One proposed mechanism of the reaction of HBr with O2 is given here. HBr + O2 → HOOBr (slow) HOOBr + HBr → 2HOBr (fast) HOBr + HBr → H2O + Br2 (fast) What is the equation for the overall reaction?

Answers

Answer 1

Answer:

4 HBr + O2  →  + 2H2O + 2Br2

Explanation:

Based on the following reaction mechanism:

HBr + O2 → HOOBr (slow)

HOOBr + HBr → 2HOBr (fast)

HOBr + HBr → H2O + Br2 (fast)

The equation for the overall reaction is the sum of the three reactions in which intermediaries of reaction (HOBr and HOOBr are canceled). That is 1 + 2 + 2*(3):

HBr + O2 + HOOBr + HBr + 2HOBr + 2HBr → HOOBr + 2HOBr + 2H2O + 2Br2

4 HBr + O2  →  + 2H2O + 2Br2

Beeing this reaction the equation of the overall reaction.


Related Questions

a binary ionic compound is made of two components name one of them​

Answers

Answer:

CATION

Explanation:

It's one is the action and the mother is a cation.

For the iodine trichloride molecule: a. Determine the number of valence electrons for each atom in the molecule b. Draw the Lewis Dot structure c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?) d. Show the polarity of each bond and for the molecule by drawing in the dipole +à

Answers

Answer:

Explanation:

a. Determine the number of valence electrons for each atom in the molecule

In this case we both atoms are halogens. Therefore we will have 7 electrons for each atom.

b. Draw the Lewis Dot structure

In this case, the formula is [tex]ICl_3[/tex], so the central atom would be "I" and the "Cl" atoms would be placed around "I". See figure 1

c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?)

In this specific case, the "I" atom don't follow the octet rule. We will have an expanded octet for iodine (more than 8 electrons).

d. Show the polarity of each bond and for the molecule by drawing in the dipole +d

The negative dipole would be placed in the atom with higher electronegativity, in this case "Cl". The positive dipole would be placed in the atom with low electronegativity, in this case "I".

I hope it helps!

What is the Arrhenius definition of an acid? A substance that increases H3O+ concentration when it is dissolved in water. A substance that increases OH– concentration when it is dissolved in water. A compound that donates protons. A compound that accepts protons.

Answers

Answer:

A substance that increases H3O+ concentration when it is dissolved in water.

Explanation:

Note that H3O+ and H+ are used quite interchangeably in chemistry.

An acid makes the H+ content higher, thereby decreasing the pH.

Answer:

a

A substance that increases H3O+ concentration when it is dissolved in water.

Explanation:

Give two examples (i.e. list 2 elements that are examples) of: a. an atom with a half-filled subshell b. an atom with a completely filled outer shell c. an atom with its outer electrons occupying a half-filled subshell and a filled subshell

Answers

Answer:

an atom with a half-filled subshell - hydrogen

an atom with a completely filled outer shell - argon

an atom with its outer electrons occupying a half-filled subshell and a filled subshell- copper

Explanation:

The outermost shell or the valence shell of the atom is the last shell in the atom. Chemical reactions occur at this outer most shell. The number of electrons on the outermost shell of an atom determines the group to which it belongs in the periodic table as well as its chemical properties.

Hydrogen has a half filled 1s sublevel. Only one electron is present in this sublevel.

Let us consider argon

1s2 2s2 2p6 3s2 3p6

The outermost ns and np levels are completely filled. Thus the outermost shell is completely filled.

In the last case; let us look at the electronic configuration of nitrogen;

1s2 2s2 2p3

The outermost 2p subshell is exactly half filled while the 2s sublevel is fully filled. The outermost shell of nitrogen is made up of 2s2 and 2p3 sublevels.

How has human action affected the
population?

Answers

Answer:

Human action has affected the population in quite a negative way. Technological innovations and urbanization gave rise to a high degree of pollution on the land, air and water.

Emission of hydrocarbons from automobiles and factories are known to pollute and cause sicknesses related to the respiratory system thereby shortening the lifespan of humans. Pollution of water bodies also cause death of sea animals and thereby reducing food availability of humans.

