It takes to break an carbon-chlorine single bond. Calculate the maximum wavelength of light for which an carbon-chlorine single bond could be broken b

Answers

Answer 1

The question is incomplete, the complete question is;

It takes 338. kJ/mol to break an carbon-chlorine single bond. Cal broken by absorbing a single photon Iculate the maximum wavelength of light for which an carbon-chiorine single bond could be Round your answer to 3 significant digits

Answer:

3.55 × 10^-7 m or 355 nm

Explanation:

Now, the energy of the photon = 338 × 10^3/6.02 × 10^23 = 5.61 × 10^-19 J

Recall that;

E= hc/λ

h= planks constant

c= speed of light

λ = wavelength

λ =hc/E

λ = 6.63 ×10^-34 × 3 × 10^8/5.61 × 10^-19

λ =3.55 × 10^-7 m or 355 nm


Related Questions

Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.

Which statement is true about the experiments? (5 points)
The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.
The same amount of heat is absorbed in both the experiments because the heat absorbed depends only on the final temperature.
The heat absorbed in Trial 2 is about 3,674 J greater than the heat absorbed in Trial 1.
The heat absorbed in Trial 2 is about 5,021 J greater than the heat absorbed in Trial 1.

Answers

Answer:

Explanation:

Using the formula below to calculate the heat absorbed in each trial:

Q = m × c × ∆T

Where;

Q = amount of heat absorbed (J)

m = mass of substance (g)

c = specific heat of water (4.184J/g°C)

∆T = change in temperature (°C)

Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.

Q = 30 × 4.184 × (40 - 0)

Q = 30 × 4.184 × 40

Q = 5,020.8J

Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.

Answer:

The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.

Explanation:

Explanation:

Using the formula below to calculate the heat absorbed in each trial:

Q = m × c × ∆T

Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.

Q = 30 × 4.184 × (40 - 0)

Q = 30 × 4.184 × 40

Q = 5,020.8J

Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.

Q=40*4.184*30

Q=5020.8J

8.7 Two products are formed in the following reaction in a 50:50 mixture. Would the resulting solution be optically active

Answers

Answer:

Yes. The solution would be optically active.

Explanation:

Diastereomer are defined as the image that is non mirror and non -identical. It is made up of two stereoisomers. They are formed when the two stereoisomers or more than two stereoisomers of the compound have the same configuration at the equivalent stereocenters.

In the given context, as the product given is a diastereomeric mixture, the product would have an optical activity in total.

So the answer is Yes.

The metal thallium becomes superconducting at temperatures below 2.39K. Calculate the temperature at which thallium becomes superconducting in degrees Celsius. Round your answer to decimal places.

Answers

Answer:

-270.76°C

Explanation:

Given that metal Thallium becomes superconducting below the temperature of 2.39 kelvin i.e. this temperature is critical temperature for Thallium and below critical temperature a metal offers no resistance to the flow of electric current. Also the metal below its critical temperature expels the magnetic field in such a way that they do not penetrate the metal and pass through its surface only.

We have the relation between kelvin scale and degree Celsius scale of temperature measurement as:

[tex]C = K - 273.15[/tex]

[tex]C=2.39-273.15\\ C=-270.76^{o}C[/tex]

The volume of an ideal gas is held constant. Determine the ratio P2/P1 of the final pressure to the initial pressure when the temperature of the gas rises (a) from 46 to 92 K and (b) from 35.4 to 69.0 oC.

Answers

Answer:

A. P₂ / P₁ = 2

B. P₂ / P₁ = 1.1

Explanation:

A. Determination of the ratio P₂/P₁

Volume = constant

Initial temperature (T₁) = 46 K

Final temperature (T₂) = 92 K

Final pressure /Initial pressure (P₂/P₁) =?

