CHEMISTRY COLLEGE

The following data is given to you about a reaction you are studying: Overall reaction: 2A  D Proposed mechanism: Step 1 A + B  C (slow) Step 2 C + A  D + B (fast) [A]o = 0.500 M [B]o = 0.0500 M [C]o = 0.500 M [D]o = 1.50 M This reaction was run at a series of temperatures and it was found that a plot of ln(k) vs 1/T (K) gives a straight line with a slope of -982.7 and a Y intercept of -0.0726. What is the initial rate of the reaction at 298K?

Answers

Answer 1

Answer:

Answer : The initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

Explanation :

The Arrhenius equation is written as:

$K=A* e^{(-Ea)/(RT)}$

Taking logarithm on both the sides, we get:

$\ln k=-(Ea)/(RT)+\ln A$ ............(1)

where,

k = rate constant

Ea = activation energy

T = temperature

R = gas constant = 8.314 J/K.mole

A = pre-exponential factor

The equation (1) is of the form of, y = mx + c i.e, the equation of a straight line.

Thus, if we plot a graph of $\ln k$ vs $(1)/(T)$ then the graph shows a straight line with negative slope. That means,

Slope of the line = $-(Ea)/(R)$

And,

Intercept = $\ln A$

As we are given that:

Slope of the line = -982.7 = $-(Ea)/(R)$

Intercept = -0.0726 = $\ln A$

Now we have to calculate the value of rate constant by putting the value of slope, intercept and temperature (298K) in equation 1, we get:

$\ln k=-(982.7)/(298)+(-0.0726)$

$\ln k=-3.37$

$k=0.0344s^(-1)$

The value of rate constant is, $0.0344s^(-1)$

Now we have to calculate the initial rate of the reaction at 298 K.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$A+B\rightarrow C$

The expression of rate law for this reaction will be,

$Rate=k[A][B]$

As we are given that:

[A] = 0.500 M

[B] = 0.0500 M

k = $0.0344s^(-1)$

Now put all the given values in the rate law expression, we get:

$Rate=(0.0344)* (0.500)* (0.0500)$

$Rate= 8.6* 10^(-4)M/s$

Therefore, the initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

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Inside a calorimeter, two solutions are mixed and result in an endothermic reaction. Which of the following best illustrates how this reaction affects the water in the calorimeter? (2 points)Select one:
a. The reaction has no effect on the water. The kinetic energy of the water molecules remains the same.
b. The reaction causes the temperature of the water to increase. The kinetic energy of the water molecules increases.
c. The reaction causes the temperature of the water to decrease. The kinetic energy of the water molecules decreases.
d. The reaction causes the temperature of the water to decrease. Then, the water gains heat from the surroundings and the kinetic energy of the water molecules increases.

Answers

The correct option is C. A calorimeter is an isolated system. In isolated systems there is no exchange whatsoever with the surrounding. In an isolated system, an endothermic reaction results in a decrease in the temperature of the system. This is in contrast to the effect of an endothermic reaction in an open system. The heat of reaction remains a positive quantity.

A student is heating a chemical in a beaker with a Bunsen burner.In a paragraph of at least 150 words, identify the safety equipment that should be used and the purpose of it for the given scenario.

Answers

When a student is warming a chemical in a container using a special burner, it is very important to focus on safety by using the right safety tools.

What is the safety equipment

First, the student needs to wear the right safety clothes like a lab coat, gloves, and goggles to protect themselves from getting splashed or hurt by chemicals. A lab coat stops chemicals from touching the skin, gloves keep the hands safe, and safety goggles protect the eyes from chemicals

and hot things.

Furthermore, using a fume hood is necessary to make sure there is enough fresh air circulating and to remove any dangerous fumes or gases that might be released while heating things up.

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Answer:The student should be wearing a lab coat or maybe an apron to prevent chemicals from spilling or exploding onto their clothes, I do recommend a lab coat better though because it can protect your skin better. Next, make sure while messing with chemicals you are always wearing goggles, if you are not wearing them there is a chance that after touching chemicals you could touch your eyes. And that brings me to washing your hands straight away after messing with chemicals. You could also wear gloves and just take them off when you're done but if you don't have clean hands afterward you could always put the chemicals all over your skin. But in case you do touch your eyes there is always an emergency eyewash station somewhere in the lab room. And if you are to get Chemicals on your skin, in your hair, on your clothes, or to be on fire, there shall be a shower somewhere to get rid of that. But if you read the instructions or listen closely to the teacher you shall have no problem.

Explanation:

I kinda got off topic

A compound contains only carbon, hydrogen, and oxygen. Combustion of 139.1 g of the compound yields 208.6 g of CO2 and 56.93 g of H2O. The molar mass of the compound is 176.1 g/mol. *Each part of this problem should be submitted separately to avoid losing your work* 1. Calculate the grams of carbon (C) in 139.1 g of the compound: grams 2. Calculate the grams of hydrogen (H) in 139.1 g of the compound. grams 3. Calculate the grams of oxygen (O) in 139.1 g of the compound. grams

Answers

Answer:

1. Mass of Carbon is 56.89g

2. Mass of Hydrogen is 6.33g

3. Mass of Oxygen is 75.88

Explanation:

The following were obtained from the question.