Calcium chloride reacts with sodium hydroxide to form solid calcium hydroxide, Ca(OH)2. The balanced net ionic equation is


Group of answer choices


Ca+2(aq) + 2OH-(aq) + 2 Na+(aq) + 2 Cl-(aq) → Ca(OH)2(s).


Ca+2(aq) + 2OH-(aq) + Na+(aq) + Cl-(aq) → Ca(OH)2(s).


Ca+2(aq) + 2OH-(aq) → Ca(OH)2(s).


CaCl2(aq) + 2NaOH(aq) → CaOH(s) + NaCl(aq).


CaCl2(aq) + 2NaOH(aq) → Ca(OH)2(s) + 2NaCl(aq).

Answers

Answer:

Ca²⁺ + 2 OH⁻ → Ca(OH)₂(s)

Explanation:

In chemistry, the net ionic equation is a way to write a chemical reaction whereas you write only the ions that are involved in the reaction.

When calcium chloride, CaCl₂ reacts with sodium hydroxide, NaOH to produce Ca(OH)₂ the only ions involved in the reaction are Ca²⁺ and OH⁻, thus, the balanced net ionic equation is:

Ca²⁺ + 2 OH⁻ → Ca(OH)₂(s)

Cl⁻ and Na⁺ are not involved in the reaction and you don't have to write them.

The balanced net ionic equation for the reaction between calcium chloride and sodium hydroxide is

Ca²⁺(aq) + 2OH⁻(aq) —> Ca(OH)₂(s)

The ionic equation for the reaction between calcium chloride and sodium hydroxide can be written as follow:

Calcium chloride => CaCl₂

Sodium hydroxide => NaOH

In solution,

CaCl₂(aq) —> Ca²⁺(aq) + 2Cl⁻(aq)

NaOH(aq) —> Na⁺(aq) + OH⁻(aq)

CaCl₂(aq) + NaOH(aq) —>

Ca²⁺(aq) + 2Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) —> Ca(OH)₂(s) + 2Cl⁻(aq) + Na⁺(aq)

Cancel the spectator ions (i.e Cl⁻ and Na⁺) and write 2 before OH⁻ to obtain the net ionic equation as shown below:

Ca²⁺(aq) + 2OH⁻(aq) —> Ca(OH)₂(s)

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A compound is known to be Na2CO3, Na2SO4, NaOH, NaCl, NaC2H3O2, or NaNO3. When a barium nitrate solution is added to a solution containing the unknown a white precipitate forms. No precipitate is observed when a magnesium nitrate solution is added to a solution containing the unknown. What is the identity of the unknown compound

Answers

Answer:

Na₂SO₄

Explanation:

Barium nitrate, Ba(NO₃)₂ produce precipitate with SO₄²⁻, CO₃²⁻. That means the precipitate could be obtained from Na₂SO₄ and Na₂CO₃.

Also, magnesium nitrate, Mg(NO₃)₂, produce precipitate just with CO₃²⁻. As the unknown solution produce no precipitate, the unknown compound is:

Na₂SO₄

The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2 and I2 are 0.453 M, what will the concentration of HI be at equilibrium

Answers

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

Interpret the following equation for a chemical reaction using the coefficients given: CO(g) Cl2(g) COCl2(g) On the particulate level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g). On the molar level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g).

Answers

Answer:

On the particulate level: 6.02 * 10²³ particles of CO(g) reacts with 6.02 * 10²³ particles of Cl₂(g) to form 6.02 * 10²³ particles of COCl2(g).

On the molar level: 1 mole of CO(g) reacts with 1 mole of Cl2(g) to form 1 mole of COCl₂(g).

Explanation:

The particulate level refers to the microscopic or atomic level of substances. It also involves the ions, protons, neutrons and molecules present in substances.

The molar level refers to the quantitative measure of substances in terms of the mole, where a mole represents the amount of substances containing the Avogadro number of particles which is equal to 6.02 * 10³ particles.

Equation of the reaction: CO(g) + Cl₂(g) ----> COCl₂(g)

From the equation above, I mole of CO gas reacts with 1 mole of Cl₂ gas to produce 1 mole of COCl₂ gas.

