Calculate the formula mass for the compound tin(IV) sulfate.

Answers

Answer 1
The molecular formula for tin(IV)  sulfate is Sn(SO₄)₂.
mass of sulfur S = 32.065
as there is 2 sulfur in tin (IV) sulfate = 32.065 x 2 = 64.13
mass of oxygen = 15.999 and there are 8 oxygen atoms in the formula; 
8 x 15.999 = 127.992
mass of tin = 118.71
now add all the masses = 64.13 + 127.992 + 118.71 = 310.832 
formula mass of tin(IV) sulfate is 310.832

Related Questions

A 60.2-ml sample of hg (density = 13.6 g/ml) contains how many atoms of hg?

Answers

density is calculated using the following rule:
density = mass / volume
therefore:
mass = density * volume
mass of Hg = 13.6 * 60.2 = 818.72 grams

From the periodic table:
molar mass of Hg = 200.59 grams

number of moles = mass / molar mass
number of moles of Hg = 818.72 / 200.59 = 4.08 moles

each mole contains Avogadro's number of atoms.
Therefore,
number of atoms in the given sample = 4.08 * 6.022 * 10^23
                                                            = 2.456976 * 10^24 atoms

Empirical formula of magnesium oxide

Answers

The empirical formula of Magnesium Oxide is MgO

In order to prove this you need to calculate how many moles of magnesium and how many moles of oxygen there are and how it all works  together. A very common example of proving it is using 0.297 g of magnesium and getting 0.493 g of the oxide after a reaction.

Mass of Mg = 0.297
Mass of MagnesiumOxide = Mass of Mg + Mass of O
0.493 g = 0.297 g + Mass of O
Mass of O = (0.493 - 0.297)g = 0.196g

Moles of Mg = 0.297 g Mg x 1 mol Mg/24.3g Mg = 0.01222 mol Mg
Moles of O = 9.196 g O x 1 mol O/16g O = 0.01225 mol O

Now you divide both the numerator and the denominator by the smaller value and you'll see that the results is MgO

Final answer:

The empirical formula for a compound with 0.130 g of nitrogen and 0.370 g of oxygen is found by converting grams to moles, getting the simplest ratio, and ensuring whole numbers, which results in N₂O₅.

Explanation:

To determine the empirical formula of a compound from the given masses of nitrogen and oxygen, we first need to convert these masses to moles. For nitrogen, 0.130 grams divided by its molar mass (approximately 14.01 g/mol) equals approximately 0.00928 moles. For oxygen, 0.370 grams divided by its molar mass (approximately 16.00 g/mol) equals approximately 0.0231 moles. Next, we divide the number of moles of each element by the smallest number of moles to find the simplest ratio of the elements.

In this case, dividing both by 0.00928 moles, we get a ratio of about 1 mole of nitrogen to 2.5 moles of oxygen. Since we want whole numbers for the empirical formula, we multiply both by 2 to get approximately 2 moles of nitrogen for every 5 moles of oxygen, giving us N₂O₅ as the empirical formula.

If the couple moment acting on the pipe has a magnitude of 400 n-m, determine the magnitude f of the vertical force applied to each wrench.

Answers

A couple moment is produced when a network of forces do not have a resultant force, but only produces moment. As a result, it induces rotation without translation. That means the two forces are equal. 

Moment = Force * Distance

In order to determine the vertical force, we must know the distance between the two forces which is represented by the length of the pipe. Suppose the pipe is 0.5 m, then the force is

400 N·m = F*0.5 m
F = 800 N

Which of the following physical properties would tell you about an element's mass per unit volume ratio? hardness boiling point crystalline shape density all of the above

Answers

Hi there, thanks for asking a question here on Brainly!

A physical property that would tell you about an element's mass per unit volume ratio would be its density.

Answer: Letter C

Hope that helps! ★ If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia 

Answer: Option (d) is the correct answer.

Explanation:

A physical property is a property that does not change the chemical composition of a substance.

For example, mass, density, boiling point etc are all physical properties.

It is known that density is  the ratio of mass per unit volume.

Mathematically,      Density = [tex]\frac{mass}{volume}[/tex]

Hence, we can conclude that out of the given options density is the physical property that would tell you about an element's mass per unit volume ratio.

You are given solutions of hcl and naoh and must determine their concentrations. you use 27.5 ml of naoh to titrate 100. ml of hcl and 18.4 ml of naoh to titrate 50.0 ml of 0.0782 m h2so4. find the unknown concentrations.

Answers

To determine the concentrations of the substances, it is important that we equate the number of equivalents of the titrant and the substance being titrated.

