Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
Answer:
He was the first scientist to observe and describe bacteria and protozoa by looking at a drop of water from a pound under a microscope. He also was the one to build the first compound microscope.
Hope this helps :)
Answer:

Explanation:
Hello,
In this case, the only source of hydrogen is in the 6 molecules of water, therefore, the atoms of hydrogen, by applying stoichiometry with the Avogadro's number is:

Best regards.
Answer:

Explanation:
You calculate the energy required to break all the bonds in the reactants.
Then you subtract the energy needed to break all the bonds in the products.
CH₃CH₂OH + 3O₂ ⟶ 2CO₂ + 3H₂O
Bonds: 5C-H 1C-C 1C-O 1O-H 3O=O 4C=O 6O-H
D/kJ·mol⁻¹: 413 347 358 467 495 799 467
