Be sure to answer all parts. Carry out the following operations as if it were a calculation of real experimental results. Express the answer with the correct number of significant figures. (3.26 × 10−3 mg) − (7.88 × 10−5 mg ) Answer Units mg Enter your answer in standard form. Do not use scientific notation.

Answer:

oxygen atoms gain electrons and hydrogen atoms lose electron

Explanation:

Hydrogen oxygen fuel cell involves redox reactions.

It is an electrochemical cell, which is used for many applications like rocket propellant.

The actual reaction is

[tex]H_{2}+\frac{1}{2}O_{2}--->H_{2}O[/tex]

Here hydrogen undergoes oxidation as it loses electrons

Oxygen undergoes reduction as it gains electrons.

The redox reactions are

At anode:

[tex]H_{2}--->2H^{+}+2e[/tex] [loss of electrons by hydrogen]

At cathode:

[tex]O_{2}+4H^{+}+4e--->2H_{2}O[/tex] [gain of electrons by oxygen]

Answer: [Ni(en)3] > [Ni(en)(H2O)4] > [Fe(NH3)6] > [FeF6]

Explanation:

Generally chelating ligands stabilize the complex more than non chelated ligands and more the no of chelated ligands more the stability.

Here en (ethylenediamine) is a chelated ligand and stabilze the complex more by chelation.

And Strong field ligand (NH3) also stabilze the complex more than weak field ligand (F).

Hence F containing complex is least stable.

The rank of the coordination compounds in the increasing order of the stability is [tex]\rm [Ni(en)_3] > [Ni(en)(H_2O)_4] > [Fe(NH_3)_6] > [FeF_6][/tex].

The coordination compound are the complexes in which the central metal atom has been bounded by the nonmetal to the complexes through the chemical bond.

The stability of the coordination complexes is attributed to the number of chelating agents to the central atoms.

The increased number of atoms to the central atoms adds to the stability of the complex. Thus, the increasing order of the stability is [tex]\rm [Ni(en)_3] > [Ni(en)(H_2O)_4] > [Fe(NH_3)_6] > [FeF_6][/tex].

Learn more about coordination complexes, here:

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Explanation:

According to Avogadro's law, equal volumes of all the gases at same temperature and pressure will also, have same number of molecules.

That is,             [tex]V \propto n[/tex]

or                      [tex]\frac{V}{n}[/tex] = k

Since, it is given that [tex]V_{1}[/tex] is 6.13 L, [tex]n_{1}[/tex] is 8.51 mol, and [tex]V_{2}[/tex] is 18.5 L.

Hence, calculate the [tex]n_{2}[/tex] as follows.

                        [tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]

                        [tex]\frac{6.13 L}{8.51 mol} = \frac{18.5 L}{n_{2}}[/tex]  

                           [tex]{n_{2}}[/tex] = 0.35 mol

Thus, we can conclude that the number of moles of gas added to the container is 0.35 mol.

Final answer:

To calculate the moles of gas added, we use the ratio of the initial and final volumes to determine the final moles, and then subtract the initial moles to find that 17.16 mol of gas were added to the container.

Explanation:

The subject of this question is Chemistry, specifically it is dealing with gas laws and the concept of molar volume at constant pressure and temperature (ideal gas law).

To find the number of moles of gas added to the container, we will apply the ideal gas law in its simplest form, which is Boyle's Law (P1V1 = P2V2), assuming constant temperature and pressure. Since P and T are constant, the relationship between volume and moles is direct according to Avogadro's Law. This means we can simply use the ratio of the initial and final volumes to find the moles added:

Initial moles (n1) = 8.51 mol

Initial volume (V1) = 6.13 L

Final volume (V2) = 18.5 L

Since the volumes are proportional to moles at constant temperature and pressure, we have:

V1 / n1 = V2 / n2

Using a simple cross-multiplication:

n2 = (V2 / V1) × n1

n2 = (18.5 L / 6.13 L) × 8.51 mol

n2 = 3.016 × 8.51 mol

n2 = 25.67 mol

The total moles in the container after addition of gas is 25.67 mol. To get the moles of gas added, we subtract the initial moles from the total moles:

Moles of gas added = 25.67 mol - 8.51 mol

Moles of gas added = 17.16 mol

Answer:

Explanation:

Writing the equation properly:

         (CH₃)₃N + H₂O ⇄ (CH₃)₃NH⁺ + OH⁻

         HNO₃ + H₂O  ⇄ H₂O + NO₃⁻

The bronsted-lowry theory defines an acid as a proton donor and a base as a proton acceptor.

