.337}{100}=3.37\times 10^{-3}

Mass of isotope Ar- 38 = 37.96 amu

% abundance of isotope 2 = 0.063 % = \frac{0.063}{100}=6.3\times 10^{-4}

Mass of isotope Ar- 40 = 39.96 amu

% abundance of isotope 2 = 99.600 % = \frac{99.600}{100}=0.996

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[(35.97\times 3.37\times 10^{-3})+(37.96\times 6.3\times 10^{-4})+(39.96\times 0.996)]

A=39.96amu

Therefore, the average atomic mass of argon is 39.96 amu

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To solve the problem, we clear ΔT of the equation and then replace our data:

Q=msΔT

ΔT=Q/ms

ΔT=\frac{-14900 J}{40,7g*0,49\frac{J}{gC} }=-747,13°C

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Thus, the change in temperature of the steel bar is -747,13°C, meaning that the temperature of the bar decreases.

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-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}

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