A gas mixture contains twice as many moles of o2 as n2. addition of 0.200 mol of argon to this mixture increases the pressure from 0.800 atm to 1.10 atm. how many moles of o2 are in the mixture?

Answers

Answer 1
number of moles of oxygen = x

number of moles of nitrogen = y

x = 2y

initial pressure, p1 = 0.8 atm

final pressure, p2 = 1.10 atm

At constant volume and temperature p1 / n1 = p2 / n2

=> p1 / p2 = n1 / n2

n1 = x + y = 2y + y = 3y

n2 = 0.2 + 3y

=> p1 / p2 = 3y / (0.2 + 3y)

=> 0.8 / 1.10 = 3y / (0.2 + 3y)

=> 0.8 (0.2 + 3y) = 1.10 (3y)

0.16 + 2.4y = 3.3y

=> 3.3y - 2.4y = 0.16

=> 0.9y = 0.16

=> y = 0.16 / 0.9

=. x = 2*0.16/0.9 = 0.356

Answer: 0.356 moles O2
Answer 2
Final answer:

The gas mixture before the addition of argon contained 1.78 moles of O2. This is calculated by first determining the combined moles of O2 and N2 using the pressure increase upon addition of argon and the information that the pressure due to moles of a gas is directly proportional to its mole count. Having the total moles, we then take the 2/3 share for O2 as stated in the problem.

Explanation:

We're dealing with a gas mixture where the total pressure of the mixture depends on the moles of each gas present. According to the ideal gas law, the total pressure exerted by the mixture is the sum of the partial pressures of each gas, with each partial pressure corresponding to the number of moles of that gas.

When the 0.200 mol of argon is added, the pressure of the gas mixture increases from 0.800 atm to 1.10 atm, a change of 0.30 atm. This change is due to the argon added, so it means that 0.200 mol of gas contributes to a pressure of 0.30 atm.

Given that 0.200 mol of argon contributes 0.30 atm pressure, and considering that initially the mixture had a pressure of 0.800 atm, we can infer that the total moles of oxygen and nitrogen before the argon addition was (0.800 atm ÷ 0.30 atm/mol) = 2.67 mol. Since the problem outlines that the gas mixture contains twice as many moles of O2 as N2, therefore the number of moles of O2 is 2/3 of the total original moles, which is (2/3 x 2.67 mol) = 1.78 mol O2.

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Related Questions

How many grams of caf2 would be needed to produce 1.29 moles of f2?

Answers

*Given: 
    1.29 moles F2

*Information needed in calculation
     CaF2 molar mass - 78.07 g/mol
     F2 molar mass - 38.00 g/mol

*Solution



Answer is: 100.72 grams of calcium fluoride.

Balanced chemical reactions:

1) CaF₂ + H₂SO₄ → 2HF + CaSO₄.

2) 2HF → F₂ + H₂.

1) From chemical reaction 2: n(F₂) : n(HF) = 1 : 2.

n(HF) = 2 · 1.29 mol.

n(HF) = 2.58 mol.

2) From chemical reaction 1: n(HF) : n(CaF₂) = 2 : 1.

n(CaF₂) = 2.58 mol ÷ 2.

n(CaF₂) = 1.29 mol; amount of substance.

m(CaF₂) = n(CaF₂) · M(CaF₂).

m(CaF₂) = 1.29 mol · 78.08 g/mol.

m(CaF₂) = 100.72 g.

If 26.35 ml of a standard 0.1650 m naoh solution is required to neutralize 35.00 ml of h2so4, what is the molarity of the acid solution?

Answers

Answer : The molarity of the [tex]H_2SO_4[/tex] is, 0.06211 M

Explanation :

Using neutralization law,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1[/tex] = basicity of an acid [tex](H_2SO_4)[/tex] = 2

[tex]n_2[/tex] = acidity of a base [tex](NaOH)[/tex] = 1

[tex]M_1[/tex] = concentration or molarity of [tex]H_2SO_4[/tex] = ?

[tex]M_2[/tex] = concentration of NaOH = 0.1650 M

[tex]V_1[/tex] = volume of [tex]H_2SO_4[/tex] = 35.00 mL

[tex]V_2[/tex] = volume of NaOH = 26.35 mL

Now put all the given values in the above law, we get the concentration of the [tex]H_2SO_4[/tex].

