Answer:
28/95 = 29.,5 % of Arsine decomposed
Explanation:A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom- poses to solid arsenic and hydrogen gas. The flask is then cooled to 273 K, at which tem- perature the pressure in the flask is 488 Torr. What percentage of arsine molecules have de- composed?
Answer in units of %.
initial pressure 332 Torr initial volume 0.46 L initial temperature 223K
final pressure 488 Torr final volume 0.46 L final 273 K
Torr is 1/760 atm 332 torr = 0.437 atm 488 Torr =0.642 atm
PV = nRT so n=RT/PV
INITIAL n= 0.082 X 223/(0.437)(0.46) = 91 moles
final n= 0.082 X 273 / (.437)(488) = 105 moles
2AsH3----------> 2As + 3H2
x moles of Arsine decomposed to make 1.5 moles of H2
the final number of moles was
(91 -X)+ 1.5 X = 105 moles
91 + 0.5 X = 105
0.5 X = 14
X =28
CHECK
if 28 moles of Arsine , then the container would have
91 --28 + 1.5(28) = 91 +14 =105 check
so 28/95 = 29.,5 % of Arsine decomposed
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The answer would be:
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The table below gives the atomic mass and relative abundance values for the three isotopes of element M.
Relative abundance (%) Atomic mass (amu)
78.99 23.9850
10.00 24.9858
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What is the average atomic mass (in amu) of element M?
2.86
5.36
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24.98
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Multiply the relative abundance by the atomic mass of each isotope, add the results together, and divide by 100. Then round to the correct amount of significant figures.
(78.99 x 23.9850) + (10.00 x 24.9858) + (11.01 x 25.9826)/100
24.30 is the average atomic mass (in amu) of element M.
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Multiply the relative abundance by the atomic mass of each isotope, add the results together, and divide by 100. Then round to the correct amount of significant figures.
(78.99 x 23.9850) + (10.00 x 24.9858) + (11.01 x 25.9826)
2430.501576 ÷ 100
24.30
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