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3 years ago

What are the radicals?Write their types with four examples.

Chemistry

2 answers:

polet [3.4K]3 years ago

4 0

Answer:

<h3>A group of atoms behaving as a unit in a number of compounds. </h3>

Explanation:

<h2>Sorry for giving half answer only .I known this much only.</h2><h2>I hope U like my answer and my answer help u.</h2>

alina1380 [7]3 years ago

3 0

Answer:

<h3>A radical is a chemical entity with an unpaired electron in it. A radical can be electrically neutral, radical cation charged positively, or radical anion charged negatively. The Radicals Examples. ... In ultraviolet light the chlorine molecule Cl2 undergoes homolysis to form two radicals of Cl.</h3>

<h3>Radicals : A radical is an atom of an element or a group of atoms of different elements that behaves as a single unit with a positive or negative charge on it. Valency : It is the number of electrons donated or accepted by the valence shell of an atom during chemical combination.</h3>

<h2>Example </h2>

<h3>: Some common examples of basic radicals are Na+, Ba2+, Fe3+ etc. Composite Radicals-Acetylimino, Cyclohexane carbonyl etc. Bivalent Acidic Radicals-The ion formed after removal of Hydrogen ion from an acid is called as Acidic Radical.</h3>

HOPE ITS CLEAR. ☺️

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Marketing affects the consumer mindset by leaving them with the decision of either purchasing a product or not and why it should be purchased or not.

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2 years ago

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

Blababa [14]

After Rounding off The percentage yield is 64%

<h3>What is Percentage Yield ?</h3>

It is the ratio of actual yield to theoretical yield multiplied by 100% .

It is given in the question

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

The reaction is

HBrO₃ (aq) + 5 HBr (aq) → 3 H₂O (l) + 3 Br₂ (aq)

Actual yield = 47.3 grams

Molecular weight of Bromic Acid is 128.91 gram

Moles of Bromic Acid = 20/128.91 = 0.155 mole

Mole fraction ratio of Bromic Acid to Bromine is 1 :3

Therefore for 0.155 mole of Bromic Acid 3 * 0.155 = 0.465 mole of Bromine is produced.

1 mole of Bromine = 159.8 grams of Bromine

0.465 of Bromine = 74.31 grams of Bromine

Percentage Yield = (47.3/74.31)*100 = 63.65 %

After Rounding off The percentage yield is 64% .

To know more about Percentage Yield

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1 year ago

The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound, C5H10O

vova2212 [387]

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2 years ago

What is the equilibrium constant expression for the ksp of sr3(po4)2?

Eddi Din [679]

The equilibrium constant expression for KSP of Sr3(PO4)2 is

KSP={(Sr^2+)^3 (PO4^3-)^2/ Sr3(PO4)2}

Explanation

write the ionic equation for Sr3(PO4)2

Sr3(PO4)2 → 3Sr^2+ + 2 PO4^3-

KSP is given by (concentration of the products raised to their coefficient /concentration of reactants raised to their coefficient)

7 0

3 years ago

Read 2 more answers

A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under

sukhopar [10]

Answer:

$\rho _2=0.22g/L$

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

$PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT$

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

$\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}$

It means we can compute the final density as shown below:

$\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}$

Now, we plug in to obtain:

$\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L$

Regards!

8 0

2 years ago

What are the radicals?Write their types with four examples.​

What are the radicals?Write their types with four examples.