g The "Coulomb barrier" is defined to be the electric potential energy of a system of two nuclei when their surfaces barely touch. The probability of a nuclear reaction greatly increases if the energy of the system is above this barrier. What is the Coloumb barrier (in MeV) for the absorption of an alpha particle by a lead-208 nucleus

Answers

Answer 1

Answer:

The Coulomb Barrier U is 25.91 MeV

Explanation:

Given that:

Atomic Mass of lead nucleus A = 208

atomic mass of an alpha particle A = 4

Radius of an alpha particle [tex]R_\alpha = R_o A^{^{\dfrac{1}{3}}[/tex]

where;

[tex]R_\alpha = 1.2 \times 10 ^{-15} \ m[/tex]

[tex]R_\alpha = R_o A^{^{\dfrac{1}{3}}[/tex]

[tex]R_\alpha = 1.2 \times 10 ^{-15} \ m \times (4) ^{^{\dfrac{1}{3}}[/tex]

[tex]R_\alpha = 1.905 \times 10^{-15} \ m[/tex]

Radius of the Gold nucleus

[tex]R_{Au}= R_o A^{^{\dfrac{1}{3}}[/tex]

[tex]R_{Au}= 1.2 \times 10 ^{-15} \ m \times (208) ^{^{\dfrac{1}{3}}[/tex]

[tex]R_{Au} = 7.11 \times 10^{-15} \ m[/tex]

[tex]R = R_\alpha + R_{Au}[/tex]

[tex]R = 1.905 \times 10^{-15} \ m + 7.11 \times 10^{-15} \ m[/tex]

[tex]R = 9.105 \times 10 ^{-15} \ m[/tex]

The electric potential energy of the Coulomb barrier [tex]U = \dfrac{Ke \ q_{\alpha} q_{Au}}{R}[/tex]

[tex]U = \dfrac{8.99 \times 10^9 \ N.m \ ^2/C ^2 \ \times 2 ( 82) \times \(1.60 \times 10^{-19} C \ \ e } {9.105 \times 10^{-15} \ m }[/tex]

U = 25908577.7eV

U = 25.908577 × 10⁶ eV

U =  25.91 MeV

The Coulomb Barrier U is 25.91 MeV


Related Questions

what is the electron configuration of the iodide ion?
A. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²
B. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶
C. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶
D. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰

Answers

Answer:

the answer to this question is C

The electron configuration of the iodide ion is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶. The correct option is option C.

What is electron configuration ?

The arrangement of an atom's or molecule's electrons in their respective atomic or molecular orbitals is known as its electron configuration; for instance, the electron configuration of the neon atom is 1s2 2s2 2p6.

According to electronic configurations, electrons move individually within each orbital while interacting with the average field produced by all other orbitals. The corrosion potential and reactivity of an atom may be calculated from its electron configuration. The electron configuration of the iodide ion is  1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶.

Therefore, the correct option is option C.

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Consider the 1H NMR spectrum for the following compound:
CH3CH2CH3
Predict the first-order splitting patterns of the indicated proton. This question uses specific splitting patterns instead of the often used generic term "multiplet."
a. doublet of quartets
b. triplet of triplets
c. septet
d. quartet of quartets
e. quintet

Answers

Answer:

See explanation

Explanation:

In this case, we have to check the neighbors of each carbon in the molecule. In propane, we have two types of carbons (see figure 1) (blue and red ones). The red carbons are equivalent. (Both have the same neighbors). Now, we can analyze each carbon:

Blue carbon

In the blue carbon, we have 6 hydrogens neighbors (three on each methyl). If we follow the n+1 rule, we will have:

6+1= 7

For the blue carbon, we will have a Septet.

Red carbons

In the red carbon, we have only 2 neighbors (the carbon in the middle only have 2 hydrogens). If we follow the n+1 rule, we will have:

2+1=3

For the red carbon, we will have a triplet.

See figure 2

I hope it helps!

HELPPP.
Which of the following is a property of matter?
O A. It takes up space.
OB. It is everywhere.
O C. It is constantly changing.
O D. It cannot be divided.

Answers

Answer:

a

Explanation:

Matter can be volume or density. So, this concludes that it is when it takes up space.

Answer: A.