[tex]2\times M_1\times 35.00mL=1\times 0.1650M\times 26.35mL[/tex]

[tex]M_1=0.06211M[/tex]

Therefore, the molarity of the [tex]H_2SO_4[/tex] is, 0.06211 M

Final answer:

The molarity of the sulfuric acid solution is calculated using the volume and molarity of a sodium hydroxide solution used in titration. After determining the moles of NaOH and using the stoichiometric relationship, we find the moles of H₂SO₄ and divide by the volume of the acid solution to get the molarity, which is approximately 0.0621 M.

Explanation:Calculating the Molarity of an H₂SO₄ Solution

To determine the molarity of the sulfuric acid solution, we'll need to use the concept of titration and the stoichiometry of the reaction that occurs between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH). Here's the balanced chemical equation for the reaction:

H₂SO₄ (aq) + 2NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)

According to the equation, one mole of H₂SO₄ reacts with two moles of NaOH. First, we calculate the moles of NaOH used:


 Moles of NaOH = Volume of NaOH × Molarity of NaOH
 Moles of NaOH = 0.02635 L × 0.1650 M = 0.00434775 moles

Since the molar ratio of NaOH to H₂SO₄ is 2:1, we divide the moles of NaOH by 2 to get the moles of H₂SO₄:


 Moles of H₂SO₄ = Moles of NaOH / 2
 Moles of H₂SO₄ = 0.00434775 moles / 2 = 0.002173875 moles

Finally, we calculate the molarity of the H₂SO₄:


 Molarity of H₂SO₄ = Moles of H₂SO₄ / Volume of H₂SO₄ in liters
 Molarity of H₂SO₄ = 0.002173875 moles / 0.03500 L = 0.0621107 M

Therefore, the molarity of the acid solution is approximately 0.0621 M.

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there are 28L of soup in a pot. Marshall serves 400mL in each bowl. If he fills 16 bowls how much soup is left in the pot? in liters

Answers

Total in pot=28 L
400 mL in each bowl
16 bowls filled
1000mL=1L
16 bowls(400mL/1 bowl)=6400mL
6400mL(1L/1000mL)=6.4L
28L-6.4L=21.6 L

What maximum amount of ammonia in kilograms can be synthesized from 5.22 kg of h2 and 31.5 kg of n2? express your answer in kilograms to one decimal place?

Answers

Answer: Equation: N2(g) + 3H2(g) → 2NH3(g) Check for limiting reactant Molar mass N2 = 28g/mol mol N2 in 31,500g = 31,500/28 = 1,125 mol N2 This will require 1,125*3 = 3,375 mol H2 Molar mass H2 = 2g/mol Mol H2 in 5,220g = 5,220/2 = 2,610 mol H2 The H2 is limiting From the equation: 3 mol H2 will produce 2mol NH3 2610 mol H2 will produce 2610*2/3 = 1,740 mol NH3 Molar mass NH3 = 17g/mol Mass of 1740 mol NH3 = 1740*17 = 29,580g NH3 produced Mass of NH3 produced = 29.6kg

In process in which An animal's cells use oxygen and digested food molecules to release the energy in food is called

Answers

In process in which An animal's cells use oxygen and digested food molecules to release the energy in food is called

cellular respiration
it's known as cellular respiration

Draw the lewis dot structure for se2−. to change the symbol of an atom, double-click on the atom and enter the letter of the new atom. show the formal charge of the atom

Answers

First of all, Se has 6 valence electrons, so its Lewis structure has 6 dots.

Symbolize the valence electrons as *, then the Lewis struecture for Se is:

     **
** Se **

Now, Se 2-, means Se atom has two more electrons, which means that the atom completes the octete rule (eight electrons in the outer shell).

This is the structe fo Se 2-

     **
** Se **   draw big square bracketts around that structre and write 2- at the upper right corner as a superscript.
     **

To begin with the Lewis dot structure,

Se possesses six valence electrons, resulting in a Lewis structure with six dots.Now, Se 2- denotes that the Se atom has two additional electrons, completing the octet rule (eight electrons in the outer shell).Make a huge square bracket around that structure and write 2- as a superscript in the upper right corner.