P₁/T₁ = P₂/T₂

P₁/46 = P₂/92

Cross multiply

46 × P₂ = P₁ × 92

Divide both side by P₁

46 × P₂ / P₁ = 92

Divide both side by 46

P₂ / P₁ = 92 / 46

P₂ / P₁ = 2

B. Determination of the ratio P₂/P₁

Volume = constant

Initial temperature (T₁) = 35.4 °C = 35.4 + 273 = 308.4 K

Final temperature (T₂) = 69.0 °C = 69 + 273 = 342 K

Final pressure /Initial pressure (P₂/P₁) =?

P₁/T₁ = P₂/T₂

P₁/308.4 = P₂/342

Cross multiply

308.4 × P₂ = P₁ × 342

Divide both side by P₁

308.4 × P₂ / P₁ = 342

Divide both side by 308.4

P₂ / P₁ = 342 / 308.4

P₂ / P₁ = 1.1

Given 200ul of a 0.5mg/ml stock solution of BSA, how much do you pipet into a test tube so that you are adding 5ug of BSA to the test tube

Answers

Answer: [tex]10\mu L[/tex] of volume needs to be pipetted out in the test tube.

Explanation:

We are given:

Mass of BSA to be formed = [tex]5\mu g=0.005mg[/tex]      (Conversion factor: [tex]1mg=1000\mu g[/tex]

Volume of stock solution = [tex]200\mu L=0.2mL[/tex]    (Conversion factor: [tex]1mL=1000\mu L[/tex]

It is also given that for the mass of BSA is 0.5 g, the volume used up is 1 mL

In order to have, 0.005 g, the volume of stock solution needed will be = [tex]\frac{1mL}{0.5g}\times 0.005g=0.01mL=10\mu L[/tex]

Hence, [tex]10\mu L[/tex] of volume needs to be pipetted out in the test tube.

If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?​

Answers

Answer:

1.88 A

Explanation:

Let's consider the reduction of copper in an electrolytic cell.

Cu²⁺ + 2 e⁻ ⇒ Cu

We can calculate the charge used to deposit 12.3 g of Cu using the following relations.

The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).

The charge used is:

[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]

We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.

5.50 h × 3600 s/1 h = 1.98 × 10⁴ s

The current used is:

I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A

The molar ratio of HPO42- to H2PO4- in a solution is 1.4. Calculate the pH of the solution. Phosphoric acid (H3PO4) is a triprotic acid with 3 pKa values: 2.14, 6.86, and 12.4.

Answers

Given is the ratio of conjugate base and conjugate acid of phosphoric acid. pH of a substance is the concentration of the hydrogen ions in its solution and higher this concentration lower is the value of pH.

pKa value is a measure of the strength of acid, it is the negative log of acid dissociation constant Ka.

The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV = nRT, where n is the number of moles of the gas and R = 0.0821 is the gas constant. Suppose that, at a certain instant, P = 8.0 atm and is increasing at a rate of 0.13 atm/min and V = 13 L and is decreasing at a rate of 0.17 L/min. Find the rate of change of T with respect to time (in K/min) at that instant if n = 10 mol.

Answers

Answer:

The rate of change of T with respect to time is 0.40 K/min

Explanation:

The gas law equation is:

[tex] PV = nRT [/tex]

We can find the rate of change of T with respect to time by solving the above equation for T and derivating with respect to time:

[tex] \frac{dT}{dt} = \frac{d}{dt}(\frac{PV}{nR}) [/tex]

[tex] \frac{dT}{dt} = \frac{1}{nR}(V\frac{dP}{dt} + P\frac{dV}{dt}) [/tex]

Where:

n: is the number of moles = 10 mol

R: is the gas constant = 0.0821

V: is the volume = 13 L

P: is the pressure = 8.0 atm

dP/dt: is the variation of the pressure with respect to time = 0.13 atm/min

dV/dt: is the variation of the volume with respect to time = -0.17 L/min

Hence, the rate of change of T is:

[tex] \frac{dT}{dt} = \frac{1}{10*0.0821}(13*0.13 - 8.0*0.17) = 0.40 K/min [/tex]    

Therefore, the rate of change of T with respect to time is 0.40 K/min

I hope it helps you!    