In a bronsted-lowry acid-base reaction, the original acid gives up its proton and becomes a conjugate base. Also, the original base accepts a proton and becomes a conjugate acid. For every acid, there is a conjugate base and for every base there is a conjugate acid.

What differentiates an acid from its conjugate base is a proton. The difference between a base and its conjugate acid is a proton.

    (CH₃)₃N          +          H₂O          ⇄         (CH₃)₃NH⁺       +         OH⁻

Bronsted-lowry base     acid                conjugate acid        conjugate base

    HNO₃                       +      H₂O       ⇄           H₃O⁺        +                NO₃⁻

Bronsted-lowry acid           base             conjugate acid      conjugate base

In the reaction (a), (CH3)3N is the Brønsted-Lowry base, H2O is the Brønsted-Lowry acid, (CH3)3NH+ is the conjugate acid, and OH- is the conjugate base. In reaction (b), HNO3 is the Brønsted-Lowry acid, H2O is the Brønsted-Lowry base, H3O+ is the conjugate acid, and NO3- is the conjugate base.

In the reaction (a) (CH3)3N(aq) + H2O(l) <-> (CH3)3NH+(aq) + OH-(aq), (CH3)3N is the Brønsted-Lowry base, H2O is the Brønsted-Lowry acid, (CH3)3NH+ is the conjugate acid, and OH- is the conjugate base.

In the reaction (b) HNO3(aq) + H2O(l) <-> H3O+(aq) + NO3-(aq), HNO3 is the Brønsted-Lowry acid, H2O is the Brønsted-Lowry base, H3O+ is the conjugate acid, and NO3- is the conjugate base.

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Answer: Option (a) is the correct answer.

Explanation:

A chemical bonding in a substance occurs when there is exchange of electrons between the atoms. These electrons are present in the outermost orbital are actually away from the nucleus of the atom.

Due to which there is no force of attraction between nucleus or valence electrons. Hence, it is easy for the outermost orbital to lose electron.

Whereas nucleus of an atom contains neutrons and protons. And, changes in the nucleus of an atom leads to nuclear reactions.

Therefore, we can conclude that the configuration of the outermost orbital determines chemical bonding characteristics.

Answer : The initial rate of the reaction is, [tex]1.739\times 10^{-3}s^{-1}[/tex]

Explanation :

First we have to calculate the rate constant of the reaction.

Expression for rate law for first order kinetics is given by :

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,

k = rate constant  = ?

t = time taken for the process  = 44 s

[tex][A_o][/tex] = initial amount or concentration of the reactant  = 0.1108 M

[tex][A][/tex] = amount or concentration left time 44 s = 0.0554 M

Now put all the given values in above equation, we get:

[tex]k=\frac{2.303}{44}\log\frac{0.1108}{0.0554}[/tex]

[tex]k=0.0157[/tex]

Now we have to calculate the initial rate of the reaction.

Initial rate = K [A]

At t = 0, [tex][A]=[A_o][/tex]

Initial rate = 0.0157 × 0.1108 = [tex]1.739\times 10^{-3}s^{-1}[/tex]

Therefore, the initial rate of the reaction is, [tex]1.739\times 10^{-3}s^{-1}[/tex]

hey There!:

Compound B is the structure that leads to Cis-nonene as major product because in this case the Leaving group i.e Br and the electrophile i.e  H is not anti-periplanar to each other .

Hope this helps!

Answer:It would depend on the eluent solvent what we use ,If we would use a polar solvent than B will be appear first followed by c and then A on the chromatography strip.

If we use non-polar eluent solvent than A will be separated first followed by C and thenB on the chromatography strip.

Explanation:

Separation using paper chromatography is dependent upon the polarity of various  pigments .

A polar pigment would move ahead in case of polar solvent used and a non-polar pigment would move ahead when we use a non-polar solvent.

So separation would occur in order of polarity  of various pigments in a given solvent.

The amount of distance travelled by each component (or pigment or spot) can be calculated using the formula for retention factor:

Rf= Distance travelled by pigment spot or solute/Distance travelled by eluent solvent

Rf= Retention factor

Retention factor is basically the ratio of distance tarvelled by the pigment or spot to the ratio of distance travelled by the solvent.

Final answer:

In a paper chromatography analysis, pigment B (highly polar) will appear closest to the origin line, followed by pigment C (moderately polar), and then pigment A (slightly polar).

