I only identify decomposition
Answer:
Acetylide , Enol ,aldehydes, tautomers, alkynes , Hydroboration, Keto
Explanation:
Reset <u>Acetylide</u> anions are strong nucleophiles that open epoxide rings by an SN2 mechanism. <u>Enol </u>tautomers have an O-H group bonded to a C=C. <u>aldehydes </u>are formed from terminal alkynes with the addition of water using BH3 then H2O2. <u>tautomers</u> are constitutional isomers that differ in the location of a double bond and a hydrogen and exist in an equilibrium with each other. <u>alkynes</u> are compounds that contain a carbon-carbon triple bond. <u>Hydroboration</u> of a terminal alkyne adds BH₂ to the less substituted, terminal carbon.<u> Keto</u> tautomers have a C=O and an additional C-H bond.
Answer:
Volume of container = 0.0012 m³ or 1.2 L or 1200 ml
Explanation:
Volume of butane = 5.0 ml
density = 0.60 g/ml
Room temperature (T) = 293.15 K
Normal pressure (P) = 1 atm = 101,325 pa
Ideal gas constant (R) = 8.3145 J/mole.K)
volume of container V = ?
Solution
To find out the volume of container we use ideal gas equation
PV = nRT
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
First we find out number of moles
<em>As Mass = density × volume</em>
mass of butane = 0.60 g/ml ×5.0 ml
mass of butane = 3 g
now find out number of moles (n)
n = mass / molar mass
n = 3 g / 58.12 g/mol
n = 0.05 mol
Now put all values in ideal gas equation
<em>PV = nRt</em>
<em>V = nRT/P</em>
V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa
V = 121.87 ÷ 101,325 pa
V = 0.0012 m³ OR 1.2 L OR 1200 ml
Answer:
P.E = 25.48 J
Explanation:
Given data:
Mass = 2 Kg
Height = 1.3 m
Potential energy = ?
Solution:
Formula:
P.E = m . g . h
P. E = potential energy
m = mass in kilogram
g = acceleration due to gravity
h = height
Now we will put the values in formula.
P.E = m . g . h
P.E = 2 Kg . 9.8 m /s² . 1.3 m
P.E = 25.48 Kg. m² / s²
Kg. m² / s² = J
P.E = 25.48 J
