Given in the problem is the mass of the liquid (500 grams) and the volume of the liquid (1000 ml = 1000 cm^3).
We can use these two givens to calculate the density of the liquid using the following rule:
density = mass / volume
density = 500 / 1000 = 0.5 grams / cm^3
Comparing the calculated density with the choices we have, we can deduce that the liquid is most likely to be propane with density 0.494 g / cm^3
The first question would be B) they are examples of cell organelles. The second question would be A) mitochondria. Hope this helps!
Answer:
1) Ethanol
Explanation:
If we will have <u>interactions</u> we will need more <u>energy</u> to break them in order to go from liquid to gas. If we need more <u>energy</u>, therefore, the <u>temperature will be higher</u>.
In this case, we can discard the <u>propanone</u> because this molecule don't have the ability to form <u>hydrogen bonds</u>. (Let's remember that to have hydrogen bonds we need to have a hydrogen bond to a <u>heteroatom</u>, O, N, P or S).
Then we have to analyze the hydrogen bonds formed in the other molecules. For ethanol, we will have only <u>1 hydrogen bond</u>. For water and ethanoic acid, we will have <u>2 hydrogen bonds</u>, therefore, we can discard the ethanol.
For ethanoic acid, we have 2 <u>intramolecular hydrogen bonds</u>. For water we have 2 <u>intermolecular hydrogen bonds</u>, therefore, the strongest interaction will be in the <u>ethanoic acid</u>.
The<u> closer boiling point</u> to the 75ºC is the <u>ethanol</u> (boiling point of 78.8 ºC) therefore these molecules would have <u>enough energy</u> to <u>break</u> the hydrogen bonds and to past from<u> liquid to gas</u>.
Answer:
Explanation:
The combustion reaction of Octane is:
To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.
We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:
Now we<u> convert 1.24 gallons to mL</u>:
- 1.24 gallon *
4693.4 mL
We <u>calculate the mass of Octane</u>:
- 4693.4 mL * 0.703 g/mL = 3.30 g Octane
Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:
- CO₂ ⇒ 3.30 g Octane ÷ 114g/mol *
* 44 g/mol = 10.19 g CO₂
- H₂O ⇒ 3.30 g Octane ÷ 114g/mol *
* 18 g/mol = 4.69 g H₂O
Answer:
X= Be
Y= B
Z=O
Explanation:
From the description of the compound XCl2, among the options listed only beryllium can form such compound with three lone pairs in the two chlorine atoms and no lone pair on the central atom X.
From the description of YCl3, only Boron among the options listed can form such a compound with no lone pair on the central atom and three lone pairs on each of the chlorine atoms.
From the description of ZCl2, only oxygen forms the compound OCl2 among the elements listed where oxygen possesses two lone pairs and each chlorine atom possesses three lone pairs each.