Since 1 mole of a substance contains 6.02 * 10²³ particles, on a particulate level, 6.02 * 10²³ particles of CO gas reacts with 6.02 * 10²³ particles of Cl₂ gas to produce 6.02 * 10²³ particles of COCl₂ gas.

Fishing trawlers in a certain bay catch a large variety of marketable fish along with a species of eel that is toxic. They normally kill the eels and throw them back into the sea. What term is used to refer to the eel? The eel species is called a(n) ______ of the fishing operation.

Answers

Answer:

1. Non-target

2. Bycatch

Explanation:

In the fishing industry, the main aim of the industry is to capture fishes that can be used or eaten and sell. A variety of fishes are captured for this purpose and since they are used therefore are known as Target catch.

But there are some species which has to be discarded because they are toxic and not useful. These non-useful species like eel which gets captured in the net while capturing other fishes are known as Non-target fish.

The eel fish which gets captured is known as bycatch fishes in the fishing operation.

Thus, Non-target and Bycatch are the correct answer.

Answer:

Its just bycatch

Explanation:

The eel species is called a(n) bycatch of the fishing operation.

Indicate whether the following represents a Chemical or Physical change: Milk sours

Answers

Answer:

Chemical Change

Explanation:

Physical change normally mean that the change can revert back to its orginal state, which in this case that is not possible therfore it is a chemical change.

Round off the following measurement to three significant digits: 29.950g

Answers

Answer:

30.0 g.

Explanation:

Hello,

In this case, for us to round the given number off to three significant figures, we firstly realize it has initially five significant figures. Thus, cutting at the third digit, which is the second nine, we will have 29.9 g, nonetheless, as a five is after such nine, we should round the nine to ten, so the result is 30.0 g.

Best regards.

If a pork roast must absorb 1500 kJkJ to fully cook, and if only 14% of the heat produced by the barbeque is actually absorbed by the roast, what mass of CO2CO2 is emitted into the atmosphere during the grilling of the pork roast?

Answers

You need to know the amount of heat generated by the combustion reaction.

Assuming propane as fuel, you can use thiis data:

C3H8(g)+5O2(g)---3CO2(g)+4H2O(g) ΔH= -2217 KJ

So when 3 moles of CO2 is emmitted 2217 kJ of heat is produced.

The molar wegiht of CO2 is 12 g/mol + 2 * 16 g/mol = 44 g/mol.

Then 3 mol * 44 g / mol = 132 g of CO2 are produced with 2217 kJ of heat.

Now you have to calculate how much energy you need to produce if only 12% is abosrbed by the pork

Energy absorbed by the pork = 12% *  total energy =>

total energy = energy absorbed by the pork / 0.12 = 1700 kJ / 0.12 = 14,166.67 kJ.

Now, state the proportion:

132 g CO2 / 2217 kJ = x / 14,166.7 kJ =>

x = 14,166.67 * 132 / 2217 = 843.48 g CO2.

Answer: 843 g of CO2

Which diagram represents this molecule?

Answers

Answer:

  C

Explanation:

The molecule has 8 carbon atoms joined by 7 C-C bonds.

The first two diagrams show 6 carbon atoms, not 8.

The last two diagrams show line segments representing C-C bonds. Only choice C shows 7 such segments.

The appropriate choice is C.

Answer:

C.

Explanation:

Which of the following reactions would be predicted by the activity series list
A. A metal ion reacts with another ion to form a precipitate.
B. A metal replaces a metallic ion below it on the list.
C. A metal replaces a metallic ion above it on the list.
D. A metal reacts with oxygen in a combustion reaction.

Answers

Answer:

The answer is B) A metal replaces a metallic ion below it on the list.

Explanation:

I just did it and got it correct, luckily I didn't use the other answer posted for this question.

A metal replaces a metallic ion below it on the list give reaction which would be predicted by the  activity series list.

So, option B is correct one.

What is Electrochemical series?

The list in which elements arranged in the increasing order of their electrode potential values is called Electrochemical series.

The Electrochemical series is also called activity series.

To learn more about  Electrochemical series here.

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A 25.0 mLsample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid in the sample was:_______,
A) 0.263
B) 0.365
C) 0.175
D) 1.83×10−4
E) 0.119

Answers

Answer:

0.263M of CH₃COOH is the concentration of the solution.

