Helium (He) is the lightest noble gas component of air, and xenon (Xe) is the heaviest. Perform the following calculations, using R = 8.314 J/(mol·K) and ℳ in kg/mol. (a) Find the rms speed of He in winter (0.°C) and in summer (30.°C). Enter your answers in scientific notation. × 10 m/s (winter) × 10 m/s (summer) (b) Find the ratio of the rms speed of He to that of Xe at 30.°C. (rate He)/(rate Xe) (c) Find the average kinetic energy per mole of He and of Xe at 30.°C. Enter your answers in scientific notation. × 10 J/mol for He × 10 J/mol for Xe (d) Find the average kinetic energy per molecule of He at 30.°C. Enter your answer in scientific notation. × 10 J/He atom

Answers

Answer 1

Answer:

Explanation:

Hello,

Data;

R = 8.314J/(mol.K)

Temp (winter) = 0°C = (0 + 273.15)K = 274.15K

Temp.(summer) = 30°C = (30 + 273.15)K = 303.15K

Molar mass of He = 4g/mol = 0.004kg/mol

Molar mass of Xe = 131.29g/mol = 0.131kg/mol

a) rms speed in winter and summer

Vrms = √(3RT/M)

R = gas constant

T = temperature of the gas

M = molar mass of the gas

In winter,

Vrms = √(3×8.314×273.15) / 0.004

Vrms = 1.30×10³m/s

In summer

Vrms = √(3×8.314×303.15) / 0.004

Vrms = 1.37×10³m/s

b) Vrms of Xe at 30°C

Vrms = √(3 × 8.314 × 303.15) / 0.131

Vrms of Xe = 240.24m/s = 2.40×10²m/s

Vrms of He = 1.37×10³m/s

Rate of He / rate of Xe = 1.37×10³ / 2.40×10²

Rate of He / rate of Xe = 5.7

c) K.E per mole

At 30°C

K.E of He = (3/2) × 8.314 × 303.15

K.E of He = 3.78×10³J/mol

K.E of Xe = (3/2) × 8.314 × 303.15

K.E of Xe = 3.78×10³J/mol

d) K.E per molecule = ½mv²

K.E of He molecule = ½ × 0.004 × 1.37×10³

K.E = 2.74J

Answer 2

The rms speed of He in winter and in summer are 1.30×10³m/s & 1.37×10³m/s respectively, ratio of the rms speed of He to that of Xe at 30 degree celsius is 5.7, average kinetic energy per mole of He and of Xe and average kinetic energy per molecule of He are discussed below.

How do we calculate the root mean square velocity?

Root mean square velocity of the gases will be calculated as:

Vrms = √(3RT/M), where

R = universal gas constant = 8.314J/(mol.K)

M = molar mass of gas in kg/mol

T = temperature in K

Root mean square velocity of Helium at 0°C or 273.15K and 30°C or 303.15K will be calculated as:

At 273.15K-

Vrms = √(3×8.314×273.15) / 0.004 = 1.30×10³m/s

At 303.15K-

Vrms = √(3×8.314×303.15) / 0.004 = 1.37×10³m/s

Vrms (winter×summer) = 1.30×10³m/s × 1.37×10³m/s = 1.78×10⁶m/s

Vrms of Xe at 303.15K

Vrms = √(3 × 8.314 × 303.15) / 0.131

Vrms of Xe = 240.24m/s = 2.40×10²m/s

Vrms of He = 1.37×10³m/s

Rate of He / rate of Xe = 1.37×10³ / 2.40×10²

Rate of He / rate of Xe = 5.7

Average kinetic energy will be calculated as:

K.E = 3RT / 2N, where

N = avgadros number = 6.022×10²³ atoms/mole

Average kinetic energy per mole of He & Xe at 303.15K is same as:

K.E = 3/2(1.38 × 10⁻²³J/K)(303.15K) = 6.2 × 10⁻²⁵J/mol

Average kinetic energy per molecule of He will be calculated as:

K.E of He molecule = ½ × 0.004 × 1.37×10³

K.E = 2.74J

Hence, calculations for the given points are described above.