Give the following reaction: ammonium nitrate—> dinitrogen monoxide + water.
a.) Write a complete balanced chemical equation.
b.) Calculate the number of molecules of water produced by 11.2g of ammonium nitrate

Answers

Answer:

a) NH₄NO₃ ⇒ N₂O + 2 H₂O

b) 1.69 × 10²³ molecules

Explanation:

Step 1: Write the balanced equation

NH₄NO₃ ⇒ N₂O + 2 H₂O

Step 2: Convert 11.2 g of NH₄NO₃ to moles

The molar mass of NH₄NO₃ is 80.04 g/mol.

11.2 g × 1 mol/80.04 g = 0.140 mol

Step 3: Calculate the moles of H₂O produced

0.140 mol NH₄NO₃ × 2 mol H₂O/1 mol NH₄NO₃ = 0.280 mol H₂O

Step 4: Calculate the number of molecules in 0.280 moles of water

We will use Avogadro's number.

0.280 mol × 6.02 × 10²³ molecules/1 mol = 1.69 × 10²³ molecules

The Bohr model of the atom explains why emission spectra are discrete. It could also be used to explain the photoelectric effect. Which is a correct explaination of the photoelectric effect according to the model

Answers

Answer:

photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.

1.rain pours from the sky
2.leaves of the plant dried
3.fluffy clouds form in the sky
4.bathing suit dries after swim
5.water puddles disappear

A.Evaporation
B.Condensation
C.Precipitation
D.Transpiration
Yan po pag pipilian

Answers

Answer:

1.Precipitation

2.Transpiration

3.Condensation

4.Evaporation

5.Evaporation

3.Condensation

Explanation:

Rain pours from the sky occurs due to the process of precipitation, leaves of the plant dried due to the process of transpiration in which the water is evaporated from the body of plant, fluffy clouds form in the sky occurs in the process of condensation, bathing suit dries after swim is due to evaporation in which water is removed and goes into the atmosphere and water puddles disappear due to the process of evaporation. Evaporation is the removal of water from the any surface whereas transpiration is the removal of water from plant body parts.

Pl hep help help me

Answers

It is true

I’ve done this before

The size of an atomic orbital is associated with:______________

a. the magnetic quantum number (ml).
b. the spin quantum number (ms).
c. the angular momentum quantum number (l).
d. the angular momentum and magnetic quantum numbers, together.
e. the principal quantum number (n).

Answers

Answer:

e. the principal quantum number (n).

Explanation:

The size of the orbital is governed and decided by the principal quantum number n, which is dependent on the overall average distance between the number of electrons as well as the nucleus. The orbital's shape is explained by the angular quantum number. The magnetic quantum number is concerned with the orbital's orientation in space. The quantum number's spin explains the spin of the electrons.

The volume of a single tantalum atom is 1.20×10-23 cm3. What is the volume of a tantalum atom in microliters?

Answers

Answer:

1.20x10⁻²⁰μL

Explanation:

1cm³ is equal to 1milliliter. As we must know, 1milliliter = 1000 microliters, 1000μL. To convert the 1.20x10⁻²³mL we need to use the conversion factor: 1mL = 1000μL.

The volume of tantalum in μL is:

1.20x10⁻²³mL * (1000μL /1L) = 1.20x10⁻²⁰μL

Can someone please please help

Answers

Answer:

oxidizer

Explanation:

an example of an oxidizers are oxygen and hydrogen peroxide

Describe the electron configuration of an atom using principal energy level, sublevels, orbitals, and periodic table. Give one example others may not think about and why you made this selection.

Silicon is not allowed.

