Answer:
option c is correct
Explanation:
the addition of catalyst does not effect the position or equilibrium constant and increase the forward and backward reaction in equal rates so no effect would be observed
<span>The molecular formula that describes the problem is
2CH3COOH (aq) + Ca(OH)2 (s) ---> Ca(CH3COO)2 (aq) + 2H2O (l)
The net equation is written as follows:
2CH3COOH- (aq) + 2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 CH3COO- (aq) + 2H2O (l)
canceling out spectator ions
2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 H2O (l)</span>
Answer:
The amount of energy liberated will be 49.38 J.
Explanation:
The amount of energy liberated (gibbs free energy) can be calculated using the following equation:
ΔG° = -nFε
n: amount of moles of electrons transfered
F: Faraday's constant
ε: cell potential
20.0 g of Zn is equal to 0.30 mol.
Two electrons are transfered during the reaction.
Therefore, n = 2x0.30 ∴ n = 0.60
ΔG° = - 0.60 x 96.485 x 0.853
ΔG° = 49.38 J
The question is incomplete. Complete question is attached below.
..........................................................................................................................
Correct Answer: <em>Option 1) 2-pentene</em>
Reason:
Following are the IUPAC rules for naming the compound
1) Select the
longest carbon chain. In present case longest carbon chain has 5 carbon atom. Hence, it is a pentane derivative.
2) In case of alkene,
replace 'e' of alkane by 'ene'3) Give
lowest number to function group. In present case, it is double bond.
Applying above rules, the IUPAC name of compound is
2-pentene
Answer:
thats cool mate
Explanation:
hope ya have a good day, im answering just for the points tbh