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Related Questions

Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant: G° for this reaction would be _________ than zero. Submit AnswerRetry Entire Group

Answers

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

The reduction of iron(III) oxide to iron metal is an endothermic process: Fe2O3(s) + 2 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = +26.3 kJ How many kilojoules of energy are required to produce 1.00 kilogram of iron metal?

Answers

Answer: Thus 234 kJ of energy are required to produce 1.00 kilogram of iron metal

Explanation:

To calculate the number of moles , we use the equation:

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]  

Putting values , we get:

[tex]\text{Moles of iron}=\frac{1000g}{56g/mol}=17.8moles[/tex]    (1.00kg=1000g)

The balanced chemical reaction is:

[tex]Fe_2O_3(s)+2CO(g)\rightarrow 2Fe(s)+3CO_2(g)[/tex]   [tex]\Delta H=+26.3kJ[/tex]

Given :

Energy released when 2 moles of [tex]Fe[/tex] is produced = 26.3 kJ

Thus Energy released when 17.8 moles of [tex]Fe[/tex] is produced =

= [tex]\frac{26.3kJ}{2}\times 17.8=234kJ[/tex]

Thus 234 kJ of energy are required to produce 1.00 kilogram of iron metal

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. What is the final level of the electron?(c=3.00×10^8m/s, h=6.63×10^-34 J·s, RH=2.179×106-18J)a. 5



b. 6



c. 8



d. 9



e. 1

Answers

Explanation:

It is given that,

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,

[tex]n_i=8[/tex]

[tex]\lambda=3745\ nm[/tex]

The amount of energy change during the transition is given by :

[tex]\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}][/tex]

And

[tex]\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}][/tex]

Plugging all the values we get :

[tex]\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5[/tex]

So, the final level of the electron is 5.

A beach has a supply of sand grains composed of calcite, ferromagnesian silicate minerals, and non-ferromagnesian silicate minerals. If it undergoes lots of chemical weathering, which sand grains will be quickly chemically weathered away?

a. Calcite
b. ferromagnesian silicate minerals
c. non-ferromagnesian silicate minerals

Answers

ferromagnesian silicate minerals (i looked it up)

The sand that grained will be quickly chemically weathered away should be option b. ferromagnesian silicate minerals.

What are ferromagnesian silicate minerals?

It should be considered as the Silicate minerals where cations of iron and the form of magnesium should be important for the chemical components. It is used for covering up the minerals. Also, calcite should be normal weather via the solution process so it required a lot of water that contains a high amount of carbonic acid.

Hence, the correct option is b.

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A 5.00g of X, the product of organic synthesis is obtained in a 1.0 dm3 aqueous solution. Calculate the mass of X that can be extracted from the aqueous solution by a 50cm3 of ethoxy ethane. (KD (X) =40.​

Answers

Answer:

mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

(m/0.05) = 200 - 40m

m = 0.05 × (200 - 40m)

m = 10 - 2m

3m = 10

m = 10/3

m = 3.33 g of X

Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Which land feature supports the theory of continental drift?

A.canyons B.volcanoes C.coal fields D.oceans

Answers

Answer:

Coals

Explanation:

The land feature that supports the theory of continental drift is ; ( C ) coal fields

Continental drift is the gradual shift in position of the earth tectonic plates ( i.e. gradual shift in the continents in relation to ocean basins) and this due to the heat from the earths' mantle.

Coal fields supports this theory because the it is an area where coal is found in large quantities and mined for commercial purposes. coal fields areas are found as a result of continental drift.

Hence we can conclude that the land feature that supports the theory of continental drift is coal fields

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Write a net ionic equation to show that benzoic acid, C6H5COOH, behaves as a Brønsted-Lowry acid in water.

Answers

Answer:

H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)

Explanation:

According to Brönsted-Lowry acid-base theory, an acid is a substance that donates H⁺. Let's consider the molecular equation showing that benzoic acid is a Brönsted-Lowry acid.

