Answer:
I think the answer would be option d.
hope it helps.
Answer:
Solution for A gas has a volume of 340.0 mL at 45.90 degree celsius. What is the new temperature of the gas, in kelvin, if the volume increased to 550.0 mL.
Missing: oC. | Must include: oC.
Explanation:
Answer:
i think the answer is C
Explanation:
not sure plsss dont bash im a beginner
Answer:
Mass of ring = 32 g
Volume of ring = 4 mL
Density of ring = 8 g/mL
Explanation:
From the question given above, the following data were obtained:
Mass of ring = 32 g
Volume of water = 64 mL
Volume of water + ring = 68 mL
Density of ring =?
Next, we shall determine the volume of the ring. This can be obtained as follow:
Volume of water = 64 mL
Volume of water + ring = 68 mL
Volume of ring =?
Volume of ring= (Volume of water + ring) – (Volume of water)
Volume of ring = 68 – 64
Volume of ring = 4 mL
Finally, we shall determine the density of the ring. This can be obtained as follow:
Mass of ring = 32 g
Volume of ring = 4 mL
Density of ring =?
Density = mass / volume
Density of ring = 32 / 4
Density of ring = 8 g/mL
Answer:
[ S2- ] = 4.0 E-47 M
Explanation:
- PbS(s) → Pb2+ + S2-
- HgS(s) → Hg2+ + S2-
∴ Ksp PbS = 3.4 E-28 = [Pb2+]*[S2-]
∴ [Pb2+] = 0.181 M
∴ Ksp HgS = 4.0 E-53 = [Hg2+]*[S2-]
∴ [Hg2+] = 0.174 M
∴ Ksp PbS > Ksp HgS ⇒ precipitate first Hg2+:
∴ [ Hg2+ ] = 1.0 E-6 M
⇒ [S2-] = 4.0 E-53 / 1.0 E-6 = 4.0 E-47 M