[tex]**\\ **Se^{2-} **\\ **[/tex]

Thus, the structure has 8 electrons represented by *.

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draw a transition state for the reaction between ethyl iodide and sodium acetate

Answers

The reaction between ethyl iodide and sodium acetate is a substitution reaction, likely involving an SN2 (nucleophilic substitution bimolecular) mechanism.

In the transition state for an SN2 reaction, the nucleophile (acetate ion) is attacking the substrate (ethyl iodide) from the backside, leading to inversion of configuration. The reaction between ethyl iodide and sodium acetate is a substitution reaction, with an SN2 (nucleophilic substitution bimolecular) mechanism most likely at work.

In the transition state:

The leaving group (iodide in this case) is partially leaving.

The nucleophile (acetate ion) is partially bonded to the carbon, approaching from the backside.

The carbon undergoing substitution is in a tetrahedral arrangement with the nucleophile, the leaving group, and two other substituents.

In comparison to molecules that interact by London dispersion forces only, the melting point of similar-sized molecules forming hydrogen bonds would most likely be
about the same.
unpredictable.
lower.
higher.

Answers

The answer is higher.

You should know that the London dispersion force is the weakest intermolercular attractive force while the hydrogen bonds form a high attractive intermolecular force, so you have to expect that in molecules forming hydrogen bonds the molecules will be more attached to each other than in similar-sized molecules that interact by London dispersion forces only, which takes you to conclude that the melting point of the molecules forming hydrogen bonds will be higher.

Calculate and compare the [h+] and ph for a 0.100 m solution of hclo4 and a 0.100 m solution of hclo (ka = 2.9 × 10–8).

Answers

[H+] for HClO(this is a weak acid so it requires an I.C.E. table to solve):


(I=initial amount. C=change in amount. E= amount at equilibrium.)
I.C.E. table
HClO. H2O. >>> H3O+ ClO4-
I. 0.100M. N/A. 0. 0
C. -x. N/A. +x. +x
E. 0.1-x. N/A. x. x.
(we don't consider water because it is not an aqueous solution)

ka=(x^2)/(0.1-x)
(2.9×10^-8)= (x^2)/(0.1-x)
(0.1-x)(2.9×10^-8) = x^2
(3.0×10^-9) - ((2.9×10^-8)x) = x^2

x^2+(2.9×10^-8)x-(3×10^-9)

a=1
b=(2.9×10^-8)
c= -(3×10^-9)

plug those values into the quadratic formula:

x= (-b +(√((b^2)-4ac))))/2a

I got x= 0.000053837

x= [H+] =0.000053837



[H+] for 0.100 M HClO4:
because HClO4 is a strong acid, it dissociates completely. Meaning that it's [H+] =0.1

HClO₄ has a higher [H⁺] than HClO (0.100 M vs. 5.4 × 10⁻⁵ M).

HClO₄ has a lower pH than HClO (1.00 vs. 4.3).

Classification of acids according to their strengthWeak acids: dissociate partially in water.Strong acids: dissociate completely in water.

HClO₄ is a strong acid. Thus, of HClO₄ is 0.100 M, H⁺ will be 0.100 M as well. We can use this value to calculate the pH for this acid.

pH = -log [H⁺] = -log 0.100 = 1.00

HClO is a weak acid. Thus, [H⁺] ≠ [HClO]. Given the acid dissociation constant (Ka) and the concentration of the acid (Ca), we can calculate [H⁺]  and pH using the following expressions.

[tex][H^{+} ] = \sqrt{Ca \times Ka } = \sqrt{0.100 \times (2.9 \times 10^{-8} ) } = 5.4 \times 10^{-5} \\\\pH = -log 5.4 \times 10^{-5} = 4.3[/tex]

HClO₄ has a higher [H⁺] than HClO (0.100 M vs. 5.4 × 10⁻⁵ M).

HClO₄ has a lower pH than HClO (1.00 vs. 4.3).