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.

In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:

0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.

As the sample of acetic acid had a volume of 25.0mL = 0.025L:

6.56x10⁻³ moles of CH₃COOH / 0.0250L =

0.263M of CH₃COOH is the concentration of the solution

A compound has an empirical formula of CHN. What is the molecular formula, if it’s molar mass is 135.13 g/mol? (C=12.01 amu, H=1.008 amu, N= 14.01)

Answers

Answer:

well the MF is 224.78 g/mol

Explanation:

just times them all by the molor mass and divide it by 3

A compound D with the molecular formula C6H12 is optically inactive but can be resolved into enantiomers. On catalytic hydrogenation, D is converted to E (C6H14) and E is optically inactive. Propose structures for D and E. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)

Answers

Answer:

D: CH2=CH-CH(CH3)-CH2-CH3 (R & S enantiomers)

E: CH3-CH2-(CH3)-CH2-CH3

(Please see the figures enclosed )

Explanation:

D is a racemic mixture (R & S) of 3-metyl-pent-1-ene, so it is optically inactive (although each of two enantiomers is optically active, the mixture is optically  inactive. The reason is that two enantiomers are present in an equal amount).

E is optically inactive, so its structure has to be symmetric.

At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=1.9×10−9 M?

Answers

Answer:

The correct answer will be "5.26 × 10⁻⁶".

Explanation:

The given values is:

[tex][H^{+}]=1.9\times 10^{-9} M[/tex]

As we know,

⇒  [tex]pH+pOH=14[/tex]

On taking log, we get

⇒  [tex]-log[H^{+}] + -log[OH^{-}] = 14[/tex]

Now,

Taking "log" as common, we get

⇒  [tex]log[H^{+}][OH^{-}]= -14[/tex]

⇒  [tex][H^{+}][OH^{-}]= 10^{-14}[/tex]

⇒  [tex][OH^{-}]=\frac{10^{-14}}{[H^{+}]}[/tex]

On putting the estimated value of "[tex][H^{+}][/tex]", we get

⇒             [tex]=\frac{10^{-14}}{1.9\times 10^{-9}}[/tex]

⇒             [tex]=5.26\times 10^{-6}[/tex]

given a k value of 0.43 for the following aqueous equilibrium suppose sample z is placed into water such that its original concentration is 0.033M assume there was zero initial concentration of either A(aq) or B(ag) once equilibrium has occured what will be the equilibrium concentration of z? K=0.43

Answers

Answer:

Less than 0.033 M:

[tex][Z]_{eq}=2.4x10^{-3}M[/tex]

Explanation:

Hello,

In this case, the described equilibrium is:

[tex]2A+B\rightarrow 2Z[/tex]

Thus, the law of mass action is:

[tex]K=\frac{[Z]^2}{[A]^2[B]}=0.43[/tex]

Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:

[tex]\frac{1}{K}=\frac{[A]^2[B]}{[Z]^2}=\frac{1}{0.43}=2.33[/tex]

Know, by introducing the change [tex]x[/tex] due to the reaction extent, we can write:

[tex]2.33=\frac{(2x)^2*x}{(0.033-2x)^2}[/tex]

Which has the following solution:

[tex]x_1=2.29M\\x_2=0.0181M\\x_3=0.0153M[/tex]

But the correct solution is [tex]x_3=0.0152M[/tex] since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:

[tex][Z]_{eq}=0.033M-2(0.0153M)[/tex]

[tex][Z]_{eq}=2.4x10^{-3}M[/tex]

Which is clearly less than 0.033 M since the addition of a product shift the reaction leftwards in order to reestablish equilibrium (Le Chatelier's principle).

Regards.

Now construct a different electrochemical cell. You put a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of the electric potential at the moment the reaction begins

Answers

Answer:

[tex]1.66~V[/tex]

Explanation:

We have to start with the half-reactions for both ions:

[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76

[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80

If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:

[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76

[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80

If we want to calculate ºE we have to add the two values, so:

ºE=0.76+0.80 = 1.56 V

Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:

[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]

On this case, Q is equal to:

[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]

Because the total reaction is:

[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]

So, the value of "Q" is:

[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]

Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:

[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]

I hope it helps!

