Answer:
Explanation:
Given data:
one side wall temperature 200°C
other side wall temperature 10°C
h = 100 W/m^2 °C
k = 2.6W/m °C
wall thickness L = 30 cm
we know that heat flux is given as
solving for temperature for cold surface is given as
Click the "draw structure" button to launch the drawing utility. under certain reaction conditions, 2,3−dibromobutane reacts wit
h two equivalents of base to give three products, each of which contains two new π bonds. product a has two sp hybridized carbon atoms, product b has one sp hybridized carbon atom, and product c has none. what is the structure of b?
1 answer:
Answer:
Explanation:
According to this. Let's analize the possible products a, b and c.
First, the problem states that we have 2 eq. of base, For this case, let's assume it's KOH. Now, As we are doing a reaction with base, means that this reaction can only take places under conditions of SN2 and E2, a fast reaction that is taking place in only 1 step.
With this in mind, let's analyze product a. This states that it has two sp hybridized carbon, in other words, a triple bond between two carbons. So the product is with no doubt, an alkyne.
Product b has only one sp hybridized carbon, which means that this carbon should cannot be an alkyne because we need two carbon atoms. The only way to have one atom of C sp hybridized, is with two double bonds, so product b would have to a alkene with two double bonds.
Product c do not have sp hybridized carbon, therefore, it only has two double bonds in two different Carbon atoms, so it's another alkene with two double bonds, but in two different atoms of carbon.
Picture attached show the product a, b and c. Hope this can help
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Explanation:
As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.
So, using ideal gas equation we will calculate the number of moles as follows.
PV = nRT
or, n =
=
= 0.0052 mol
Also, No. of moles =
0.0052 mol =
mass = 0.0104 g
As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at
= 24 mm Hg
= atm
= 0.03158 atm
Now, P =
= 0.963 atm
Hence, n =
= 0.0056 mol
So, mass of = 0.0056 mol × 2
= 0.01013 g (actual yield)
Therefore, calculate the percentage yield as follows.
Percent yield =
=
= 97.49%
Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.
Answer:
450. g of 0.173 % KCN solution contains 779 mg of KCN.
Explanation:
Mass of the solution = m
Mass of the KCN in solution = 779 mg
Mass by mass percentage of KCN solution = 0.173%
1 mg = 0.001 g
m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g
450. g of 0.173 % KCN solution contains 779 mg of KCN.