(1) 27.5 mL of NaOH in 100 mL of HCl
(2) 18.4 mL of NaOH to titrate 50 mL of 0.0782 M of H2SO4

Of the two items, the one that can be easily answered is concentration of NaOH. Let x be the concentration of NaOH.
       (18.4 mL)(x) = (50 mL)(0.0782 mol/L)(2 equivalents / 1 mol of H2SO4)
The value of x from the equation is 0.425 M of NaOH

Then, for the first item, 
      (27.5 mL)(0.425) = (100)(y)

The value of y from the equation is 0.116875 M

ANSWERS: The concentrations of HCl and NaOH are 0.12 M and 0.425 M. 
             

The half life of a certain radioactive element is 1,250 years what percent of the atoms remain after 7500 years

Answers

Each half-life results in ~50% (1/2) of the original element remaining.

7500/1250 = 6 half-lives, so 100(1/2)^6
= 100(0.015625)
= 1.5625% of the original element would remain
Answer:

The amount of element after 7500 years will be 1.607% of initial amount.

Explanation:

The half life of certain elements shows the time after which half of the amount   of the element will disintegrate.

Half life of given atom = 1250 years.

The amount of element after 7500 years:

Total number of half life during this period = 7500/1250 = 6.

The amount of element after 1 half life = 50%

                            After 2nd half life = 25%

                            After 3rd half life = 12.5%

                             After 4 half life = 6.25%

                             After 5 half life = 3.125

                             After 6 half life = 1.607%

So the amount of element after 7500 years will be 1.607% of initial amount.




State the law of conservation of mass. Then apply the law to this question: what would be the total mass of the products of a reaction in which 10 grams of water decomposes into elements hydrogen and oxygen?

Answers

The law of conservation of mass states that:
"For any closed system, mass is preserved. This means that the total amount of mass remains constant as mass can neither be created nor destroyed".

Applying this to the decomposition we have,
since the total mass of reactants (water) is 10 grams, therefore, the total mass of products (hydrogen and oxygen) is also 10 grams

Hey there!

The law of conservation of mass express  that in any physical change or chemical reaction, mass is neither created or destroyed. This was according to the law. The total mass the products in the decomposition of water would be 10 grams. The total mass of the products would be the same mass as the total mass of the reactants.



Hope this helped you


Tobey

Which of the following is a unit of volume of liquids?
1. Gram
2. Liters
3. Liters per cubic gram
4. Gram per cubic centimeter
PLEASE HELP!

Answers

Your answer would be B. Liters
its liters hope this helped u

a student 600 ml cup of cocoa gets cold. It was only 25°C. He put it in the microwave. How many calories must transfer to the cocoa to bring it up to 70°C

Answers

To get the answer, you must know first the formula for heat.

Heat = mass X specific heat X change in temp 

We know that the mass is 600

Specific heat is 1

And the change in temperature is 70 – 25 which is 45.

Now plugging in those values:

 

Heat = 600 x 1 x 45 = 27,000 calories


Note: calorie is a unit of heat

Write the word and formula equations representing the chemical reaction that occurred between the. barium chloride solution , BaCl2(aq) , and the sodium sulfate solution Na2SO4(aq).

Answers

Word Equation: Barium chloride solution reacts with sodium sulfate solution to produce sodium chloride and barium sulfate.

Final answer:

Mixing solutions of barium chloride and sodium sulfate results in a precipitation reaction, forming insoluble barium sulfate and soluble sodium chloride. The balanced chemical equation for this reaction is BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq). The net ionic equation simplifies to Ba2+(aq) + SO42−(aq) → BaSO4(s).

Explanation:

Barium Chloride and Sodium Sulfate Reaction

When aqueous solutions of barium chloride (BaCl2(aq)) and sodium sulfate (Na2SO4(aq)) are mixed, a precipitation reaction occurs producing barium sulfate and sodium chloride.

The word equation for this reaction is:

Barium chloride + Sodium sulfate → Barium sulfate + Sodium chloride

The balanced chemical equation is:

BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)

In the chemical equation, solid barium sulfate (BaSO4) is the precipitate, indicating a chemical change has occurred. Understanding solubility rules is key to predicting that barium sulfate is insoluble in water and will precipitate from the mixture.

The complete ionic equation for this reaction would show all the ions that are present in solution separately. Since BaSO4 is insoluble, it will not dissociate into ions:

Ba2+(aq) + 2Cl−(aq) + 2Na+(aq) + SO42−(aq) → BaSO4(s) + 2Na+(aq) + 2Cl−(aq)

The net ionic equation omits the spectator ions (in this case, Na+ and Cl−) and includes only the ions involved in the formation of the precipitate:

Ba2+(aq) + SO42−(aq) → BaSO4(s)

Which answer describes what happens during a chain reaction? Protons released during a fission reaction trigger other fission reaction. Electrons and protons released during a fission reaction trigger other fission reactions. Neutrons released during a fission reaction trigger other fission reactions

Answers

Neutrons released during a fission reaction trigger other fission reactions. 

