That is only the combustion of a hydrocarbon. Rust is a combustion reaction because oxygen is added.
Fe(s) + O2(g) => FeO2(s)
Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Answer:
just wait 3 minutes ill tell you the answer cause its in my book
Explanation:
Answer:
C. Particle size
Explanation:
The sand, which has smaller particles, will go through the sieve, while the rice (with a larger particle size) will not
Answer is: the maximum concentration of Pb²⁺ is 6.8·10⁻³ M.
Chemical reaction 1: PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq).
Chemical reaction 2: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Ksp(PbCl₂) = 1.7·10⁻⁵.
c(NaCl) = c(Cl⁻) = 0.0500 M.
Ksp(PbCl₂) = c(Pb²⁺) · c(Cl⁻)².
c(Pb²⁺) = Ksp(PbCl₂) ÷ c(Cl⁻)².
c(Pb²⁺) = 1.7·10⁻⁵ M³ ÷ (0.0500 M)².
c(Pb²⁺) = 0.000017 M³ ÷ 0.0025 M².
c(Pb²⁺) = 0.0068 M = 6.8·10⁻³ M.