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2.5 million atoms of a particular element have a mass of 8.33 x 10-16 grams. what is this element

Answers

To determine the element, the mass in grams is converted to atomic mass units using the known mass of an atomic mass unit. The calculated atomic mass matches the approximate atomic masses of elements like Neon or Calcium. However, a precise identification may require isotopic composition.

The student is asking about the identity of an element based on a given mass and number of atoms. To find the answer, we use the concept of atomic mass units (u) and Avogadro's number. The mass of a single atomic mass unit is 1.661 × 10-24 grams. With 2.5 million atoms having a mass of 8.33 × 10-16 grams, we can calculate the average atomic mass of an individual atom.

First, we divide the total mass by the number of atoms:

8.33 × 10-16 g / 2.5 million atoms = 3.332 × 10-22 g/atom.

Next, we convert this mass into atomic mass units by dividing by the mass of one atomic mass unit:

3.332 × 10-22 g/atom / 1.661 × 10-24 g/u = 20.04 u/atom.

This calculated value can be compared to the atomic mass or atomic weight of elements listed in the periodic table to identify the element. The mass is approximately 20 u, which suggests the element could be Neon (Ne) with an atomic mass of approximately 20.18 u or Calcium (Ca) with an atomic mass of 40.08 u considering the natural abundance of isotopes. For a more precise identification, additional information such as isotopic composition would be needed.

Calculate the frequency of the n = 6 line in the Lyman series of hydrogen.

Answers

Lyman Series Working Formula:

1/λ = RH (1-(1/n^2))

Given:

n = 6
RH = Rydberg's constant = 1.0968x10^7 m^-1
c = speed of light = 3x10^8 m/s

Required: 


Frequency (Hertz or cycles per second)

Solution:

To solve for the wavelength λ, we substitute the given in the working formula
1/λ = RH (1-(1/n^2))
1/λ = 1.0968x10^7 m^-1 (1-(1/6^2))
λ = 0.0000000938 m or 93.8 nm 

To get the frequency, we will use the formula below. 
f = c/
λ

We then substitute c or the speed of light,
= (3x10^8 m/s) / 0.0000000938 m 

Therefore,
= 3.2x10^15 s^-1

ANSWER: Frequency = 3.2x10^15 s^-1

The frequency of the n = 6 line in the Lyman series of hydrogen is [tex]\boxed{{\mathbf{3}}{\mathbf{.196 \times 10 }}{{\mathbf{s}}^{{\mathbf{ - 1}}}}}[/tex].

Further Explanation:

The given problem is based on the concept of the emission spectrum of a hydrogen atom.

Lyman series:

When an electron undergoes transition from any higher energy orbit [tex]\left({{{\text{n}}_{\text{f}}}=2,{\text{ }}3,{\text{ }}4,...}\right)[/tex] to first energy orbit [tex]\left({{{\text{n}}_{\text{i}}} = 1}\right)[/tex] then it emits the energy to complete the process. This spectral lines formed due to this emission is known as the Lyman series of hydrogen atom.

The formula to calculate the energy of transition in the hydrogen atom is,

[tex]\Delta E={R_{\text{H}}}\left({\frac{1}{{{{\left({{{\text{n}}_{\text{i}}}}\right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}}\right)}^2}}}}\right)[/tex]

Where,

[tex]\Delta E[/tex] is the energy difference between two energy levels.

[tex]{R_{\text{H}}}[/tex] is a Rydberg constant and its value is [tex]2.179 \times {10^{ - 18}}{\text{ J}}[/tex].

[tex]{{\text{n}}_{\text{i}}}[/tex] is the initial energy level of transition.

[tex]{{\text{n}}_{\text{f}}}[/tex] is the final energy level of transition.

Substitute the 1 for [tex]{{\text{n}}_{\text{i}}}[/tex] , 6 for [tex]{{\text{n}}_{\text{f}}}[/tex] and [tex]2.179\times {10^{ - 18}}{\text{ J}}[/tex] for [tex]{{\text{R}}_{\text{H}}}[/tex] in the above formula to calculate the value of energy of the given transition.