C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The complete ionic equation includes all the ions and molecular species.

C₆H₅COO⁻(aq) + H⁺(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The net ionic equation includes only the ions that participate in the reaction and the molecular species.

H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)

Why at night, under the mercury or sodium vapor lights in a mall parking lot, do cars seem to be peculiar colors?

Answers

Answer:

Here's what I find  

Explanation:

Sodium vapour produces mostly yellow light.

Mercury vapour produces mostly blue light.

A white car reflects all colours back to our eyes.

A coloured car, say green, absorbs all the other colours and reflects the green.

Under a sodium light, a white or a yellow car will appear yellow. A red car will be nearly black because there is no red light to reflect.

Under a mercury light, a white or a blue car will appear blue. A yellow car will be quite dark, but there will be a yellowish tinge because there is some yellow in the Hg spectrum for it to reflect.  

The picture below shows a red car and a black car under sodium light. Can you tell which is which?

the iupac name of the compound​

Answers

Answer:

3-Pentyn-1-ol

Explanation:

tripple bond is at 3 postion from alochol

carbon are 5 atoms so pent

yn becauese its alkyne

Why can liquids change shape but solids cannot

Answers

Answer:

Explanation:

the forces between the molecules are stronger in solid than in liquids

Which best describes the relationship between heat internal energy, and thermal energy?
O Internal energy is heat that flows, and heat is the part of thermal energy that can be transferred.
O Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred.
O Thermal energy is heat that flows, and heat is the part of internal energy that can be transferred.
O Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred.

Answers

Answer:

Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred

Explanation:

structure and correct name for 5-octyne​

Answers

Answer:

Explanation:

(R)-5-octyne-4-ol

C8H140

Correct name for 5-octyne is 3-octane.

Structure is attached below.

3-Octane:

Octane is a hydrocarbon and an alkane with the chemical formula C₈H₁₈, and the condensed structural formula CH₃(CH₂)₆CH₃. Octane has many structural isomers that differ by the amount and location of branching in the carbon chain. One of these isomers, 2,2,4-trimethylpentane (commonly called iso-octane) is used as one of the standard values in the octane rating scale.

The structure for 5-octyne is given below.

Correct name for it is 3-octane.

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A chemistry student weighs out of ascorbic acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the final equivalence point. Round your answer to significant digits.

Answers

Answer:

14.3 mL  

Explanation:

Assume the student used 0.113 g ascorbic acid  and 0.0900 mol·L⁻¹ NaOH.

1. Balanced chemical equation.

The formula of ascorbic acid is H₂C₆H₆O₆ (MM = 176.12 g/mol).

However, for the balanced equation, let's write it as H₂A.

[tex]\rm H_{2}A + 2NaOH \longrightarrow Na_{2}A + 2H_{2}O[/tex]

2. Moles of ascorbic acid

[tex]\text{Moles of H$_{2}$A} =\text{0.113 g H$_{2}$A} \times \dfrac{\text{1 mmol H$_{2}$A}}{\text{0.176 12 mg H$_{2}$}A} = \text{0.6416 mmol H$_{2}$A}[/tex]

3. Moles of NaOH

The molar ratio is 2 mmol NaOH:1 mmol H₂A.

[tex]\text{Moles of NaOH}= \text{0.6416 mmol H$_{2}$A} \times \dfrac{\text{2 mmol NaOH}}{\text{1 mmol H$_{2}$A}} =\text{1.283 mmol NaOH}[/tex]

4. Volume of NaOH

[tex]V = \text{1.283 mmol NaOH}\times \dfrac{\text{1 mL NaOH}}{\text{0.0900 mmol NaOH}} = \textbf{14.3 mL NaOH}\\\\\text{The student will need $\large \boxed{\textbf{14.3 mL NaOH}}$}[/tex]

which process is used to produce gases from solutions of salts dissolved in water or another liquid?