Explanation:

In a paper chromatography analysis, the pigments A, B, and C will appear on the surface of the chromatography strip in the following order:

Pigment B (highly polar)

This order is determined by the polarity of the pigments. In chromatography, more polar substances tend to travel more slowly through the stationary phase, while less polar substances travel faster. Therefore, pigment B, being highly polar, will have the slowest migration and appear closest to the origin line, followed by pigment C and then pigment A.

Answer:

That iron atom is oxidized. It loses two electrons.

Explanation:

Compare the formula of an iron atom and an iron(II) ion:

The superscript [tex]+2[/tex] in the iron(II) ion is the only difference between the two formulas. This superscript indicates a charge of [tex]+2[/tex] on each ion. Atoms and ions contain protons. In many cases, they also contain electrons. Each proton carries a positive charge of [tex]+1[/tex] and each electron carries a charge of [tex]-1[/tex]. Atoms are neutral for they contain an equal number of protons and electrons.

Protons are located at the center of atoms inside the nuclei. They cannot be gained or lost in chemical reactions. However, electrons are outside the nuclei and can be gained or lost. When an atom loses one or more electrons, it will carry more positive charge than negative charge. It will becomes a positive ion. Conversely, when an atom gains one or more electrons, it becomes a negative ion.

An iron atom [tex]\mathrm{Fe}[/tex] will need to lose two electrons to become a positive iron(II) ion [tex]\mathrm{Fe^{2+}}[/tex] with a charge of [tex]+2[/tex] on each ion. That is:

[tex]\rm Fe \to Fe^{2+} + 2\;e^{-}[/tex].

This definition can be written as the acronym OILRIG. (Khan Academy.)

In this case, each iron atom loses two electrons. Therefore the iron atoms here are oxidized.

Final answer:

When an iron atom becomes an ion Fe+2, it loses two electrons, increasing its oxidation number to +2. This is oxidation and is common for transition metals like iron which can create ions of different charges.

Explanation:

During a chemical reaction, when an iron atom becomes the ion Fe+2, it means that the iron atom has lost two electrons. This process is known as oxidation and results in an increase in the oxidation number of iron from 0 to +2. Iron atoms can lose electrons from their outermost shell when they react, and transition metals like iron can have more than one possible charged state, commonly forming Fe2+ or Fe3+ ions. In the provided reaction, 4 Fe + 3O2 → 2 Fe2O3, iron is in the +3 oxidation state, indicating it has lost three electrons. Therefore, in different chemical environments, iron atoms can lose varying numbers of electrons, resulting in various oxidation states such as Fe2+ or Fe3+.

Answer: The spontaneous cell reaction having smallest [tex]E^o[/tex] is [tex]I_2+Cu\rightarrow Cu^{2+}+2I^-[/tex]

Explanation:

We are given:

[tex]E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V[/tex]

The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

[tex]E^o_{cell}=E^o_{oxidation}+E^o_{reduction}[/tex]

The combination of the cell reactions follows:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

[tex]I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)[/tex]

Oxidation half reaction:  [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-[/tex]   [tex]E^o_{oxidation}=0.45V[/tex]

Reduction half reaction:  [tex]I_2(s)+2e^-\rightarrow 2I_-(aq.)[/tex]   [tex]E^o_{reduction}=0.54V[/tex]

[tex]E^o_{cell}=0.45+0.54=0.99V[/tex]

Thus, this cell will not give the spontaneous cell reaction with smallest [tex]E^o_{cell}[/tex]

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

[tex]I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)[/tex]

Oxidation half reaction:  [tex]Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-[/tex]   [tex]E^o_{oxidation}=-0.34V[/tex]

Reduction half reaction: [tex]I_2(s)+2e^-\rightarrow 2I_-(aq.)[/tex]   [tex]E^o_{reduction}=0.54V[/tex]

[tex]E^o_{cell}=-0.34+0.54=0.20V[/tex]

Thus, this cell will give the spontaneous cell reaction with smallest [tex]E^o_{cell}[/tex]

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

[tex]Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)[/tex]

Oxidation half reaction:  [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-[/tex]   [tex]E^o_{oxidation}=0.45V[/tex]

Reduction half reaction:  [tex]Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)[/tex]   [tex]E^o_{reduction}=0.34V[/tex]

[tex]E^o_{cell}=0.45+0.34=0.79V[/tex]

Thus, this cell will not give the spontaneous cell reaction with smallest [tex]E^o_{cell}[/tex]

Hence, the spontaneous cell reaction having smallest [tex]E^o[/tex] is [tex]I_2+Cu\rightarrow Cu^{2+}+2I^-[/tex]

Answer:

Explanation:

People who use marble to carve monuments and sculptures are always concerned because it can easily chemically weather.

