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3 years ago

2 KClO2 (s) 0 2 KCl (s) + 3 02 (g)

Chemistry

1 answer:

Nonamiya [84]3 years ago

5 0

Volume of O2 : 168 L

<h3>Further explanation</h3>

Given

Reaction

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

5 moles of KCIO3

Required

volume of O2

Solution

From the equation, mol O2 :

= 3/2 x moles KClO3

= 3/2 x 5 moles

= 7.5 moles

Assumed at STP( 1 mol = 22.4 L) :

= 7.5 x 22.4 L

= 168 L

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The tabulated data were collected for this reaction:

erastovalidia [21]

Answer:

ai) Rate law, $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

$CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)$

If $A + B \rightarrow C + D$ , the rate of backward reaction is given by:

$Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

$0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1$

Using the information provided in lines 3 and 4 of the table and insering the value of a:

$0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\$

$0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5$

The rate law is: $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

The rate constant $k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$ then becomes:

$k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25$

b) Overall order of reaction = a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

3 0

3 years ago

Why does a pan of warm water eventually become the same temperature as the room it is in

rewona [7]

the room will either cool down or warm up the water until it is the same temperature as the room around it.

4 0

3 years ago

what is the maximum number of covalent bonds that an oxygen atom with atomic number 8 can make with hydrogen? question 4 options

Oliga [24]

Maximum number of covalent bonds that an oxygen atom can make with hydrogen is 2.

the ground state electronic configuration of oxygen is 2s² 2p⁴ that means it has 6 electrons in its valence shell and require two electrons are required to complete its octate.
Two bonds are created when an electron donor atom shares the two needed electrons with oxygen. The ability of two oxygen atoms to share valence electrons results in the creation of a double bond between the two atoms.
There are no longer any empty orbitals in the octet of oxygen after it is complete. As a result, it is unable to accept more electrons or create more bonds.

Therefore, Oxygen can only generate two bonds because it needs two additional electrons to complete its octet, after which it will run out of empty orbitals in which to receive additional electrons and create additional bonds.

learn more about octate here:

https://brainly.in/question/24161245

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2 years ago

Is there a phobia for chemical reactions specifically?

gizmo_the_mogwai [7]

Answer:

chemonoia

Explanation:

Many people have a strong fear of “chemicals” that belies the evidence. It's called chemonoia – and it may be damaging their health. If you were paranoid, you might think there really was something in the water that's damaged our sense of reason.

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3 years ago

The Ka for acetic acid (HC2H3O2) is 1.80 x 10-5 . Determine the pH of a 0.0500mol/L acetic acid solution.

TiliK225 [7]

Answer:

pH = 3.02

Explanation:

Acetic Acid is a weak acid (HOAc) that ionizes only ~1.5% as follows:

HOAc ⇄ H⁺ + OAc⁻.

In pure water the hydronium ion concentration [H⁺] equals the acetate ion concentration [OAc⁻] and can be determined* using the formula [H⁺] = [OAc⁻] = SqrRt(Ka·[acid]) = SqrRt(1.8x10⁻⁵ x 0.0500)M = 9.5x10⁻⁴M.

By definition, pH = -log[H⁺] = -log(9.5x10⁻⁴) = 3.02

______________________________________________________

*This formula can be used to determine the [H⁺] & [Anion⁻] concentrations for any weak acid in pure water given its Ka-value and the molar concentration of acid in solution.

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3 years ago