[tex]\begin{aligned}\Delta E &={R_{\text{H}}}\left({\frac{1}{{{{\left( {{{\text{n}}_{\text{i}}}}\right)}^2}}} \frac{1}{{{{\left({{{\text{n}}_{\text{f}}}}\right)}^2}}}}\right)\\&=\left({2.179 \times {{10}^{ - 18}}{\text{ J}}} \right)\left({\frac{1}{{{{\left( {\text{1}}\right)}^2}}}-\frac{1}{{{{\left({\text{6}} \right)}^2}}}}\right)\\&={\mathbf{2}}{\mathbf{.118 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 18}}}}{\mathbf{ J}}\\\end{aligned}[/tex]

Now we can calculate the frequency of the given transition by using the following formula.

[tex]\Delta E = hv[/tex]

Where,

[tex]\Delta E[/tex] is the energy difference between two energy levels.

[tex]h[/tex] is a Plank’s constant and its value is [tex]6.626\times {10^{ - 34}}{\text{ Js}}[/tex].

v is a frequency of the transition.

Rearrange the above formula to calculate the frequency of the given transition and substitute the value of [tex]\Delta E[/tex].

[tex]\begin{aligned}\Delta E&=hv\hfill\\v&=\frac{{\Delta E}}{h}\hfill\\&=\frac{{{\text{2}}{\text{.118}} \times {\text{1}}{{\text{0}}^{ - {\text{18}}}}{\text{ J}}}}{{6.626 \times {{10}^{ - 34}}{\text{ Js}}}} \hfill\\&={\mathbf{3}}{\mathbf{.196 \times 10 }}{{\mathbf{s}}^{{\mathbf{ - 1}}}} \hfill \\\end{aligned}[/tex]

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of atom

Keywords: Lyman series of hydrogen, hydrogen, n=6 transition, emission spectra, Plank’s constant, transition, energy orbits.

Will bromine react with sodium? Explain your answer.

Answers

Sodium metal reacts vigorously with all the halogens to form sodium halides. So, it reacts with fluorine, then forms form respectively sodium
Hope this helped :)

Which of the following tasks should be performed before measuring the mass of any object on a triple-beam balance? A. Make sure that all the counterweights are set to zero. B. Make sure that the balance is on a level surface. C. Make sure that the balance pan is clean and free of debris. D. all of these

Answers

B. make sure that the balance is on level surface

its all of these trust me

Which unit abbreviation is a measurement of force?



m/s



m/s2



N



N/s

Answers

So,

Recall Newton's Second Law of Motion: force = mass x acceleration

Also recall:

acceleration = velocity/time
velocity = distance/time

Distance is expressed in meters.
Time is expressed in seconds.
Mass is expressed in kilograms.

v = m/s
a = m/s^2
F = m * a
F = kg * m/s^2

[tex]F= \frac{kg*m}{s^2} [/tex]

The unit for force is the Newton (N).

Answer:

Newton is the unit of force, abbreviated as "N"

Explanation:

Let us understand the given units:

a) m/s: it is meter per second. It is SI unit of speed or velocity.

b) m/s: it is meter per second square. It is SI unit of acceleration (velocity per second).

c) N : it is unit of force. It stands for Newton.It is equal to 1 kilogram meter per second squared.

d) N/s: it is newton per second. It is unit of momentum.

PLEASE HELP, 30 POINTS!

Suppose 1 kg each of water (4.19 J/g⋅∘C), brick (0.90 J/g⋅∘C), iron (0.46 J/g⋅∘C), and plastic (1.01 J/g⋅∘C) were held at the same initial temperature and heated for an equivalent amount of time. Indicate the relative final temperatures by ordering from lowest to highest resulting temperature (if using the Intro tab of the PhET to help visualize the temperature changes, assume the water iron, and plastic are 1 kg in mass, and the brick has a mass of 0.5 kg). Assume no heat is lost to the surroundings.
Rank from lowest resulting temperature to highest resulting temperature. To rank items as equivalent, ove

Answers

Scenario 1:

Mass of  each of water, brick, iron, and plastic = 1 kg = 1000 g

Specific heat of water = 4.19 J/g⋅∘C

To raise the temperature of  1 g of water by 1∘C the heat required is 4.19 J

Therefore, to raise the temperature of  1000 g of water by 1∘C the heat required is 4.19  x 1000 = 4190 J