A.Electrolysis
B.Metallic bonding
C.Ionic bonding
D. Polar covalent bonding

Answers

Answer:

A.Electrolysis

Explanation:

A.Electrolysis

For example,  electrolysis of solution of NaCl in water gives H2 and O2.

A student mixed 115 g of sugar, 350 g of water and 5 g of spices. What will be the mass of the solution?
these are the options
470g

465g

350g

120g

Answers

Answer:

[tex]m_{solution}=470g[/tex]

Explanation:

Hello,

In this case, a solution is formed when a solute is completely dissolved in a solvent, thus, for this situation, the sugar is the solute and the water the solvent but in addition to them we find spices which are also considered in the total mass of the solution. In such a way, for computing the total mass we must add the mass of three constituents (115 g sugar, 350 g water and 5 g spices) as shown below:

[tex]m_{solution}=115g+350g+5g\\\\m_{solution}=470g[/tex]

Best regards.

Aqueous solutions of sodium sulfate and potassium chloride are mixed. What is the precipitate and how many molecules are formed?

Answers

Answer:

The two products of this reaction, Sodium Chloride and Potassium Sulfate, are both soluble in water, hence, there's no precipitate formed from this reaction.

Explanation:

Sodium sulfate is Na₂So₄

Potassium Chloride is KCl

When they both react, theres a double displacement where ions and radicals are exchanged

Na₂SO₄ + KCl → NaCl + K₂SO₄

The products are

NaCl - Sodium Chloride

K₂SO₄ - Potassium Sulfate

The two products are soluble in water, hence, there's no precipitate formed from this reaction.

And we would need numerical values of the number of moles of the two reactants mixed to obtain the number of moles of products formed and then the number of molecules from Avogadro's constant.

Hope this Helps!!!

100. mLmL of 0.200 MMHClHCl is titrated with 0.250 MMNaOHNaOH. Part A What is the pH of the solution at the equivalence point? Express the pH numerically.

Answers

Answer:

pH = 7.0

Explanation:

When HCl reacts with NaOH, H₂O and NaCl are produced, thus:

HCl + NaOH → H₂O + NaCl

At equivalence point, all HCl reacts with NaOH. The only you will have is water.

Equilbrium of water is:

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

K = 1x10⁻¹⁴ = [H⁺] [OH⁻]

As H⁺ = OH⁻ because both are produced from the same water-

1x10⁻¹⁴ = [H⁺]²

1x10⁻⁷M = [H⁺]

As pH = -log= [H⁺]

pH = 7.0

-The pH at equivalence point in the titration of a strong acid with a strong base is always 7.0-

8. A 25.0 mL sample of an H2SO4 solution is titrated with a 0.186 M NaOH solution. The equivalence point is reached with 12.9 mL of base. The concentration of H2SO4 is ________ M. (Hint: write a balanced chemical equation first!)

Answers

Answer:

  0.0480 M

Explanation:

The reaction is ...

  H₂SO₄ + 2NaOH ⇒ Na₂SO₄ +2H₂O

That is, 2 moles of NaOH react with each mole of H₂SO₄. Then the molarity of the H₂SO₄ is ...

  moles/liter = (0.186 M/2)(12.9 mL)/(25.0 mL) ≈ 0.0480 M

c) What is the pH of the buffer system in part a when 0.030 moles of strong acid are added (without a change in volume)

Answers

Answer:

remain the same

Explanation:

The pH of the buffer system remain the same when 0.030 moles of strong acid are added because buffer system has the property to resist any change in the pH  when acid or base is added to the solution. In buffer system, one molecule is responsible for neutralizing the pH of the solution by giving H+ or OH-.This molecule is known as buffer agent. If more base is added, the molecule provide H+ and when more acid is added to the solution, then the molecule add OH- to the solution.

Balance the chemical equation
Fe2O3 (s) + CO (g) 2 Fe(s) + CO2 (g)
Express your answer as a chemical equation. Identify all of the phases in your answer.

