Answer:
The number of moles of potassium hydroxide, KOH required to make 4 moles of K₂SO₄ is 8 moles of KOH
Explanation:
2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
From the above reaction, we have 2 moles of KOH combining with 1 mole of H₂SO₄ to produce 1 mole of K₂SO₄ and 2 moles of H₂O.
Therefore the number of moles of potassium hydroxide that will be needed to make 4 moles of K₂SO₄ is;
8KOH + 4H₂SO₄ → 4K₂SO₄ + 8H₂O
8 moles of KOH is required to make 4 moles of K₂SO₄.
Answer:
sp2
Explanation:
Orbital hybridization is defined as the changes in the orbitals of atoms when they pair with electrons of other atoms to form chemical bonds.
C atoms in C2Cl4 or tetrachloroethene have a total of three bond pairs, two with each chlorine atom and one with other carbon atom and zero lone pairs. Thus, C atoms in C2Cl4 is sp2 hybridized.
Hence, the correct answer is "sp2".
The oxidizing agent is the one that is reduced in the reaction. In this reaction, the charge of Cu falls from +2 to zero charge (neutral atom in the right side). Hence, CuO is the oxidizing agent. The reducing agent, the one being oxidized is carbon from zero charge to +4. The answer is CuO.
Answer:
6.24 x 10-3 M
Explanation:
Hello,
In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:
![Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BBrO%5E-%5D%7D%7B%5BHBrO%5D%7D)
Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change 
, we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:
Thus, we obtain a quadratic equation whose solution is:
Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.
Best regards.
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)
