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.
Solution :
An 
reaction is a type of 
in organic chemistry where the enol or an enolate ion reacts with the carbonyl compound and forms a 
-hydroxyaldehyde or a -hydroxyketone, and then followed by a dehydration to give conjugated enone.
Benzaldehyde reacts with methylketone and forms two products:
Answer:
106 mL
Explanation:
In order to be able to answer this question, you must understand what the density of a substance tells you.
The density of a substance is nothing more than the mass of that substance that occupies one unit of volume.
In your case, the density of ethanol is given in Grams per milliliter, which means that one unit of volume will be
1 mL
.
So, ethanol has a density of
0.785 g mL
−
1
, which is equivalent to saying that if you take exactly
1 mL
of ethanol and weigh it, you will end up with a mass of
0.785 g
.
Now, you know that the volume you're using has a mass of
83.3 g
. Well, if you get
0.785 g
for every
1 mL
of ethanol, it follows that this much mass will correspond to a volume of
83.3
g ethanol
⋅
ethanol's density
1 mL
0.785
g ethanol
=
106.11 mL
Rounded to three sig figs, the answer will be
V
ethanol
=
106 mL
Hope this helps
Answer:
74.9%.
Explanation:
Relative atomic mass data from a modern periodic table:
- Ca: 40.078;
- C: 12.011;
- O: 15.999.
What's the <em>theoretical</em> yield of this reaction?
In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?
Molar mass of CaCO₃:
.
Number of moles of CaCO₃ available:
.
Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.
.
Molar mass of CO₂:
.
Mass of the 0.400655 moles of 
expected for the 40.1 grams of CaCO₃:
.
What's the <em>percentage</em> yield of this reaction?
.
Answer:
Balanced.
Explanation:
Both sides are pulling with equal force.