Answers

Answer:

[tex]Fe_2O_3+3CO\Rightarrow \:2Fe+3CO_2[/tex]

Explanation:

[tex]Fe_2O_3+CO\Rightarrow \:2Fe+CO_2\\\\Fe_2O_3+3CO\Rightarrow \:2Fe+3CO_2[/tex]

Best Regards!

"The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb" for NH3

Answers

Answer:

Kb = 1.77x10⁻⁵

Explanation:

When NH₃, a weak base, is in equilibrium with waterm the reaction that occurs is:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

And the dissociation constant, Kb, for this equilibrium is:

Kb = [NH₄⁺] [OH⁻] / [NH₃]

To find Kb you need to find the concentration of each species. The equilibrium concentrations are:

[NH₃] = 0.950M - X

[NH₄⁺] = X

[OH⁻] = X

Where X is reaction coordinate.

You can know [OH⁻] and, therefore, X, with pH of the solution, thus:

pH = -log [H⁺] = 11.612

[H⁺] = 2.4434x10⁻¹²

As 1x10⁻¹⁴ = [H⁺] [OH⁻]

1x10⁻¹⁴ / 2.4434x10⁻¹² = [OH⁻]

4.0926x10⁻³ = [OH⁻] = X

Replacing, concentrations of the species are:

[NH₃] = 0.950M - X

[NH₄⁺] = X

[OH⁻] = X

[NH₃] = 0.9459M

[NH₄⁺] = 4.0926x10⁻³M

[OH⁻] = 4.0926x10⁻³M

Replacing in Kb expression:

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Kb = [4.0926x10⁻³M] [4.0926x10⁻³M] / [0.9459M]

Kb = 1.77x10⁻⁵

The branch of science which deals with the chemical bond is called Chemistry.

The correct answer to the question is  [tex]Kb = 1.77*10^{-5[/tex]

Explanation:

When NH₃, is acts as a  weak base it forms an equilibrium with water the reaction occurs is:

[tex]NH_3(aq) + H_2O(l) ---><NH_4^+(aq) + OH^-(aq)[/tex]

The formula we gonna use is as follows:-

[tex]Kb = \frac{[NH_4^+] [OH^-]}{[NH_3]}[/tex]

The data is given in the question is as follows:-

[NH₃] = 0.950M - X [NH₄⁺] = X [OH⁻] = X

Where X stands for reaction coordinate.

After solving the ph of the compound the value is as follows:-

[NH₃] = [tex]0.9459M[/tex] [NH₄⁺] = [tex]4.0926*10^{-3}M[/tex] [OH⁻] = [tex]4.0926*10^{-3}M[/tex]

Putting the value in the formula.

[tex]Kb = \frac{[4.0926*10^{-3}M] [4.0926*10^{-3}M]}{[0.9459M]}[/tex]

After solving the equation the value of Kb is [tex]1.77*10^{-5[/tex]

Hence, the correct answer is [tex]1.77*10^{-5[/tex]

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A correct name for the following compound is:_________.a) 4-bromo-3,8-dimethylbicyclo[5.2.2]nonane b) 3,8-dimethyl-4-bromo-bicyclo[5.2.O)nonane c) 4-bromo-3,8-dimethylbicyclo[5.2.1]decane d) 7-bromo-2,6-dimethylbicyclo[5.2.0]nonane e) 4-bromo-3,8-dimethylbicyclo[5.2.0]nonane

Answers

Answer:

e) 4-bromo-3,8-dimethylbicyclo[5.2.0]nonane

Explanation:

The missing image of the the  compound we are to name is attached below.

Before we can name an organic compound; It is crucial we know the guiding rules in naming them.

1. Select the longest continuous carbon chain as the root hydrocarbon and name according to the number of carbon atoms it contains, adding appropriate suffix to indicate the principal substituent group.

2. Number the carbon atoms in the root hydrocarbon from the end which gives the lowest number to the substituents.

3. If the same substituent is present two  or more times  in a molecule; the number of this substituent is indicated by the prefix di -(2), tri - (3) , tetra - (4) etc attached to the substituent name.