Marble is a metamorphic rock derieved from limestone. The minerals that makes up limestone are calcite and dolomite. When a metamorphic transformation occurs, the rock is subjected to intense temperature and pressure. Marble is made of calcite and dolomite minerals.

Rain water dissolves carbon dioxide to form weak carbonic acid. This weak carbonic acid can easily dissolve marble thereby defacing the monument and the sculpture.

Acid rain can cause damage to marble monuments and sculptures because it reacts with the carbonate minerals in marble, leading to their dissolution and degradation over time.

People who care for monuments and sculptures made of marble are concerned about acid rain because marble is primarily composed of carbonate minerals such as calcite (CaCO3) and dolomite (Ca,Mg(CO3)2). When acid rain falls on marble, the weak acid reacts with the carbonate minerals, causing them to dissolve. This can lead to the degradation and erosion of the marble over time.

The reaction between acid rain and marble can be represented by the following equation: CaCO3(s) + H₂SO4 (aq) → CaSO4(s) + H₂O(1) + CO2(g). The reaction produces calcium sulfate, water, and carbon dioxide gas, which contribute to the dissolution of the marble.

Therefore, the concern for acid rain arises from the fact that it can cause significant damage to marble monuments and sculptures, affecting their aesthetic value and structural integrity.

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Answer: The concentration of [tex]HNO_3[/tex] comes out to be 0.088 M.

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]

We are given:

Conversion factor: 1 L = 1000 mL

[tex]n_1=1\\M_1=?M\\V_1=0.105L=105mL\\n_2=2\\M_2=0.100M\\V_2=46.5mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 105=2\times 0.100\times 46.5\\\\M_1=0.088M[/tex]

Hence, the concentration of [tex]HNO_3[/tex] comes out to be 0.088 M.

The concentration of the HNO3 solution is 0.443 M.

To determine the concentration of the HNO3 solution, we can use the equation, concentration = (volume of solution titrated)/(volume of solution required for complete neutralization). In this case, the volume of solution titrated is 0.105 L and the volume of 0.100 M Ba(OH)2 solution required for complete neutralization is 46.5 mL (or 0.0465 L). Plugging these values into the equation, we get: concentration = 0.0465 L / 0.105 L = 0.443 M.

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Hey there!:

* S,and Ne are non-metals.

* Si is a metalloid.Generally metalloids acts as semiconductor.

* Ag and Hg are metals because metals exhibits lustre property,  melleable and ductility  properties.

Hope this helps!

Explanation:

Metals are the substances which lose electrons to attain stability and hence they form cations.

Metals are good conductors of heat and electricity and generally they are solid at room temperature. Metals have a shiny surface and they are also malleable and ductile.

For example, Ag: Silver Shiny solid at room temperature that is also a good conductor of heat and electricity. Hence, silver is a metal.

Non-metals are the substances which gain electrons to gain stability and hence they form anions.

Non-metals are bad conductors of heat and electricity. They are brittle and non-shiny in nature.  

For example, S: Sulfur Powdery yellow solid is a non-metal.

Metalloids are the substances which show properties of both metals and non-metals.

Metalloids are moderately able to conduct heat and electricity. Generally metalloids are solid at room temperature. They are also able to react with other molecules.

For example, Si: Silicon Semiconductor is a metalloid.

Hence, the given substances are classified as follows.

Answer: The concentration of [tex]COF_2[/tex] remains at equilibrium is 0.40 M.

Explanation:-

Initial concentration of [tex]COF_2[/tex] = 2 M

The given balanced equilibrium reaction is,

                       [tex]2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g[/tex]

Initial conc.                      2 M             0             0            

At eqm. conc.           (2-2x) M            x M        x M

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[CO_2]\times [CF_4]}{[COF_2]^2}[/tex]

Now put all the given values in this expression, we get :

[tex]4.20=\frac{(x)\times (x)}{(2-2x)^2}[/tex]

By solving the term 'x', we get :

x =  0.80 M

Thus, the concentrations of [tex]COF_2[/tex] at equilibrium is : (2-2x) = 2-2(0.80)=0.40 M

The concentration of [tex]COF_2[/tex] remains at equilibrium is 0.40 M.

