Answer:
alkenes and alkynes are considered as unsaturated hydrocarbons because they are capable of reacting with hydrogen in the presence of a catalyst to form a saturated hydrocarbon (alkanes)
Explanation:
A saturated compound contains a chain of carbon atoms joined by single bonds, with hydrogen atoms attached to each carbon atom.
Propane for example, CH₃-CH₂-CH₃.
It is saturated because all the carbon atoms holds maximum number of hydrogen atoms.
Alkenes such as Propene (CH₂=CH-CH₃) and alkynes like Propyne,
(CH₃-C≡C-H) are unsaturated because the central carbon atoms contain less number of hydrogen atoms than they possibly could.
2.67×
g of
form when 6. 00 moles of bromine react completely.
As per the chemical equation 2 moles of Aluminum react with 3 moles of Bromine to generate 2 moles of Aluminum Bromide
Molecular Weight of Al= 27 g/mol
Molecular Weight of Bromine= 159.8 g/mol
Molecular Weight of Aluminum Bromide= 266.7 g/mol
So,
Moles of Br = 6 moles
Moles of Al =
= 4moles
similarly,
moles of
=
= 1mole
Moles of
=
1 mole = ![\frac{weight}{266.7}](https://tex.z-dn.net/?f=%5Cfrac%7Bweight%7D%7B266.7%7D)
weight = 2.67×
g
Learn more about Aluminum Bromide here;
brainly.com/question/19459538
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A. 6.59kj is the answer.
There you go!
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15