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V125BC [204]
2 years ago
14

In a chemical process, you need to force a compound to bond with a halogen, causing it to lose hydrogen. Which type of reaction

do you need to perform?
Chemistry
1 answer:
Kazeer [188]2 years ago
5 0
Answer is Halogenation because Halogenation is a type of substitution reaction in which a hydrogen atom is replaced by a halogen atom in a molecule.
The molecule looses its hydrogen atom as the halogen is introduced into the molecule. This sort of reaction is very common in organic chemistry. Many hydrocarbons can be halogenated in the presence of light.
You might be interested in
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules. Several other gases make up the remaining 1% of air molecules.
kherson [118]

Answer:

The partial pressure of the other gases is 0.009 atm

Explanation:

Step 1: Data given

Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.

The atmospheric pressure = 0.90 atm

Step 2: Calculate mol fraction

If wehave  100  moles of air, 78 moles will be nitrogen,

21  moles will be  oxygen, and  1  mol will be other gases.

Mol fraction = 1/100 = 0.01

Step 3: Calculate the partial pressure of the other gases

Pgas = Xgas * Ptotal

⇒ Pgas = the partial pressure = ?

⇒ Xgas = the mol fraction of the gas = 0.01

⇒Ptotal = the total pressure of the pressure = 0.90 atm

Pgas = 0.01 * 0.90 atm

Pgas = 0.009 atm

The partial pressure of the other gases is 0.009 atm

4 0
3 years ago
What elements are being reduced and oxidized in this chemical formula?
dangina [55]

Answer: The element Na (Sodium) is getting oxidized and Hydrogen is getting reduced.

Explanation:

Oxidation reactions are the reactions in which addition of oxygen takes place.

Reduction reactions are the reactions in which loss of oxygen takes place.

For a given reaction:

2Na(s)+2H_2O(l)\rightarrow H_2(g)+2NaOH(aq.)

Sodium is getting oxidized because there is an addition of reaction with that element.

Hydrogen is getting reduced because there is a removal of oxygen with that element.

3 0
3 years ago
2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
2 years ago
A star is mainly made of hydrogen gas, but most classes of stars include other elements such as helium, calcium, and other heavi
Xelga [282]
That would be true, since most other stats include helium, calcium, etc
6 0
2 years ago
Read 2 more answers
When 12 moles of methanol (ch3oh) and 24 moles of oxygen gas react according to the chemical equation below, what is the limitin
Nataliya [291]
The balanced reaction equation for the reaction between CH₃OH and O₂ is 
                        2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)
Initial moles        12                24
Reacted moles   12                18
Final moles          -                   6              12               24
 
The stoichiometric ratio between CH₃OH and O₂ is 2 : 3
Hence,
   reacted moles of O₂ = reacted moles of CH₃OH x (3/2)
                                      = 12 mol x 3 / 2  
                                      = 18 mol

All of CH₃OH moles react with O₂.

Hence, the limiting agent is CH₃OH. 

Excess reagent is O₂.
Amount of moles of excess reagent left = 24 - 18 mol = 6 mol
                                            
5 0
3 years ago
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