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3 years ago

Examples of the types of organic reactions.

Chemistry

1 answer:

svp [43]3 years ago

5 0

Answer:

substitution ,elimination,addition organic reactions

Explanation:

In a substitution reaction, one atom or a group of atoms is substituted by another atom or a group of atoms to form a new substance.Elimination reaction

There are some reactions which involve the elimination or removal of the adjacent atoms. After these multiple bonds are formed and there is a release of small molecules as products.

One of the examples of elimination reaction is the conversion of ethyl chloride to ethylene.

Addition reaction is nothing but just the opposite of elimination reaction. In an addition reaction, the components A and B are added to the carbon-carbon multiple bonds and this is called addition reaction. In the reaction given below when HCl is added to ethylene, it will give us ethylene chloride.

HCl + CH2=CH2→ CH3CH2Cl

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Which of these is an example of a chemical reaction that occurs at a very fast rate?

Aloiza [94]

I believe the answer is an explosion

7 0

3 years ago

Read 2 more answers

Is combining water and powdered drink mix a chemical reaction?

VikaD [51]

Answer:

No, it's a physical reaction.

Explanation:

A chemical change produces new chemical compounds, but combining water and powder is just mixing the powder with the water. It's not a new compound.

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3 0

2 years ago

Balance the following equation:

Firdavs [7]

Answer: a) $2K_2CrO_4+3Na_2SO_3+10HCl\rightarrow 4KCl+3Na_2SO_4+2CrCl_3+5H_2O$

b) 1 mole of $SO_2$ is produced.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The skeletal equation is:

$K_2CrO_4+Na_2SO_3+HCl\rightarrow KCl+Na_2SO_4+CrCl_3 +H_2O$

The balanced equation will be:

$2K_2CrO_4+3Na_2SO_3+10HCl\rightarrow 4KCl+3Na_2SO_4+2CrCl_3+5H_2O$

Thus the coefficients are 2, 3 , 10 , 4 , 3 , 2 and 5.

b) Oxidation: $2I-^-\rightarrow I_2+2e-^-$

Reduction: $SO_4^{2-}+2e^-+4H^+\rightarrow SO_2+2H_2O$

Net reaction: $2I-^-+SO_4^{2-}+4H^+\rightarrow I_2+SO_2+2H_2O$

When 1 mole of $I_2$ is produced, 1 mole of $SO_2$ is produced.

8 0

3 years ago

1) A potassium carbonate hydrate has a formula K2CO3.XH2O. 10g of the hydrate leave 7.83g of anhydrous salt upon heating. Deduce

OLEGan [10]

The formula of the hydrated potassium carbonate salt is K₂CO₃.2H₂O

Based on the calculated mass ratio of carbon and oxygen in carbon dioxide, carbon has a fixed composition.

<h3>What are hydrated compounds?</h3>

Hydrated compounds are compounds that contain one or more molecules of water physically combined with a molecule of the compound.

The formula of the hydrated potassium carbonate salt is determined as follows:

Mass of the hydrated sample = 10.0 g

Mass of anhydrous salt = 7.83

mass of water = 10 - 7.83

mass of water = 2.17 g

Molar mass of water = 18.0 g

Molar mass of anhydrous potassium carbonate = 138 g

moles of anhydrous potassium carbonate in sample = 7.83/138

moles of anhydrous potassium carbonate = 0.056 moles

moles of water in the hydrated salt = 2.17/18

moles of water in the hydrated salt = 0.12

Mole ratio of water to anhydrous salt = 0.12/0.056

Mole ratio of water to anhydrous salt = 1 : 2

Formula of hydrated salt = K₂CO₃.2H₂O

The mass ratio of carbon to oxygen in the compounds is given below:

Sample 1:

Mass ratio = 3.62 / (13.26 - 3.62)

Mass ratio = 0.38 : 1

Sample 2:

Mass ratio = 5.91 / (21.66 - 5.91)

Mass ratio = 0.38 : 1

Sample 3:

Mass ratio = 7.07 / (25.91 - 7.07)

Mass ratio = 0.38 : 1

Carbon has a fixed composition.

Learn more about hydrated compounds at: brainly.com/question/11112492

#SPJ1

6 0

1 year ago

Calculate the ph of a 0.021 m nacn solution. [ka(hcn) = 4.9  10–10]

ohaa [14]

Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).
Chemical reaction 2: CN⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.021 M.
Ka(HCN) = 4.9·10⁻¹⁰.
Kb(CN⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵.
Kb = [HCN] · [OH⁻] / [CN⁻].
[HCN] · [OH⁻] = x.
[CN⁻] = 0.021 M - x..
2.04·10⁻⁵ = x² / (0.021 M - x).
Solve quadratic equation: x = [OH⁻] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.

4 0

3 years ago

Examples of the types of organic reactions. ​

Examples of the types of organic reactions.