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3 years ago

Is carbon or a fluorine atom more reactive?

Chemistry

2 answers:

AveGali [126]3 years ago

8 0

I Believe it's Carbon

katrin [286]3 years ago

7 0

Fluorine,definitely.

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What happens when hydrogen reacts with nitrogen at necessary conditions.

Orlov [11]

Answer:

When hydrogen gas combines with nitrogen to form Ammonia the following chemical reaction will take place. Our equilibrium reaction will be N2(g) + 3H2(g) ⇔ 2NH3(g) + Heat. In this case, Hydrogen and nitrogen react together to form ammonia.

Explanation:

3 0

3 years ago

Read 2 more answers

State three differences between an electron and a proton

NNADVOKAT [17]

An electron is smaller than a proton. An electron has the opposite charge of a proton. An electron is outside the nucleus, while a proton is part of the nucleus.

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3 years ago

barium chloride (Ba(CIO3)2) breaks down to form barium chloride and oxygen. what is the balanced equation for this equation?

kompoz [17]

Answer:

Ba(ClO₃)₂ → BaCl₂ + 3 O₂

Explanation:

When exposed to heat, barium chlorate (Ba(ClO₃)₂ breaks down into an inorganic compound (Barium chloride - BaCl₂) and a molecule (Oxygen - O₂).

6 0

4 years ago

Write a balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air.

Lelechka [254]

Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

Explanation:

When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.

For example, chemical equation for oxidation of methane is as follows.

$CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O$

Number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 2

Number of atoms present on product side are as follows.

C = 1
H = 2
O = 3

To balance this equation, multiply $O_{2}$ by 2 on reactant side. Also, multiply $H_{2}O$ by 2 on product side. Hence, the equation can be rewritten as follows.

$CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$

Now, the number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 4

Number of atoms present on product side are as follows.

C = 1
H = 4
O = 4

Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.

Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

4 0

3 years ago

Question 3 0

Tanya [424]

Answer:

Option D. KBr < KCl < NaCl

Explanation:

We'll begin by calculating the number of mole of each sample.

This can be obtained as follow:

For NaCl:

Mass = 1 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 1/58.5

Mole of NaCl = 0.0171 mole

For Kbr:

Mass = 1 g

Molar mass of KBr = 39 + 80 = 119 g/mol

Mole of KBr =?

Mole = mass /Molar mass

Mole of KBr = 1/119

Mole of KBr = 0.0084 mole

For KCl:

Mass = 1 g

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1/74.5

Mole of KCl = 0.0134 mole

Summary

Sample >>>>>>>> Number of mole

NaCl >>>>>>>>>> 0.0171

KBr >>>>>>>>>>> 0.0084

KCl >>>>>>>>>>> 0.0134

Arranging the number of mole of the sampl in increasing order, we have:

KBr < KCl < NaCl

5 0

3 years ago