5.58 X 
Litres is the volume, in liters, occupied by 0.015 molecules of oxygen at STP.
Explanation:
Data given:
molecules of oxygen = 0.015
number of moles of oxygen =?
temperature at STP = 273 K
Pressure at STP = 1 atm
volume = ?
R (gas constant) = 0.08201 L atm/mole K
to convert molecules to moles,
number of moles = 
number of moles = 2.49 x 
Applying the ideal gas law since the oxygen is at STP,
PV = nRT
rearranging the equation:
V = 
putting the values in the rearranged equation:
V = 
V = 5.58 X Litres.
Answer:
the awnser is 22
Explanation:
9+6 = 15 + 7 = 22 have a great day
<span>The oxidation number that is given to chlorine in a chemical combination indicates how many electrons are gained or lost because it is a neutral compound, perchloric acid has a total charge of 0. Hydrogen has an oxidation number of +1,whereas oxygen has an oxidation number of -2. So for the total charge to be 0, 1 hydrogen + 4 oxygen of negative charge added together must equal 0, which will make the oxidation number of chlorine, solving this algebraically +7.</span>
Based on the assumptions made about the type of car, the distance Honda Insight with EPA gas mileage rating of 57mi/gal will travel on 355 mL of gasoline is 8.6 km
<h3>What is the distance in kilometres a car can travel on 355 mL of gasoline?</h3>
The distance a car will travel on gasoline will depend on mileage rating of the car.
- Assuming that the car is a Honda Insight with EPA gas mileage rating of 57mi/gal.
The units are converted to convert Km/mL as follows:
- 1 mile = 1.609344 Km
- 1 gallon = 3785.411 mL
Then;
57 mi/gal = 57 × 1.609344 km/3785.411 mL
57 mi/gal = 0.0242 km/mL
Then;
Distance travelled on 355 mL = 0.0242 × 355
Distance travelled = 8.59 km
Therefore, the distance Honda Insight with EPA gas mileage rating of 57mi/gal will travel on 355 mL of gasoline is 8.6 km
Learn more about distance and car mileage at: brainly.com/question/17146782
PH = pKa + log
![\frac{[base]}{[Acid]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Bbase%5D%7D%7B%5BAcid%5D%7D%20)
Acid is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
so pH = 4.74 + log (0.2/0.2) = 4.74
This is called maximum buffer capacity (when acid conc. and base conc. are equal) the pH = pKa in this case
