In an alkene, cis and trans isomers are possible because the double band is rigid, cannot rotate, has groups attached to the carbons of the double bond that are fixed relative to each other, and only occurs with double bonds-possibility that molecule will have different geometries; two different molecules with slightly different properties.
-Trans-2 ends of chain across the double bond.
While naming Cis-Trans isomers the prefix cis or trans are placed in front of the alkene name when there are cis-trans isomers.
The molarity of solution made by dissolving 15.20g of i2 in 1.33 mol of diethyl ether (CH3CH2)2O is =0.6M
calculation
molarity =moles of solute/ Kg of the solvent
mole of the solute (i2) = mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol
moles is therefore= 15.2 g/253.8 g/mol = 0.06 moles
calculate the Kg of solvent (CH3CH2)2O
mass = moles x molar mass
molar mass of (CH3CH2)2O= 74 g/mol
mass is therefore = 1.33 moles x 74 g/mol = 98.42 grams
in Kg = 98.42 /1000 =0.09842 Kg
molarity is therefore = 0.06/0.09842 = 0.6 M
The reaction given above is a combustion reaction. All combustion reactions are exothermic, meaning they give off heat when they react,
Answer:
2578.99 years
Explanation:
Given that:
100 g of the wood is emitting 1120 β-particles per minute
Also,
1 g of the wood is emitting 11.20 β-particles per minute
Given, Decay rate = 15.3 % per minute per gram
So,
Concentration left can be calculated as:-
C left = ![[A_t]=\frac{11.20\ per\ minute}{15.3\ per\ minute\ per\ gram}\times [A_0]= 0.7320[A_0]](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5Cfrac%7B11.20%5C%20per%5C%20minute%7D%7B15.3%5C%20per%5C%20minute%5C%20per%5C%20gram%7D%5Ctimes%20%5BA_0%5D%3D%200.7320%5BA_0%5D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Also, Half life of carbon-14 = 5730 years
Where, k is rate constant
So,
The rate constant, k = 0.000120968 year⁻¹
Time =?
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
So,
![\frac {[A_t]}{[A_0]}=e^{-0.000120968\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-0.000120968%5Ctimes%20t%7D)
![\frac {0.7320[A_0]}{[A_0]}=e^{-0.000120968\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B0.7320%5BA_0%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-0.000120968%5Ctimes%20t%7D)
<u>t = 2578.99 years</u>
It takes 21.3 days
<h3>Further explanation</h3>
Given
5 hr = 8 kg Alcohol
Required
Days to consume 1000 kg of glucose
Solution
Alcoholic fermentation⇒ glucose produces ethanol and carbon dioxide,
C₆H₁₂O₆ → 2 C₂H₅OH + 2CO₂
mol ethanol :
moles of glucose to produce 108.7 moles ethanol :
54.35 moles = 5 hours
moles of 1000 kg of glucose :
So for 5555.5 moles, it takes :