Mass of the compound = 139.1g

Mass of CO2 produced = 208.6g

Mass of H2O produced = 56.93

1. Determination of mass of Carbon (C). This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 208.6

Mass of C = 56.89g

2. Determination of the mass of Hydrogen (H). This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 56.93

Mass of H = 6.33g

3. Determination of the mass of oxygen (O).

This is illustrated below:

Mass of the compound = 139.1g

Mass of C = 56.89g

Mass of H = 6.33g

Mass of O = Mass of compound - (mass of C + Mass of H)

Mass of O = 139.1 - (56.89 + 6.33)

Mass of O = 139.1 - 63.22

Mass of O = 75.88

Rank the following compounds in order of decreasing acid strength using periodic trends. Rank the acids from strongest to weakest. To rank items as equivalent, overlap them. H2Se

HBr

H2O

HI

Answers

Explanation:

It is known that acidic strength of hydrides of same group tends to increase when we move from top to bottom in a group. On the other hand, acidic strength of hydrides of same period elements increases when we move from left to right in a period.

As both bromine and iodine belongs to the same group. Also, selenium and oxygen are same group elements. Therefore, their acidic strength increases on moving down the group.

Therefore, we can conclude that acidic strength of given compounds from strongest to weakest is as follows.

HI > HBr > $H_(2)Se$ > $H_(2)O$

Final answer:

To rank the acids in decreasing acid strength using periodic trends, consider the size, electronegativity, and presence of lone pairs of electrons. HI is the strongest acid, followed by HBr, H2O, and H2Se.

Explanation:

To rank the acids in order of decreasing acid strength using periodic trends, we need to consider the size and electronegativity of the atoms. The larger the atom, the weaker the acid, and the more electronegative the atom, the stronger the acid. Additionally, we can consider the presence of lone pairs of electrons, as they increase the acidity.

HI - Iodine (I) is larger and less electronegative than the other halogens. It also has a lone pair of electrons, making it the strongest acid.
HBr - Bromine (Br) is larger and less electronegative than chlorine (Cl), and it also has a lone pair of electrons. It is the second strongest acid.
H2O - Oxygen (O) is smaller and more electronegative than the halogens. It does not have a lone pair of electrons, making it a weaker acid than the halogens.
H2Se - Selenium (Se) is larger and less electronegative than sulfur (S). However, it does not have a lone pair of electrons, making it the weakest acid.

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What is the isoelectric point of proteins?

Answers

Isoelectric point. The isoelectric point (pI, pH(I), IEP), is the pH at which a particular molecule carries no net electrical charge in the statistical mean. The standard nomenclature to represent the isoelectric point is pH(I), although pI is also commonly seen, and is used in this article for brevity.

The isoelectric point (pI, pH(I),IEP), is the pH at which a particular molecule carries no net electrical charge in the statistical mean.

Identify two ions that have the following ground-state electron configurations Part B
[Ar]3d^5
Check all that apply.
A- Fe2+
B- Fe3+
C- Mn2+
D- V+
E- Sc2+

Answers

Answer: $Fe^(2+):24:[Ar]3d^5$

$Mn^(2+):23:[Ar]3d^5$

Explanation:

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom. The electrons are filled according to Afbau's rule in order of increasing energies.

The electronic configuration for given elements is as follows:

$Fe:26:[Ar]3d^64s^2$

$Fe^(2+):24:[Ar]3d^5$

$Fe^(3+):23:[Ar]3d^4$

$Mn:25:[Ar]3d^54s^2$

$Mn^(2+):23:[Ar]3d^5$

$V:23:[Ar]3d^34s^2$

$V^+:22:[Ar]3d^34s^1$

$Sc:21:[Ar]3d^14s^2$

$Sc^(2+):19:[Ar]3d^1$

Final answer:

The ions Fe2+ and Mn2+ have the ground-state electron configuration [Ar]3d^5.

Explanation:

The ground-state electron configuration [Ar]3d^5 indicates a level of electrons in 3d subshell after the Argon core electron configuration. Now, iron (Fe) has a base atomic configuration of [Ar]3d^6 4s2. When it loses 2 electrons (to form Fe2+), it tends to lose from both the 3d and the 4s sublevels, giving [Ar]3d^5 (which is our required configuration).

However, it's also important to consider Manganese (Mn), which has a base configuration of [Ar]3d^5 4s2. It usually loses 2 electrons from the 4s sublevel first when it forms Mn2+ which results in a configuration [Ar]3d^5.

So, the two ions with the electron configuration [Ar]3d^5 are Fe2+ and Mn2+.

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