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miss Akunina [59]
3 years ago
15

How many degrees of unsaturation are lost or gained in the base-induced transformation from mito-ph to its photoactive ring-open

ed form?
Chemistry
1 answer:
Fudgin [204]3 years ago
6 0

Unsaturation (IHD) 2 hydrogen Needed

IHD = [(2n+2) -H]/2
(H: X=1, N=-1, O= zero)

Unsaturation:
Double bonds = 1
Rings = 1
Triple Bonds = 2

The degrees of unsaturation in a molecule are additive — a molecule with one double bond has one degree of unsaturation, a molecule with two double bonds has two degrees of unsaturation, and so forth.



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1 lb of CO2 occupies 0.6 ft^3 at a pressure of 200 psi. Determine the temperature of the system.
Anarel [89]

<u>Answer:</u> The temperature of the system is 273 K

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

Given mass of carbon dioxide = 1 lb = 453.6 g   (Conversion factor: 1 lb = 453.6 g)

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation, we get:

\text{Moles of carbon dioxide}=\frac{453.6g}{44g/mol}=10.31mol

To calculate the temperature of gas, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of carbon dioxide = 200 psia = 13.6 atm   (Conversion factor:  1 psia = 0.068 atm)

V = Volume of carbon dioxide = 0.6ft^3=16.992L    (Conversion factor:  1ft^3=28.32L )

n = number of moles of carbon dioxide = 10.31 mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the system = ?

Putting values in above equation, we get:

13.6atm\times 16.992L=10.31mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times T\\\\T=273K

Hence, the temperature of the system is 273 K

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