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Novay_Z [31]
4 years ago
9

An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the

course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically.
The reaction of compound A forming compound B was studied and the following data were collected:

Time (s) [A](M)
0. 0.184
200. 0.129
500. 0.069
800. 0.031
1200. 0.019
1500. 0.016

a.) What is the average reaction rate between 0. and 1500. s?
b.) What is the average reaction rate between 200. s and 1200. s?
c.) What is the instantaneous rate of the reaction at t=800 s?
Chemistry
1 answer:
Sauron [17]4 years ago
8 0

Answer:

a) 0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.

b) 0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.

c) Instantaneous rate of the reaction at t=800 s :

Instantaneous rate  : \frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s

Explanation:

Average rate of the reaction is given as;

R_{avg}=-\frac{\Delta A}{\Delta t}=\frac{A_2-A_1}{t_2-t_1}

a.) The average reaction rate between 0.0 s and 1500.0 s:

At 0.0 seconds the concentration was  = A_1=0.184 M

t_1=0.0s

At 1500.0 seconds the concentration was  = A_2=0.016 M

t_2=1500 s

R_{avg]=-\frac{0.016 M-0.184 M}{1500.0 s-0.0 s}=0.000112 M/s

0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.

b.) The average reaction rate between 200.0 s and 1200.0 s:

At 0.0 seconds the concentration was  = A_1=0.129 M

t_1 =200.0 s

At 1500.0 seconds the concentration was  = A_2=0.019M

t_2=1200 s

R_{avg]=-\frac{0.019 M-0.129M}{1200.0s-200.0s}=0.00011 M/s

0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.

c.) Instantaneous rate of the reaction at t=800 s :

At 800 seconds the concentration was  = A=0.031 M

t =800.0 s

Instantaneous rate  : \frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s

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What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0×1015Hz?
RideAnS [48]

The kinetic energy of the emitted electrons of cesium when it is exposed to UV rays of frequency 1.0 \times {10^{15}}\;{\text{Hz}}  is  \boxed{6.63 \times {{10}^{ - 19}}\;{\text{J}}}

Further Explanation:

Photoelectric effect:

When light is made to fall on any substance, electrons are emitted from it. This is known as the photoelectric effect and the emitted electrons are called photoelectrons. The electrons are emitted because of the transference of energy from light to the electrons.

Cesium is a member of the alkali metal group so it is highly reactive and shows photoelectric effect to the maximum extent. It can remove its electron so easily because of its atomic size. Due to large atomic size of cesium, its outermost electrons are held very less tightly to the nucleus and therefore removed easily.

According to the Planck-Einstein equation, the energy is proportional to the frequency and is expressed as follows:

{\mathbf{E=h\nu }}                                   ......(1)

Here,

{\text{E}} is the energy.

h is the Plank’s constant.

\nu is the frequency.

The frequency of UV rays is 1.0 \times {10^{15}}\;{\text{Hz}} or 1.0 \times {10^{15}}\;{{\text{s}}^{ - 1}}

The value of Planck’s constant is 6.626 \times {10^{ - 34}}\;{\text{J}}\cdot{\text{s}} .

Substitute these values in equation (1)

\begin{aligned}{\text{E}}&=\left( {6.626 \times {{10}^{ - 34}}\;{\text{J}}\cdot{\text{s}}}\right)\left( {1.0 \times {{10}^{15}}\;{{\text{s}}^{ - 1}}}\right)\\&=6.63\times {10^{ - 19}}\;{\text{J}}\\\end{aligned}

But when electrons are ejected out from the surface of the substance, all of its energy is considered as kinetic energy.

So the kinetic energy of the electrons is {\mathbf{6}}{\mathbf{.63 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 19}}}}\;{\mathbf{J}} .

Learn more:

1. Statement about subatomic particle: brainly.com/question/3176193

2. The energy of a photon in light: brainly.com/question/7590814

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: kinetic energy, frequency, energy, photoelectric effect, Planck's constant, light, electrons, photoelectrons, proportional, transference, reactive, cesium.

7 0
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How many grams of lithium sulfide must be dissolved in 1600.0 g water to make a 2.0 molal solution?
klasskru [66]
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the molality of solution to be prepared is 2.0 molal. 
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Answer:

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At STP, 1 mol of any given mass occupies 22.4 L. With that information in mind we <u>calculate the volume that 0.025 moles of O₂ gas would occupy</u>:

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Thus the answer is 0.56 liters.

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