Answer:
2.1 M is the molarity of the HCl solution.
Explanation:
![HCl+NaOH\rightarrow H_2O+NaCl](https://tex.z-dn.net/?f=HCl%2BNaOH%5Crightarrow%20H_2O%2BNaCl)
Molarity of HCl solution = ![M_1=?](https://tex.z-dn.net/?f=%20M_1%3D%3F)
Volume of HCl solution = ![V_1=50.0mL](https://tex.z-dn.net/?f=V_1%3D50.0mL)
Ionizable hydrogen ions in HCl = ![n_1=1](https://tex.z-dn.net/?f=n_1%3D1)
Molarity of NaOH solution = ![M_2=1.05 M](https://tex.z-dn.net/?f=%20M_2%3D1.05%20M)
Volume of NaOH solution = ![V_2=100.0 mL](https://tex.z-dn.net/?f=V_2%3D100.0%20mL)
Ionizable hydroxide ions in NaOH = ![n_2=1](https://tex.z-dn.net/?f=n_2%3D1)
(neutralization )
![M_1=\frac{M_2V_2}{V_1}=\frac{1.05M\times 100.0 mL}{50.0 mL}](https://tex.z-dn.net/?f=M_1%3D%5Cfrac%7BM_2V_2%7D%7BV_1%7D%3D%5Cfrac%7B1.05M%5Ctimes%20100.0%20mL%7D%7B50.0%20mL%7D)
![M_1=2.1 M](https://tex.z-dn.net/?f=M_1%3D2.1%20M)
2.1 M is the molarity of the HCl solution.
<span>The theoretical yield for a reaction is calculated based on the limiting reagent. This allows researchers to determine how much product can actually be formed based on the reagents present at the beginning of the reaction.</span>
<span>The actual yield will never be 100 percent due to limitations.</span>
Answer:
Cu(s) in Cu(NO₃)₂(aq)
Explanation:
The standard reduction potential (E°) is the energy necessary to reduce the atom in a redox reaction. When an atom reduces it gains electrons from other than oxides. As higher is E°, easily it will reduce. The substance that reduces is at the cathode of a cell, where the electrons go to, and the other that oxides are at the anode of the cell.
The standard reduction potentials from Al(s) and Cu(s) are, respectively, -1.66V and +0.15V, so the half-cell of Cu(s) in Cu(NO₃)₂(aq) is the cathode.
2H2+O2----->2H2O. I hope this helps