Answers

Explanation:

The electron density number as well as the sublevel letter are used to describe valence electrons in an atom. The third total energy and subbasement p, for example, is denoted by 3p. The electron configuration of oxygen, for example, is 1s^2 2s^2 2p^4, which means the first two electrons will couple up in the 1s orbital, while the following two protons will pair up in the 2s orbital.

The sample atom is Carbon with electron configuration; 1s² 2s² 2p².

The principal energy level of an electron refers to the shellp in which the electron is located relative to the atom's nucleus. In this case only 2 energy levels exist in a carbon atom; which are energy level 1 and 2

The sublevels exist within a principal energy and the electron configuration of an atom is described with consideration of energy sublevels. The sublevels in a carbon atom are;

s and p energy sublevels.

The orbitals in this configuration are: 1s 2s 2px 2py 2pz in which case; each orbital can accommodate 2 electrons each.

Ultimately, the location of an element on the periodic table with respect to group and period are used to determine the valency and no. of energy levels in the atom of that Element.

Read more:

https://brainly.com/question/13749106

b) What is the change in entropy of the reaction if ΔH° = -3.2 kJ mol-1?

Answers

I would go w A I just took the test it was very way I got straight b on it

Based upon the intermolecular forces present, rank the following substances according to the expected boiling point for the substance.

a. HCl
b. NaCl
c. N2
d. H2O

Answers

It would be N2!!!!!!!!!!!!!!!!!!

Calculate the percent dissociation of benzoic acid C6H5CO2H in a 1.3M aqueous solution of the stuff. You may find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits.

Answers

Answer:

the percent dissociation is  0.69 %

Explanation:

Given the data in the question;

benzoic acid C₆H₅CO₂H

C₆H₅COOH[tex]_{(aq)[/tex]  ⇔ C₆H₅COO[tex]_{-(aq)[/tex] + H[tex]_{+(aq)[/tex]

Ka =  [C₆H₅COO- ][ H+ ] / [ C₆H₅COOH ] = 6.28 × 10⁻⁵

given that it dissociated in a 1.3 M  aqueous solution.

so Initial concentration is;  

[ C₆H₅COOH ] = 1.3

[C₆H₅COO- ] = 0

[ H+ ] = 0

Change in concentration

[ C₆H₅COOH ] = -x

[C₆H₅COO- ] = +x

[ H+ ] = +x

Concentration equilibrium

[ C₆H₅COOH ] = 1.3 - x

[C₆H₅COO- ] = +x

[ H+ ] = +x

Hence,

x² / ( 1.3 - x ) = 6.28 × 10⁻⁵

6.28 × 10⁻⁵( 1.3 - x ) = x²

8.164 × 10⁻⁵ - 6.28 × 10⁻⁵x = x²

x² + 6.28 × 10⁻⁵x - 8.164 × 10⁻⁵ = 0

solve for x

ax² + bx - c = 0

x = [ -b ± √( b² - 4ac ) ] / [ 2a ]

we substitute  

x = [ -6.28 × 10⁻⁵ ± √( (6.28 × 10⁻⁵)² - (4 × 1 × -8.164 × 10⁻⁵ ) ) ] / [ 2 × 1 ]  

x = [ -6.28 × 10⁻⁵ ± 0.01807 ] / [ 2]

x = [ -6.28 × 10⁻⁵ - 0.01807 ] / [ 2]  or [ -6.28 × 10⁻⁵ + 0.01807 ] / [ 2]

x = -0.0090664 or 0.0090036

so x = 0.0090036

hence

[ H+ ] = +x = 0.0090036 M

[C₆H₅COO- ] = +x = 0.0090036 M

Initial concentration of [ C₆H₅COOH ] = 1.3 M

concentration of C₆H₅COOH dissociated = 0.0090036 M

percent dissociation of C₆H₅COOH will be;

⇒ ( 0.0090036 M / 1.3 M ) × 100 = 0.69 %

Therefore, the percent dissociation is  0.69 %

Ammonia is produced by the reaction of nitrogen and hydrogen: N2(g) + H2(g)  NH3(g)
(a) Balance the chemical equation.
(b) Calculate the mass of ammonia produced when 35.0g of nitrogen reacts with hydrogen.

