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Anni [7]
3 years ago
12

The reaction between SO2 and O2 to give SO3 was studied to determine the rate law for the reaction. The following results were o

btained The rate law is a) k[SO2][O2] b) k[SO2]2[O2] c) k[SO2]2 d) k[SO2][ O2]−2 e) k[SO2][ O2]−1
Chemistry
1 answer:
Tom [10]3 years ago
8 0

Answer : (b) The rate law expression for the reaction is:

\text{Rate}=k[SO_2]^2[O_2]

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The general reaction is:

A+B\rightarrow C+D

The general rate law expression for the reaction is:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

R = rate  law

k = rate constant

[A] and [B] = concentration of A and B reactant

Now we have to determine the rate law for the given reaction.

The balanced equations will be:

2SO_2+O_2\rightarrow 2SO_3

In this reaction, SO_2 and O_2 are the reactants.

The rate law expression for the reaction is:

\text{Rate}=k[SO_2]^2[O_2]^1

or,

\text{Rate}=k[SO_2]^2[O_2]

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3 years ago
An individual is hospitalized and the initial blood work indicates high levels of hco3- in the blood and a ph of 7. 47. This wou
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An individual is hospitalized and the initial blood work indicates high levels of HCO_{3} ^{-} in the blood and a pH of 7. 47. This would indicate the individual probably has compensated respiratory acidosis.

A chronic illness usually leads to compensated respiratory acidosis because the kidneys have time to adjust to the delayed onset. Even if the PCO_{2} is elevated in a compensated respiratory acidosis, the pH is within the usual range.

The kidneys counteract a respiratory acidosis by increasing the amount of HCO_{3} that tubular cells reabsorb from the tubular fluid, the amount of H^{+} that collecting duct cells secrete while also producing HCO_{3} , and the amount of NH_{3} buffer that is formed through ammoniagenesis.

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5 0
9 months ago
Under certain conditions, the equilibrium constant of the reaction below is Kc=1.7×10−3. If the reaction begins with a concentra
svetoff [14.1K]

Answer:

[Cl2] equilibrium = 0.0089 M

Explanation:

<u>Given:</u>

[SbCl5] = 0 M

[SbCl3] = [Cl2] = 0.0546 M

Kc = 1.7*10^-3

<u>To determine:</u>

The equilibrium concentration of Cl2

<u>Calculation:</u>

Set-up an ICE table for the given reaction:

               SbCl5(g)\rightleftharpoons SbCl3(g)+Cl2(g)

I                 0                    0.0546     0.0546

C              +x                        -x               -x

E               x                  (0.0546-x)    (0.0546-x)

Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M

The equilibrium concentration of Cl2 is:

= 0.0546-x = 0.0546-0.0457 = 0.0089 M

5 0
3 years ago
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What would a scientist’s next steps be if his/her data failed to support their hypothesis?
Paul [167]
To change only one variable which is very important than to test the experiment to match the hypothesis again, I think. It’s been a while since I was on that lesson‍♀️
6 0
3 years ago
Calcule la densidad del hidrógeno H2 en g/L a 327 mm Hg y 48ºc
tankabanditka [31]
<h3>The density of H₂ = 0.033 g/L</h3><h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm , N/m²

V = volume, liter  

n = number of moles  

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)

T = temperature, Kelvin  

n = N / No  

n = mole  

No = Avogadro number (6.02.10²³)  

n = m / MW

m = mass  

MW = molecular weight

For density , can be formulated :

\tt \rho=\dfrac{P\times MW}{R\times T}

P = 327 mmHg = 0,430263 atm

R = 0.082 L.atm / mol K

T = 48 ºC = 321.15 K

MW of H₂ =  2.015 g/mol

The density :

\rho=\dfrac{0,430263\times 2.015 }{0.082\times 321.15}\\\\\rho=0.033~g/L

4 0
2 years ago
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