Answer:

D)the first ionization energy (IE1) of B is most probably lower than the first ionization energy (IE1) of A

Explanation:

As mentioned the possible chemical formula for a compound can be predicted from electron configurations of the elements.

The first element 'A' has electronic configuration as "[core]ns²

It means it has two valence electrons and it can give those valence electrons to attain noble gas or full filled stability.

So the element is most likely to form a dipositive ion.

Now for element B the configuration is :[Core]ns²np⁵

thus it has seven valence electrons and it will most likely to accept an electron to gain full filled stability. So B will form a mono negative ion.

The possible formula will be AB₂ [Like MgCl₂]

So the false statements are:

D)the first ionization energy (IE1) of B is most probably lower than the first ionization energy (IE1) of A : the first ionization energy of A will be lower than B as B cannot give an electron easily.

The other statements are correct:

A) element A serves as the reducing agent in the reaction to form AxBy : As A will lose electrons so it will acts as reducing agent and itself gets oxidized

B) yes it is whole number as calculated above.

C) x + y = 1+2 = 3

E) Already calculated that B has oxidation state of -1.

The false statement is that; "the first ionization energy (IE1) of B is most probably lower than the first ionization energy (IE1) of A"

The first ionization energy is the energy required to remove the first electron from an atom. This is a periodic trend that increases across the period but decreases down the group.

Looking at the electron configurations of A and B it is clear that A is an alkaline earth metal while B is a halogen. As such, A is a reducing agent and B will have an oxidation state of -1 as expected of halogens.

The compound will have the formula AX2 hence x + y = 3. However, since ionization energy increases across the period, the first ionization energy of B must certainly be greater than that of A.

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Answer:

Calculate the steric energy or heat of formation for one single bond isomel of trans-benzAlacetone using the usual energy minimization procedure. The result should be a planar molecule, Then deliberately hold the dihedral angle defined by atoms 1, 2, 3, and 4 at 0 90, and 180 and again calculate the energies of the molecule.

Answer:

Explanation:

It's false, but not because it is not a decomposition equation. It does resemble one.

As near as I can tell, it should be

2H2O >>> 2H2 + O2

Answer:

a. NaCl

b. Dissociation (the compound breaking down into ions)

To express the equilibrium concentration of F2, the equation [F2] = (k -1[NO2F][F]) / (k1[NO2]) is used, based on the principle that the forward reaction rate equals the reverse reaction rate at equilibrium.

To write an expression for the equilibrium concentration of F2 in terms of the rate constants k1 and k -1, and the equilibrium concentrations of NO2, NO2F, and F, one can use the principle that at equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. For the given reaction NO2(g) + F2(g) → NO2F(g) + F(g), the forward rate is given by k1[NO2][F2] and the reverse rate by k -1[NO2F][F]. Setting these equal to each other gives us:

k1[NO2][F2] = k -1[NO2F][F]

Solving for the equilibrium concentration of F2, we have:

[F2] = (k -1[NO2F][F]) / (k1[NO2])

This expression relates the equilibrium concentration of F2 to the rate constants and the equilibrium concentrations of NO2, NO2F, and F.

By calculating the number of moles of each element (carbon, hydrogen, nitrogen, and oxygen) and determining the empirical formula as C9H15N1O11, and seeing that the molar mass of this empirical formula is close to the given molar mass of procaine, we can conclude that its molecular formula is indeed C9H15N1O11.

Given CO2 mass = 8.58 g, H2O mass = 2.70 g, and N mass = 0.279 g.

To find the amount of C and H, we'll use the molar ratios derived from the formulas of CO2 and H2O respectively. For CO2, it’s 1 mole of Carbon per mole of CO2 (MW of C = 12.01 g/mol, MW of CO2 = 44.01 g/mol). Hence moles of C = (8.58 g CO2)*(1 mole C/44.01 g CO2) = 0.195 mol C. Similar calculation for H from H2O yields 0.300 mol H. Given the mass of nitrogen (0.279 g), we find 0.02 mol N. We then subtract the masses of C, H and N from the total mass (3.54 g procaine) to get the mass of O, and consequently, 0.218 mol of O.

Now, to find the empirical formula, divide the obtained moles of each element by the smallest mole value, which is N = 0.02. Thus, the empirical formula is C9H15N1O11.

Last step for finding the molecular formula involves dividing the given molar mass of procaine (236 g/mol) by the molar mass of the above empirical formula (~207 g/mol). Getting approximately 1 as a result, the molecular formula will be the same as the empirical, that is C9H15N1O11.

