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Fantom [35]
3 years ago
9

A chef in restaurant baking 2 different desserts in the oven . He removes a pumpkin and some chocolate chip cookies at the same

time from the oven . The smell of cookies reaches you a minute before you smell the pie. Since the particles have the same temperature , the smell of cookies reaches you first because____________________________.​
Chemistry
1 answer:
valentina_108 [34]3 years ago
5 0

Answer:

the air molecules from the cookies diffused faster than the air molecules from the pumpkin

Explanation:

The smell of cookies reaches first because the air molecules from the cookies were able to diffuse faster than the air molecules from the pumpkin.

<em>At constant temperature, the diffusion rates of two molecules would be dependent on the size of the molecules. The bigger the size of the molecule, the more time it will take for the molecule to diffuse and vice versa. </em>

Thus, <u>since the pumpkin and the cookies were baked at the same temperature, the smell of the cookies was perceived first because the air molecules of the cookies are smaller in size than the air molecules of the pumpkin.</u>

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Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:
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Answer:

60.42% is the percent yield of the reaction.

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Moles of methane gas at 734 Torr and a temperature of 25 °C.

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Temperature of the methane gas = T = 25 °C = 298.15 K

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Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of  water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

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CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

\frac{3}{1}\times 0.6402 mol=1.9208 mol of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

\frac{1}{22.4 L}\times 26 L= 1.1607 mol

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

\frac{Experimental}{Theoretical}\times 100

\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%

60.42% is the percent yield of the reaction.

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