Neutrons released during a fission reaction trigger other fission reactions.


When an atom undergoes nuclear fission it releases free neutrons.


These free neutrons then trigger the next fission reaction by reacting with the next atom.

Thus a chain reaction is formed by neutrons released dusing a fission reaction.

if a man has brown eye with the recessive gene for blue eyes (Bb), each of his sex cells will have

Answers

B - brown eyes
b - blue eyes

If we were to fill out a punette square each of his sex cells would contain at least one brown eyed gene. 

one gene or the other, B or b that is the answer for study island


Name of molecule formed upon oxidation of aldehyde to carbonyl group

Answers

Aldehyde is a group of organic compounds with the general form of RCHO. Oxidation of aldehyde can lead to the formation of the carboxylic acids, with the general form of RCO₂H. 

With this oxidation, the typical reagents promoting the oxidation are chromic acids, chromate salts, permanganate, etc. 
Final answer:

Upon oxidation, an aldehyde is converted into a carboxylic acid. This process involves the addition of an oxygen atom to the carbon-oxygen double bond of the aldehyde, transforming it into a carboxyl group while preserving the carbon chain length.

Explanation:

The oxidation of an aldehyde leads to the formation of a carboxylic acid. When an aldehyde, which contains at least one hydrogen atom bonded to the carbonyl carbon, undergoes oxidation, a carboxylic acid is produced. This transformation occurs because the oxidation process increases the oxidation state of the carbonyl carbon by incorporating an additional oxygen atom, converting the carbon-oxygen double bond into a carboxyl group (COOH). This chemical reaction is significant in both laboratory and biological contexts, as it transforms a relatively simple organic molecule into a more complex and functional one.

The process can be summarized as the conversion of an aldehyde group (-CHO) into a carboxyl group (-COOH), retaining the same number of carbon atoms in the backbone of the molecule. This reaction is facilitated by various oxidizing agents, and it underscores the chemical reactivity of aldehydes compared to other functional groups, such as ketones, which do not further oxidize under mild conditions.

A saturated solution of barium chloride at 30 degrees Celsius contains 150 g water. How much additional barium chloride can be dissolved by heating this solution to 60 degrees Celsius?

Answers

Solubility of barium chloride at 30 degree Celsius is 38.2g /100 g water and solubility of barium chloride at 60 degree Celsius is 46.6 g / 100 g water.
The quantity of barium chloride that is dissolved in water at 30 degree Celsius = 38.2 * [150/100] = 57.30 g.
The quantity of barium chloride that will be dissolved in water at 60 degree Celsius = 46.6 * [150/100] = 69.90 g
 The difference between these quantities is the amount of barium chloride that can be dissolved by heating the barium chloride to 60 degree Celsius.
69.90 - 57.30 = 12.60 g. Therefore, 12.60 g of barium chloride can still be dissolved in the water by heating the water to 60 degree Celsius. 

The melting of ice cream is an example of which of the following?

Chemical change

Chemical property

Physical change

Physical property

Answers

Answer:

Answer is physical change

Explanation:

I took the test and got the answer right

No new substance is created during a physical change, and the substance's chemical makeup stays the same. Physical characteristics, such as shape and size, vary. Here the melting of ice cream is a Physical change. The correct option is C.

The physical state of ice cream has only changed from solid to liquid (a physical change), and this change only lasts until the melted ice cream begins to cool (a temporary change). This means that melting ice cream is a reversible, temporary, and physical change.

Ice dissolves by an endothermic mechanism. Ice needs some heat to melt because the warmth will cause connections to dissolve when absorbed. When a solid reaches its freezing point, its ions or molecules disintegrate, causing its molecules to become loosely packed.

Thus the correct option is C.

To know more about physical change, visit;

https://brainly.com/question/28742279

#SPJ6

Which alkyl chloride, through primary, is essentially unreactive in sn2 reactions?

Answers

Final answer:

In SN2 reactions, primary alkyl chlorides are generally reactive, but steric hindrance can cause some to be less reactive. Alkyl chlorides with neighboring bulky groups can be unreactive in SN2 due to restricted access for nucleophiles. Tertiary alkyl halides are typically unreactive in SN2 reactions, preferring SN1 or E1 mechanisms.

Explanation:

The student's question asks which alkyl chloride is essentially unreactive in SN2 reactions when considering primary alkyl chlorides. SN2 reactions involve a nucleophile attacking the carbon centre and displacing the leaving group, typically a halide, through a concerted mechanism with inversion of stereochemistry. This mechanism is characteristic of primary and methyl alkyl halides, with important factors being the steric accessibility of the central carbon and the strength of the leaving group.