4. If there is more than one type of substituent in the molecule ; the substituents are named according to the alphabetical order but where  there are mixed substituents ; the inorganic are named first.

5. In selecting and numbering the longest continuous chain, the functional groups are given preference over substituents., i.e the functional group is given the smallest possible number.

In the light of the above guiding rules; we were able to name the given compound because the compound contains nine carbons in the ring form which result to root name nonane. The two methyl are on the third and eight carbon; bromine is on the fourth carbon ; there are two cyclic ring present in the compound where we have 5 carbons in one structure, another 2 carbons in the second structure and zero carbon in the bridge structure which eventually result to the correct  name:

4-bromo-3,8-dimethylbicyclo[5.2.0]nonane

Elvira Walks 4 miles to the west from school and stops at the store. She then walks 3 miles south.


What is the Distance and the displacement?

Answers

Answer:

distance = 7 miles

displacement = 5 miles

Explanation:

Distance is a scalar quantity as it takes account of magnitude traveled but not the direction traveled from starting point.

The distance traveled is the sum total of distances moved

distance = 4 + 3 = 7 miles

Displacement however, is a vector and measure the shortest possible distance traveled in a given direction from the starting point.

The path of Elvis' walking forms a right-angle triangle with the hypotenuse being the displacement and the other two sides being the distance traveled west and south.

Using Pythagoras' theorem; c² = a² + b²

where c = hypotenuse and a and b are the other two sides

c² = 4² + 3²

c² = 16 + 9 = 25

√c² = √25

c = 5

Therefore, displacement = 5 miles

For the reaction of nitrogen gas and hydrogen gas to make methane, describe FOUR stresses that would shift the equilibrium to the left toward the reactants? N2(g) + 3 H2(g) ⇌ 2 NH3(g) + Energy

Answers

Answer:

- Addition of NH₃(g)

- Removal of N₂(g)

- Increase of temperature

- Pressure decrease

Explanation:

According to Le Chatelier's principle, if we apply an stress to a reaction at equilibrium, the system will try to shift the equilibrium in order to decrease the stress. If we add reactants, the equilibrium will shift toward the formation of more products (to the consumption of reactants) and vice versa.

The stresses we can apply to this equilibrium are the following:

- Addition of NH₃(g) : it is a product, thus its addition will result in a shift toward reactants.

- Removal of N₂(g): it is a reactant, thus its removal from the reaction mixture will result in a shift toward reactants.

- Increase of temperature: the reaction is exothermic, so it releases energy. Energy is a product. If we add energy (increase the temperature), we are adding a product, so the equilibrium will shift toward the reactants.

- Pressure decrease: because both reactants and products are in the gas phase. A decrease in pressure shifts an equilibrium to the side of the reaction with greater number of moles of gas. In this case, the reactants side has greater number of moles of gas (1 mol + 3 moles= 4 moles) than the products side (2 moles). Thus, the equilibrum will shift toward reactants.

A volume of 0.800 L of a 2 x 10–4 M Ba(NO3)2 solution is added to 0.200 L of 5 x 10–4 M Li2SO4. Does BaSO4 precipitate? Explain your answer

Answers

Answer:

The BaSO₄ precipitates as a solid.

Explanation:

In order to determine the precipitation, we need to look at the reactants, we have: Ba(NO₃)₂ and Li₂SO₄

Salts from nitrate are all soluble and sulfates can make precipitate if they react to elements from group 2. (Solubility rules) So this reaction, has a precipitate but we can predict it, if we look at the reaction and we take acount the Kps:

Ba(NO₃)₂ (aq) + Li₂SO₄ (aq)  →  2LiNO₃ (aq)  +  BaSO₄ (s) ↓

Let's find out the moles of each ion:

Ba(NO₃)₂ (aq) →  Ba²⁺  (aq)  +  2NO₃⁻  (aq)

