How to cook brainly?<br> ps:e

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

A flashlight battery is hooked to a toy motor, and then the battery and the connections are sprayed with a water-proof coating.

denis-greek [22]

Explanation:

Formula to calculate work done by motor is as follows.

Work done by motor = $mass \times g \times height$

where, g = gravitational constant = 10 $m/s^{2}$

Therefore, work done by motor is as follows.

Work done by motor = $1.00 kg \times 10 m/s^{2} \times 10.0 m$

= 100.0 J

Now, heat lost by water will be calculated as follows.

q = $mC \times \Delta T$

= $g \times 4.184 J/g^{o}C \times 0.024^{o}C$

= 10.0 J

Hence, heat gained by motor = heat lost by water

As, heat gained by motor = 10.0 J

So, change in energy = heat gained - work done

Therefore, change in energy will be calculated as follows.

Change in energy = heat gained - work done

= (10.0 J) - (100.0 J)

= -90.0 J

Thus, we can conclude that change in the energy of the battery contents is -90.0 J.

4 0

3 years ago

What is the molarity of a solution that contains 1000.0 mg of AgNO3 that has been dissolved in 500 mL of water

Vaselesa [24]

0.012moldm⁻³

Explanation:

Given parameters:

Mass of AgNO₃ = 1000mg

Volume of water = 500mL

Unknown:

Molarity of solution = ?

Solution:

The molarity of a solution is the number of moles of a solute dissolved in volume of solvent.

Molarity = $\frac{xnumber of moles}{Volume}$

Number of moles of AgNO₃ = ?

Number of moles = $\frac{mass}{molar mass}$

Molar mass of AgNO₃ = 108 + 14 + 3(16) = 170g/mol

convert mass to g;

1000mg = 1g

Number of moles = $\frac{1}{170}$ = 0.00588moles

convert the given volume to dm³;

1000mL = 1dm³;

500mL = 0.5dm³

Now solve;

Molarity = $\frac{0.00588}{0.5}$ = 0.012moldm⁻³

learn more:

Molarity brainly.com/question/9324116

#learnwithBrainly

4 0

3 years ago

At 25°C and constant pressure, carbon monoxide gas combines with oxygen gas to give carbon dioxide gas with the evolution of 10.

stealth61 [152]

Answer : The value of $\Delta H$ for the reaction is, -565.6 kJ

Explanation :

First we have to calculate the molar mass of CO.

Molar mass CO = Atomic mass of C + Atomic mass of O = 12 + 16 = 28 g/mole

Now we have to calculate the moles of CO.

$\text{Moles of }CO=\frac{\text{Mass of }CO}{\text{Molar mass of }CO}=\frac{1g}{28g/mole}=\frac{1}{28}mole$

Now we have to calculate the value of $\Delta H$ for the reaction.

The balanced equation will be,

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

From the balanced chemical reaction we conclude that,

As, $\frac{1}{28}mole$ of CO release heat = 10.1 kJ

So, 2 mole of CO release heat = $2\times 28\times 10.1=565.6kJ$

Therefore, the value of $\Delta H$ for the reaction is, -565.6 kJ (The negative sign indicates the amount of energy is released)

4 0

3 years ago

(04.03 LC)

pishuonlain [190]

Carrying capacity, D
Emigration, C
Limiting factor, A
Population dynamics, B

4 0

3 years ago

How to cook brainly? ps:e