Specific heat of brick =  0.90 J/g⋅∘C

To raise the temperature of  1 g of brick by 1∘C the heat required is 0.9J

Therefore, to raise the temperature of 1000 g of brick by 1∘C the heat required is 0.90  x 1000 = 900 J

Specific heat of iron = 0.46 J/g⋅∘C

To raise the temperature of  1 g of iron by 1∘C the heat required is 0.46 J

Therefore, to raise the temperature of  1000 g of iron by 1∘C the heat required is 0.46  x 1000 = 460 J

Specific heat of plastic = 1.01 J/g⋅∘C

To raise the temperature of  1 g of plastic by 1∘C the heat required is 1.01 J

Therefore, to raise the temperature of  1000 g of plastic by 1∘C the heat required is 1.01  x 1000 = 1010 J

So it takes the greatest amount of heat in Joules to raise the temperature of water, followed by plastic, followed by brick and iron.

Now all of them are heated for an equivalent amount of time as a result water will have the lowest resulting or final temperature. The resulting or final temperature of plastic will be greater than water. The resulting or final temperature of brick will be greater than plastic. . The resulting or final temperature of iron will be greater than brick.

Water                   plastic    brick    iron

(lowest resulting temperature) -->  (highest resulting temperature)



How does most of the water in the water cycle move from lakes and rivers directly back into the atmosphere?

Answers

It goes from lakes and as the temp rises, it evaporates into clouds. from there, the clouds get heavy, and it produces rain, and the rain forms rivers and then back into lakes.
It happens through precipitation when water evaporates from the suns rays and rises into the atmosphere

How many liters of water are required to dissolve 1.00 g of barium chromate? express your answer in liters to three significant figures?

Answers

Assuming the water is present at 20 °C, the solubility of barium chromate is 0.2275 mg/100 mL

Converting this to mg/L, we get:

2.275 mg/L

Converting this value again, to grams per liter, we get:

0.002275 g/L

To dissolve one gram, we require:

1/0.002275 = 439.56044

440 liters of water.

The solubility  Barium chromate in grams per liter = 2.78 . 10⁻³ grams/L

359 liters of water are required to dissolve 1.00 g of Barium chromate

Further explanation

Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility

1. Temperature: 2. Surface area: 3. Solvent type: 4. Stirring process:

Ksp is an ion product in equilibrium

Solubility relationships and solubility constants (Ksp) of the AxBa solution can be stated as follows.

AₓBₐ (s) ← ⎯⎯⎯⎯ → x Aᵃ⁺ (aq) + a Bˣ⁻ (aq)

s                             xs               as

Ksp = [Aᵃ⁺] ˣ [Bˣ⁻] ᵃ

Ksp = (xs) ˣ (as) ᵃ

Solubility units in the form of mol / liter or gram / liter

At 25.°C, the molar solubility of Barium chromate  BaCrO₄ in water is 1.10. 10⁻⁵M.

to change units to grams / liter, we multiply by molar mass:

M BaCrO₄ = Ba + Cr + 4. Ar O

M BaCrO₄ = 137 + 52 + 4.16

M BaCrO₄ = 253

So the solubility is in grams / liter

= 1.10 . 10⁻⁵ mol / liter x 253 grams / mol

=  278.3 .10⁻⁵ = 2.78 . 10⁻³ grams/L

(3 significant numbers, 2.7 and 8)

If we dissolve 1 gram of Barium chromate into the solution, we need water :

= 1 grams / 2.78 . 10⁻³ grams / liter

= 359 liters

(3 significant numbers, 3.5 and 9)

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Keywords: solubility, silver chromate, a significant number

The name of the compound whose formula is aucl3 is

Answers

Gold III Or Gold trichloride |

Part b an element belongs to group 9. what common name might be used to describe the group to which this element belongs?

Answers

Group 9 of the periodic table is comprised of cobalt (Co), Rhodium (Rh), Iridium (Ir), and one which is still chemically uncharacterized Meitnerium (Mt). These elements are all metals. These metals are termed Transition metals. 

These metals are in the d-block because the last elements are in the d-orbitals. 

Group 9 of the Periodic Table does not have a widely recognized common name but is part of the broader category known as transition elements or transition metals.

