Answer:
0.1 M is the answer and I have explained it in the attachment below.
Hope this helped!
The <span>reactions are an explosion hope it helps.</span>
Answer: The electronic configuration of Chlorine will be ![[Ne]3s^23p^5](https://tex.z-dn.net/?f=%5BNe%5D3s%5E23p%5E5)
Explanation: Chlorine is an element which belongs to the P-block of the periodic table. Its atomic number is 17.
Atomic Number = Number of electrons
Number of electrons = 17
The nearest noble gas which lies to Chlorine is Neon which has 10 electrons. Rest of the 7 electrons are filled in the 3s and 3p orbitals. Hence, the electronic configuration of Chlorine is
![[Cl]=[Ne]3s^23p^5](https://tex.z-dn.net/?f=%5BCl%5D%3D%5BNe%5D3s%5E23p%5E5)
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)