Explanation:

it takes up space

Given the information below, which is more favorable energetically, the oxidation of succinate to fumarate by NAD+ or by FAD? Fumarate + 2H+ + 2e- → Succinate E°´ = 0.031 V NAD+ + 2H+ + 2e- → NADH + H+ E°´ = -0.320 FAD + 2H+ + 2e- → FADH2 E°´ = -0.219

Answers

Answer:

Oxidation by FAD  

Explanation:

1. Oxidation by NAD⁺

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻;                  E°´ =  -0.031 V  

NAD⁺ + 2H⁺ + 2e⁻ ⇌ NADH + H⁺;                      E°´ = -0.320 V

Succinate + NAD⁺ ⇌ Fumarate  + NADH + H⁺; E°' =  -0.351 V

2. Oxidation by FAD

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻;        E°´ = -0.031 V  

FAD + 2H⁺ + 2e⁻ ⇌ FADH₂;                    E°´  = -0.219 V

Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V

Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.

FAD is the stronger oxidizing agent.

The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.  

 

The gas evolved in the metal carbonate reaction with acid turns limewater milky The milky substance formed is​

Answers

Answer:

The gas evolved because of reaction of acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas. This happens because of formation of white precipitate of calcium carbonate.

Hope it helps you!

The milky substance formed is CO₂ gas.

What leads to the formation of white precipitate of calcium carbonate ?

The gas evolved because of reaction of acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas.

This happens because of formation of white precipitate of calcium carbonate.

Lime water turns milky when the gas liberated from an acidified carbonate solution is passed into it.

The liberated CO₂ reacts with lime water to give calcium bicarbonate as the precipitate.

Hence, the milky substance formed is CO₂ gas.

Learn more about white precipitate, here:

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The substance nitrogen has the following properties: normal melting point: 63.2 K normal boiling point: 77.4 K triple point: 0.127 atm, 63.1 K critical point: 33.5 atm, 126.0 K At temperatures above 126 K and pressures above 33.5 atm, N2 is a supercritical fluid . N2 does not exist as a liquid at pressures below atm. N2 is a _________ at 16.7 atm and 56.5 K. N2 is a _________ at 1.00 atm and 73.9 K. N2 is a _________ at 0.127 atm and 84.0 K.

Answers

Answer:

- N2 does not exist as a liquid at pressures below 0.127 atm.

- N2 is a solid at 16.7 atm and 56.5 K.

- N2 is a liquid at 1.00 atm and 73.9 K

- N2 is a gas at 0.127 atm and 84.0 K.

Explanation:

Hello,

At first, we organize the information:

- Normal melting point: 63.2 K.

- Normal boiling point: 77.4 K.

- Triple point: 0.127 atm and 63.1 K.

- Critical point: 33.5 atm and 126.0 K.

In such a way:

- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).

- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.

- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.

- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.

Best regards.

which conditions make snow

Answers

Answer:

It depends on the weather.

Explanation:

Like rain and hail, snow comes from the water vapor that forms the clouds, but obviously its formation is different: snow forms when the temperature is low in the atmosphere. In these conditions the water vapor drops freeze and when they collide form tiny crystals that join together to form snowflakes, which fall to Earth when they are heavy enough.

1.60 mL of a suspension of 320.0 mg/5.00 mL aluminum hydroxide is
added to 2.80 mL of hydrochloric acid. What is the molarity of the
hydrochloric acid?

Answers

Answer:

1.41 M

Explanation:

First we must use the information provided to determine the concentration of the aluminum hydroxide.

Mass of aluminum hydroxide= 320mg = 0.32 g

Molar mass of aluminum hydroxide= 78 g/mol

Volume of the solution= 5.00 ml

From;

m/M= CV

Where;

m= mass of aluminum hydroxide= 0.32 g

M= molar mass of aluminum hydroxide = 78 g/mol

C= concentration of aluminum hydroxide solution = the unknown

V= volume of aluminum hydroxide solution = 5.0 ml

0.32 g/78 g/mol = C × 5/1000

C = 4.1×10^-3/5×10^-3

C= 0.82 M

Reaction equation;

Al(OH)3(aq) + 3HCl(aq) -----> AlCl3(aq) + 3H2O(l)

Concentration of base CB= 0.82 M

Volume of base VB= 1.60 ml

Concentration of acid CA= the unknown

Volume of acid VA= 2.80 ml

Number of moles of acid NA = 3

Number of moles of base NB= 1

Using;

CA VA/CB VB = NA/NB

CAVANB = CBVBNA

CA= CB VB NA/VA NB

CA= 0.82 × 1.60 × 3/ 2.80 ×1

CA= 1.41 M

Therefore the concentration of HCl is 1.41 M

Note the dynamic equilibrium in the opening photo which solution changes color when the pH of both solutions is increased explain?

