Answer: Heat of vaporization is 41094 Joules
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:

where,
= initial pressure at 429 K = 760 torr
= final pressure at 415 K = 515 torr
= enthalpy of vaporisation = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 429 K
= final temperature = 515 K
Now put all the given values in this formula, we get
![\log (\frac{515}{760}=\frac{\Delta H}{2.303\times 8.314J/mole.K}[\frac{1}{429K}-\frac{1}{415K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B515%7D%7B760%7D%3D%5Cfrac%7B%5CDelta%20H%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B429K%7D-%5Cfrac%7B1%7D%7B415K%7D%5D)

Thus the heat of vaporization is 41094 Joules
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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Answer:
The available amounts of each reactant
Explanation:
The limiting reactant is the reactant that will limit the equation.
For example, I have 2N2 and H2O. In this situation, H2O would be the limiting reactant since there is only one atom, while N2 has two atoms.
Think of it like you're making peanut butter and jelly. You have 50 cans of peanut butter and only 10 jelly. Which means you can only make sandwiches using 10 jars of jelly and 10 jars of peanut butter.
Answer:
d = 0.992 g/L
Explanation:
Data Given:
Pressure of nitric oxide (NO) = 0.866 atm
Temperature of a gas = 46.2° C
Convert the temperature to kelvin = 46.2° C + 273
temperature in kelvin = 319.2 K
density of nitric oxide (NO) = ?
Solution:
Density of a gas can be calculated by
d = PM /RT
Where
d = density
P = Pressure
M = molar mass of gas
R = ideal gas constant = 0.0821 L atm mol⁻¹ K⁻¹
T = temperature
So,
Molar mass of NO = 30 g/mol
Put values in the formula:
d = PM /RT
d = 0.866 atm × 30 g/mol / 0.0821 L atm mol⁻¹ K⁻¹ × 319.2 K
d = 25.98 atm. g/mol / 26.2 L atm mol⁻¹
d = 0.992 g/L
Answer:
Zymase is acting as a catalyst
Explanation:
Zymase is an enzyme that is naturally produced in yeast. Enzymes are biological catalysts that speed up the rates of reactions in living things.
Zymase catalyses the breakdown of glucose into ethanol and carbon dioxide, which causes the bread to rise.
Zymase speeds up this reaction, but is not physically changed itself. Therefore, it is a catalyst.
Actually, zymase represents a collection of enzymes in yeast, not just one!