Answers

Answer:

a) N2 (g) + H2 = 2 NH3

b) You have to state the mass of hydrogen

Inbox
Question 1
Flagged
III
Initial Knowledge Check
Drafts
Sent
Write the symbol for every chemical element that has atomic number less than 14 and atomic mass greater than 23.2 u.

Answers

Answer:

that symbol less than atomic number 14 and greater than mass number 23.2 is mg

Who knows Cameron Herrin?​

Answers

Explanation:

Cameron Herrin has killed a mother and her baby on a highway in Tampa, Florida

on 2018 on a illegal race

Each of the following sets of quantum numbers is supposed to specify an orbital. Choose the one set of quantum numbers that does NOT contain an error.

a. n = 4, l = 3, ml =-4
b. n = 2, l = 2, ml =0
c. n = 3, l = 2, ml =-2
d. n = 2, l = 2, ml =+1

Answers

Answer:

n = 3, l = 2, ml =-2

Explanation:

Quantum numbers are a set of values which can be used to describe the energy and position of an electron in space.

There are four sets of quantum numbers;

1) principal quantum number

2) orbital quantum number

3) spin quantum number

4) magnetic quantum number.

The values of orbital quantum number include; -l to +l;

The set of quantum numbers without error is ; n = 3, l = 2, ml =-2

PLEASE HELP FAST!!

Which of the following ions is formed when an acid is dissolved in a solution?
H+
O−
OH−
SO42+

Answers

Answer:

H+

Explanation:

acid produce H+ when it is dissolved in a solution

300.0 mL of a 0.335 M solution of NaI is diluted to 700.0 mL. What is the new concentration of the solution?

Answers

Answer: The new concentration of the solution is 0.143 M.

Explanation:

Given: [tex]V_{1}[/tex] = 300.0 mL,    [tex]M_{1}[/tex] = 0.335 M

[tex]V_{2}[/tex] = 700.0 mL,         [tex]M_{2}[/tex] = ?

Formula used is as follows.

[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]

Substitute values into the above formula as follows.

[tex]M_{1}V_{1} = M_{2}V_{2}\\0.335 M \times 300.0 mL = M_{2} \times 700.0 mL\\M_{2} = 0.143 M[/tex]

Thus, we can conclude that the new concentration of the solution is 0.143 M.

Which pairs of aqueous solutions will not produce a precipitate when mixed AgNo3(aq) and NaCl(aq)?

Answers

Answer:

CHCI3

Explanation:

there are no free CI ions hence it doesnt precipitate with an aqeous solution of AQUO33

Write the balanced reaction for the methanol cannon demo that includes their Lewis structures . The reaction is the combustion of methanol (CH3OH). Include the states (s, l, g) in your balanced equation as well.

Answers

Answer:

The reaction is the combustion of methanol (CH3OH).

Write the balanced chemical equation.

Draw Lewis structures for each structure.

Explanation:

The balanced chemical equation for the combustion of methane is shown below:

[tex]2CH_3OH(g)+3O_2(g)->2CO_2(g)+ 4 H_2O(g)[/tex]

Lewis structures of the given molecules are shown below:

Please can someone please help me !!

Answers

Answer:

False

Explanation:

3. Does entropy increase or decrease in the following processes?
A. Complex carbohydrates are metabolized by the body, converted into simple sugars.
Answer: Increase
es-lesund
B. Steam condenses on a glass surface.
Answer:
decreare
-->
MgCl2(s)
C. Mg(s) + Cl2(g)
correct
Answer:

Answers

Answer:

HOPE IT helps much as you can

Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points from highest to lowest. CoCl3, NH4Cl, Li2SO4

Answers

Answer:

NH4Cl > Li2SO4 > CoCl3

Explanation:

Let us recall that the freezing point depression depends on the molality of the solution and the number of particles present.

