Answer: The standard entropy of vaporization of ethanol is 0.275 J/K
Explanation:
Using Gibbs Helmholtz equation:
For a phase change, the reaction remains in equilibrium, thus 
Given: Temperature = 285.0 K
Putting the values in the equation:
Thus the standard entropy of vaporization of ethanol is 0.275 J/K
Answer:
<u>2-chlorohexane</u>
Explanation:
<u>In this figure</u> :
- There are 6 carbon atoms
- The Cl atom is bonded to the 2nd carbon atom
⇒ The Cl is a substituent group, termed as -chloro
⇒ Based on IUPAC nomenclature, the 6 atom chain starts with hex
⇒ There are only single bonds present, so it is an alkane
<u>The name is</u> :
Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole
Answer:
<em>Weather forecasters often discuss the models they use to help predict the weather. ... Weather observations (pressure, wind, temperature and moisture) obtained from ground sensors and weather satellites are fed into these equations. The observations are brought into the models in a process known as data assimilation.</em>
Explanation:
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<em>I </em><em>hope</em><em> this</em><em> helps</em><em>!</em></h2>
Answer:
44.8 L of O2 will react (option D)
Explanation:
Step 1: Data given
Number of moles of SO2 = 4.00 moles
STP = Pressure = 1 atm and temperature = 273 K
Step 2: The balanced equation
2 SO2(g) + O2(g) → 2 SO3(g)
Step 3: Calculate moles of O2
For 2 moles SO2, we need 1 mol O2 to produce 2 moles SO3
For 4.00 moles SO2 we need 4.00 / 2 = 2.00 moles O2
Step 4: Calculate volume of O2
For 1 mol we have a volume of 22.4 L
V = (n*R*T)/ p
V = (2.00 * 0.08206 * 273)/p
V = 44.8 L
For 2.00 moles we have a volume of 2*22.4 = 44.8 L
44.8 L of O2 will react (option D)
