which of the following compounds would be considered an electrolyte? a. c6h12o6 b. naoh c. co2 d. agcl

The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.

Brilliant_brown [7]

Answer : The rate of reaction is,

$Rate=4.77\times 10^{-19}M/s$

The appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$

The rate law expression will be:

$Rate=k[NO_2][O_3]$

Given:

Rate constant = $k=1.69\times 10^{-4}M^{-1}s^{-1}$

$[NO_2]$ = $1.77\times 10^{-8}M$

$[O_3]$ = $1.59\times 10^{-7}M$

$Rate=k[NO_2][O_3]$

$Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})$

$Rate=4.77\times 10^{-19}M/s$

The expression for rate of appearance of $NO_3$ :

$\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}$

As, $\text{Rate of reaction}=4.77\times 10^{-19}M/s$

So, $\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s$

Thus, the appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

7 0

2 years ago

. In a titration, a 25.0 mL sample of 0.150 M HCl is neutralized with 44.45 mL of Ba(OH)2. a. Write the balanced molecular equat

choli [55]

Answer:

Equation of reaction:

a) 2HCl + Ba(OH)2 ==> CaCl2 + 2H2O

b) Molarity of base = 0.042 M.

Explanation:

Using titration equation

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2

NB is the number of mole of base = 1

CA is the molarity of acid =0.15M

CB is the molarity of base = to be calculated

VA is the volume of acid = 25 ml

VB is the volume of base = 44.45mL

Substituting

0.15×25/CB×44.45 = 2/1

Therefore CB =0.15×25×1/44.45×2

CB = 0.042 M.

5 0

2 years ago

Describe the electron sharing between 2 oxygen atoms

erastova [34]

When two oxygen atoms get close to each other, the attractions from the nucleus of both atoms attract the outer electrons.

(BRAINLIEST PLEASE!!!)

3 0

2 years ago

For the hypothetical compound AX2, which of the following statements is true?. a.) A is a nonmetal from Group 3A, and X is a met

Ierofanga [76]

The cations has positive charges that are metals while the anions have negative charges that are non-metals. Upon reaction, there is an exchange in charges that are reflected in the subscripts of the atoms. In this case, compound AX2 must have a cation, A belonging to group 2 A with +2 charge and anion, X belonging to Group 7A with -1 charge. Answer is D.

4 0

2 years ago

Read 2 more answers

What element in the second period has the largest atomic radius?

Allushta [10]

Answer:

Na is the element from the second period that has the largest atomic radius

Explanation:

The atomic radius is a chemist property from the periodic table. It is decreased when we move in a period from the periodic, so the element in the second period that has the largest radius is Na, and the shortest, the Ar

The atomic radius indicates the distance between the nucleus and the outermost valence layer. In the periods it decreases with increasing Atomic Number, to the right, due to the attraction that the nucleus exerts on the electrons of the outermost orbitals, thus decreasing the core-electron distance.

3 0

2 years ago