I can tell you're not very educated because everyone knows that breathing pure oxygen for long periods of time can sometimes hurt us. Oxygen in lower levels, such as levels found in atmosphere are just right for us to breathe. Get a life and stop trying to scare young kids that just want help on their homework.
Answer: 26.5 mm Hg
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
where,
= initial pressure at 
= ?
= final pressure at 
= 100 mm Hg
= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature =
Now put all the given values in this formula, we get
![\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BP_1%7D%7B100%7D%29%3D%5Cfrac%7B28000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B299.5%7D-%5Cfrac%7B1%7D%7B267.9%7D%5D)
Thus the vapor pressure of 
in mmHg at 26.5 ∘C is 26.5
Answer:
0.486 L
Explanation:
Step 1: Write the balanced reaction
2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)
Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃
The molar mass of KCIO₃ is 122.55 g/mol.
1.52 g × 1 mol/122.55 g = 0.0124 mol
Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃
The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol
Step 4: Calculate the volume corresponding to 0.0186 moles of O₂
0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L
Answer:
It is necessary to use models to study sub- microscopic objects such as atoms and molecules because they are too small to be seen.
