Which type of bond is present within a water molecule

If 133mL of a 1.2 M glucose solution is diluted to 41.0L, what is the molarity of the diluted solution?

WITCHER [35]

the formula we is as follows:-

M1V1= M2V2

where

M1=1.2

V1=0.133l

V2=41l

M2=?

1.2 × 0.133 = 41 × M2

0.1596 = 41 × M2

M2 = 0.15960/41

M2 = 0.0038926829

7 0

3 years ago

__________ is important to a system because each part of a system relies on the previous work of another part.

Debora [2.8K]

I’m guessing A. Order

3 0

2 years ago

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En un recipiente colocamos unos cubos de hielo luego tapamos y observamos que sucede despues de un tiempo

SSSSS [86.1K]

Three questions come along with the given statement. It is in Spanish language:

a) Por qué se humedeció la parte exterior del frasco?

b) Por qué el hielo disminuyó su volumen y ahora es agua?

c) Por qué puede haber agua en el exterior?

These are the three answers (in English).

First question:

a) Por qué se humedeció la parte exterior del frasco?

The question is Why did the outside of the bottle get wet?

Answer:

The outside of the bottle get wet because the ice cubes cooled the walls of the bottle, so the air surrounding the bottle also cooled.

The air contains humidity (water) in gas phase. The hotter the air the more the amount of humidity it can retain, the cooler the air the less the amount of humidity it can retain.

Then, when the air close to the walls of the bootle got cooler some of the water in the air became liquid and those are the drops of water that you see in the outside of the bottle.

Second question

b) Por qué el hielo disminuyó su volumen y ahora es agua?

The question is Why did the ice diminish its volume and now it is water?

Answer:

The ice diminished its volume and now it is water, becasue the ice, which is cooler than the surroundings, received heat energy (from the surroundings) and then its temperature increased. At some moment, this temperature reached the melting point of the ice (water) and it started to become liquid.

Third question

c) Por qué puede haber agua en el exterior?

The question is: Why can there be water outside?

Answer:

The water outside is outside since the beginning: it is in the air. You do not see it because it is gas state. When the air close to the walss of the bottle got cooler, part of the water in the air became liquid.

4 0

2 years ago

Which statement is the best example of an object and motion that would make it hard for people to accept Newton’s first law? A r

djverab [1.8K]

Answer:

The best example of an object and motion that would make it hard for people to accept Newtons first law is, "A rolling ball eventully slows down and comes to a stop".

Explanation:

7 0

3 years ago

In each reaction box place the best reagent and conditions from the list below benzene 3 boxes

den301095 [7]

Answer : The correct answer is 1) AlCl₃ - CH₃Cl 2) HNO₃ -H₂SO₄ at room temperature 3) Fuming HNO₃ -H₂SO₄ at 90-100 ⁰ C heat .

I think this reaction is forming 2,4,6- trinitrotoluene from benzene, since the product is not mentioned. Following are the steps to convert Benzene to 2,4,6 trinitrotoluene .

Step 1: Conversion of Benzene to Toluene .

Benzene can be converted to toluene by Friedel Craft Alkylation of benzene . In this reaction reagent AlCl₃ and Ch3Cl is used . Electrophile CH³⁺ is produced which attached on carbon of benzene and formation of Toluene and HCl occur.

$Benzene \frac{AlCl3}{Ch3Cl}> Toulene + HCl$

Step 2 : Conversion of Toluene to dinitrotoluene.

Dinitritoluene is prepared from toluene by Nitration . This reaction uses Electrophilic substitution mechanism . The reagents used are HNO₃ and H₂SO₄ at room temperature . These reagents produces NO₂⁺ ( nitronium ion ), a electrophile which attacks on C2 and C4 Carbon atoms of Toluene.

Toluene $Tolune \frac{HNO3 -H2SO4}{30-40 degree C} -> 2,4- dinitrotoluene$

Step 3) Conversion of Dinitro toluene to trinitrotoluene.

This reaction is extended nitration of toluene . Further nitration is done in extreme condition . The temperature of reaction is increased to 90- 100 ⁰ C . Due to which there is more production of NO²⁺ ion occurs from HNO₃ -H₂SO₄ and they attack on C6 carbon atom of dinitrotoluene which forms 2,4,6- trinitrotoluene.

Dinitrotoluene $2,4 -dinitrotoluene \frac{fuming HNO3-H2So4}{90-100 C} -> 2,4,6-trinitrotoluene.$

So over all reaction uses three reagents in order :

$Benzene \frac{AlCl3}{CH3Cl} -> Toluene \frac{HNO3-H2So4}{room temp} -> 2,4-dinitrotoluene \frac{Fuming HNO3 -H2SO4}{Heating at 90-100 C} -> 2,4,6-trinitrotoluene .$

3 0

3 years ago

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