An element that belongs to Group 9 is part of the transition metals on the Periodic Table. Unfortunately, there isn't a widely used common name for Group 9 specifically, unlike other groups with well-known common names such as the alkali metals for Group 1 or the halogens for Group 17. However, elements in Group 9 are part of the larger family known as the transition elements, which share similar properties such as conducting electricity and heat, possessing a high density, and having a high melting and boiling points.

A student determines the acetic acid concentration of a sample of distilled vinegar by titration of 25.00 ml of the vinegar with standardized sodium hydroxide solution using phenolphthalein as an indicator. which error will give an acetic acid content for the vinegar that is too low

Answers

Inaccuracy of the determined value for the acetic acid concentration of the sample can arise in many mishandling and error sources. One of the sources of error is the inaccuracy in the reading of through the glasswares used such as burette. 

Inaccuracy may also arise from the too much addition of sodium hydroxide to the solution, this means that endpoint had already been reached, however, the standard is still added. 
Final answer:

Adding a too small volume of sodium hydroxide solution during titration will give an acetic acid content for the vinegar that is too low.

Explanation:

The error that will give an acetic acid content for the vinegar that is too low is adding a too small volume of sodium hydroxide solution during titration. In this case, the student titrated 25.00 ml of the vinegar, and if the volume of sodium hydroxide solution added is insufficient, it will not neutralize all the acetic acid present in the vinegar. This will result in a lower calculated concentration of acetic acid in the vinegar.

Name two elements in which the last electrons to be added are placed into s subshells

Answers

Lithium and Sodium both have their last elections added to s subshells. All alkali and alkali earth metals have their last electrons placed in s subshells.

Dinosaur fossils are often dated by using an element other than carbon, like potassium-40, that has a longer half life (in this case, approximately 1.25 billion years). suppose the minimum detectable amount is 0.1% and a dinosaur is dated with 40k to be 67 million years old. is this possible?

Answers

The equation relating radioactive decay is:

N(i) = N(o) * exp(-0.693*t / t(half))

The fraction:
N(i)/N(o) represents the fraction of original substance present after t years and a half-life of t(half). Substituting the values,

N(i)/N(o) = exp((-0.693 * 67)/(1,250))

N(i)/N(o) = 0.964 or 96.4%

This means that the amount is far above the minimum detectable amount, so this method is feasible.

Using enthalpies of formation, calculate the quantity of heat produced when 14.0 g of propane is completely combusted in air under standard conditions. Assume that liquid water is forming

Answers

The balanced chemical reaction here is:

C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(l)

 

Calculating for the molar change in enthalpy:

Enthalpy change = 3(-393.5) + 4(-285.83) - (-103.85) = -2220.0 kJ/mole 

 

Now, propane has a molar mass of 44g/mol. So, for us to calculate the heat released by burning 14 grams of propane, we have: 
-2220 kJ/mole * ( 14g / 44g/mol) = -706.36 kJ 

 

Answer:

-706.36 kJ 

(negative means heat is released)

Heating oxygen difluoride gas yields oxygen gas and fluorine gas? Whats the balance chemical equation for this?

Answers

2OF2(g) —> O2(g) + 2F2(g) hope that helps

Heating oxygen difluoride gas yields oxygen gas and fluorine gas

2OF2(g) —> O2(g) + 2F2(g)

what are the properties of oxygen ?

Oxygen atom belongs to group of chalcogen represented in  the periodic table which is one of the abundant and essential element and participate in combustion reaction abundantly found in the Earth’s crust.

oxygen is colorless, odorless whereas liquid oxygen is paramagnetic, forms oxides with element except helium, neon, krypton, and argon.

Dioxygen form of oxygen is allotropic where as Trioxygen is the most reactive allotrope of oxygen which can damage to lung tissue and it is called as ozone.

These oxygen are used in the production and manufacturing of glass and stone products, it also be used in the process of melting, refining, and manufacturing of steel along with other metals.

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The statement "matter can be neither created nor destroyed by chemical means, but it can be changed from one form to another" is the chemical law of

Answers

The law of the conservation of matter.

It the chemical law of conservation of matter


How many moles of oxygen are required to produce 37.15 g CO2?