However, among primary alkyl chlorides, alkyl chlorides with increased steric hindrance due to neighbouring substituents can be less reactive in SN2 reactions. While a standard primary alkyl chloride like ethyl chloride is typically reactive in SN2 reactions, bulkier substrates that are still considered primary may react slower or not at all. An example could be a primary alkyl chloride with adjacent bulky groups that restrict the approach of the nucleophile, leading to steric hindrance. However, without further structural information about specific alkyl chlorides, standard primary alkyl chlorides like methyl chloride or ethyl chloride do not exhibit this unreactivity in SN2 reactions.

In contrasting SN1 and SN2 mechanisms, the ease of nucleophilic attack plays a significant role. Tertiary alkyl halides are more prone to SN1 and E1 mechanisms due to their highly hindered nature which impedes nucleophilic attack, making them unsuitable for SN2 reactions. Conversely, primary alkyl halides are typically more reactive in SN2 reactions. It is important to note that in SN1 reactions, the rate-determining step is unimolecular, involving the formation of a carbocation intermediate. This intermediate is planar, allowing for nucleophilic attack on either side which can lead to a racemic mixture if the original molecule was chiral.

Neopentyl chloride, despite being a primary alkyl chloride, is unreactive in SN2 reactions due to significant steric hindrance from adjacent methyl groups. This obstructs the nucleophile's approach.

In SN2 reactions, the rate of reaction is significantly influenced by steric hindrance around the electrophilic carbon. Although primary alkyl chlorides are generally highly reactive due to minimal steric hindrance, there are exceptions. One notable exception is neopentyl chloride (1-chloro-2,2-dimethylpropane), which is essentially unreactive in SN2 reactions despite being classified as a primary alkyl chloride. This lack of reactivity is due to the significant steric hindrance created by the methyl groups attached to the adjacent carbon atoms, which obstruct the nucleophile's approach.

Key Factors Affecting SN2 Reactions:

Steric Hindrance: Increased steric hindrance around the electrophilic carbon hinders the nucleophile from attacking effectively.Alkyl Group Structure: Simple primary alkyl chlorides work well in SN2 reactions, but bulky primary structures like neopentyl chloride do not.

In summary, while most primary alkyl halides are reactive in SN2 mechanisms, the presence of bulky groups adjacent to the reaction site can significantly reduce their reactivity.

how many grams oxygen are present in 76.8g of CO2

Answers

there would be 55.8 g present

Anwer: 21.0g

Explanation:

But I noticed you got another.  I wanted to see what I did wrong. Can you show me how you got 55.8g

Ionization involves completely removing an electron from an atom. light of a particular wavelength can cause ionization to occur if it has the required energy. the energy to ionize a certain element is 342 kj/mol. what wavelength contains enough energy in a single photon to ionize one atom of this element?

Answers

350 nanometers We first need to calculate how much energy is needed to ionize a single atom. So divide the kilojoules/mol energy value by avogadro's number. 342 kj/mol / 6.0221409x10^23 = 5.67904x10^-19 Joules So we need a photon with 5.67904x10^-19 Joules of energy. The equation to express the energy of a photon is E = hc/w where h = Planck constant (6.626070040Ă—10â’34 Js) c = speed of light (299792458 m/s) w = wavelength of photon Solving for w E = hc/w Ew = hc w = hc/E Substituting known values w = 6.626070040Ă—10^â’34 Js * 299792458 m/s / 5.67904x10^-19 J w = 1.9864458x10^-25 Jm / 5.67904x10^-19 J w = 3.497855x10^-7 m w = 3.50x10^-7 m w = 350 nm

Final answer:

The wavelength containing enough energy in a single photon to ionize one atom of the given element is approximately 182 nm.

Explanation:

Ionization involves completely removing an electron from an atom. The energy required to ionize a certain element is 342 kJ/mol. The energy of a photon is given by the equation E = hc/λ, where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. To calculate the wavelength, we can rearrange the equation to find λ = hc/E. Substituting the energy required for ionization, we have λ = (6.63x10^-34 J.s * 3x10^8 m/s) / (342x10^3 J/mol). Converting the result to nanometers, we find that the wavelength contains enough energy in a single photon to ionize one atom of the element is approximately 182 nm.

An average copper penny minted in the 1960s contained about 3.000 g of copper. how much chalcopyrite had be mined to produce 100 pennies?