A sample of chloroform, CHCl 3 , , was determined to have a molecular mass of 112.3g / (mol) . Its molecular mass is known to be 119.5g / (mol) . Calculate the absolute error and the percent error

Answers

Answer:

Explanation:

in your case ,

Meaured value = 112.3

actual value = 119.5

Absolute error= measured value - actual value

Percent error = [measured value - actual value  / actual value ] x 100

Hope this help you to find the answer

Determine the amount of heat (in kJ) associated with the production of 5.71 × 104 g of ammonia according to the following equation. N2(g) + 3H2(g) 2NH3ΔH°rxn = −92.6 kJ Assume that the reaction takes place under standard-state conditions at 25°C.

Answers

Answer:

[tex]Q=-3.11x10^5kJ[/tex]

Explanation:

Hello,

In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:

[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]

Best regards.

The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ

We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃

Mass of NH₃ = 5.71×10⁴ g

Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mole of NH₃ =?

Mole = mass / molar mass

Mole of NH₃ = 5.71×10⁴ / 17

Mole of NH₃ = 3358.82 moles

Finally, we shall determine the heat required to produce 3358.82 moles (i.e 5.71×10⁴ g) of NH₃. This can be obtained as follow:

N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ

Since reaction took place at standard conditions, it means:

1 moles of NH₃ required −92.6 kJ

Therefore,

3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ

Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ

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Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. HF (aq) + SO32- ⇌ F- + HSO3- Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ In this reaction: The formula for the conjugate _____ of HF is The formula for the conjugate _____ of SO32- is

Answers

Explanation:

A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.

On the other hand;

Bronsted-Lowry acid is the substance that donates the proton.

HF (aq) + SO32- ⇌ F- + HSO3-

In the forward reaction;

Bronsted-Lowry acid : HF

Bronsted-Lowry base: SO32-

In the backward reaction;

Bronsted-Lowry acid : HSO3-

Bronsted-Lowry base: F-

The conjugate base of HF is F-

The conjugate acid of SO32- is HSO3-

A solution of pentane and ethanol (CH3CH2OH)that is 50.% pentane by mass is boiling at 57.2°C. The vapor is collected and cooled until it condenses to form a new solution.
Calculate the percent by mass of pentane in the new solution. Here's some data you may need:

normal boiling point density vapor pressure at
57.2°C
pentane 36.°C 0.63gmL 1439.torr
ethanol 78.°C 0.79gmL 326.torr
Be sure your answer has 2 significant digits.
dont round during math only for answer!
Note for advanced students: you may assume the solution and vapor above it are ideal.

Answers

Answer:

The correct answer is 81.52 percent.

Explanation:

Based on the given information, the boiling point of pentane is 36 degree C and the boiling point of ethanol is 78 degree C. The density of pentane and ethanol is 0.63 g/ml and 0.79 g/ml. The vapor pressure of pentane at 57.2 degree C is 1439 torr and the vapor pressure of ethanol at 57.2 degree C is 326 torr.  

In the given case, 50 percent pentane by mass signifies that mass of pentane is 50 grams. Thus, the mass of ethanol will be 100-50 = 50 grams.  

The moles or n can be calculated by using the formula,  

n = weight/molecular mass

The molecular mass of pentane is 72.15 g per mol and the molar mass of ethanol is 46.07 g/mol.  

The moles of pentane is,  

= 50 g/72.15 g/mol = 0.6930 mol

The moles of ethanol is,  

= 50 g/46.07 g/mol = 1.0853 mol

The mole fraction of pentane is,  

= 0.6930 mol / (0.6930 + 1.0853) mol = 0.3897  

The mole fraction of ethanol is,  

= 1.0853 mol / (0.6930 + 1.0853) mol = 0.6103

Now the vapor pressure of solution will be,  

= pressure of pentane * mole fraction of pentane + pressure of ethanol * mole fraction of ethanol

= (1439 * 0.3897) + (326 * 0.6103)

= 759.736 torr

The vapor pressure of pentane within the solution,  

= vapor pressure of pentane * mole fraction of pentane

= 1439 torr * 0.3897

= 560.778 torr

The fraction of pentane is,  

= 560.778 / 759.736 = 0.738

Let us assume that the total mole is 1, the mole fraction of pentane is 0.738, so the mole fraction of ethanol will become, 1-0.738 = 0.262