Answers

Answer:

The colour of the orange solution becomes yellow.  

Explanation:

1. Before adding NaOH

Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.

Both solutions are involved in the same equilibrium:

[tex]\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O[/tex]

The first beaker contains mostly chromate ions with a few dichromate ions.

The position of equilibrium lies to the left and the solution is yellow.

The second beaker contains mostly dichromate ions with a few chromate ions.

The position of equilibrium lies to the right and the solution is orange.

2. After adding NaOH

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Beaker 1

If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).

The system responds by having the dichromate react with water to replace the H⁺.  

At the same time, the system forms more of the yellow chromate ion.

The position of equilibrium shifts to the left.

However, the solution is already yellow, so you see no change in colour.

Beaker 2

The reaction is the same as in Beaker 1.

This time, however, as the dichromate ion disappears, do does its orange colour.

Also, the yellow chromate is being formed and its yellow colour appears .

The colour changes from orange to yellow.

Two moles of copper (II) sulfate,CuSO4,contains blank moles of O

Answers

Answer:

8 mol

Explanation:

Step 1: Given data

Moles of copper (II) sulfate: 2 mol

Chemical formula of copper (II) sulfate: CuSO₄

Step 2: Establish the molar ratio of copper (II) sulfate to oxygen

According to the chemical formula, the molar ratio of copper (II) sulfate to oxygen is 1:4.

Step 3: Calculate the moles of O in 2 mol of CuSO₄

[tex]2molCuSO_4 \times \frac{4molO}{1molCuSO_4} = 8molO[/tex]

Name the compound Ga S3​

Answers

Answer:

Gallium(III) sulfide

What is the core charge of helium and why?

Answers

Answer:

Formula for effective nuclear charge is as follows. So, for He atom value of S = 0.30 because the electrons are present in 1s orbital. Therefore, calculate the effective nuclear charge for helium as follows. Thus, we can conclude that the effective nuclear charge for helium is 1.7

Explanation:

The effective nuclear charge experienced by a 1s electron in helium is +1.70.

neeeeed helpppppppppp

Answers

Answer:

Option C. Will always.

Explanation:

A spontaneous reaction is a reaction that occurs without an external supply of heat.

This implies that spontaneous reaction will always occur as no external supply of heat is needed.

You use 10.0 mL of solution A, 10.0 mL of solution B, and 70.0 mL of water for your first mixture. What is the initial concentration of KIO3

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  initial concentration is  [tex]C_f = 0.0022 \ M[/tex]

Explanation:

From the question we are told that

    The volume of solution A is  [tex]V_i = 10.0 mL[/tex]

    The concentration of A is [tex]C_i = 0.0200 \ M[/tex]

    The volume of solution B  is  [tex]V_B = 10.0mL[/tex]

    The volume of water is  [tex]V_{w } = 70.0 mL[/tex]

Generally the law of dilution is mathematically represented as

             [tex]C_i * V_i = C_f * V_f[/tex]

Where  [tex]C_f[/tex] is the concentration of  the mixture

            [tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as

            [tex]V_f = 10 + 10 + 70[/tex]

           [tex]V_f = 90mL[/tex]

So  

      [tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]

substituting values

       [tex]C_f = \frac{0.0200 * 10 }{90}[/tex]

       [tex]C_f = 0.0022 \ M[/tex]

Note the mixture obtained is  [tex]KIO_3[/tex]

A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of moles of an unknown gas requires 6.34 minutes to effuse through the same barrier. The molar mass of the unknown gas is:________.
g/mol.