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Answer: 48 Liters.

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 745 mmHg

[tex]P_2[/tex] = final pressure of gas =360 mmHg

[tex]V_1[/tex] = initial volume of gas = 26.5 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]23.3^oC=(273+23.3)K=296.3K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-13.7^oC=273+(-13.7)K=259.3K[/tex]

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]\frac{745mmHg\times 26.5L}{296.3K}=\frac{360mmhg\times V_2}{259.3K}[/tex]

[tex]V_2=48L[/tex]

Therefore, the volume of the gas will be 48 Liters.

To produce 8 moles of ammonia, 4 moles of nitrogen and 12 moles of hydrogen would have been reacted, according to the stoichiometry of the reaction.

The balanced chemical reaction for the production of ammonia (NH3) is given by: N2(g) + 3H2(g) → 2NH3(g). This implies that for every 2 moles of ammonia produced, 1 mole of nitrogen (N2) and 3 moles of hydrogen (H2) are reacted. Therefore, if 8 moles of NH3 were produced, the amount of nitrogen reacted would be 8/2 moles of N2, which is 4 moles of N2. Using the stoichiometric ratio, the moles of hydrogen reacted would be 3 x 4, which is 12 moles of H2.

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To produce 8.00 moles of NH3, the reaction would consume 4.00 moles of N2 and 12.00 moles of H2, as inferred from the stoichiometric ratios provided by the balanced equation.

The chemical equation for the creation of ammonia, a principle nitrogen fertilizer, from nitrogen and hydrogen is shown as: N2(g) + 3H2(g) → 2NH3(g). To find out how many moles of H2 and N2 were reacted to produce 8.00 moles of NH3, we use the stoichiometric ratios provided by the balanced equation. The balanced equation shows that one mole of N2 reacts with three moles of H2 to produce two moles of NH3. Therefore, if 8.00 moles of NH3 were produced, we can infer that the reaction consumed 4.00 moles of N2 and 12.00 moles of H2. This is because the ratio of N2 to NH3 in the reaction is 1:2 and the ratio H2 to NH3 is 3:2, so we divide 8 by 2 to find the moles of N2, and multiply 8 by 3/2 to find moles of H2.

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Answer: Correct answer is both A and B.

Explanation:

Chemistry is a branch of science that deals with different substances in recognizing the matter present in them.

For example, chemistry deals with states of matter like solid, liquid or gas.

Also, it tells us about how different substances present in nature react or what is their composition and properties.

Basically, chemistry helps us knowing nature of a matter and how it can be transformed.

Thus, we can conclude that chemistry is the study of matter and its transformation.

Final answer:

Chemistry is the study of matter and its properties, including macroscopic and microscopic phenomena.

Explanation:

Chemistry is the study of matter and the changes it undergoes, encompassing both macroscopic and microscopic phenomena. It focuses on substances, their properties, and how they interact and transform during chemical reactions. There is a vast array of specializations within chemistry, including but not limited to physical chemistry, organic chemistry, inorganic chemistry, analytical chemistry, and biochemistry.

Matter consists of anything that has mass and occupies space. This broad definition includes the air we breathe, the food we eat, and even materials like plastics and metals. Understanding the changes matter undergoes, whether through physical or chemical processes, is central to this science. Moreover, chemistry is often known as the central science because it intersects with other scientific disciplines such as physics, biology, and environmental science.

Answer:

[tex]\boxed{\text{0.50 mol/L}}[/tex]

Explanation:

The balanced equation is

2COF₂ ⇌ CO₂+CF₄; Kc = 9.00

1. Set up an ICE table

[tex]\begin{array}{ccccc}\rm 2COF_{2} & \, \rightleftharpoons \, & \rm CO_{2} & +&\rm CF_{4}\\2.00& & 0& & 0\\-x& & +x & & +x\\2.00 - x& & x & &x \\\end{array}[/tex]

2. Solve for x

[tex]K_{c} = \dfrac{[\rm CO][ \rm CF_{4}]}{[\rm COF_{2}]^{2}} = 9.00\\\\\begin{array}{rcl}\dfrac{x^{2}}{(2.00 - x)^{2}} & = & 9.00\\\dfrac{x}{2.00 - x} & = & 3.00\\x & = &3.00(2.00 - x)\\x & = & 6.00 - 3.00x\\4.00x & = & 6.00\\x & = & \mathbf{1.50}\\\end{array}[/tex]

3. Calculate the equilibrium concentration of COF₂

c = (2.00 - x) mol·L⁻¹ = (2.00 - 1.50) mol·L⁻¹ = 0.50 mol

[tex]\text{The equilibrium concentration of COF$_{2}$ at equilibrium is $\boxed{\textbf{0.50 mol/L}}$}[/tex]

Check:

[tex]\begin{array}{rcl}\dfrac{1.50^{2}}{0.50^{2}} & = & 9.00\\\\\dfrac{2.25}{0.25}& = & 9.00\\\\9.00 & = & 9.00\\\end{array}[/tex]

OK.

