Hello!
<span>We have the following statement data:
</span>
Data:
<span>As the percentage is the mole fraction multiplied by 100:
</span>
<span>The mole fraction will be the percentage divided by 100, thus:
</span><span>What is the partial pressure of oxygen in this mixture?
</span>
<span>To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:
</span>
In relation to
:
<span>
Answer:
</span><span>
b. 320.0 mm hg </span>
Answer:
no because it already had it in there and it exploded by a fuse
Explanation:
Hope this helps?? :))
Answer:
to collect liquid ethanol and leave ethene as a gas because ethanol has hydrogen bonds
Explanation:
The chemist would be lesser than the temperature of the mixture as to collect the liquid ethanol and then leave ethene as a gas since the ethanol is a bond that should be hydrogen. Also -OH that available in the ethanol would be responsible for the hydrogen bonds also it is the main and significant molecular forice
So as per the given situation the above represent the answer
Answer:
Explanation:
At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).
Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.
Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).
A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.
<u>a. C₂H₄:</u>
- C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)
Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.
The following analysis just shows that the other options are not right.
<u>b. C₂H₂:</u>
- 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)
The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.
<u>с. С₃Н₈</u>
- C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)
The mole ratio is 1 mol C₃H₈ : 5 mol O₂
<u>d. C₂H₆</u>
- 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)
The mole ratio is 2 mol C₂H₆ : 7 mol O₂
Answer:
D. 7
Explanation:
The halogens are found in Group 7 of the Periodic Table. If Bromine is a halogen, then that means it would be found in Group 7. Also, elements in Group 7 have 7 electrons in its outer shell, Bromine would have 7 electrons in its outer shell.
