Any physical matter that is obtained or made from plants, animals, or the ground without undergoing a chemical process is called
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While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.
A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.
The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.
<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm
<u>Explanation:</u>
We are given:
Initial partial pressure of ethylene gas = 1.8 atm
Initial partial pressure of water vapor = 4.7 atm
Equilibrium partial pressure of ethylene gas = 1.16 atm
Equilibrium partial pressure of water vapor = 4.06 atm
The chemical equation for the reaction of ethylene gas and water vapor follows:
<u>Initial:</u> 1.8 4.7
<u>At eqllm:</u> 1.8-x 4.7-x
Evaluating the value of 'x'
The expression of for above equation follows:
Putting values in above expression, we get:
When more ethylene is added, the equilibrium gets re-established.
Partial pressure of ethylene added = 1.2 atm
<u>Initial:</u> 2.36 4.06 0.64
<u>At eqllm:</u> 2.36-x 4.06-x 0.64+x
Putting value in the equilibrium constant expression, we get:
Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.
So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm
Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm
Answer: The molecular formula will be
Explanation:
Mass of = 17.95 g
Mass of = 4.87 g
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 17.95 g of carbon dioxide, = of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 4.87 g of water, = of hydrogen will be contained.
Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g
Mass of C = 4.89 g
Mass of H = 0.541 g
Mass of O = 6.57 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H=
Moles of O=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =
For H =
For O =
The ratio of C : H : O = 1: 1 : 1
Hence the empirical formula is .
Hence the empirical formula is
The empirical weight of = 1(12)+1(1)+1(16)= 29 g.
If 0.25 moles has mass of 44.0 g
Thus 1 mole has mass of =
Thus molecular mass is 176 g
Now we have to calculate the molecular formula.
The molecular formula will be=