The equilibrium concentration of COF2 can be found by solving a quadratic equation derived from the equilibrium constant expression and initial concentration, and then subtracting the amount that reacted from the initial concentration.

The student asked about the equilibrium concentration of carbonyl fluoride, COF2, in a specific chemical reaction where it is converted to carbon dioxide and carbon tetrafluoride. To determine the concentration of COF2 that remains at equilibrium starting with an initial concentration of 2.00 M, we must use the equilibrium constant (Kc) value provided, which is 9.00.

The equilibrium expression for the reaction 2COF2(g) ⇌ CO2(g) + CF4(g) is Kc = [CO2][CF4] / [COF2]2. Let the change in concentration of COF2 at equilibrium be 'x'. Thus, at equilibrium, we'll have [COF2] = 2.00 - 2x, [CO2] = x, and [CF4] = x. Plugging these into the equilibrium expression, we have 9.00 = x2 / (2.00 - 2x)2. Solving for x will give us the change in concentration of COF2, from which we can determine the equilibrium concentration of COF2 by subtracting 'x' from the initial concentration.

Therefore, to find the equilibrium concentration of COF2, one would need to solve the quadratic equation resulting from the substitution of equilibrium concentrations in terms of 'x' into the Kc expression, then calculate 2.00 M - 2x.

Hey there!:

the chemical reaction between 2-iodoctane and sodium nitrite is as follows:

Answer:

On the attached picture.

Explanation:

Hello,

In the case, the reaction between 2-iodooctane and sodium nitrite, leads to the formation of an alkyl nitrite and a nitro alkane as shown on the attached picture. Once the reaction began, the salt breaks and the sodium bonds with the iodine from the 2-iodooctane to form sodium iodide, in such a way, a free radical in the second carbon is formed so the NO₂ could bond both as a nitrite and as a nitro radical; therefore, the formed species are octyl 2-nitrite and 2-nitrooctane.

Best regards.

The problem posed involves stoichiometry, requiring calculations related to reactants and the products in the given chemical reaction. Utilizing molar masses and the balanced chemical equation, we identify chlorine as the limiting reactant. Therefore, the maximum mass of aluminum chloride that can be produced when reacting 15.0g of aluminum with 20.0g of chlorine is approximately 25.0g.

The subject of this question relates to an area of chemistry known as stoichiometry, which involves calculating the amounts of reactants and products in chemical reactions. To determine the maximum mass of aluminum chloride that can be produced from 15.0g of aluminum and 20.0g of chlorine, we first need to convert these amounts to moles. We can do this by dividing the mass of each compound by its molar mass. The molar mass of aluminum (Al) is approximately 27 g/mol, and the molar mass of chlorine (Cl2) is approximately 71 g/mol. Therefore, we have approximately 0.56 moles of aluminum and 0.28 moles of chlorine.

Looking at the balanced chemical equation, we see that the reaction ratio between Al and Cl2 is 2:3. We have more than twice the number of moles of Al as Cl2, which means that Cl2 is the limiting reactant – it will be completely used up before all of the aluminum is reacted. Therefore, the amount of product (AlCl3) formed is determined by the amount of Cl2 present.

The balanced chemical equation also shows that the reaction ratio between Cl2 and AlCl3 is 3:2, so for every 3 moles of Cl2 reacted, we form 2 moles of AlCl3. Therefore, we can multiply the number of moles of Cl2 (0.28 moles) by 2/3 to find the number of moles of AlCl3 produced. We then convert this number of moles back to grams of AlCl3 by multiplying by the molar mass of AlCl3 (approximately 133 g/mol). This gives us approximately 25.0g of AlCl3 as the maximum amount that can be formed under these conditions.

