according to the bohr model of an atom, what happens when an electron moves from the second energy level to the third energy level and then back to the second energy level?​

Answer: The empirical formula is [tex]C_3H_3O_2[/tex].

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2[/tex] = 0.8635

Mass of [tex]H_2O[/tex]= 0.1767 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.8635 g of carbon dioxide, =[tex]\frac{12}{44}\times 0.8635=0.2355g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1767 g of water, =[tex]\frac{2}{18}\times 0.1767=0.0196g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (0.4647) - (0.2355+0.0196) = 0.2096 g

Mass of C = 0.2355 g

Mass of H = 0.0196 g

Step 1 : convert given masses into moles.

Moles of C =[tex] \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{0.2355g}{12g/mole}=0.0196moles[/tex]

Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.0196g}{1g/mole}=0.0196moles[/tex]

Moles of O=[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.2096g}{16g/mole}=0.0131moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =[tex]\frac{0.0196}{0.0131}=1.5[/tex]

For H =[tex]\frac{0.0196}{0.0131}=1.5[/tex]

For O = [tex]\frac{0.0131}{0.0131}=1[/tex]

The ratio of C : H: O = 1.5 : 1 .5 : 1

Converting to simple whole number ratio:

The ratio of C : H: O = 3 : 3: 2

Hence the empirical formula is [tex]C_3H_3O_2[/tex].

The original concentration : 800%

Given

Diluted 1/40

Final concentration 20%

Required

The original concentration

Solution

Dilution is the process of adding solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

 n₁ = n₂

M₁.V₁ = M₂.V₂

diluted 1/40(dilution factor)⇒ V₁/V₂=1/40⇒V₂/V₁=40

M₂ = 0.2(20%)

Input the value for M₁ :

M₁=M₂ x (V₂/V₁)

M₁ = 0.2 x 40

M₂ = 8(800%)

=

Answer:

0.768g of Fe are in the tablet = 38.4% by mass of Fe in the tablet

Explanation:

The balanced redox equation of Fe²⁺ with KMnO₄ in acidic medium is:

5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Where 5 moles of Fe²⁺ react per mole of KMnO₄

The moles required of KMnO₄ are:

27.50cm³ * (1dm³ / 1000cm³) = 0.0275dm³ * (0.100mol / dm³) = 0.00275mol KMnO₄

Moles of Fe²⁺ are:

0.00275mol KMnO₄ * (5 moles Fe²⁺ / 1mol KMnO₄) = 0.01375moles of Fe²⁺

In grams (Molar mass Fe: 55.845g/mol):

0.01375moles of Fe²⁺ * (55.845g/mol) =

And percentage is:

0.768g Fe / 2.000g * 100 =

Answer:

Hey

Explanation:

The normal range for urine specific gravity is 1.005 to 1.030.

Answer: Vibråtory movement.

Explanation: when particles bounce against each other the friction creates thermal energy. Think about what happens when you rub your hands together and they get warmer, that the friction between your hands making thermal energy.

Answer:

19.47%

Explanation:

[tex]\frac{1.29}{1.29+5.35}[/tex]

Please recheck, for this answer may not be correct

4 in hydrogen 4 in oxygen

Explanation:

hope this helps

Answer:

A. Theoretical yield of C is 9.03 g

B. Mass of B consumed is 5.53 g

Explanation:

A. Determination of the theoretical yield of C.

Actual yield of C = 6.5 g

Percentage yield of C = 72.%

Theoretical yield of C =?

Percentage yield = Actual yield /Theoretical yield × 100

72% = 6.5 / Theoretical yield

72 / 100 = 6.5 / Theoretical yield

Cross multiply

72 × Theoretical yield = 100 × 6.5

72 × Theoretical yield = 650

Divide both side by 72

Theoretical yield = 650 / 72

Theoretical yield = 9.03 g

Therefore, the theoretical yield of C is 9.03 g

B. Determination of the mass of B consumed.

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

A + B —> C

Mass of A = 3.5 g

Mass of C = 9.03 g

Mass of B =?

