Using accurate measurements, using pure chemicals and performing the reaction under the most ideal conditions is important to get a valuable percent yield.
<h3>How we calculate the percent yield?</h3>
Percent yield of any chemical reaction is define as the ratios of the actual yield to the theoretical yield of the product and multiply by the 100.
To get the high percent yield or actual yield of any reaction, we have to perform the reaction under ideal condition because if we not use the standard condition then we get the low rate of reaction. Reactants should be present in the pure form as impurity make unwanted products and reduce the productivity of main product and accurate amount of reactants also important for the spontaneous reaction.
Hence, options (a), (b) & (c) are correct.
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Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
Just find the energy of the <span>blueviolet light with a wavelength of 434.0 nm using the formula:
E = hc / lambda
E = energy
c= speed of light = 3 x 10^8 m/s
h = planck's constant = 6.6 x 10^{-34} m^2 kg / s
lambda = 434 nm = 434 x 10^{-9} m
Putting these values (with appropriate units) in the above formula :
we get: Energy, E = 4.5 x 10^{-19} J
E = 0.45 x 10^{-18} J
Now, the </span>minimum energy is 2.18×10^-{18} J but our energy is 0.45 x 10^{-18} J which is less.
<span>Means the electron will not be removed
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