Answer : The volume of oxygen at STP is 112.0665 L
Solution : Given,
The number of moles of 
= 5 moles
At STP, the temperature is 273 K and pressure is 1 atm.
Using ideal gas law equation :
where,
P = pressure of gas
V = volume of gas
n = the number of moles
T = temperature of gas
R = gas constant = 0.0821 L atm/mole K (Given)
By rearranging the above ideal gas law equation, we get
Now put all the given values in this expression, we get the value of volume.
Therefore, the volume of oxygen at STP is 112.0665 L
Answer: A golfer hitting a golf ball.
Explanation:
The atomic particles move more in this option than the others.
The answer that correctly describes a reaction that forms a disaccharide from two monosaccharides is Galactose + glucose = lactose.
Cellulose is not a disaccharide, glycogen is not a monosaccharide, and sucrose isn't either.
The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>
The equation for the complete combustion of C2H2 is as below:
The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Reacting to produce hydrogen gas is a chemical property
