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lawyer [7]
2 years ago
15

Gases emitted during volcanic activity often contain high concentrations of hydrogen sulfide and sulfur dioxide. these gases may

react to produce deposits of sulfur according to the equation: 2 h2s(g) so2(g) → 3 s(s) 2 h2o(g) for the complete reaction of 6.41 mol of hydrogen sulfide:
Chemistry
1 answer:
kompoz [17]2 years ago
7 0
There can be three possible answers to this question: the amount of moles of SO₂ gas needed to react with 6.41 mol H₂S, and the amount of S and H₂O gas produced.

Amount of SO₂:
6.41 mol H₂S (1 mol SO₂/2 mol 2 mol H₂S) = <em>3.205 moles SO₂ gas</em>

Amount of S:
6.41 mol H₂S (3 mol S/2 mol 2 mol H₂S) =<em> 9.615 moles S solid</em>

Amount of H₂O:
6.41 mol H₂S (2 mol H₂O/2 mol 2 mol H₂S) = <em>6.41 moles H₂O gas</em>
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The ________ orbital is degenerate with 5py in a many-electron atom.
arsen [322]

The ________ orbital is degenerate with 5py in a many-electron atom.


<h2>5px is the correct answer</h2>
5 0
2 years ago
What will be the pressure in atmospheres in a 750.0 mL vessel containing 5.00 g of argon gas at 15°C?
alukav5142 [94]
It is 5.00 and thats the answer
8 0
2 years ago
At 73.0 ∘c , what is the maximum value of the reaction quotient, q, needed to produce a non-negative e value for the reaction so
damaskus [11]
Here we will use the general formula of Nernst equation:

Ecell = E°Cell - [(RT/nF)] *㏑Q

when E cell is cell potential at non - standard state conditions

E°Cell is standard state cell potential = - 0.87 V

and R is a constant = 8.314 J/mol K

and T is the temperature in Kelvin = 73 + 273 = 346 K

and F is Faraday's constant = 96485 C/mole

and n is the number of moles of electron transferred in the reaction=2  

and Q is the reaction quotient for the reaction 
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔  Br2(aq) + SO2(g) +2H2O(l)

so by substitution :

0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q      → solve for Q 


∴ Q = 4.5 x 10^-26 
6 0
2 years ago
What is the stoichiometric ratio between BaCl2 and NaCl
bixtya [17]
<span>BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
7 0
2 years ago
Tritium 3 1H decays to 3 2He by beta emission. Find the energy released in the process. Answer in units of keV.
-Dominant- [34]

Answer:

The energy released in the decay process = 18.63 keV

Explanation:

To solve this question, we have to calculate the binding energy of each isotope and then take the difference.

The mass of Tritium = 3.016049 amu.

So,the binding energy of Tritium =  3.016049 *931.494 MeV

= 2809.43155 MeV.

The mass of Helium 3 = 3.016029 amu.

So, the binding energy of Helium 3 = 3.016029 * 931.494 MeV

= 2809.41292 MeV.

The difference between the binding energy of Tritium and the binding energy of Helium is: 32809.43155 - 2809.412 = 0.01863 MeV

1 MeV = 1000keV.

Thus, 0.01863 MeV = 0.01863*1000keV = 18.63 keV.

So, the energy released in the decay process = 18.63 keV.

7 0
2 years ago
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