and .
Assuming complete decomposition of both samples,
First compound:
; of the first compound would contain
Oxygen and mercury atoms seemingly exist in the first compound at a ratio; thus the empirical formula for this compound would be where the subscript "1" is omitted.
Similarly, for the second compound
; of the first compound would contain
and therefore the empirical formula
.
Answer:
Vapour pressure of benzene over the solution is 253 torr
Explanation:
According to Raoult's law for a mixture of two liquid component A and B-
vapour pressure of a component (A) in solution =
vapour pressure of a component (B) in solution =
Where are mole fraction of component A and B in solution respectively
are vapour pressure of pure A and pure B respectively
Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr
So, vapour pressure of benzene in solution =
= 253 torr
Answer:
0.861 L
Explanation:
We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:
(P₁V₁)/T₁ = (P₂V₂)/T₂
Convert the temperatures to degrees Kelvin.
25.0°C -> 298 K, 100.0°C -> 373 K
Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:
(165(2.5))/298 = (600(V₂))/373
V₂ = (165(2.5)(373))/(298(600))
V₂ = 0.861 L
The answer would be A) sand, it is not soluble in water
Answer:
900 K
Explanation:
Recall the ideal gas law:
Because only pressure and temperature is changing, we can rearrange the equation as follows:
The right-hand side stays constant. Therefore:
The can explodes at a pressure of 90 atm. The current temperature and pressure is 300 K and 30 atm, respectively.
Substitute and solve for <em>T</em>₂:
Hence, the temperature must be reach 900 K.