A + B = C

3.5 + B = 9.03

Collect like terms

B = 9.03 – 3.5

B = 5.53 g

Thus, the mass of B consumed in the reaction is 5.53 g

Answer:

a solution color becoming less intense  due to dilution- is not an evidence of a chemical reaction

bubbles (gas formation) - evidence of a chemical reaction

explosion or fire - evidence of a chemical reaction

changes in color- evidence of a chemical reaction

precipitation- evidence of a chemical reaction

changes in temperature - evidence of a chemical reaction

a solid liquifying - is not an evidence of a chemical reaction

solution colors mixing - is not an evidence of a chemical reaction

Explanation:

A chemical change is not easily reversible and yields new substances. It is often accompanied by a loss or gain of heat.

In the answer section, i have shown some evidences that lead us to conclude that a chemical reaction has taken place. The occurrence of a chemical change often goes with the formation of new substances as earlier stated and any of these signs may accompany the process.

For instance, when a metal is dropped in dilute acid solution, bubble of hydrogen gas indicates that a chemical reaction has taken place.

Answer:

Incomplete combustion produces carbon monoxide (CO), a toxic gas.

Explanation:

Combustion is the rapid chemical reaction of oxygen in the air or direct oxygen, which is defined as oxidizer, with the different elements that make up the fuel (mainly carbon (C) and hydrogen (H)). These chemical reactions release energy producing local increases in temperature, which causes a flow of heat to the outside.

In other words, on the one hand you have the fuel, which is the substance that oxidizes and that you can find in a solid, liquid or gaseous state and which is made up mostly of carbon and hydrogen.

The oxidizer, for its part, is the substance that oxidizes the fuel, and it is generally the oxygen found in the air.

Certain amounts of oxidizer and fuel must come into contact for combustion to occur.

Incomplete combustion occurs when part of the fuel does not fully react. Incomplete combustion produces carbon monoxide (CO), a toxic gas.

Answer:

One electron per single hydrogen atom.

Explanation:

Hello!

In this case, given the reaction:

[tex]H_2+O_2\rightarrow H_2O[/tex]

Whereas we can identify the following half-reaction for hydrogen:

[tex]H_2^0\rightarrow H_2^+[/tex]

Whereas we see that each single hydrogen atom gains one electron in order to go from 0 to +1, which is also related to an oxidation half-reaction.

Best regards!

2.51 mol Cu

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Step 1: Define

1.51 × 10²⁴ atoms Cu

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

2.50747 mol Cu ≈ 2.51 mol Cu

Answer:

2.343 g of HEPES sodium salt is needed.

Explanation:

So, from the question above we have the following parameters which is going to help in solving this particular question.

The number of moles = 90mM, the pH = 7.50, FW for the HEPES sodium salt = 260.31, and the volume of HEPES sodium salt required = 100mL.

Therefore, the amount of HEPES sodium salt required to make the buffer = 260.31 × 9 × 0.001 = 2.343 grams of HEPES sodium salt is needed in 10mL.

Answer:

3,4,5,6,7,8

Explanation:

Groups 13  - 18 are the main group on the periodic table

The group number corresponds the number of valence electrons they contains.

Valence electrons are the number of electrons in the outermost shell of an atom.

So group 13 - 18 corresponds to a valency of 3,4,5,6,7,8.

Groups with a valency of 8 are the noble gases.

Answer:

C = 18.29 g  

Explanation:

Given data:

Mass of beryllium needed = ?

Mass of nitrogen = 18.9 g

Solution:

Chemical equation:

3Be + N₂    →    Be₃N₂

now we will calculate the number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 18.9 g/ 28 g/mol

Number of moles = 0.675 mol

Now we will compare the moles of nitrogen and Be from balance chemical equation.

                     N₂        :       Be        

                       1          :       3

                  0.675       :      3/1×0.675 = 2.03 mol

Number of moles of Be needed are 2.03 mol.

Mass of Beryllium:

Mass = number of moles × molar mass

Mass = 2.03 mol ×   9.01 g/mol

Mass = 18.29 g  

Answer:

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Explanation:

)??

