The hydrogen deficiency index( HDI) of strigol is = 10
<h3>How to calculate HDI:</h3>
The hydrogen deficiency index is used to measure the number of degree of unsaturation of an organic compound.
Strigol is an example of an organic compound because it contains carbons and hydrogen.
To calculate the HDI using the molecular formula given (C19H20O6) the formula for HDI is used which is:

where C = number of carbon atoms = 19
n= number of nitrogen atoms = 0
h= number of hydrogen atoms = 20
X = number of halogen atoms = 0
Note that oxygen was not considered because it forms two bonds and has no impact.
There for HDI =

HDI=

HDI =

HDI = 10
Therefore, the hydrogen deficiency index of strigol is = 10
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Answer:
54g
Explanation:
Given parameters:
Number of moles of H₂O = 3 moles
Unknown:
mass of water = ?
Solution:
To solve this problem, we use the expression below:
mass = number of moles x molar mass
Molar mass of H₂O = 2(1) + 16 = 18g/mol
Mass of water = 3 x 18 = 54g
A magnet because it will remove only metal and not the saw dust
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Answer:
Longer hydrocarbon molecules have a stronger intermolecular force. More energy is needed to move them apart so they have higher boiling points . This makes them less volatile and therefore less flammable
The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>
the diagram of the emission spectrum has been added.
<em>From the given</em><em> chart;</em>
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- <em>c is the speed of light = 3 x 10⁸ m/s</em>
- <em>f is the frequency of the wave</em>
- <em>λ is the wavelength</em>

The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
<em />
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
Learn more here:brainly.com/question/15962928