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STALIN [3.7K]
2 years ago
9

Part A: Calculate the mass of butane needed to produce 75.6g of carbon dioxide.

Chemistry
1 answer:
lisabon 2012 [21]2 years ago
5 0

Answer:

Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g.

Explanation:

Part B:

The mass of water produced when 4.86 g of butane(C4H10) react with excess oxygen is calculated as below

calculate the moles of C4H10 used = mass/molar mass

moles = 4.86g/58 g/mol =0.0838 moles

write a balanced equation for reaction

2 C4H10 + 13 O2 = 8 CO2 + 10 H2O

by use of mole ratio between C4H10 to H2O which is 2:10 the moles of

H20= 0.0838 x10/2 = 0.419 moles of H2O

mass = moles x molar mass

=0.419 molx 18 g/mol = 7.542 grams of water is formed

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When optically active (S)-2-methylcyclopentanone is treated with aqueous base, the compound loses its optical activity. Explain
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Answer:

See explanation below

Explanation:

In this case, we need to see which is the structure of this compound. Now, racemization occurs basically because we are in an aqueous basic medium, and the ketone can reacts again with water in the medium to form the starting reagent.

First, the base will take out the Alpha hydrogen from the ketone, then, the negative charge goes down and opens up the carbonile group, forming a double bond in there. Later, with the water of the medium, it reacts and substract a proton, and then, with keto enolic equilibrium, forms again the ketone, but this ketone is different from the start, it will be the  R isomer which is not optically active.

See picture below for mechanism

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Which of the following would dissolve in water, and why?
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CH3 + HCl <=> CH3Cl + H2O
dmitriy555 [2]

Answer:

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

Explanation:

Step 1: Data given

Kp = 4.7 x 10^3 at 400K

Pressure of CH3OH = 0.250 atm

Pressure of HCl = 0.600 atm

Volume = 10.00 L

Step 2: The balanced equation

CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)

Step 3: The initial pressure

p(CH3OH) = 0.250atm

p(HCl) = 0.600 atm

p(CH3Cl)= 0 atm

p(H2O) = 0 atm

Step 3: Calculate the pressure at the equilibrium

p(CH3OH) = 0.250 - X atm

p(HCl) = 0.600 - X atm

p(CH3Cl)= X atm

p(H2O) = X atm

Step 4: Calculate Kp

Kp = (pHO * pCH3Cl) / (pCH3* pHCl)

4.7 * 10³ =  X² /(0.250-X)(0.600-X)

X = 0.249962

p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm

p(HCl) = 0.600 - 0.249962 = 0.350038 atm

p(CH3Cl)= 0.249962 atm

p(H2O) = 0.249962 atm

Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)

Kp = 4.7 *10³

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

4 0
3 years ago
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