Answer:
C. 0.20 M Mg ion & 0.40 M Cl ion
Explanation:
MgCl₂ is a ionic salt which is dissociated as this
MgCl₂ → Mg²⁺ + 2Cl⁻
First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M
Molarity . volume = moles.
0.6 mol/l . 0.2l = 0.12 mol
MgCl₂ → Mg²⁺ + 2Cl⁻
0.12mol 0.12 0.24
This moles are also in 400mL of water, so the new concentration is
[Mg²⁺] = 0.12 m/0.6L = 0.2M
[Cl⁻] = 0.24 m/0.6L = 0.4M
Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)
The balanced chemical reaction describing this decomposition is as follows:
<span>4c3h5n3o9 .............> 6N2 + 12CO2 +10H2O + O2
From the periodic table:
mass of oxygen = 16 grams
mass of nitrogen = 14 grams
mass of hydrogen = 1 gram
mass of carbon = 12 grams
Therefore:
mass of </span><span>C3H5N3O9 = 3(12) + 5(1) + 3(14) + 9(16) = 227 grams
mass of O2 = 2(16) = 32 grams
From the balanced chemical equation:
4(227) = 908 grams of </span>C3H5N3O9 produce 32 grams of O2. Therefore, to know the amount of oxygen produced from 4.5*10^2 grams <span>C3H5N3O9, all we need to do is cross multiplication as follows:
amount of oxygen = (4.5*10^2*32) / (908) = 15.859 grams</span>
Answer:
ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.
Explanation:
Answer:
The molar mass of copper (II) nitrate is 187.5 g/mol.
Explanation:
The molar mass is the mass of all the atoms in a molecule in grams per mole. To calculate the molar mass of a molecule, we first obtain the atomic weights from the individual elements in a periodic table. We then count the number of atoms and multiply it by the individual atomic masses.
A binary compound of oxygen with another element is called oxide. An oxide is a binary compound of oxygen and another element. Oxygen combines with metals and non-metals to form respective oxides.