0.8 L . 0.0002 M = 1.6×10⁻⁴ moles. These are the moles of Ba²⁺ in 1 L of solution, so [Ba²⁺] = 1.6×10⁻⁴ M

Li₂SO₄ (aq) →  2Li⁺ (aq)  +  SO₄⁻² (aq)

0.2 L . 0.0005 M = 1×10⁻⁴ M. These are the moles of sulfate, in 1 L of solution so [SO₄⁻²] =  1×10⁻⁴ M

BaSO₄ (s)  ⇆   Ba²⁺  (aq) + SO₄⁻² (aq)      Kps = 1.08×10⁻¹⁰

We need to compare Kps to [ Ba²⁺] . [SO₄⁻²]

[ Ba²⁺] . [SO₄⁻²] = 1.6 ×10⁻⁸ >  Kps →  Precipitate.

If  [Ba²⁺] . [SO₄⁻²] = Kps  → Solution is saturated

If [ Ba²⁺] . [SO₄⁻²] < Kps → There is no precipitate formed, or if it is formed, it will be quickly dissolved to reach saturation.

What is the name of CaCl2 7H2o

Answers

Answer:

calcium chloride dihydrate

Consider the addition of an electron to the following atoms from the fourth period. Rank the atoms in order from the most negative to the least negative electron affinity values based on their electron configurations.

Atom or ion Electron configuration
Br 1s22s22p63s23p64s23d104p5
Ge 1s22s22p63s23p64s23d104p2
Kr 1s22s22p63s23p64s23d104p6

Answers

Answer:

Ge: [Ar] 3d10 4s2 4p2 => 6 electrons in the outer shell

Br: [Ar] 3d10 4s2 4p5 => 7 electrons in the outer shell

Kr: [Ar] 3d10 4s2 4p6 => 8 electrons in the outer shell

Explanation:

The electron affinity or propension to attract electrons is given by the electronic configuration. Remember that the most stable configuration is that were the last shell is full, i.e. it has 8 electrons.

The closer an atom is to reach the 8 electrons in the outer shell the bigger the electron affinity.

Of the three elements, Br needs only 1 electron to have 8 electrons in the outer shell, so it has the biggest electron affinity (the least negative).

Ge: needs 2 electrons to have 8 electrons in the outer shell, so it has a smaller (more negative) electron affinity than Br.

Kr, which is a noble gas, has 8 electrons and is not willing to attract more electrons at all, the it has the lowest (more negative) electron affinity of all three to the extension that really the ion is so unstable that it does not make sense to talk about a number for the electron affinity of this atom.

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Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have reacted.

Answers

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

A chemistry student weighs out of hypochlorous acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.

Answers

Answer:

Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.

A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.

Explanation:

1 mole HClO = 74.44g

0.0941g = [tex]\frac{0.0941}{74.44}[/tex] = 0.00126 moles

Concentration = no. of moles/volume in L

Hence, Concentration of HClO = 0.00126/ 0.250L

= 0.005M.

C1V1 =C2V2

0.005 × 250 mL = 0.2 × V2

Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.

An atom of element number 33 (As) is in its ground electronic state. Which one of the following sets quantum numbers could not apply to any of its electrons?
A) n=2 l=1 ml= -1 ms= 1/2
B) n=3 l=0 ml=0 ms= -1/2
C) n=3 l=2 ml=-2 ms= -1/2
D) n=4 l=0 ml=0 ms= -1/2
E) n=4 l=2 ml=1 ms= 1/2

Answers

Answer:

E) n=4 l=2 ml=1 ms= 1/2

Explanation:

Arsenic is a member of group 15 in the periodic table. Its electronic configuration is;

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3. Its condensed electronic configuration can be written as [Ar]4s2 3d10 4p3. This electronic configuration shown here can now enable us to consider each option given in the question in order to meaningfully arrive at a logical answer.

If we look at option E, the data given for that electron is; n=4 l=2 ml=1 ms= 1/2. This refers to an electron in a 4d orbital. In the ground state configuration of arsenic shown above, there is no 4d orbital, hence option E must be the correct answer.

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