If the dehydration reaction of an alcohol is successful. The changes would be seen in the IR spectrum for the product compared to the starting material are as,
- The O-H and C-O band is disappear from stating material
- The addition of a C-C double bond band in the product.
In dehydration reaction of alcohol ( O-H and C-O bond ) contain , the water molecule ( ) is release from the reactant and C-C double bond is form which is known as alkene in the product .
The reactant and product have different structure. To determine the structure of the compound IR spectroscopy is used. In IR spectrum the peak corresponds to 3400-3600 cm is missing in the product of dehydration reaction of an alcohol. It means O-H band is disappear from stating material.
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Answer:
\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}
Explanation:
0.030 cm³ × ? = x m³
You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.
For example, you know that centi means "× 10⁻²", so
1 cm = 10⁻² m
If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).
If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.
So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.
We choose the former because it has the desired units on top.
The "cm" is cubed, so we must cube the conversion factor.
The calculation becomes
The correct answer is BeCl_2(l)+2Cl^-(solvated)→BeCl_4^2-.
Evaluating be behavior to see :
how it differs from the other Group 2A (2) members.
In this reaction Be behaves like other alkaline earth metals
The complete equation can be given as
BeCl_2(l)+2Cl^-(solvated)→BeCl_4^2-
BeCl_2 tends to form a chloro bridged dimer in the vapour state, however at high temperatures of the order of 1200K, this dimer dissociates into the linear monomer.
BeCl_2 has a chain structure in its solid form. Each Be atom in this structure is surrounded by chlorine atoms, two of which are connected by conversion bonds and the remaining two by covalent coordinate connections. This chain structure is displayed.
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Answer:
I’m not a expert on it, but I think B or D
Explanation:
The rheo leaf peel will lose water by osmosis and become flaccid. If the rheo leaf was boiled, this would not happen.
<h3>What is osmosis?</h3>
Across living membranes, water moves from a region of high to a region of low water potential.
Concentrated sugar solutions have low water potential when compared to the cell sap of the cells of rheo leaf peel. Thus, water will move from the rheo leaf peel to the sugar solution.
This will result in the wilting of the leaf peel.
If the leaf peel is boiled, the membranes in the cells die and can no longer function as a selectively permeable interface.
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