Answer:
Chlorine has higher ionization energy.
Explanation:
Sodium is present on left side of periodic table in period 3 while chlorine is present on right side of periodic table in period 3.
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required. Thus chlorine have higher ionization energy as compared to the sodium.
Answer:
hope this helped :( got this from my friend a long time ago but still not sure if it's really correct this
Answer:
We'll have 1 mol Al2O3 and 3 moles H2
Explanation:
Step 1: data given
Numer of moles of aluminium = 2 moles
Number of moles of H2O = 6 moles
Step 2: The balanced equation
2Al + 3H2O → Al2O3 + 3H2
Step 3: Calculate the limiting reactant
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
Aluminium is the limiting reactant. It will completely be consumed (2 moles).
H2O is in excess. There will react 3/2 * 2 = 3 moles
There will remain 6 - 3 = 3 moles
Step 4: Calculate moles products
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
For 2 moles Al we'll have 2/1 = 1 mol Al2O3
For 2 moles Al We'll have 3/2 * 2 = 3 moles H2
We'll have 1 mol Al2O3 and 3 moles H2
Answer:
El potencial celular estándar, 
is +0.46 V
Explanation:
Las reacciones de media célula son;
Media reacción del ánodo Cu²⁺ + 2e⁻ ↔ Cu, E ° = 0.34 V
Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V
Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V
y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V
que es más alta que la del cobre presente, por lo tanto, el cobre se oxidará en el ánodo
Por lo tanto, en el ánodo, tendremos
Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)
En el cátodo
2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)
El potencial celular estándar, = +0.46 V
