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nikdorinn [45]
1 year ago
12

The interhalogen ClF₃ is prepared in a two-step fluorination of chlorine gas:Cl₂(g) + F₂(g) ⇄ CIF(g) CIF(g) + F₂(g) ⇄ CIF₃(g)(a)

Balance each step and write the overall equation.
Chemistry
1 answer:
bogdanovich [222]1 year ago
8 0

The interhalogen CIF3 is prepared in a two-step fluorination of chlorine gas : Cl₂(g) + F₂(g) ⇄ CIF(g) CIF(g) + F₂(g) ⇄ CIF₃(g). The balanced equation will be

           Cl₂(g) + 2F₂(g) ⇌ ClF(g) + CIF₃(g)

Solution:

These are the two equation

Cl 2(g)+F 2(g) ⇌ ClF(g)                             (1)

ClF(g)+F 2 (g) ⇌ ClF 3 (g)                         (2)

Balance equation one

      Cl 2(g)+F 2(g) ⇌ ClF(g)                        

This will be the result

      Cl 2 (g)+F 2 (g)⇌2ClF(g)

Balance equation two

      ClF(g)+F 2 (g)⇌ClF 3 (g)

The equation is already balanced

On adding both the equation the result will be

Cl 2 (g)+2F 2(g) ⇌ ClF(g)+ClF 3 (g)

​

To learn more click the given link

brainly.com/question/11904811

#SPJ4

​

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Answer:

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5 0
3 years ago
If 42.8 mL of 0.204 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solut
Aneli [31]

Hey There!

At neutralisation moles of H⁺ from HCl  = moles of OH⁻ from Ca(OH)2  so :

0.204 * 42.8 / 1000  => 0.0087312 moles

Moles of Ca(OH)2 :

2 HCl + Ca(OH)2 = CaCl2 + 2 H2O

0.0087312 / 2 => 0.0043656 moles (  since each Ca(OH)2 ives 2 OH⁻ ions )

Therefore:

Molar mass Ca(OH)2 = 74.1 g/mol

mass = moles of Ca(OH)2 * molar mass

mass =  0.0043656 * 74.1

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Hope that helps!

6 0
3 years ago
How many moles of methane are in 20.32 x 10^16 molecules
Marizza181 [45]

Taking into account the definition of avogadro's number, 3.37×10⁻⁷ moles of methane are 20.32×10¹⁶ molecules.

First of all, you have to know that Avogadro's number indicates the number of particles of a substance (usually atoms or molecules) that are in a mole.

Its value is 6.023×10²³ particles per mole and it applies to any substance.

Then you can apply the following rule of three: if 6.023×10²³ molecules are contained in 1 mole of methane, then 20.32×10¹⁶ molecules are contained in how many moles of methane?

amount of moles of methane= (20.32×10¹⁶ molecules × 1 mole)÷ 6.023×10²³ atoms

Solving:

<u><em>amount of moles of methane= 3.37×10⁻⁷ moles</em></u>

Finally, 3.37×10⁻⁷ moles of methane are 20.32×10¹⁶ molecules.

Learn more about Avogadro's Number:

  • <u>brainly.com/question/11907018?referrer=searchResults </u>
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  • <u>brainly.com/question/1528951?referrer=searchResults</u>
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