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Final answer:

Sponging down with isopropanol can help lower the body temperature through rapid evaporation, which can assist in preventing dehydration. Nevertheless, athletes should prioritize proper hydration by consuming adequate fluids, particularly sports drinks that contain a balance of water, electrolytes, and sugar, to maintain endurance and performance.

Explanation:

To avoid dehydration during or after a long-distance athletic event, athletes may sponge down with isopropanol because it accelerates the evaporation process when applied to the skin. This rapid evaporation causes a cooling effect which can help to lower the body temperature. However, it is important to note that the best way to prevent dehydration is through proper hydration, which involves drinking enough fluids, such as water or sports drinks, before, during, and after exercise. These fluids should contain the necessary electrolytes like sodium, which is crucial for fluid absorption and replacing what is lost through sweat.

The American College of Sports Medicine recommends that the goal of drinking fluids during exercise is to prevent dehydration without over-hydrating. Replacing lost fluids without overconsumption is key to maintaining optimal endurance and performance. To ensure proper rehydration, it's essential to consume sports drinks that contain the correct proportions of sugar, water, and sodium. Sodium in sports drinks not only enhances fluid absorption but also aids in maintaining blood-glucose levels for muscle fuel and replaces some of the electrolytes lost in sweat.

Answer:

The expected ratio of half-lives for a reaction will be 5:1.

Explanation:

Integrated rate law  for zero order kinetics is given as:

[tex]k=\frac{1}{t}([A_o]-[A])[/tex]

[tex][A_o][/tex] = initial concentration

[A]=concentration at time t

k = rate constant

if, [tex][A]=\frac{1}{2}[A_o][/tex]

[tex]t=t_{\frac{1}{2}}[/tex], the equation (1) becomes:

[tex]t_{\frac{1}{2}}=\frac{[A_o]}{2k}[/tex]

Half life when concentration was 0.05 M=[tex]t_{\frac{1}{2}}[/tex]

Half life when concentration was 0.01 M=[tex]t_{\frac{1}{2}}'[/tex]

Ratio of half-lives will be:

[tex]\frac{t_{\frac{1}{2}}}{t_{\frac{1}{2}}'}=\frac{\frac{[0.05 M]}{2k}}{\frac{[0.01 M]}{2k}}=\frac{5}{1}[/tex]

The expected ratio of half-lives for a reaction will be 5:1.

Answer:

The correct option is d. Membranes, the brain and the nervous system.

Explanation:

The sphingolipids are a class of lipids which contain a sphingosine amino alcohol.

The main function of sphingolipid is the protection of the cells against harmful environment by forming a chemically resistant and mechanically stable membrane.

Sphingolipids are present in abundance in the brain and in the central nervous system where they are the important constituents of the plasma membranes. They are also important for the proper functioning, development and maintenance of the nervous system.

Therefore, Sphingolipids are particularly present in the membranes, the brain and the nervous system.

Calcium carbide (CaC₂) reacts with water to produce acetylene (C₂H₂): CaC₂ (s) + 2 H₂O (g) → Ca (OH)₂ (s) + C₂H₂ (g) Production of 6.5 g of C₂H₂ (MW = 26.036 g/mol) requires consumption of 2.3 gm. Hence, option B is the correct option.

The balanced chemical equation given is:

CaC₂ (s) + 2 H₂O (g) → Ca (OH)₂ (s) +C₂H₂ (g)

The molar ratio between CaC₂ and C₂H₂ is 1:1, and the molar ratio between H₂O and C₂H₂ is 2:1.

To determine the mass of H₂O required to produce 6.5 g of C₂H₂, one can use the following steps:

The number of moles of C₂H₂:

  Moles of C₂H₂ = Mass / Molar mass = 6.5 g / 26.036 g/mol

The molar ratio between  H₂O and C₂H₂ is 2:1, the number of moles of  H₂O required is half of the moles of C₂H₂:

  Moles of  H₂O = Moles of C₂H₂ / 2

The mass of  H₂O using its moles and molar mass:

  Mass of  H₂O = Moles of  H₂O × Molar mass of H₂O

1. Moles of C₂H₂ = 6.5 g / 26.036 g/mol ≈ 0.2494 mol

2. Moles of  H₂O = 0.2494 mol / 2 ≈ 0.1247 mol

3. Mass of  H₂O = 0.1247 mol × 18.016 g/mol ≈ 2.244 g

So, the consumption of H₂O required to produce 6.5 g of C₂H₂ is approximately 2.244 g.

The closest option to this value is 2.3 g, so the correct answer is:

B. 2.3

Learn more about water consumption here.

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