Answers

To know the answer, determine first the mass of the copper used:

Copper used = 100 pennies x 3.0g Cu per penny = 300.0 g Cu 



then, determine the pathway and molar ratio from Cu formed back to CuFeS2 required using the balanced reactions:

1 Cu2S from 2 CuS; 2Cu from 1 Cu2S; 2CuS from 2CuFeS2 
Therefore 2Cu from 2CuFeS2, they are in a one to one molar ratio. 



then, convert g Cu to moles and g of CuFeS2:
= 300.0 g Cu * 1 mol Cu/63.546g Cu * 2 mol CuFeS2/2 moles Cu

= 4.72 moles CuFeS2 



chalcopyrite had to be mined = 4.72 moles CuFeS2 * 183.54 g CuFeS2/1 mole CuFeS2 = 866.49 g CuFeS2

The amount of chalcopyrite that had be mined to produce 100 pennies is  [tex]\boxed{{\text{866}}{\text{.49g}}}[/tex].

Further explanation:

Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.

Consider the general reaction,

[tex]{\text{A}}+2{\text{B}}\to3{\text{C}}[/tex]

Here,

A and B are reactants.

C is the product.

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

The elemental copper that exists in its native state can be mined from its naturally occurring oxide or sulfide minerals. To extract copper from chalcopyrite certain reactions sequence is followed. First, the chalcopyrite is converted to its copper sulfide by the following reaction:

[tex]2{\text{CuFe}}{{\text{S}}_2}+3{{\text{O}}_2}\to2{\text{CuS}}+2{\text{FeO}}+2{\text{S}}{{\text{O}}_2}[/tex]

In the second step, the ferrous oxide thus formed is then treated with silica as follows:

[tex]2{\text{FeO}}+{\text{Si}}{{\text{O}}_2}\to2{\text{FeSi}}{{\text{O}}_3}[/tex]

In the third step, the copper sulfide decomposes as shown below:

[tex]2{\text{CuS}}\to{\text{C}}{{\text{u}}_2}{\text{S}}+{\text{S}}[/tex]

In the final sequence of the reaction, the [tex]{\text{C}}{{\text{u}}_2}{\text{S}}[/tex] formed undergoes following oxidation to form elemental copper.

[tex]{\text{C}}{{\text{u}}_2}{\text{S}}+{\text{S}}\to{\text{2Cu}}+2{\text{S}}{{\text{O}}_2}[/tex]

From these reactions, it is the evident that 2 moles of [tex]{\text{CuFe}}{{\text{S}}_2}[/tex] forms 2 moles of elemental copper. Therefore the molar ratio of the combination of Cu and [tex]{\mathbf{CuFe}}{{\mathbf{S}}_{\mathbf{2}}}[/tex] is 1:1.

Since one penny contains 3 g of Cu hence the mass of copper in 100 pennies is calculated as follows:

[tex]\begin{gathered}{\text{Mass of Cu}}=\left({\frac{{3\;{\text{g}}}}{{1\,{\text{penny}}}}}\right)\left({100\;{\text{pennies}}}\right)\\=300\;{\text{g}}\\\end{gathered}[/tex]

Using the mass of copper and the molar mass we would determine the number of moles of copper by applying the following formula.

[tex]{\text{Number of moles}}=\frac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}[/tex]                             …… (1)

The given mass is [tex]300{\text{g}}[/tex].

The molar mass is [tex]{\text{63}}{\text{.546g/mol}}[/tex].

Substitute the values in equation (1).

 [tex]\begin{gathered}{\text{Number of moles}}=\frac{{300.0{\text{g}}}}{{{\text{63}}{\text{.546g/mol}}}}\\=4.72099{\text{mol}}\\\end{gathered}[/tex]

The molar ratio of combination of Cu and [tex]{\text{CuFe}}{{\text{S}}_2}[/tex] is 1:1, so the number of moles of [tex]{\mathbf{CuFe}}{{\mathbf{S}}_{\mathbf{2}}}[/tex] is 4.72099 mol.

The moles of [tex]{\text{CuFe}}{{\text{S}}_2}[/tex] calculated can be used to determine the corresponding mass in grams. The formula to calculate the mass is as follows:

[tex]{\text{Mass of CuFe}}{{\text{S}}_2}=\left({{\text{moles of CuFe}}{{\text{S}}_2}}\right)\left({{\text{molar mass of CuFe}}{{\text{S}}_2}}\right)[/tex]  …… (2)

The moles of [tex]{\text{CuFe}}{{\text{S}}_2}[/tex]is [tex]4.72099{\text{ mol}}[/tex].

The molar mass of [tex]{\text{CuFe}}{{\text{S}}_2}[/tex]is[tex]183.54{\text{g/mol}}[/tex].

Substitute the values in equation (2).

 [tex]\begin{gathered}{\text{Mass of CuFe}}{{\text{S}}_2}=\left({4.72099{\text{ mol}}}\right)\left({183.54{\text{g/mol}}}\right)\\=866.49{\text{g}}\\\end{gathered}[/tex]

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2. The net ionic equation for the reaction of [tex]{\text{MgS}}{{\text{O}}_{\text{4}}}[/tex] with [tex]{\text{Sr}}{\left({{\text{N}}{{\text{O}}_3}}\right)_2}[/tex] : https://brainly.com/question/4357519

Answer details:

Grade: College

Subject: Chemistry

Chapter: Mole concept and stoichiometry

Keywords: copper, penny, pennies, CuFeS2, stoichiometry, O2, moles, mass, S and Cu2S.

