Answer:
Initially 
of nitrogen dioxide were in the container .
Explanation:
Volume of the container at low pressure and at room temperature =
Number of moles in the container = 
After more addition of nitrogen gas at the same pressure and temperature.
Volume of the container after addition = 
Number of moles in the container after addition=
Applying Avogadro's law:
(at constant pressure and temperature)
Initially of nitrogen dioxide were in the container .
Answer:
Theoretical yield = 2.5 g
Explanation:
Given data:
Mass of sodium = 79.7 g
Mass of water = 45.3 g
Theoretical yield of hydrogen gas = ?
Solution:
Chemical equation:
2Na + 2H₂O → 2NaOH + H₂
Number of moles of sodium:
Number of moles = mass/ molar mass
Number of moles = 79.7 g / 23 g/mol
Number of moles = 3.5 mol
Number of moles of water:
Number of moles = mass/ molar mass
Number of moles = 45.3 g / 18g/mol
Number of moles = 2.5 mol
Now we will compare the moles of hydrogen gas with water and sodium.
H₂O : H₂
2 : 1
2.5 : 1/2×2.5 =1.25 mol
Na : H₂
2 : 1
3.5 : 1/2×3.5 =1.75 mol
water will be limiting reactant.
Theoretical yield:
Mass = number of moles × molar mass
Mass = 1.25 mol × 2 g/mol
Mass = 2.5 g
Divide the difference by the accepted value and <span> Multiply times 100 to make the value a percent. Use significant figures in all your calculations. When you subtract (Step #1) round your answer to the correct number of significant figures.</span>
Answer:
1= K⁺= (Z=19) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰
2 = Zn²⁺= (Z = 30) =1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰, 3d¹⁰
Explanation:
When an atom lose or gain the electron ions are formed. There are two types of ions cation and anion.
Cation are formed when atom lose the electron.
Anion are formed when an atom gain the electron.
In given question potassium loses its valance electron and form K⁺ cation. Thus its electronic configuration will be written as,
₁₉K⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰
While the electronic configuration of potassium in neutral form is:
₁₉K = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
The atomic number of zinc is 30 and its electronic configuration is:
₃₀Zn= 1s² 2s² 2p⁶ 3s² 3p⁶ 4s², 3d¹⁰
When zinc atom loses its 2 valance electrons the electron configuration will be,
₃₀Zn²⁺= 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰, 3d¹⁰
