1) Write the balaced chemical equation:
H2 + 2O2 → 2H2O
2) Infere the molar ratios:
1 mol H2 : 2 mol of water
3) Make the calculus as the direct proportion relation:
[2 mol H2O] / [1 mol H2] * 7 mol H2 = 14 mol H2
As you see you produce the double number of moles of H2O than number of moles of H2 used.
Answer: 14 moles
Answer:
6.68 X 10^-11
Explanation:
From the second Ka, you can calculate pKa = -log (Ka2) = 6.187
The pH at the second equivalence point (8.181) will be the average of pKa2 and pKa3. So,
8.181 = (6.187 + pKa3) / 2
Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11
Answer:
1.6 L
Explanation:
Using Charle's law
Given ,
V₁ = 1.5 L
V₂ = ?
T₁ = 12 °C
T₂ = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (12 + 273.15) K = 285.15 K
T₂ = (32 + 273.15) K = 305.15 K
Using above equation as:
New volume = 1.6 L
Is true. Nitrogen gas behaves more like an ideal gas as the
temperature increases. Under normal conditions such as normal pressure and temperature
conditions , most real gases behave qualitatively as an ideal gas. Many
gases such as air , nitrogen , oxygen ,hydrogen , noble gases , and some heavy
gases such as carbon dioxide can be treated as ideal gases within a reasonable tolerance. Generally,
the removal of ideal gas conditions tends to be lower at higher temperatures and lower density (that is at lower pressure ), since the work made by the intermolecular
forces is less important compared to the kinetic energy<span> of the particles, and the size of the molecules is less important
compared to the empty space between them. </span><span>The ideal gas model
tends to fail at lower temperatures or at high pressures, when intermolecular
forces and intermolecular size are important.</span>
(1,0)n +(235,92)U --->(91,36)Kr + (142,56) Ba + 3(1,0)n