Answer: a) [tex]1.7\times 10^{-4}[/tex]

b) [tex]3.4\times 10^{-4}[/tex]

Explanation:

The reaction is :

[tex]2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)[/tex]

Rate = Rate of disappearance of [tex]N_2O_5[/tex] = Rate of appearance of [tex]NO_2[/tex]

Rate =  [tex]-\frac{d[N_2O_5]}{2dt}[/tex] = [tex]\frac{d[NO_2]}{4dt}[/tex]

Rate of disappearance of [tex]N_2O_5[/tex] = [tex]\frac{\text {change in concentration}}{time}[/tex] = [tex]\frac{0.100-0.066}{200-0}=1.7\times 10^{-4}[/tex]

a) Rate of disappearance of [tex]N_2O_5[/tex] = [tex]-\frac{d[N_2O_5]}{2dt}[/tex]

Rate of appearance of [tex]NO_2[/tex] = [tex]\frac{d[NO_2]}{4dt}[/tex]

b) Rate of appearance of [tex]NO_2[/tex] =  [tex]\frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}[/tex]

A) Find the rate of disappearance of [tex]N_2O_5[/tex] from t = 0 s to t = 200s

[tex]Rate = \frac{1}{2}(\frac{-\delta N_2O_5}{\delta t})\\\\Rate = -\frac{1}{2}(\frac{0.066 - 0.100}{200 - 0})\\\\Rate = 8.5*10^{-5}[/tex]

B) Find the rate of appearance of [tex]NO_2[/tex] from t = 0 s to t = 200s

According to rate law,

[tex]\frac{1}{2}(\frac{-\delta N_2O_5}{\delta t}) = \frac{1}{4}(\frac{\delta NO_2}{\delta t})\\\\8.5*10^{-5} = \frac{1}{4}(\frac{\delta NO_2}{\delta t})\\\\\frac{\delta NO_2}{\delta t} = 4 * 8.5*10^{-5}\\\\Rate = 3.4*10^{-4}[/tex]

For more information on rate of disappearance

https://brainly.com/question/20708865?referrer=searchResults

Answer:

Fluorite

What is Fluorite? Fluorite is an important industrial mineral composed of calcium and fluorine (CaF2). It is used in a wide variety of chemical, metallurgical, and ceramic processes

Which of the statements about liquids and gases are true?

The particles in liquids and gases can move past one another freely.

Which of the statements about liquids and gases are true?

Both liquids and gases have a definite volume.

The particles in liquids and gases can move past one another freely

Both liquids and gases have a definite shape.

Neither liquids nor gases have a definite shape.

The particles in liquids and gases can move past one another freely.

Answer:

a. True.

Explanation:

Hello!

In this case, since dullness and smoothness are characteristic of materials that are not able to conduct the electricity, as they do not provide an appropriate pathway for the electrons to move around because of their amorphous structure, we realize the answer is a. True.

On the contrary, a lustrous and robust solid, is actually able to conduct electricity because it provides the pathway for the electron movements and therefore to conduct electricity.

Best regards!

Considering the solution of mole fraction, molality and molarity, you obtain that:

(a) the mole fraction of thiophene in the solution is 0.039.

(b) the molality of thiophene in the solution 0.4429 [tex]\frac{moles}{kg}[/tex].

(c)  the molarity of thiophene in the solution is 0.373[tex]\frac{moles}{L}[/tex].

You know that:

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

Being the molar mass of each compound equals to:

the number of moles of each compound can be calculated as:

The total moles is obtained from the addition of the moles of the solute (C₄H₄S) and the solvent (C₇H₈):

total moles = moles C₄H₄S + moles C₇H₈ = 0.096 moles + 2.35 moles = 2.45 moles

Then, the mole fraction of thiophene in the solution can be calculated as:

[tex]mole fraction of thiophene=\frac{0.096 moles}{2.45 moles}[/tex]

Solving:

mole fraction of thiophene= 0.039

Finally, the mole fraction of thiophene in the solution is 0.039.

Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.

The Molality of a solution is determined by the expression:

[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]

In this case, you know:

Replacing:

[tex]Molality C_{4} H_{4}S =\frac{0.096 moles}{0.21675 kg}[/tex]

molality C₄H₄S= 0.4429 [tex]\frac{moles}{kg}[/tex]

Finally, the molality of thiophene in the solution 0.4429 [tex]\frac{moles}{kg}[/tex].