Let us also recall that freezing point depression is a colligative property. It depends on the number of particles present in solution.

Usually, the more the number of particles present, the lower the freezing point. Hence, NH4Cl which has only two particles will have the highest freezing point while CoCl3 which has four particles will have the lowest freezing point.

Other Questions
QUICK PLZ!!!!! Which graph shows the result of dilating this figure by a factor of One-third about the origin? On a coordinate plane, triangle A B C has points (negative 6, 6), (6, 6), (6, negative 6). On a coordinate plane, triangle A prime B prime C prime has points (negative 2, 2), (2, 2), (2, negative 2). On a coordinate plane, triangle A prime B prime C prime has points (negative 3, 3), (3, 3), (3, negative 3). On a coordinate plane, triangle A prime B prime C prime has points (Negative 18, 18), (18, 18), (18, negative 18). On a coordinate plane, triangle A prime B prime C prime has points (negative 12, 12), (12, 12), (12, negative 12). If a cube has an edge of length e, then the lateral surface area is: Researchers note that a plant closes its stomata on a hot, dry, windy day. What statement describes the plant response to theseconditions?A)The stomatal closings and the environmental conditions are unrelated for this plant.B)The stomatal closings and the environmental conditions are unrelated forthis plant.Water loss and water stress is highest on hot, dry, windy days, so the plantcloses stomata to conserve water.Hot, dry, windy days are not conducive to pollen spread, so the plant closesits stomata to prevent pollination.Photosynthesis is decreased on hot, dry, windy days, so the plant closes itsstomata to prevent carbon dioxide diffusion.aD) The Assertion that, non formal institutions have no role to play in the processing stage at the system theory is an empty rhetoric 93Find the value of y. if x=-1 , y=1 and z =2 then find the value of. a . 2x+3y +4z b. 5x2 +3y2+7z2 (I/not/meet) you at the park this afternoon. I have to see the doctorA. haven't to meetB. can't meetC. don't have to meet Which word correctly completes the sentence?Esto est arriba de la cabeza y puede ser de diferentes colores.Es _____.la bocael cabelloel estmagolos dientes Write the chemical formula for the following: a. The conjugate acid of amide ion, NH-b. The conjugate base of nitric acid, HNO 1. Ill tell you about this tomorrow, Mary. said Tom.A. Tom said to Mary that he will tell her about that the next day.B. Tom told Mary that I would tell you about that the next day.C. Tom told Mary that he would tell her about that the next day.D. Tom told Mary that she would tell him about that the next day. Hypevitaminosis is largely attributed to which of the following 15. Moment actuel ou pass ? Choisissez l'expression correcte. 1. (L'anne dernire - Cette anne), nous sommes alls en vacances en Italie et (l'anne dernire cette anne) nous retournons Venise. 2. Nouveau programme : (cette semaine - la semaine dernire), les enfants partent en vacances avec leur pre. 3. (Cette semaine - La semaine dernire), a va, mais (cette semaine - la semaine dernire), je n'ai pas dormi pendant plusieurs nuits. Find the missing side round to the nearest tenth write a story started i couldn't belive what my dad had just said The cross-section of a searchlight mirror is shaped like a parabola. The light bulb is located 3 centimeters from the base along the axis of symmetry. If the mirror is 20 centimeters across at the opening, find its depth in centimeters. (Round your answer to the nearest tenth if necessary.) Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speaker O b) Interrupting the speaker to share their opinion c) Avoiding eye contact d) Asking questions to make sure they understand what's being said Lesson 1 Skills Practice Lines For Exercises 1-12, use the figure at the right. In that figure, line m is parallel.Classify each pair of angles as alternate interior, alternate exterior, or corresponding. Pictures Below. Given a dilation around the origin, what is the scale factor K?D o, K = (2,4) (3,6) Dont just say the letter.What is it? look below for the image