Which hybridization scheme occurs about nitrogen when nitrogen forms a double bond?

Answers

The electronic configuration of nitrogen is

1s2 2s2 2p3

Now during formation of double bond there will be formation of a pi bond

The nitrogen will undergo sp2 hybridization. One extra p orbital will form pi bond by side ways overlapping

the two sp2 hybridized orbital will form sigma bond with two atoms

the third sp2 hybridized orbital will have one lone pair of electrons .

The hybridization around nitrogen when it forms a double bond is [tex]\boxed{s{p^2}}[/tex] .

Further explanation:

Prediction of hybridization:

The hybridization can be determined by calculating the number of hybrid orbitals (X) which is formed by the atom. The formula to calculate the number of hybrid orbitals (X) is as follows:

 [tex]\boxed{{\text{X = }}\frac{1}{2}\left[ {{\text{VE}} + {\text{MA}} - c + a}\right]}[/tex]

Where,

VE is a total number of valence electrons of the central atom. MA is the total number of monovalent atoms/groups surrounding the central atom. c is the charge on the cation if the given species is a polyatomic cation. a is the charge on the anion if the given species is a polyatomic anion.

Note: In MA only monovalent species should be considered and for divalent atoms or groups MA is equal to zero.

Generally, the least electronegative atom is considered as the central atom. Calculate the hybridization as follows:

1. If the value of X is 2 then it means two hybrid orbitals are to be formed and thus the hybridization is sp.

2. If the value of X is 3 then it means three hybrid orbitals are to be formed and thus the hybridization is [tex]s{p^2}[/tex] .

3. If the value of X is 4 then it means four hybrid orbitals are to be formed and thus the hybridization is [tex]s{p^3}[/tex].

4. If the value of X is 5 then it means five hybrid orbitals are to be formed and thus the hybridization is [tex]s{p^3}d[/tex].

5. If the value of X is 6 then it means six hybrid orbitals are to be formed and thus the hybridization is [tex]s{p^3}{d^2}[/tex] .

The ground state electronic configuration for nitrogen (N) is,

 [tex]1{s^2}2{s^2}2{p^3}[/tex]

Therefore, the valence electrons associated with nitrogen (N) atom are 5.

The two electrons of nitrogen are involved in the formation of the double bond and the electron present in 2s subshell will remain as the lone pair on nitrogen atom. There is only one electron left on the nitrogen atom out of five, and therefore the total number of monovalent atoms that can surrounding the central atom (MA) is 1.

Since the molecule is a neutral species and thus the value of a and c is 0.

Substitute these values in the above formula.

 [tex]\begin{aligned}\text{X}&=\dfrac{1}{2}[5+1-0+0]\\&=\dfrac{1}{2}[6]\\&=\boxed{3}\end{aligned}[/tex]

Since the value of X is 3, it means 3 hybrid orbitals are to be formed and therefore the hybridization of nitrogen is [tex]{\mathbf{s}}{{\mathbf{p}}^{\mathbf{2}}}[/tex] .

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Covalent bonding and molecular structure

Keywords: hybridization, nitrogen, ground state electronic configuration, sp2, valence electrons, monovalent atoms, VE, MA, a, c, X, 5, 0, N, least electronegative, central atom.

How much heat is absorbed during production of 112 g of NO by the combination of nitrogen and oxygen?
N2(g)+O2(g)→2NO(g), ΔH = + 43.0 kcal

Answers

From the given chemical equation we see that 43 kcal of energy is needed for every 2 moles of NO. First let us calculate the moles of NO with a molar mass of 30 g/mol.

moles NO = 112 g / (30 g/mol) = 3.73 mol

 

So the total heat absorbed is:

heat = (43 kcal / 2 mol) * 3.73 mol

heat = 80.195 kcal

if a sample of gas is intially at 1.8 atm,22.0 l, and 26.4 c, what will be the volume if the pressure is reduced by 0.8 atm and the temperature is lowered to 20.3 c?

Answers

Final answer:

To find the new volume of the gas, we can use Boyle's Law, which states that if temperature and amount are constant, pressure and volume are inversely proportional. Using the equation P1V1 = P2V2 and substituting the given values, we find that the final volume will be approximately 39.6 L.

Explanation:

To solve this problem, we can use Boyle's Law, which states that if the temperature and amount of gas are kept constant, the pressure and volume of a gas are inversely proportional.