The mass of pentane = 0.738 * 72.15 = 53.2467

The mass of ethanol = 0.262 * 46.07 = 12.07034

The percent by mass of pentane in new solution will be,  

Mass% = mass of pentane/Total mass * 100%

= 53.2467/(53.2467 + 12.07034) * 100%

= 53.2467/65.31704 * 100 %

= 81.52 %

Of the following three atoms, one tends to lose three electrons, another tends to lose two electrons, and another tends to lose one electron. Rank these atoms in order of the number of electrons they tend to lose, from most to fewest: potassium, K; calcium, Ca; gallium, Ga. Rank from most to fewest.

Answers

Answer: Ga > Ca > K

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 metals tend to get stable by losing electrons to attain noble gas configuration.

[tex]K:19:1s^22s^22p^63s^23p^64s^1[/tex]

[tex]K^+:18:1s^22s^22p^63s^23p^6[/tex]

[tex]Ca:20:1s^22s^22p^63s^23p^64s^2[/tex]

[tex]Ca^{2+}:18:1s^22s^22p^63s^23p^6[/tex]

[tex]Ga:31:1s^22s^22p^63s^23p^64s^23d^{10}4s^24p^1[/tex]

[tex]Ga^{3+}:28:1s^22s^22p^63s^23p^64s^23d^{10}[/tex]

Thus gallium (Ga) loses three electrons, Calcium (Ca) loses 2 electrons and Potassium (K) loses one electron.

What are extensive properties of Oxygen?

Answers

the extensive property of Oxygen is Volume, Enthalpy, Entropy.

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 5.00 mL of 0.0100 M HCl to 25.0 mL of this solution is __________ M.

a. 0.0980
b. 0.0817
c. 0.0167
d. 0.0850
e. 0.00253

Answers

Answer:

The answer is "Option b"

Explanation:

In this question first we calculates the moles in F-, HF, and in HCL, which can be defined as follows:

Formula:

[tex]\ Number \ of \ moles\ = \ Molarity \times \ Volume \ in \ litter[/tex]

[tex]\ moles \ in\ F- = 0.100 \ M \times 0.0250 L\\\\[/tex]

                     [tex]=\ 0.0025 \ moles[/tex]

[tex]\ moles \ in \ HF \ = 0.126M \times 0.0250 L[/tex]

                       [tex]= 0.00315 \ moles[/tex]

[tex]\ moles \ in \ HCl = 0.0100M \times 0.00500 L[/tex]

                       [tex]= 0.00005 \ moles[/tex]

[tex]\ Reaction: \\\\F - + H+ \rightarrow HF[/tex]

[tex]\Rightarrow \ moles \ in \ F- = 0.0025 \\\\\Rightarrow \ moles \ in \ H+ = 0.00005 \\\\ \Rightarrow \ moles \ in \ HF = 0.00315\\\\ \ total \ moles = 0.00250 -0.0000500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00315 + 0.00005\\\\\ total \ moles =0.00245 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00245[/tex]

[tex]\ total \ volume \ in \ the \ solution = \ V = \ 0.0300 L\\\\ after \ addition \ of \ HCl \ the \ concentration \ of \ F- \ = 0.00245\ moles \div V[/tex]

                                                                                [tex]=\frac{ 0.00245 \ moles }{0.0300L}\\\\= \frac{245 \times 10^4}{300 \times 10^5} \\\\= \frac{245}{3000} \\\\ = 0.0817 M[/tex]

What are plastic bottles made of?
Polyethylene

halogen

silicon

Alkyl groups

Answers

Answer:

polyethylenes

Explanation:

the plastic bottles used to hold potable water and other drinks are made from polyethylene because, the material is both strong and light.

hope this helped!

Answer: Polyethylenes

Explanation: I got 100% on the test :)

what is the balanced equation for calcium sulfate?​

Answers

Answer:

CaSO4

Explanation:

Calcium sulfate (or calcium sulphate) is the inorganic compound with the formula CaSO4 and related hydrates.

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