Molarity is the number of moles of solute that are dissolved in a certain volume and is determined by the following expression:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

Assuming that the volumes of solute and solvent are additive, you can add the volume of C₄H₄S and C₇H₈.

But first, you need yo know the volume of C₄H₄S, which can be calculated from the mass and density:

[tex]Volume C_{4} H_{4}S =8.10 grams\frac{1 mL}{1.065 grams}[/tex]= 7.606 mL= 0.007606 L

Then, the total volume of the solution is calculated as:

total volume of the solution= volume C₇H₈ + volume C₄H₄S

total volume of the solution= 0.250 L + 0.007606 L = 0.257606 L

So, the molarity of thiophene in the solution can be calculated as:

[tex]Molarity C_{4} H_{4} S=\frac{number of moles of C_{4} H_{4} S}{totalvolumeof the solution}[/tex]

Replacing:

[tex]Molarity C_{4} H_{4} S=\frac{0.096 moles}{0.257606 L}[/tex]

Solving:

Molarity C₄H₄S= 0.373 [tex]\frac{moles}{L}[/tex]

Finally, the molarity of thiophene in the solution is 0.373[tex]\frac{moles}{L}[/tex].

In summary, you get:

(a) the mole fraction of thiophene in the solution is 0.039.

(b) the molality of thiophene in the solution 0.4429 [tex]\frac{moles}{kg}[/tex].

(c)  the molarity of thiophene in the solution is 0.373[tex]\frac{moles}{L}[/tex].

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Answer:

H₂SO₄

have a great day

Answer:

Explanation:

The chemical formula for Sulfuric acid​ is H₂SO₄

Answer:

436.55841 grams.

Explanation:

The molecular formula for Iron(III) Iodide is FeI3. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Iron(III) Iodide, or 436.55841 grams.

You use the 1 / 436.55841 conversion rate. This is different for each element, according to the stoichiometry tables. Iron(II) oxide would be 2/3 of the 1/436.55841 conversion rate, that is, multiplying the fraction to denominator.

1. The mole ratio of H2SO4 and H2O in the equation = 3:3

2. The mole ratio of Mg and Mn in the equation = 3:2

Given

Reaction

1. Fe2O3 + H2SO4 → Fe2(SO4)3 + H2O

2. Mg + Mn2O3 → MgO + Mn.

Required

Balanced equation

Solution

1. Fe₂O₃ + H₂SO₄ → Fe₂(SO₄)₃ + H₂O

aFe₂O₃ + bH₂SO₄ → Fe₂(SO₄)₃ + cH₂O

Fe, left = 2a, right=2⇒2a=2⇒a=1

S, left=b, right=3⇒b=3

H, left = 2b, right = 2c⇒2b=2c⇒2.3=2c⇒c=3

The equation becomes :

Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O

2. Mg + Mn₂O₃ → MgO + Mn.

aMg + Mn₂O₃ → bMgO + cMn.

Mn, left=2, right=c⇒c=2

O₂, left=3, right=b⇒b=3

Mg, left = a, right=b⇒a=b⇒a=3

The equation becomes :

3Mg + Mn₂O₃ → 3MgO + 2Mn.

2265 g Fe₃O₄

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

[RxN - Balanced] Fe₃O₄ + 4H₂ → 3Fe + 4H₂O

[Given] 705.0 g H₂O

Step 2: Identify Conversions

[RxN] 4 mol H₂O → 1 mol Fe₃O₄

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Fe - 55.85 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Molar Mass of Fe₃O₄ - 3(55.85) + 4(16.00) = 231.55 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 4 sig figs.

2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄

Answer:

Gravitational force between two objects depends on: a. The weight of the objects c. The mass of each object b. The distance between the objects d. Both b and c Please select the best answer from the choices provided A B C D

✓d

A scientific control is an experiment or observation designed to minimize the effects of variables other than the independent variable. This increases the reliability of the results, often through a comparison between control measurements and the other measurements.