Answer:
The solution is basic.
Explanation:
We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):
- If no. of millimoles of acid > that of base; the solution is acidic.
- If no. of millimoles of acid = that of base; the solution is neutral.
- If no. of millimoles of acid < that of base; the solution is basic.
- We need to calculate the no. of millimoles of acid and base:
no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.
no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.
<em>∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).</em>
<em>So, the solution is: basic.</em>
Answer:
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
Explanation:
Sodium is present in group 1.
It is alkali metal.
It has one valence electron.
The atomic number of sodium is 11.
Its atomic mass is 23 amu.
The longhand notation of electronic configuration of sodium can be written as,
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
The electronic configuration in shorthand notation( noble gas) would be written as,
Na₁₁ = [Ne] 3s¹
Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.
It react with halogen and form salt. Such as sodium chloride.
2Na + Cl₂ → 2NaCl
<span>the sharing of electron pairs between atoms</span>
The chemical equation would be:
2NO(g) + O2(g) --> 2NO2 (g)
<span>At equilibrium state, the partial pressure of the gases would be as follows : </span>
<span>NO = 522 - 2x </span>
<span>O2 = 421 - x </span>
<span>NO2 = 2x </span>
<span>- - - - - - - - - - - - -</span>
<span>943 - x = 748 </span>
<span>x = 195</span>
Calculating for Kp,
<span>Kp = (NO2)^2/ ((NO)^2 * (O2)) </span>
<span>Kp = (2 * 195)^2/ ((522 - 2 * 195)^2 * (421 - 195)) </span>
<span>Kp = 0.0386 </span>
