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Lelu [443]
1 year ago
14

Which element in Group 17 is the most active nonmetal? (1) Br (2) I (3) Cl (4) F

Chemistry
1 answer:
Maksim231197 [3]1 year ago
7 0

The element of the group 17 that is most active non metal is fluorine.

The group 17 of the periodic table contains bromine(Br), iodine(I), Chlorine(Cl) and fluorine(F).

Among all the elements of the group 17. Fluorine is the smallest in size.

Because of the small size of fluorine it has the highest electronegativity in group 17.

This high electronegativity makes it a very active non metal. It provides a very high oxidizing power and low dissociation energy to the fluorine atom.

Also because of the very small size the source of attraction between the nucleus and the electrons is very high in floor in atom.

It reacts readily to form oxides and hydroxides.

So, we can conclude here that fluorine is the most active non metal of group 17.

To know more about group 17, visit,

brainly.com/question/26440054

#SPJ4

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Answer:

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3 years ago
1. How many molecules are found in 13.7 moles of CuNO3? <br><br> Please explain step by step
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Answer:

1 moles

Explanation:

The answer is 0.0011897028836018. We assume you are converting between moles CuNo3 and gram. You can view more details on each measurement unit: molecular weight of CuNo3 or grams The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles CuNo3, or 840.546 grams.

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In the sentence "After conducting the science experiment,we recorded and analyzed the results," what does the word conducting me
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Anything related to the word "performing/conducting"
4 0
4 years ago
Read 2 more answers
1. How is the genome like a stack of papers?
Ivanshal [37]

Answer:

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2 years ago
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When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from
IgorLugansk [536]

The question is incomplete, here is the complete question.

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from a mixture of 3.0g Ag and 3.0g S_8.

Answer : The mass of silver sulfide is produced from a mixture is 3.44 grams.

Explanation : Given,

Mass of Ag = 3.0 g

Mass of S_8 = 3.0 g

Molar mass of Ag = 107.8 g/mole

Molar mass of S_8 = 256 g/mole

Molar mass of Ag_2S = 247.8 g/mole

First we have to calculate the moles of Ag and S_8.

\text{ Moles of }Ag=\frac{\text{ Mass of }Ag}{\text{ Molar mass of }Ag}=\frac{3.0}{107.8g/mole}=0.0278moles

\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=\frac{3.0g}{256g/mole}=0.0117moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

16Ag(s)+S_8(s)\rightarrow 8Ag_2S(s)

From the balanced reaction we conclude that

As, 16 mole of Ag react with 1 mole of S_8

So, 0.0278 moles of Ag react with \frac{0.0278}{16}=0.00174 moles of S_8

From this we conclude that, S_8 is an excess reagent because the given moles are greater than the required moles and Ag is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Ag_2S

From the reaction, we conclude that

As, 16 mole of Ag react to give 8 mole of Ag_2S

So, 0.0278 moles of Ag react to give \frac{0.0278}{16}\times 8=0.0139 moles of Ag_2S

Now we have to calculate the mass of Ag_2S

\text{ Mass of }Ag_2S=\text{ Moles of }Ag_2S\times \text{ Molar mass of }Ag_2S

\text{ Mass of }Ag_2S=(0.0139moles)\times (247.8g/mole)=3.44g

Therefore, the mass of silver sulfide is produced from a mixture is 3.44 grams.

4 0
3 years ago
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