Non-valence electrons: 1s22s22p6. Therefore, we write the electron configuration for Na: 1s22s22p63s1. What is the highest principal quantum number that you see in sodium's electron configuration? It's n = 3, so all electrons with n = 3 are valence electrons, and all electrons with n < 3 are non-valence electrons.
Answer:
H+(aq) + OH-(aq) → H2O(l)
Explanation:
Step 1: Data given
nitrious acid = HNO3
sodium hydroxide = NaOH
Step 2: The unbalance equation
HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:
H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)
This separation technique is a 4-step procedure. First, add H₂SO₄ to the solution. Because of common ion effect, BaSO₄ will not react, only Mg(OH)₂.
Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2 H₂O
The aqueous solution will now contain MgSO₄ and BaSO₄. Unlike BaSO₄, MgSO₄ is soluble in water. So, you filter out the solution. You can set aside the BaSO₄ on the filter paper. To retrieve Mg(OH)₂, add NaOH.
MgSO₄ + 2 NaOH = Mg(OH)₂ + Na₂SO₄
Na₂SO₄ is soluble in water, while Mg(OH)₂ is not. Filter this solution again. The Mg(OH)₂ is retrieved in solid form on the filter paper.
Answer:
3.676 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have different values of V and T:
(V₁T₂) = (V₂T₁)
V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,
V₂ = ??? L, T₂ = 40°C + 273 = 313 K,
- Applying in the above equation
(V₁T₂) = (V₂T₁)
∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.
