<span>0.0165 m
The balanced equation for the reaction is
AgNO3 + MgCl2 ==> AgCl + Mg(NO3)2
So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights.
Atomic weight silver = 107.8682
Atomic weight chlorine = 35.453
Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol
Now how many moles were produced?
0.1183 g / 143.3212 g/mol = 0.000825419 mol
So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division.
0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m
Rounding to 3 significant figures gives 0.0165 m</span>
Hey there!:
HCl + MnO2 → MnCl2 + H2O + Cl2
* in HCl the oxidation state of Cl is -1 .
* on the product side the oxidation state is 0 .
* therefore Cl gains electrons .
* in MnO2 the oxidation state of Mn is +4
* in MnCl2 the oxidation state of Mn is +2
Therefore Mn loses electrons
Answer A
Hope That helps!
Q1. They are highly reactive. Q2. High reactivity, nonmetallic. Q3. Oxygen has an ion charge of -2. Q4. LiCl I believe. Q5. How electrons are shared. Q6 1. Q7. Share 2 valence electrons, I believe.
<span>Mass of the solution = 0.17m
Kb for C6H5NH2 = 3.8 x 10^-10
We know Ka for C6H5NH2 = 1.78x10^-11
We have Kw = Ka x Kb => Ka = Kw / Kb
=> (C2H5NH2)(H3O^+)/(C2H5NH3^+) => 1.78x10^-11 = K^2 / 0.17
K^2 = 3 x 10^-12 => K = 1.73 x 10^-6.
pH = -log(Kw(H3O^+)) = -log(1.73 x 10^-6) = 5.76</span>
