A car accelerates from rest to a speed of 24 m/s in 8 seconds. What is the car's average acceleration

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(c) The units of rate constant is: $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

(c) The units of the rate constant can be obtained from (2).

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

3 years ago

Consider the five balanced chemical reactions listed below, all using O2 as a reactant. Normally, O2 is an excess reagent for re

ad-work [718]

Answer: option E. None because in all the reactions O2 is in excess

Explanation:

7 0

3 years ago

Given that the formula of butane C4H10 the accepted value for the molar mass should be

deff fn [24]

Answer:

carbon mass = 12.01g/mol

hydrogen mass = 1.01g/mol

4 carbon atoms and 10 hydrogen so

12.01 x 4 + 1.01 x 10

48.04g/mol + 10.10g/mol

= 58.14g/mol

4 0

2 years ago

What did Josef Loschmidt and Amedeo Avogadro Contribute to our understanding of basic molecular numbers, sizes, and reaction rat

mario62 [17]

From Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.

<h3>What is Avogadro's number?</h3>

Avogadro's number, or Avogadro's constant, is the number of particles found in one mole of a substance.

The Avogadro's number is given as 6.02 x 10²³.

Summary of Josef Loschmidt and Amedeo Avogadro Contribution to chemistry.

Equal volumes of gas contain equal numbers of molecules,
Elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.

Thus, from Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.

Learn more about Avogadro's here: brainly.com/question/1581342

#SPJ1

4 0

2 years ago

Question 12 of 25

NISA [10]

Answer:

B. Equal amounts of all gases have the same volume at the same

conditions

Explanation:

Amedo Avogadro found the relationship between volume of a gas and the number of molecules contained in the volume.

The law states that "equal volumes of all gases at the same temperature and pressure contains equal number of molecules or moles".

The law describes the behavior of gases when involve in chemical reactions. It enables one to change over at will in any statement about gases from volumes to molecules and vice versa.

So, the right option is B which implies that equal amounts of all gases have the same volume at the same conditions.

8 0

3 years ago