Answers

Answer:

The molar mass of the unknown gas is [tex]\mathbf{ 51.865 \ g/mol}[/tex]

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

[tex]R \ \alpha \ \dfrac{1}{\sqrt{d}}[/tex]

[tex]R = \dfrac{k}{d}[/tex]  where K = constant

If we compare the rate o diffusion of two gases;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}[/tex]

Since the density of a gas d is proportional to its relative molecular mass M. Then;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

Rate is the reciprocal of time ; i.e

[tex]R = \dfrac{1}{t}[/tex]

Thus; replacing the value of R into the above previous equation;we have:

[tex]\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}[/tex]

We can equally say:

[tex]{\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

[tex]{\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}[/tex]

[tex]M_2 = 32 \times ( \dfrac{6.34}{4.98})^2[/tex]

[tex]M_2 = 32 \times ( 1.273092369)^2[/tex]

[tex]M_2 = 32 \times 1.62076418[/tex]

[tex]\mathbf{M_2 = 51.865 \ g/mol}[/tex]

The rate at which two methyl radicals couple to form ethane is significantly faster than the rate at which two tert-butyl radicals couple. Offer two explanations for this observation.

Answers

Answer:

1. stability factor

2. steric hindrance factor

Explanation:

stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.

ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive

List the following compounds in order from strongest acid to weakest acid. Rank the acids from strongest to weakest.
CH2CHCH2COOH CH2CH2CH2COOH CH3CHCH2COOH CH3CH2CH2COOH
Strongest Weakest

Answers

Answer:

CH3CH2CH2COOH<CH2(F)CH2CH2COOH<CH3CH(F)CH2COOH<CH2(F)CH(F)CH2COOH

Explanation:

We know that the presence of highly electronegative elements in carboxylic acid molecules lead to -I inductive effect. This implies that electrons are withdrawn along the chain towards the electronegative element. As electrons are withdrawn towards the electronegative element, the electron cloud of the carbonyl- hydrogen bond in the acid weakens and the hydrogen can now be easily lost as a proton, that is , the molecule becomes more acidic.

The -I inductive effect increases with increase in the number of electronegative elements present in the molecule and the proximity of the electronegative element to the carbonyl group. The closer the electronegative element is to the carbonyl group, the greater the acidity of the molecule since the -I inductive effect dies out with increasing distance from the carbonyl group. Also, the more the number of electronegative elements in the molecule, the greater the - I inductive effect and the greater the acidity of the molecule, hence the answer.

A 1.00 liter solution contains 0.31 M sodium acetate and 0.40 M acetic acid. If 0.100 moles of barium hydroxide are added to this system, indicate whether the following statements are TRUE or FALSE . (Assume that the volume does not change upon the addition of barium hydroxide.)

a. The number of moles of CH3COOH will remain the same.
b. The number of moles of CH3COO- will increase.
c. The equilibrium concentration of H3O+ will decrease.
d. The pH will decrease.
e. The ratio of [CH3COOH] / [CH3COO-] will remain the same.

Answers

Answer and Explanation:

The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.

If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:

Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻

The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.

Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol

Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol

moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-

According to this, the following are the answers to the sentences:

a. The number of moles of CH₃COOH will remain the same ⇒ FALSE

The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions

b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE

Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)

c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE

The equilibrium concentration of OH⁻ is increased

d. The pH will decrease⇒ FALSE

pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:

[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]

pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15

Thus, the pH will increase.

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the surroundings

Answers

Answer:

[tex]V=27992L=28.00m^3[/tex]

Explanation:

Hello,

In this case, the combustion of methane is shown below:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]

Best regards.

3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 mL of pure cyclohexene product. Calculate the student’s percent yield for this reaction. Show your work.

Answers

Answer:

[tex]49.45~%[/tex]

Explanation:

In this case, we have to start with the chemical reaction:

[tex]C_6H_1_2O~->~C_6H_1_0~+~H_2O[/tex]

So, if we start with 10 mol of cyclohexanol ([tex]C_6H_1_2O[/tex]) we will obtain 10 mol of cyclohexanol ([tex]C_6H_1_0[/tex]). So, we can calculate the grams of cyclohexanol if we calculate the molar mass:

[tex](6*12)+(10*1)=82~g/mol[/tex]

With this value we can calculate the grams:

[tex]10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0[/tex]

Now, we have as a product 500 mL of [tex]C_6H_1_0[/tex]. If we use the density value (0.811 g/mL). We can calculate the grams of product:

[tex]500~mL\frac{0.811~g}{1~mL}=405.5~g[/tex]

Finally, with these values we can calculate the yield:

[tex]%~=~\frac{405.5}{820}x100~=~49.45%[/tex]%= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

Based on the data given, the percentage yield of the student's work is 49.45 %.

