First, let's write the givens in the form of a chemical equation:
3A + B ...................> 4X + 2Y
Now we find that this equation implies the following:
For every 4X and 2Y formation, 3A and 1B must disappear (react).
Comparing this implication to the above choices, we find that the right answer is: <span>The rate of formation of X is four times the rate of disappearance of B.</span>
Answer:
There is nothing to answer lol
Answer:
The 12L helium tank pressurized to 160 atm will fill <em>636 </em>3-liter balloons
Explanation:
It is possible to answer this question using Boyle's law:
Where P₁ is the pressure of the tank (160atm), V₁ is the volume of the tank (12L), P₂ is the pressure of the balloons (1atm, atmospheric pressure) And V₂ is the volume this gas will occupy at 1 atm, thus:
160atm×12L = 1atm×V₂
V₂ = 1920L
As the tank will never be empty, the volume of the gas able to fill balloons is the total volume minus 12L, thus the volume of helium able to fill balloons is:
1920L - 12L = 1908L
1908L will fill:
1908L×
= <em>636 balloons</em>
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I hope it helps!
Compounds. They both have more than one element in them
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
