Thank you for posting your question here. Below is the solution:
HNO3 --> H+ + NO3-
<span>HNO3 = strong acid so 100% dissociation </span>
<span>** one doesn't need to find the molarity of water since it is the solvent </span>
<span>0M HNO3 </span>
<span>1x10^-6M H3O+ </span>
<span>1x10^-6M NO3- </span>
<span>1x10^-8M OH-.....the Kw = 1x10^-14 = [H+][OH-] </span>
<span>you have 1x10^-6M H+ so, 1x10^-14 / 1x10^-6 = 1x10^-8M OH- </span>
<span>1x10^-6 Ba(OH)2 = strong base, 100% dissociation </span>
<span>1x10^-6M Ba2+ </span>
<span>2x10^-6M OH- since there are 2 OH- / 1 Ba2+ </span>
<span>0M Ba(OH)2 </span>
<span>5x10^-9M H3O+</span>
_Mg + _HCL = _MgCl2 + H2
Separate the terms on each side:
_Mg + _HCl = _MgCl2 + H2
Mg- 1 Mg-1
H-1 H-2
Cl-1 Cl-2
Mg is balanced on both sides so move on to the next (put a 1 in the space).
1Mg
There are two H's and two Cl's on the results side, so to balance the equation put a 2 as a coefficient for HCl and it'll all balance out.
2HCl
Balamced equation will be:
1Mg + 2HCL = 1MgCl2 + H2
Explanation:
Magnesium has atomic number 12. It will be distributed in K, L, M shell in the following way:
K shell can accommodate a maximum of 2 electrons.
L shell can accommodate a maximum of 8 electrons and
M shell will accommodate 2 electrons
So the configuration becomes 2, 8,3
Oxygen had 6 valence electrons
Answer:

Explanation:
Hello!
In this case, since the decomposition of sodium hydrogen carbonate is:

Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:

Best regards!