The volume of carbon dioxide produced is : 5.4 L
<u>Given data:</u>
moles of propane = 0.080 moles
<h3>Determine the volume of Carbon dioxide produced </h3>
The chemical reaction
C₃H₈ + 5O₂ ---- > 3CO₂ + 4H₂O
From the reaction
I mole of C₃H₈ = 3CO₂
0.080 moles of C₃H₈ = 3 * 0.080 = 0.24
Applying the equation below to determine the volume of CO₂
Pv = nRT
= 0.24 * R * T
v = ( 0.24 * 8.314 * 273 ) / 1 atm
= 544.7 ml = 5.4 L
Hence we can conclude that The volume of carbon dioxide produced is : 5.4 L
Learn more about Propane burning at STP : brainly.com/question/11903456
Answer : The amount of codeine in one teaspoonful is, 30 mg
Explanation : Given,
Amount of codeine per fluid ounce = 60 mg
Now we have to determine the amount of codeine present in one teaspoonful.
As we know that:
1 teaspoonful = 0.5 fluid ounce
As, the amount of codeine per fluid ounce = 60 mg
So, the amount of codeine 0.5 fluid ounce = 
Thus, the amount of codeine in one teaspoonful is, 30 mg
Answer: The red blood cell will shrink due to water loss
Explanation:
Because 1% NaCl solution is slightly more concentrated than its isotonic 0.9% form, the red blood cell will lose its cell containing fluids such as water to the MORE concentrated environment.
This water loss after a prolonged period will result in shrinking of the red blood cell.
Answer:
D, if the cliff is higher than the refrigerator
Explanation:
Answer:

Explanation:
Hello there!
In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

And the concentrations are:
![[acid]=0.000855mol/0.025L=0.0342M](https://tex.z-dn.net/?f=%5Bacid%5D%3D0.000855mol%2F0.025L%3D0.0342M)
![[base]=0.000781mol/0.025L=0.0312M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.000781mol%2F0.025L%3D0.0312M)
Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

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