Answers

Carbon = 12.010. Oxygen = 15.999 x 2 15.999 x 2 = 31.998 + 12.010 = 44.008 \frac{37.15 grams * 1 mole CO2}{44.008 grams}
1.407.........................................................

When a hammer strikes a compound formed by covalent bonds, what will most likely happen to the compound?

It will break into many pieces
It will reform into a new shape
It will spread out and then return to it's original shape
It will stay solid and resist the force of the hammer

Answers

Answer: Option (a) is the correct answer.

Explanation:

A compound formed by sharing of electrons is known as a covalent compound.

A covalent compound has weak intermolecular forces due to which it is brittle in nature. Hence, when we hit hammer on a covalent compound then it breaks easily.

Thus, we can conclude that when a hammer strikes a compound formed by covalent bonds, then most likely it will break into many pieces.

Answer:

option A.

Explanation:

have a good day.

:)

Calculate the ph for a 0.3 m solution of (a) sodium leucinate

Answers

To get the pH value for a certain aqueous solution, the concentration of the hydronium ions (in moles per liter) should be know. This concentration is also known as the "molarity".
pH = -log [H3O+]

We are given that the molarity = 0.3 m
Therefore,
pH = -log[0.3] = 0.5228787

Manganese(iv) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3mno2+4al→3mn+2al2o3part awhat mass of al is required to completely react with 30.0 g mno2?

Answers

Final answer:

To completely react with 30.0 g of MnO2, 107.92 g of aluminum is required.

Explanation:

To determine the mass of aluminum required to completely react with 30.0 g of MnO2, we need to use the balanced chemical equation and the molar mass of MnO2. In the balanced equation, the coefficient of MnO2 is 3, which means that 3 moles of MnO2 react with 4 moles of Al.

First, calculate the molar mass of MnO2:

Molar mass of Mn = 54.94 g/molMolar mass of O = 16.00 g/molMolar mass of MnO2 = (54.94 g/mol) + 2(16.00 g/mol) = 86.94 g/mol

Next, convert 30.0 g of MnO2 to moles:

Moles of MnO2 = Mass of MnO2 / Molar mass of MnO2 = 30.0 g / 86.94 g/mol = 0.344 moles

Using the mole ratio from the balanced equation, we can calculate the moles of Al required:

Moles of Al = (3/3) x 4 moles of Al = 4 moles of Al

Finally, convert moles of Al to mass:

Mass of Al = Moles of Al x Molar mass of Al = 4 moles x 26.98 g/mol = 107.92 g

The mass of aluminum required to completely react with 30.0 g of MnO2 is 107.92 g.

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Final answer:

To completely react with 30.0 g of MnO2, 12.4 g of Al is required.

Explanation:

To determine the mass of aluminum required to completely react with 30.0 g of MnO2, we need to use the balanced equation for the reaction:

3MnO2 + 4Al -> 3Mn + 2Al2O3

From the equation, we can see that the mole ratio of MnO2 to Al is 3:4. We can use this ratio to calculate the mass of Al.

First, convert the mass of MnO2 to moles using its molar mass. Then, use the mole ratio to find the moles of Al. Finally, convert the moles of Al back to mass using its molar mass.

Let's calculate:

Convert the mass of MnO2 to moles: (30.0 g MnO2) / (86.94 g/mol MnO2) = 0.345 mol MnO2Using the mole ratio, calculate the moles of Al: 0.345 mol MnO2 * (4 mol Al / 3 mol MnO2) = 0.460 mol AlConvert the moles of Al to mass: 0.460 mol Al * (26.98 g/mol Al) = 12.4 g Al

Therefore, 12.4 grams of Al are required to completely react with 30.0 grams of MnO2.

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A 108 49in source emits a 633-kev gamma photon and a 606-kev internal-conversion electron from the k shell. what is the binding energy of the electron in the k shell?

Answers

Binding energy is the energy needed to emit the electron from the shell. Using the formula below to compute for BE. Binding Energy BE = Energy of photon - Kinetic energy electron
where 
Energy proton= 633 keV
KE electron = 606 keV
Binding energy BE = 27 keVThe binding energy of the k subshell is equal to 27 keV.

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