The first word in the name of an ester is derived from the alcohol used in the esterification.
<h3>What is esterification?</h3>
Esterification is a chemical process where an organic acid with the formula is combined with an alcohol molecule having the chemical formula (ROH).
The process of esterification is known to produce an ester molecule and during this phenomenon is released water (H2O).
An example of an esterification reaction occurs when ethanoic acid (i.e., the active ingredient of vinegar) can react with C2H5OH (i.e., ethanol) in order to form the ethyl ethanoate molecule, which is a well-known ester molecule.
In conclusion, the first word in the name of an ester is derived from the Alcohol used in the esterification.
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Answer:
P = 164 Atm
Explanation:
PV = nRT => P = nRT/V
n = 10.0 moles
R = 0.08206 L·Atm/mol·K
T = 27.0°C = 300 K
V = 1.50 Liters
P = (10.0 mol)(0.08206 L·Atm/mol·K )(300 K)/(1.50 Liters) = 164.12 Atm ≅ 164 Atm (3 sig. figs.)
Answer:
The answer would be C 214g
Explanation:
890j of heat causes 4.6°c increase in temperature
specific heat of aluminium after is o.9022 j /g°c
now by using the formula .The mass of aluminium would be c that is 214 g
This doesn't need an ICE chart. Both will fully dissociate in water.
Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.
Step 1:
write out balanced equation for the reaction
HClO4+KOH ⇔ KClO4 + H2O
the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4
Step 2:
Determining the number of moles present in HClO4 and KOH
Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4
Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L
Remember:
M = moles/L so we have 0.025 L of 0.723 moles/L HClO4
Multiply the volume in L by the molar concentration to get:
0.025L x 0.723mol/L = 0.0181 moles HClO4.
Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH
Step 3:
Determine how much HClO4 remains after reacting with the KOH.
Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:
moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0
This means all of the HClO4 is used up in the reaction.
If all of the acid is fully reacted with the base, the pH will be neutral = 7.
Determine the H3O+ concentration:
pH = -log[H3O+]; [H3O+] = 10-pH = 10-7
The correct answer is 1.0x10-7.
Answer:
You
Explanation:
Will have to fill in the graph organizer with a story