We can use the formula:

P1V1 = P2V2

Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Substituting the given values, we have:

(1.8 atm)(22.0 L) = (1.0 atm + (-0.8 atm))(V2)

Solving for V2, we find that the final volume will be approximately 39.6 L.

Final answer:

To find the new volume after a pressure reduction and temperature decrease, convert the temperatures to Kelvin and use the combined gas law, showing that the initial and final states of the gas can be related by (P1V1T2) = (P2V2T1). Insert the known values into the equation and solve for the new volume, V2.

Explanation:

To calculate the new volume of a gas when the pressure and temperature change, we can use the combined gas law which relates the pressure, volume, and temperature of a gas. The combined gas law is expressed as (P1V1T2) = (P2V2T1) where P is pressure, V is volume, and T is temperature in Kelvin.

To solve the student's question, first convert the temperatures from Celsius to Kelvin by adding 273.15: 26.4°C + 273.15 = 299.55 K and 20.3°C + 273.15 = 293.45 K. Now, insert the values into the combined gas law equation:

 (P1V1T2) = (P2V2T1),

 (1.8 atm × 22.0 L × 293.45 K) = ((1.8 atm - 0.8 atm) × V2 × 299.55 K),

 Solve for V2, the new volume of the gas after pressure and temperature changes.

Evaluate the probability of finding an electron in a small region of a hydrogen 1s orbital

Answers

Use the equation for the wavefunction for a 1 s H atom (page 25 of the 5th edition of the textbook, page 21 of the 4th edition of textbook) and square it: Ψ2r,Θ,Φ( ) =1πa03e−2ra0to get the probability density.

Then take the ratio of the probability density at r=0.55 ao and at r=0.Ψ 20.55ao,Θ, Φ( )Ψ20,Θ,Φ( )=1πao3e−2 0.55ao( )ao1πao3e−2 0( )ao=e−2 0.55ao( )aoe−2 0( )ao==e−2(0.55)1=0.3329 = 0.33

The ratio of the probability density at r = 0.55 a0 to that at r = 0 is 0.33.

See attachment for better understanding.

A compound has the formula c8h8 and does not contain any double or triple bonds. all the carbon atoms are chemically identical and all the hydrogen atoms are chemically identical. draw a lewis structure for this molecule.

Answers

Final answer:

The molecular formula C8H8 might represent a cyclic compound where carbon atoms are linked in an 8-membered ring, with each carbon atom being bonded to two other carbon atoms and a hydrogen atom, maintaining the tetravalency of carbon.

Explanation:

The compound you mentioned, C8H8, does not fit with the typical hydrocarbon pattern which typically results in hydrocarbons like alkanes having a general formula of CnH2n+2. However, since the question stated clearly that there are no double or triple bonds and all carbon atoms are identical, there's a possibility that the compound is a cyclic structure. Let's consider a structure where carbon atoms are linked in a cyclic form, forming an 8-membered ring. Each carbon atom in this ring will form two bonds with two other carbon atoms and the remaining two bonds will be with hydrogen atoms. In this way, each carbon atom will maintain its tetravalency (4 bonds per carbon atom), and there will be 8 hydrogen atoms, exactly fitting the given formula C8H8. To draw the Lewis structure, arrange 8 carbon atoms in a ring and attach a hydrogen atom to each carbon atom. Remember, each line represents a single bond between atoms.

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Final answer:

The request for a Lewis structure of C8H8 without double or triple bonds and with all carbon and hydrogen atoms being chemically identical suggests an unusual compound that contradicts common organic chemistry principles, hence a conventional structure cannot be provided.

Explanation:

The student is asking for a Lewis structure of a compound with the formula C8H8 that does not contain any double or triple bonds. Given that all carbon and hydrogen atoms in this compound are chemically identical, this suggests a cyclic structure where each carbon atom is bound to two other carbons and a single hydrogen atom, forming a ring. The best structure fitting this description would be cyclooctatetraene, although it has alternating single and double bonds, which contradicts the no double or triple bonds condition.

However, considering the compound has no double or triple bonds, the compound would be highly unusual and not in line with standard organic chemistry principles. All carbon atoms making four bonds is a fundamental concept in organic chemistry, which leads to the conclusion that either the information given in the question is incorrect or it is a highly unconventional compound not typically covered in standard chemistry curricula. Therefore, without double or triple bonds, the compound's structure cannot be satisfactorily drawn while adhering to typical valency rules.

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The abundance of 12c is 98.9%. 13c is another naturally occurring isotope. what is the percent abundance of 13c?

Answers

Actually the total abundance of the isotopes of any element in this world must sum up to 100%. So we initially know that 12 C is 98.9 percent abundant, therefore the remaining of the 100 percent must be of 13 C, that is:

 

13 C = 100% - 98.9%

13 C = 1.1% 

Final answer:

The percent abundance of carbon-13 (13C) is 1.1%, which is calculated by subtracting the abundance of carbon-12 (12C) from 100%.

