Answer:
0.453 moles
Explanation:
The balanced equation for the reaction is:
2Fe(s) + 3O2(g) ==> 2Fe2O3
From the equation, mass of O2 involved = 16 x 2 x 3 = 96g
mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2
= 100g
Therefore 96g of O2 produced 100g of Fe2O3
32.2g of O2 Will produce 100x32.2/96
= 33.54g of Fe2O3
Converting it to mole using number of mole = mass/molar mass
but molar mass of Fe2O3 = 26 + (16 X 3)
= 74g/mole
Therefore number of mole of 33.54g of Fe2O3 = 33.54/74
= 0.453 moles
The three sub-atomic particles that made up an atom are electron, proton and neutron. The proton and neutron are present in the nucleus of an atom whereas electrons revolve around the nucleus in the orbits. The atoms are electrically neutral that is they have no charge due to the presence of equal number of protons (which is positively charged) and electrons (which is negatively charged).
All the atoms have negatively charged electrons which are counter balanced by the presence of equal number of protons in the nucleus of an atom and thus, the element is neutral in nature and possess no charge.
Answer:
well it turns into N3H12PO4
Explanation:
Answer:
The answer is "Option B".
Explanation:
![\to C \ CH_3COONa = \frac{(0.01\ mol + 5 \ E-4\ mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\](https://tex.z-dn.net/?f=%5Cto%20C%20%5C%20CH_3COONa%20%3D%20%20%5Cfrac%7B%280.01%5C%20%20mol%20%2B%205%20%5C%20E-4%5C%20%20mol%29%7D%7B%280.105%5C%20L%20%29%7D%5C%5C%5C%5C%5Cto%20C%20%5C%20CH_3COONa%20%3D%200.1%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20Ka%20%3D%20%28%5BH_3O%5E%7B%2B%7D%5D%5Ctimes%20%5Cfrac%7B%280.1%20%2B%20%5BH_3O%5E%2B%5D%29%29%7D%7B%280.0905%20-%20%5BH_3O%5E%2B%5D%29%7D%20%3D%201.75%5C%20E-5%5C%5C%5C%5C%5Cto%200.1%5BH_3O%5E%2B%5D%20%2B%20%5BH_3O%5E%2B%5D%5E2%20%3D%20%281.75%20E-5%29%5Ctimes%20%280.0905%20-%20%5BH_3O%5E%2B%5D%29%5C%5C%5C%5C)
![\to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\ E-6 - 1.75\ E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to [H_3O^+] = 1.5835\ E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7](https://tex.z-dn.net/?f=%5Cto%20%5BH_3O%5E%2B%5D%5E2%20%5C%200.1%5BH_3O%5E%2B%5D%20%3D%201.584%5C%20%20E-6%20-%201.75%5C%20%20E-5%5BH_3O%5E%2B%5D%5C%5C%5C%5C%5Cto%20%5BH_3O%5E%2B%5D%5E2%20%2B%200.1000175%5BH_3O%5E%2B%5D%20-%201.584%20%5C%20E-6%20%3D%200%5C%5C%5C%5C%5Cto%20%20%5BH_3O%5E%2B%5D%20%3D%201.5835%5C%20%20E-5%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20pH%20%3D%20-%20%5Clog%20%5BH_3O%5E%2B%5D%5C%5C%5C%5C%5Cto%20%20pH%20%3D%20-%20%5Clog%20%281.5835%20%5C%20E-5%29%5C%5C%5C%5C%20%5Cto%20pH%20%3D%204.8004%20%3E%204.7)
