Here we have to get the alcohols which are formed in the reaction of an alkene with H₂SO₄ and H₂O.
The alcohols which are formed in the reaction between alkene and H₂SO₄ and H₂O are majorly secondary and tertiary alcohol. As primary alcohol only ethanol can be produced in this reaction.
The addition of H₂SO₄/H₂O generates alcohol from alkene. The general reaction is shown in the figure. In this reaction methanol (CH₃OH) cannot be prepared and in primary alcohol only ethanol (C₂H₅OH) can be prepared from ethene.
In case of propene and other higher alkene always secondary or tertiary alcohols are produced as the addition of sulphuric acid and water depends on the Markonikov's rule of addition to the less hindered position.
Answer:
Explanation:
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In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:
In terms of mass, specific heat and temperature change is:
Now, solve for the final temperature, as follows:
Then, plug in the masses, specific heat and temperatures to obtain:
Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.
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11.48-gram of are needed to produce 6.75 Liters of
gas measured at 1.3 atm pressure and 298 K
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
First, calculate the moles of the gas using the gas law,
PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.
Given data:
P= 1.3 atm
V= 6.75 Liters
n=?
R=
T=298 K
Putting value in the given equation:
Moles = 0.3588 moles
Now,
Mass= 11.48 gram
Hence, 11.48-gram of are needed to produce 6.75 Liters of
gas measured at 1.3 atm pressure and 298 K
Learn more about the ideal gas here:
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