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:
The empirical formula is CH2O, and the molecular formula is some multiple of this
Explanation:
In 100 g of the unknown, there are 40.0⋅g12.011⋅g⋅mol−1 C; 6.7⋅g1.00794⋅g⋅mol−1 H; and 53.5⋅g16.00⋅g⋅mol−1 O.
We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers. Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.Clearly n=2, and the molecular formula is 2×(CH2O) = CxHyOz.
Answer:
water, lead, and wood
Explanation:
All are correct on Edg 2020
Oceans :unusable
Rivers :usable
Glaciers :usable
Freshwater: usable
Lakes :usable
Groundwater :usable
