The periodic table is based on an element's

There are always attractive forces between a collection of atoms or molecules which may not be negligible as suggested by the ki

Harrizon [31]

Answer:

Negligible

Explanation:

According to the kinetic theory of gases, the degree of intermolecular interaction between gases is minimal and gas molecules tend to spread out and fill up the volume of the container.

If the attraction between gas molecules increases, then the volume of the gas decreases accordingly. This is because, gas molecules become highly attracted to each other.

This intermolecular attractive force may be so strong, such that the actual volume of the gas become negligible compared to the volume of the container.

7 0

3 years ago

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 gas has a pressure of 5.7 atm at 100.0°C. What is its pressure at20.0°C (Assume volume is unchanged)

son4ous [18]

Answer:

$\large \boxed{\text{4.5 atm}}$

Explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ =5.7 atm; T₁ = 100.0 °C

p₂ = ?; T₂ = 20.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ = (100.0 + 273.15) K = 373.15

T₂ = (20.0 + 273.15) K = 293.15

2. Calculate the new pressure

$\begin{array}{rcl}\dfrac{5.7}{373.15} & = & \dfrac{p_{2}}{293.15}\\\\0.0153 & = & \dfrac{p_{2}}{293.15}\\\\0.0153\times 293.15 &=&p_{2}\\p_{2} & = & \textbf{4.5 atm}\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.5 atm}}$}$

6 0

3 years ago

Question 2(Multiple Choice Worth 4 points) (01.01 MC) Dominic and Eva are using the same type of stopwatches to measure the time

LekaFEV [45]

The results of Dominic and Eva's experiment is unreliable and can lead to a pseudoscientific claim primarily because they did not repeat their tests multiple times. Although, they used the same type of stopwatch in recording the time it takes for the chemical reaction to occur, they have different reactions times. Thus, it would have been better if they conducted several trials then obtained the average of their results.

6 0

3 years ago

Read 2 more answers

NEED HELPPP PLSSSSSSSSSS

Leto [7]

I would pick the first option in the third option

4 0

3 years ago