A solid calcium metal reacts with molecular oxygen. WORD EQUATION = CHEMICAL EQUATION =

Why can't methanol, CH3OH, be used as a solvent for sodium amide, NaNH2? Sodium amide is nonpolar and methanol is polar. Sodium

Elodia [21]

Answer: sodium amide undergoes an acid -base reaction

Explanation:

sodium amide is a ionic compound and basically exists as sodium cation and amide anion. Amide anion is highly basic in nature and hence as soon as there is amide anion generated in the solution , Due to its very pronounced acidity it very quickly abstracts the slightly acidic proton available on methanol.

This leads to formation of ammonia and sodium methoxide.

Hence sodium amide reacts with methanol and abstracts its only acidic proton and form ammonia and sodium Methoxide.

Hence the 3rd statement is a corrects statement.

So we cannot use methanol for sodium amide because sodium amide itself would react with methanol and the inherent molecular natur of sodium amide would then change.

The 1st and 2nd statements both are incorrect because both the compounds methanol as well as sodium amide have dipole moments and hence are polar molecules.

The 4th statement is also incorrect as both the molecules have dipole moment and hence there would be ion-dipole forces operating between them.

The following reaction occurs:

NaNH₂+CH₃OH→NH₃+CH₃ONa

4 0

3 years ago

The Lucas test has _______ results based on the type of alcohol present because the reaction involves a _________, which is ____

Olin [163]

Answer:

1) positive

2) carbocation

3) most stable

4) faster

Explanation:

A common test for the presence of alcohols can be achieved using the Lucas reagent. Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride.

The reaction of Lucas reagent reacts with alcohols leading to the formation of an alkyl chloride. Since the reaction proceeds via a carbocation mechanism, tertiary alcohols give an immediate reaction. Once a tertiary alcohol is mixed with Lucas reagent, the solution turns cloudy almost immediately indicating an instant positive reaction.

Secondary alcohols may turn cloudy within five minutes of mixing the solutions. Primary alcohols do not significantly react with Lucas reagent obviously because they do not form stable carbocations.

Therefore we can use the Lucas reagent to distinguish between primary, secondary and tertiary alcohols.

5 0

3 years ago

Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Det

mixer [17]

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

8 0

3 years ago

3.79 x 10^8 = .mol 2 2 kg.m g.cm

yuradex [85]

Answer:

sorry I have no idea what the answer is

3 0

3 years ago

What are three limitations of current cloaking technology.

kirill [66]

Explanation:

1.undetectable to electromagnetic waves

2.hiding an object from an illumination containing diffre t wave lengths become difficult as the object sizes grow.

3. reduce the scattering by two orders.

4 0

2 years ago