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3 years ago

How is the periodic law demonstrated in halogens

Chemistry

2 answers:

Zina [86]3 years ago

4 0

Answer:

The periodic law demonstrated in halogens by:

All halogens have the same number of valence electrons and similar chemical properties

Explanation:

The halogen are located in the group of seven A of the periodic table example of halogen are iodine and chlorine.

The halogens are the reactive non-metals in the periodic table, it reacts with alkali metals.

Halogen responds with iron, iron wool outbreak into flames when it responds with fluorine gas.

Halogen range from solid to liquid to gaseous at room temperature.

inn [45]3 years ago

3 0

Halogens is defined as the group of 7 periodic table. As, every periodic table contains 7 valence electrons and they only need 1 more to complete an outer shell, that is why they are extremely reactive. And according to the law that recurring patterns of the properties of elements arise when they are arranged in order of increasing atomic number. As the halogen all act very similarly with each other in chemical reaction, it is true.

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The Wolff-Kishner reaction involves the reaction of an aldehyde/ketone with hydrazine in the presence of KOH. The process is use

Ivan

Answer:

in the attached image is the reaction mechanism.

Explanation:

The first reaction (reaction 1) shown in the attached image is the Wolff-Kishner reduction, which is characterized when the carbonyl is reduced to an alkane in the presence of a hydrazine and a base. In reaction 1, the aldehyde reacts with hydrazine to produce oxime. This mechanism begins with the attack of the amine on the carbonyl group. Proton exchange happens and the water leaves the molecule.

In reaction 2, the KOH is deprotoned in nitrogen and organized to form the bond between the nitrogen molecule. this deprotonation releases the nitrogen gas

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3 years ago

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tatiyna

Answer:

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3 years ago

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zimovet [89]

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3 years ago

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Wewaii [24]

Answer:

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3 years ago

What mass of iron is formed when 10 grams of carbon react with 80 grams of iron iii oxide?

yKpoI14uk [10]

Answer:

55.85 grams of Fe is formed.

Explanation:

Identify the reaction:

2Fe₂O₃ + 3C → 4Fe + 3CO₂

Identify the limiting reactant, previously determine the mol of each reactant

(mass / molar mass)

10 g / 12 g/m = 0.83 moles C

80 g / 159.7 g /m = 0.500 moles Fe₂O₃

2 moles of oxide need 3 moles of C, to react

0.5 moles of oxide, will need ( 0.5 . 3)/ 2 = 0.751 mol

I have 0.83 moles of C, so C is the excess.

The limiting is the oxide.

3 mol of C need 2 mol of oxide to react

0.83 mol of C, will need (0.83 . 2)/ 3 = 0.553 mol of oxide, and I only have 0.5 (That's why Fe₂O₃ is the limiting)

Ratio is 2:4 (double)

If I have 0.5 moles of oxide, I will produce the double, in the reaction.

1 mol of Fe, will be produce so its mass is 55.85 g

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3 years ago