What is the equation of the reaction?

The equation of the clay-catalyzed dehydration of cyclohexanol is given below:

C₆H₁₂O ----> C₆H₁₀ + H₂O

From the equation of the reaction, 1 mole of cyclohexanol yields 1 mole of cyclohexene.

Therefore 10 moles of cyclohexanol should yield 10 moles of cyclohexene.

What is the moles of cyclohexene obtained?

First we determine the mass of cyclohexene obtained.

Mass = density * volume

volume of cyclohexene = 500 mL

density = 0.811 g/mL

mass of cyclohexene = 500 * 0.811

mass of cyclohexene = 405.5 g

Number of moles of cyclohexene = mass/molar mass

molar mass of cyclohexene = 82 g

moles of cyclohexene = 405.5 g/82

moles of cyclohexene = 4.945 moles

What is the percentage yield?

Percentage yield = actual yield /expected yield * 100%

Percentage yield = 4.945/10 * 100%

Percentage yield = 49.45%

Therefore, the percentage yield of the student's work is 49.45 %.

learn more about percentage yield at: https://brainly.com/question/8638404

How fast are the atoms moving if the temperature of a gas is cold?

A. very, very slowly

B. they are stagnant

C. very, very quickly

Answers

Answer:

i think option a is correct answer because when there is low temperature then the kinetic enegry will be very less and the atoms moves very slowly.

Answer:

A. very, very slowly

Explanation:

A is the answer because atoms will move faster in hot gas than in cold gas.

an ideal gas is at a pressure 1.00 x 10^5 N/m^2 and occupies a volume 11.00 m^3. If the gass is compressed to a volume of 1.00 m^3 while the temperature remains constant, what will be the new pressure in the gas.

Answers

Answer:

[tex]P_2=1.1x10^6Pa[/tex]

Explanation:

Hello.

In this case, we can solve this problem by applying the Boyle's law which allows us to understand the pressure-volume behavior as a directly proportional relationship:

[tex]P_1V_1=P_2V_2[/tex]

In such away, knowing the both the initial pressure and volume and the final volume, we can compute the final pressure as shown below:

[tex]P_2=\frac{P_1V_1}{V_2}[/tex]

Consider that the given initial pressure is also equal to Pa:

[tex]P_2=\frac{1.00x10^5Pa*11.00m^3}{1.00m^3}\\ \\P_2=1.1x10^6Pa[/tex]

Which stands for a pressure increase when volume decreases.

Regards.

If the particles of matter that make up a substance are relatively far apart and can move freely, the substance is in what state?
gaseous
liquid
solid​

Answers

Answer:

Gaseous

Explanation:

Gasses can move freely and do not form the shape of their containers

Liquids are more free than solids, but they conform to the shape of their container

Solids are not free

The element nitrogen forms a(n) _______ with the charge . The symbol for this ion is , and the name is . The number of electrons in this ion is .

Answers

Answer:

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10.

Explanation:

The element nitrogen is in the Group 15 in the Periodic Table, so it tends to gain 3 electrons (3 negative charges) to fill its valance shell with 8 electrons.

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10 (the original 7 plus the 3 gained). It is isoelectronic with the gas Neon, which accounts for its stability.

The pressure in an automobile tire is 2.0 atm at 27°C. At the end of a journey on a hot summer day the pressure has risen to 2.2 atm. What is the temperature of the air in the tire? a. 272.72 K b. 330 K c. 0.014 K d. 175 K

Answers

Hey there!

For this we can use the combined gas law:

[tex]\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}[/tex]

We are only working with pressure and temperature so we can remove volume.

[tex]\frac{P_{1} }{T_{1}} = \frac{P_{2} }{T_{2}}[/tex]

P₁ = 2 atm

T₁ = 27 C

P₂ = 2.2 atm

Plug these values in:

[tex]\frac{2atm}{27C} = \frac{2.2atm}{T_{2}}[/tex]

Solve for T₂.

[tex]2atm = \frac{2.2atm}{T_{2}}*27C[/tex]

[tex]2atm * T_{2}={2.2atm}*27C[/tex]

[tex]T_{2}={2.2atm}\div2atm*27C[/tex]

[tex]T_{2}=1.1*27C[/tex]

[tex]T_{2}=29.7C[/tex]

Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.

