Answer:
The partial pressure of the other gases is 0.009 atm
Explanation:
Step 1: Data given
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.
The atmospheric pressure = 0.90 atm
Step 2: Calculate mol fraction
If wehave 100 moles of air, 78 moles will be nitrogen,
21 moles will be oxygen, and 1 mol will be other gases.
Mol fraction = 1/100 = 0.01
Step 3: Calculate the partial pressure of the other gases
Pgas = Xgas * Ptotal
⇒ Pgas = the partial pressure = ?
⇒ Xgas = the mol fraction of the gas = 0.01
⇒Ptotal = the total pressure of the pressure = 0.90 atm
Pgas = 0.01 * 0.90 atm
Pgas = 0.009 atm
The partial pressure of the other gases is 0.009 atm
Answer:
2.8 x 10²³ molecules H₂O
1.4 x 10²³ molecules O₂
Explanation:
First, you will need the balanced chemical equation for the formation of water:
2H₂ + O₂ -> 2H₂O
This will help in determining the mole ratios between water and oxygen, which we will need later.
Let's first calculate the number of H₂O (water) molecules. This will require stoichiometry. We are also given the mass, so we must convert mass into moles, then moles into molecules. mass -> moles -> molecules
8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (6.02 x 10²³ molecules H₂O/1 mol H₂O) = 2.8404 x 10²³ molecules H₂O
Rounded to 2 significant digits: 2.8 x 10²³ molecules H₂O
Now, to find the molecules of water, we can begin with the same stoichiometric equation, but before we convert to molecules, we will have to convert moles of water to moles of oxygen. This is where we will use the mole ratio of water to oxygen we got from the balanced chemical equation earlier. 2H₂O:1O₂
8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (1 mol O₂/2 mol H₂O) x (6.02 x 10²³ molecules O₂/1 mol O₂) = 1.4202 x 10²³ molecules O₂
Rounded to 2 significant digits: 1.4 x 10²³ molecules O₂
Answer:
1. Butyraldehyde; 2. but-3-en-1-ol
Explanation:
1. Peak ≥ 160 ppm
The formula C₄H₈O shows that the Index of Hydrogen Deficiency = 1.
It could be caused by either a ring or a double bond.
A peak at ≥ 160 ppm strongly indicates a C=O group, so the rest of the molecule can contain no rings or double bonds.
There are no other heteroatoms, so the compound most be an aldehyde or a ketone.
One compound that meets these criteria is butyraldehyde, CH₃CH₂CH₂CH=O (see Fig. 1.)
2. No rings; all peaks < 160 ppm
If all peaks are < 160 ppm, there can no C=O group.
There is no ring, so there must be a C=C double bond.
There is no other unsaturation, so the O atom must be present as an alcohol or an ether.
One compound that meets these criteria is but-3-en-1-ol, CH₂=CHCH₂CH₂OH (see Fig. 2).
