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s344n2d4d5 [400]
3 years ago
6

Which corresponds to a pressure of 1.23 atm? 8.73 mmHg 9.23 mmHg 125 mmHg 935 mmHg

Chemistry
2 answers:
Sergeu [11.5K]3 years ago
7 0

 The value  which  corresponds to  pressure  of 1.23  atm    is 935  mm hg

 <u><em>calculation</em></u>

convert  1.23 atm  to  mmhg


that is   1 atm = 760   mm hg

            1.23 atm =?    mm hg

 by cross  multiplication

={( 1.23 atm  x 760  mm hg) / 1 atm} =935 mm hg

 

rosijanka [135]3 years ago
5 0
(935mmHg/760mmHg)=1.23 atm
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Iridium has only two naturally occurring isotopes. The mass of iridium-191 is 190.9605 amu and the mass of iridium-193 is 192.96
cluponka [151]

Answer:

Abundance of Iridium-193 is 62.75%

Explanation:

From the question given above, the following data were obtained:

Isotope A (Iridium-191):

Mass of A = 190.9605 amu

Abundance of A = A%

Isotope B (Iridium-193):

Mass of B = 192.9629 amu

Abundance B = (100 – A) %

Relative atomic mass of Iridium = 192.217 amu

Next, we shall determine the abundance of isotope A (Iridium-191). This can be obtained as follow:

Relative atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

192.217 = [(190.9605 × A%)/100] + [(192.9629 × (100 – A)%)/100]

192.217 = 1.909605A% + 1.929629(100 – A)%

192.217 = 1.909605A% + 192.9629 – 1.929629A%

Collect like terms

192.217 – 192.9629 = 1.909605A% – 1.929629A%

–0.7459 = –0.020024A%

Divide both side by –0.020024

A% = –0.7459 / –0.020024

A% = 37.25 %

Finally, we shall determine the abundance of Isotope B (Iridium-193).

This can be obtained as follow:

Abundance of A (Iridium-191) = 37.25 %

Abundance of B (Iridium-193) =?

Abundance B = 100 – A%

Abundance B = 100 – 37.25 %

Abundance of B (Iridium-193) = 62.75%

4 0
3 years ago
Consider the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) Using standard thermodynamic data at 298K, calculate the entropy chang
balandron [24]

Answer:

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∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants

∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]

∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K

∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.

For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,

then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.

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