Answer:
3.18 L
Explanation:
Step 1: Given data
- Initial pressure (P₁): 0.985 atm
- Initial volume (V₁): 3.65 L
- Final pressure (P₂): 861.0 mmHg
Step 2: Convert P₁ to mmHg
We will use the conversion factor 1 atm = 760 mmHg.
0.985 atm × 760 mmHg/1 atm = 749 mmHg
Step 3: Calculate the final volume of the gas
Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L
PH = 0.1289<span> for </span>1.50<span> M solution of weak acid with Ka value of </span><span>.73</span>
The answer is Photosphere Apex.
Answer:
Mass of heptane = 102g
Vapor pressure of heptane = 454mmHg
Molar mass of heptane = 100.21
No of mole of heptane = mass/molar mass = 102/100.21
No of mole of heptane = 1.0179
Therefore the partial pressure of heptane = no of mole heptane *Vapor pressure of heptane
Partial pressure of heptane = 1.0179*454mmHg
Partial pressure of heptane = 462.1096 = 462mmHg
the partial pressure of heptane vapor above this solution = 462mmHg
Answer:
The first thing that you need to do here is to figure out the mass of the sample.
To do that, you can use its volume and the fact that aluminium is said to have a density of
2.702 g cm
−
3
, which implies that every
1 cm
3
of aluminium has a mass of
2.702 g
.
Explanation:
