Answer:
it will turn blue black to any substance
Answer:
(edit: nvm I figured it out, here is the answer)
Explanation:
<h3>Answer:</h3>
2.55 × 10²² Na Atoms
<h3>Solution:</h3>
Data Given:
M.Mass of Na = 23 g.mol⁻¹
Mass of Na = 973 mg = 0.973 g
# of Na Atoms = ??
Step 1: Calculate Moles of Na as:
Moles = Mass ÷ M.Mass
Moles = 0.973 g ÷ 23 g.mol⁻¹
Moles = 0.0423 mol
Step 2: Calculate No, of Na Atoms as:
As 1 mole of sodium atoms counts 6.022 × 10²³ and equals exactly to the mass of 23 g. So, we can write,
Moles = No. of Na Atoms ÷ 6.022 × 10²³ Na Atoms.mol⁻¹
Solving for No. of Na Atoms,
No. of Na Atoms = Moles × 6.022 × 10²³ Na Atoms.mol⁻¹
No. of Na Atoms = 0.0423 mol × 6.022 × 10²³ Na Atoms.mol⁻¹
No. of Na Atoms = 2.55 × 10²² Na Atoms
<h3>Conclusion: </h3>
2.55 × 10²² sodium atoms are required to reach a total mass of 973 mg in a substance of pure sodium.
When Ksp = [A2+] [S2-]
when A is the metal: Fe, Ni, Pb, and Cu
When we have [S2-] = 0.1 m and we have Ksp for each metal So by substitution in Ksp formula we can get [A2+] for each metal and compare its value with solution concentration 0.01 M, when we have a concentration more than 0.01 M So there are no sulfides precipitates
- [Fe2+] = Ksp/[S2-]
by substitution with Fe2+ Ksp value:
= 6x10^2 / 0.1
= 6x10^3 M
when [Fe2+] > 0.01 M
∴ no precipitate- [Ni2+] = Ksp /[S2-]
by sustitution with Ni Ksp value :
= 8x10^-1 / 0.1
= 8 M
When [Ni2+] > 0.01 M
∴ no precipitate-[Pb2+] = Ksp / [S2-]
by substitution with Pb Ksp value:
= 6x10^-7 / 0.1
= 6 x 10^-6 M
when [Pb2+] < 0.01 M
∴PbS will be precipited-[Cu2+] = Ksp / [S2-]
by substitution with Cu2+ Ksp value:
= 6x10^-16 / 0.1
= 6x10^-15 M
when [Cu2+] < 0.01 M
∴ CuS will be precipited∴The sulfides precipitates are CuS & PbS
Copper.
Easy to look up, why use brainly? ;P