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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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san4es73 [151]

The number of moles of the magnesium (mg) is 0.00067 mol.

The number of moles of hydrogen gas is 0.0008 mol.

The volume of 1 more hydrogen gas (mL) at STP is 22.4 L.

<h3>Number of moles of the magnesium (mg)</h3>

The number of moles of the magnesium (mg) is calculated as follows;

number of moles = reacting mass / molar mass

molar mass of magnesium (mg) = 24 g/mol

number of moles = 0.016 g / 24 g/mol = 0.00067 mol.

<h3>Number of moles of hydrogen gas</h3>

PV = nRT

n = PV/RT

Apply Boyle's law to determine the change in volume.

P1V1 = P2V2

V2 = (P1V1)/P2

V2 = (101.39 x 146)/(116.54)

V2 = 127.02 mL

Now determine the number of moles using the following value of ideal constant.

R = 8.314 LkPa/mol.K

n = (15.15 kPa x 0.127 L)/(8.314 x 290.95)

n = 0.0008

<h3>Volume of 1 mole of hydrogen gas at STP</h3>

V = nRT/P

V = (1 x 8.314 x 273) / (101.325)

V = 22.4 L

Learn more about number of moles here: brainly.com/question/13314627

#SPJ1

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How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mmhg and a temperature of 21.0°c, can be made
Shalnov [3]

The answer is V = 0.6 L


The explanation:


1- First we will get P (H) :


p (H) = p total - p (H2O)


= 752 mmHg - 18.65 mmHg


= 733 mmHg


= 733 mmHg X (1 atm / 760 atm) = 0.964 atm


2- then when the reaction equation is:


Zn(s)+2HCl(aq) -> H2(g)+ZnCl2(aq)


so, then 1 mole of Zn will produce 1 mole of H2.


so, we need to get first the mole of Zn:


mole Zn = mass / molar mass


= 1.566g / 65.38g/mol

= 0.02395mol


∴ 0.02395 mol of Zn will produce 0.02395 mol of H2


by using the ideal gas equation we can get the volume V :


when :


PV = nRT


when T is the temperature = 21 + 273 = 294 K


and R is the ideal gas constant = 0.0821 L atm /K mol


and n is the number of moles = 0.2395 mol


and P is the pressure = 0.964 atm


so, by substitution:


V = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm)


= 0.6 L

3 0
3 years ago
Read 2 more answers
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