How many grams of water are needed to form 21.8 grams of oxygen gas?
1 answer:
Answer:
24.5g
Explanation:
Given parameters:
Mass of oxygen gas = 21.8g
Unknown:
Mass of water = ?
Solution:
This problem deals with the decomposition of water;
Reaction equation;
2H₂O → 2H₂ + O₂
2 mole of water will produce 1 mole of oxygen gas
First find the number of mole of the oxygen gas given;
Number of moles =
molar mass of O₂ = 2(16) = 32g/mol
Number of moles of O₂ = = 0.68moles
1 mole of oxygen gas will be produced from 2 mole of water
0.68moles of oxygen gas will be produce from 0.68 x 2 moles of water; = 1.36moles of water
Mass of water = number of moles x molar mass
Molar mass of water = 2(1) + 16 = 18g/mol
Mass of water = 1.36 x 18 = 24.5g
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Answer:
Explanation:
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In this case, the Boyle's is mathematically defined via:
Which stands for an inversely proportional relationship between volume and pressure, it means the higher the volume the lower the pressure and vice versa. In such a way, since the volume is decreased to one quarter, we can write:
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Answer:
The factor of increasing reaction rate is 1,85x10¹².
Explanation:
Using arrhenius formula:
Where k is rate constant; A is frecuency factor; Eₐ is activation energy; R is gas constant (0,008134 kJ/molK); T is temperature 25°C = 298,15K
Thus, replacing for an activation energy of 125 kJ/mol assuming A as 1:
k = 1,25x10⁻²²
When activation energy is 55kJ/mol:
k = 2,31x10⁻¹⁰
Thus, the factor of increasing reaction rate is:
2,31x10⁻¹⁰/1,25x10⁻²² =<em> 1,85x10¹²</em>
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I hope it helps!