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Murljashka [212]
4 years ago
12

The Reaction between ketones and alcohol

Chemistry
1 answer:
Colt1911 [192]4 years ago
7 0
It has been demonstrated that water adds rapidly to the carbonyl function of aldehydes and ketones to form geminal-diol. In a similar reaction alcohols add reversibly to aldehydes and ketones to form hemiacetals (hemi, Greek, half). This reaction can continue by adding another alcohol to form an acetal.
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How can you separate a liquid from a liquid?
Yakvenalex [24]
We can use distillation to separate liquid to liquid
7 0
4 years ago
Explain how the Rutherford gold foil experiment correctly advanced atomic theory away from the plum pudding model?
Rudiy27

Answer is: Rutherford demonstrate that J.J Thompson's Plum Pudding model was not accurate.

Rutherford theorized that atoms have their charge concentrated in a very small nucleus.

This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).  

Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.

That is different from Plum Pudding model, because it shows that most of the atom is empty space.

7 0
3 years ago
Read 2 more answers
Consider the reaction: 2BrF3(g) --> Br2(g) + 3F2(g)
riadik2000 [5.3K]

Answer : The entropy change of reaction for 1.62 moles of BrF_3 reacts at standard condition is 217.68 J/K

Explanation :

The given balanced reaction is,

2BrF_3(g)\rightarrow Br_2(g)+3F_2(g)

The expression used for entropy change of reaction (\Delta S^o) is:

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]

where,

\Delta S^o = entropy change of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

\Delta S_f^0_{(Br_2)} = 245.463 J/mol.K

\Delta S_f^0_{(F_2)} = 202.78 J/mol.K

\Delta S_f^0_{(BrF_3)} = 292.53 J/mol.K

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]

\Delta S^o=268.74J/K

Now we have to calculate the entropy change of reaction for 1.62 moles of BrF_3 reacts at standard condition.

From the reaction we conclude that,

As, 2 moles of BrF_3 has entropy change = 268.74 J/K

So, 1.62 moles of BrF_3 has entropy change = \frac{1.62}{2}\times 268.74=217.68J/K

Therefore, the entropy change of reaction for 1.62 moles of BrF_3 reacts at standard condition is 217.68 J/K

3 0
4 years ago
Prove that PV = nRT.​
qaws [65]

Find your answer in the explanation below.

Explanation:

PV = nRT is called the ideal gas equation and its a combination of 3 laws; Charles' law, Boyle's law and Avogadro's law.

According to Boyle's law, at constant temperature, the volume of a gas is inversely proportional to the pressure. i.e V = 1/P

From, Charles' law, we have that volume is directly proportional to the absolute temperature of the gas at constant pressure. i.e V = T

Avogadro's law finally states that equal volume of all gases at the same temperature and pressure contain the same number of molecules. i.e V = n

Combining the 3 Laws together i.e equating volume in all 3 laws, we have

V = nT/P,

V = constant nT/P

(constant = general gas constant = R)

V = RnT/P

by bringing P to the LHS, we have,

PV = nRT.

Q.E.D

6 0
3 years ago
Express 0.0004120 in scientific notation.
Montano1993 [528]

Answer:

4.120 × 10^-4

Explanation:

6 0
3 years ago
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