Answer:
4.68x10⁹
Explanation:
Kp is the equilibrium constant based on presure, and depends only on the gas substances. For a generic equation:
aA(g) + bB(g) ⇄ cC(g) + dD(g)
The reaction given can be summed to form the third one:
C(s) + CO₂(g) ⇄ 2CO (g) K'p = 1.30x10¹⁴
CO(g) + Cl₂(g) ⇄ COCl₂ (g) K''p = 6.00x10⁻³
We need to multiply the second reaction by 2, so CO will be simplified. If we multiplied a reaction for n, the new Kp will be (Kp)ⁿ, so:
C(s) + CO₂(g) ⇄ 2CO (g) K'p = 1.30x10¹⁴
2CO(g) + 2Cl₂(g) ⇄ 2COCl₂(g) (K''p)²= (6.00x10⁻³)²
The Kp of the reaction resulted by the sum will be: Kp = K'p*K''p
C(s) + CO₂(g) + 2Cl₂(g) ⇄ 2CO(g) + 2COCl₂(g)
Kp = 1.30x10¹⁴ * (6.00x10⁻³)²
Kp = 1.30x10¹⁴*3.60x10⁻⁵
Kp = 4.68x10⁹
Lead(II) nitrate will react with iron(III) chloride to produce the precipitate lead(II) chloride as shown in the balanced reaction
2FeCl3(aq) + 3Pb(NO3)2(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)
Calculating the amount of the precipitate lead(II) chloride each reactant will produce:
mol PbCl2 = 0.050L Pb(NO3)2 (0.100mol/1L)(3mol PbCl2/3mol Pb(NO3)2)
= 0.00500mol PbCl2
mol PbCl2 = 0.050L FeCl3 (0.100mol FeCl3/1L)(3mol PbCl2/2mol FeCl3) = 0.00750mol PbCl2
The reactant Pb(NO3)2 produces a lesser amount of the precipitate PbCl2, therefore, the lead(II) nitrate is the limiting reagent for this reaction.
Answer:Chromatography technique that uses paper sheets or strips as the adsorbent being the stationary phase through which a solution is made to pass is called paper chromatography. It is an inexpensive method of separating dissolved chemical substances by their different migration rates across the sheets of paper.
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Answer:
265 mL is the new volume for the gas
Explanation:
We decompose the Ideal Gases Law in order to find the answer of this question: P . V = n . R . T
We can propose the formula for the 2 situations, where n remains constant.
R refers to 0.082 L.atm/mol.K which is physic constant.
We convert the temperature to Absolute value:
67.5°C + 273 = 340.5 K
80°C + 273 = 353 K
We convert the volume to L → 242.2 mL . 1 L/1000 mL = 0.2422 L
We convert the pressure values to atm:
882 Torr . 1 atm/ 760 Torr = 1.16 atm
840 Torr . 1atm / 760 Torr = 1.10 atm
P₁. V₁ / T₁ = P₂ . V₂ / T₂ → Let's replace data:
1.16 atm . 0.2422L / 340.5K = 1.10 atm . V₂ / 353 K
(1.16 atm . 0.2422L / 340.5K) . 353K = 1.10 atm . V₂
V₂ = 0.291 L.atm / 1.10 atm → 0.2647 L ≅ 265 mL
