Answer: 8.28g Na
Explanation: use ideal gas law
PV= nRT
Solve for moles of Cl2
n= PV/ RT
Substitute:
= 1 atm x 4.0 L / 0.08205 L.atm/ mol. K x 273 K
= 0.18 moles Cl2
Do stoichiometry to solve for m of Na
2 Na + Cl2 => 2 NaCl2
=0.18 moles Cl2 x 2 mol Na/ 1 mol Cl2 x 23g Na / 1 mol Na
= 8.28 g Na.
Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
When it passes from one medium
Answer:
592 K or 319° C
Explanation:
From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;
V1/T1= V2/T2
Initial volume V1 = 1.75 L
Initial temperature T1= 23.0 +273 = 296 K
Final volume V2= 3.50 L
Final temperature T2 = the unknown
T2= V2T1/V1= 3.50 × 296 / 1.75
T2 = 592 K or 319° C
V1 = 445ml V2 = 499ml
T1 = 274 K T2 = ?
By Charles Law,
V1/T1 = V2/T2
445/274 = 499/T2
By solving we get,
T2 = 307.25 K