Explanation:

The percent abundance of carbon-13 (13C) can be calculated by subtracting the abundance of carbon-12 (12C) from 100%. If carbon-12 comprises 98.9% of natural carbon, the calculation would be 100% - 98.9% = 1.1%. Therefore, the percent abundance of carbon-13 is 1.1%.

For the reaction shown here, 2.3 mola is mixed with 1.6 molb and 3.1 molc. what is the limiting reactant? 3a+b+2c→3d

Answers

We know that the reactant ratios

a : b : c

are

3 : 1 : 2

So we will check the amount of the other two reactants required by taking one as the base.

With a as the base, 

2.3 : 0.77 : 1.53

The amounts of b and c required for 2.3 moles of a are present

With b as the base,


4.8 : 1.6 : 3.2

Both a and c are present in amounts less than that required for 1.6 moles of b.


With c as the base, 

4.65 : 1.55 : 3.1

Only a is limiting in this case

From the above considerations, we see that a is always in quantities less than that which would be required, so a is the limiting reactant.
Final answer:

By comparing the relative amounts required of each reactant according to the stoichiometric coefficients in the chemical equation, 'b' is identified as the limiting reactant as it will be consumed first.

Explanation:

In order to determine the limiting reactant in a chemical reaction, we need to take into account the stoichiometric coefficients in the balanced chemical equation. In the equation given, 3a + b+ 2c→3d, for every 3 moles of 'a', 1 mole of 'b' and 2 moles of 'c' are needed. Given the amounts of each, we have 2.3 mol 'a', 1.6 mol 'b', and 3.1 mol 'c'. For 'a' and 'c' to completely react, there should be 1 'b' for every 3 'a' and 1 'b' for every 2 'c'. Therefore, 'a' requires 2.3/3 = 0.77 mol 'b' and 'c' requires 3.1/2 = 1.55 mol 'b'. However, only 1.6 mol 'b' is available. Therefore, 'b' will be consumed before either 'a' or 'c', making 'b' the limiting reactant.

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A compound composed of only c and f contains 17.39 c by mass. what is its empirical formula

Answers

Since the Carbon C is 17.39% by mass hence the Fluorine F is 82.61% by mass. Divide each mass % by the respective molar masses, that is:

 

C = 17.39 / 12 = 1.45

F = 82.61 / 19 = 4.35

 

Divide the two by the smaller number, so divide by 1.45

 

C = 1.45 / 1.45 = 1

F = 4.35 / 1.45 = 3

 

So the empirical formula is:

CF3 

) how would your results have been affected if you had used water-insoluble pens, instead of water-soluble ones, in part 2 of this experiment

Answers

The experiment is based on how paper chromatography can be used to separate inks into their dye components. 
If water-insoluble pens are used in the reaction, the dyes components of the ink will not dissolve in water. Because the components of the ink did not dissolve in water, when the solution is spotted on the chromatography paper, the components of the ink will not be separated out.

The energy released during a nuclear reaction in a power plant is directly used to __________

A. create electric current
B. fuse hydrogen into helium
C. heat water
D. energize control rods

Answers

Final answer:

In a nuclear power plant, the energy released from the fission of uranium-235 is used to heat water, producing steam that drives turbines to generate electricity.

Explanation:

The energy released during a nuclear reaction in a power plant is directly used to heat water. In a nuclear power plant, a nuclear reactor uses the energy produced in the fission of uranium-235 to produce electricity. The fission reaction generates heat, which is transferred to water surrounding the fuel rods causing it to turn into steam. This steam then drives turbines connected to generators, thereby generating electric current. Control rods made out of materials like boron, cadmium, or hafnium, which absorb neutrons, are used to control the rate of the fission reaction, but they are not directly energized by the reaction.

Completing a Punnett Square

Suppose you are tutoring a classmate on how to complete a Punnett square.

How would you explain to a classmate the process of using a Punnett square for a cross between one pea plant with genotype Tt and a second pea plant with genotype Tt, where T is the allele for a tall pea plant, and t is the allele for a short pea plant? Make sure to describe how you would find the probability that a given offspring of the two parents is tall or short.

Answers

wow long time since I've done this but I believe the answer would be... Since T is the dominant gene that would mean being tall is a dominant gene. well one of the parents of the plants have to tt and one would have to be TT for all of the offspring to have the genes Tt. So to find if the pea plants will be tall or short all you have to do is take each of the parents genes and combines them. If there is a T in any of the answers that means that they offspring has a dominant gene off being tall. As for the outlook of what the offspring will be they will all be Tt.

(In the picture it shows your question just with different letters.)

Hint: The father gene is always on the top and the mother gene is on the left hand side.
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