Hope this helps!

which resonance form would be the most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene

Answers

The question is incomplete as the options are missing, however, the correct complete question is attached.

Answer:

The correct answer is option A. ( check image)

Explanation:

The most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene in given options is option a due to the fact that this resonating form follows the octet rule is satisfied for all atoms and additional π bond is present in between C-O that makes it more stable, while in other options there are positive charge which means they do not follows octet rule completely.

Thus, the correct answer is option A. ( check image)

The reaction rate is k[Ce4+][Mn2+] for the following reaction: 2Ce4+(aq) + Tl+(aq) + Mn2+(aq) → 2Ce3+(aq) + Tl3+(aq) + Mn2+(aq What is the catalyst?

Answers

Answer:

Manganese (II) ion, Mn²⁺

Explanation:

Hello,

In this case, given the overall reaction:

[tex]2Ce^{4+}(aq) + Tl^+(aq) + Mn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Tl^{3+}(aq) + Mn^{2+}(aq)[/tex]

Thus, since manganese (II) ion, Mn²⁺ is both at the reactant and products, we infer it is catalyst, since catalysts are firstly consumed but finally regenerated once the reaction has gone to completion. Moreover, since inner steps are needed to obtain it, we can infer that the given rate law corresponds to the slowest step that is related with the initial collisions between Ce⁴⁺ and Mn²⁺

Best regards.

A student followed the procedure of this experiment to determine the percent NaCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of the commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-ml aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na S,O, solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g ml.
(1) Calculate the mass of commercial bleaching solution titrated. 1.1 450=559
(2) Determine the percent NaOCl in the commercial bleaching solution.
(3) Calculate the mass of one gallon of the commercial bleaching solution.

Answers

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Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation of the solute. Express your answer in atmospheres to three significant figures. Pvap = atm

Answers

The question is incomplete, the solute was not given.

Let the solute be K₂CrO₄ and the solvent be water

Complete Question should be like this:

The density of a 0.438 M solution of potassium chromate (K₂CrO₄) at 298 K is 1.063 g/mL.

Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation.  

Pvap = ________atm

Answer:

Pvap (of water above the solution) = 0.0306 atm

Dissolution of the solute

K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻

Explanation:

Given

volume of solution = 1 Litre = 1000 mL of the solution

density of the solution = 1.063 g/mL

concentration of the solution= 0.438M

temperature of the solution= 298 K

vapour pressure of pure water = 0.0313atm

Recall: density = mass/volume

∴mass of solution = volume x density

m = 1000 x 1.063 = 1063 g

To calculate the moles of K₂CrO₄ = volume x concentration

= 1 x 0.438 = 0.438 mol

Mass of K₂CrO₄ = moles x molar mass = 0.438 x 194.19 = 85.055 g

Mass of water = mass  of solution - mass of K₂CrO₄

= 1063 - 85.055 = 977.945 g

moles of water = mass/molar mass

∴ moles of water = 977.945/18.02 = 54.27 mol

 Dissolution of the solute

K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻

(dissolution is the process by which solute(K₂CrO₄) is passed into solvent(H₂O) to form a solution

moles of ions = 3 x moles of K₂CrO₄

= 3 x 0.438 = 1.314 mol

Vapor pressure of solution = mole fraction of water x vapor pressure of water  

= 54.27/(54.27 + 1.314) x 0.0313 = 0.0306 atm

A sample of a compound is made up of 57.53 g C, 3.45 g H, and 39.01 g F. Determine the empirical formula of this compound.

Answers

Answer:

C7H5F3

Explanation:

The following data were obtained from the question:

Mass of Carbon (C) = 57.53g

Mass of Hydrogen (H) = 3.45g

Mass of Fluorine (F) = 39.01g

The empirical formula of the compound can be obtained as follow:

C = 57.53g

H = 3.45g

F= 39.01g

Divide each by their molar mass

C = 57.53/12 = 4.79

H = 3.45/1 = 3.45

F = 39.01/19 = 2.05

Divide each by the smallest

C = 4.79/2.05 = 2.3

H = 3.45/2.05 = 1.7

F = 2.05/2.05 = 1

Multiply through by 3 to express in whole number

C = 2.3 x 3 = 7

H = 1.7 x 3 = 5

F = 1 x 3 = 3

Therefore, the empirical formula for the compound is C7H5F3

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