Answer:
The amount in grams of hydrogen gas produced is 0.551 grams
Explanation:
The parameters given are;
Number of atoms of potassium, aₙ = 3.289 × 10²³ atoms
Chemical equation for the reaction is given as follows;
2K + 2H₂O
KOH + H₂
Avogadro's number,
, regarding the number of molecules or atom per mole is given s follows;
= 6.02 × 10²³ atoms/mole
Therefore;
The number of moles of potassium present = 3.289 × 10²³/(6.02 × 10²³) = 0.546 moles
2 moles of potassium produces one mole of hydrogen gas, therefore;
1 moles of potassium produces 1/2 mole of hydrogen gas, and 0.546 moles of potassium will produce 0.546/2 moles of hydrogen which is 0.273 moles of hydrogen gas
The molar mass of hydrogen gas = 2.016 grams
Therefore, 0.273 moles will have a mass of 0.273×2.016 = 0.551 grams.
The amount in grams of hydrogen gas produced = 0.551 grams.
Carbon will have 4 valence electrons. It will have 2 in the p orbital and 2 in the s orbital. You can see this when you find the noble gas configuration of carbon which is [He]2s²2p² showing that carbon has 4 valence electrons.
I hope this helps. Let me know if anything is unclear.
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Electrons are orbiting around the nucleus in a specific energy level as described in Bohr's atomic model. There are 7 energy levels all in all; 1 being the strongest and nearest to the nucleus, and 7 being the weakest and farthest away from the nucleus. Electron can transfer from one energy level to another. If it increases energy, it absorbs energy. If it goes down an energy level, it emits energy in the form of light. This light can be measure in wavelength through the Rydberg equation:
1/λ =R(1/n₁² -1/n₂²), where
λ is the wavelength
R is the Rydberg constant equal to 1.097 × 10⁻7<span> per meter
n</span>₁ and n₂ are the energy levels such that n₂>n₁
In the Paschen series is an emission spectrum of hydrogen when the energy level is at least n=4. So, this covers n=4 to n=7.
1/λ =(1.097 × 10⁻7)(1/4² -1/7²)
λ = 216.57 ×10⁻⁶ m or 216.57 μm
The balanced equation for the above reaction is;
CH₄ + 2O₂ ---> CO₂ + 2H₂O
Stoichiometry of CH₄ to O₂ is 1:2
The number of methane moles present - 1.44 g/ 16 g/mol = 0.090 mol
Number of oxygen moles present - 9.5 g/ 32 g/mol = 0.30 mol
If methane is the limiting reagent,
0.090 moles of methane react with 0.090x 2 = 0.180 mol
only 0.180 mol of O₂ is required but 0.30 mol of O₂ has been provided therefore O₂ is in excess and CH₄ is the limiting reactant.
Number of moles of water that can be produced - 0.180 mol
Therefore mass of water produced - 0.180 x 18 g/mol = 3.24 g
Therefore mass of 3.24 g of water can be produced