Give the formula for the alkene containing 15 carbons.

Answer:

Rocky.

Explanation:

Jovian planets are giant gas balls not unlike the SUN although they have a small rocky central core. Pluto is a rock ice planet---more like Europa, a satellite of Jupiter. In fact, Pluto is probably the largest of the so-called KUIPER BELT objects

Answer:

Defective collagen proteins affect an individuals joints, as collagen is an important component protein in body connective tissues

Explanation:

Collagen is a crucial protein in the extracellular matrix of various connective tissues in body (tendons, ligaments, muscles). It constitutes major protein component in mammals body.

Collagen Vascular disease is a group of diseases affecting 'collagen' connecting tissue. Apart from inheritance, they can con occur due to autoimmune disease. This implies that body's immune system fights against itself & mistakenly damages its own tissues. So, these diseases have a huge bearing on body joints components. Eg : Lupus , Arthritis are common collagen vascular diseases. Arthritis mostly affects adults above 30 years of age, lupus can be diagnosed in younger people above 15 years of age also.

Answer:

pH in the [tex]H_{2}/H^{+}[/tex] half cell is 0.84.

Explanation:

Oxidation: [tex]Zn(s)-2e^{-}\rightarrow Zn^{2+}(aq.)[/tex]

Reduction: [tex]2H^{+}(aq.)+2e^{-}\rightarrow H_{2}(g)[/tex]

-------------------------------------------------------

Overall: [tex]Zn(s)+2H^{+}(aq.)\rightarrow Zn^{2+}(aq.)+H_{2}(g)[/tex]

[tex]E_{cell}^{0}=E_{H^{+}\mid H_{2}}^{0}-E_{Zn^{2+}\mid Zn}^{0}[/tex]  = (0.00 V) + (0.76 V) = 0.76 V

According to Nernst equation for this cell reaction at room temperature (298 K):              

                           [tex]E_{cell}=E_{cell}^{0}-\frac{0.0592}{n}log\frac{[Zn^{2+}].P_{H_{2}}}{[H^{+}]^{2}}[/tex]

where, [tex]E_{cell}[/tex] is cell potential, n is number of electron exchanged, [tex]P_{H_{2}}[/tex] is pressure of [tex]H_{2}[/tex] in atm and species under third bracket represent molarity of the respective species.

So, [tex]0.68V=0.76V-\frac{0.0592}{2}log\frac{(1.3M)\times (8atm)}{[H^{+}]^{2}}V[/tex]

   [tex]\Rightarrow[/tex] [tex][H^{+}]=0.1436M[/tex]

pH = [tex]-log[H^{+}][/tex] = -log(0.1436) = 0.84

Final answer:

The pH in the H+/H2 half-cell of the given galvanic cell is 8.5.

Explanation:

In the given galvanic cell, the reduction half-reaction is 2H+ (aq) + 2e → H₂(g), and the overall reaction is Zn(s) + 2H+ (aq) -> Zn²+ (aq) + H₂(g). The reduction potential for the H2(g)/H+(aq) half-reaction is 0.00 V, and for the Zn(s)/Zn²+(aq) half-reaction is -0.76 V.

To find the pH in the H+/H2 half-cell, we can use the Nernst equation:

Ecell = E°cell - (0.0592 V / n) * log[H+]

Using the given cell potential of 0.68 V and plugging in the values, we can calculate the pH in the H+/H2 half-cell to be 8.5.

Answer:

a) p1v1= p2v2

852 x 40.0= 4260 x v2

v2= 8 L

b) 852 x 40 = p2 x 69.0

p2= 494 torr

Answer: c: The temperature decreases because the most energetic molecules escape

Explanation:

Evaporation is surface phenomenon in which liquid molecules gain energy from surrounding molecules and thus these high energy molecules escape from the surface in the form of vapors thus leaving low energy molecules in the system. As the kinetic energy of the left molecules decreases, the temperature of the molecules decreases as kinetic energy is directly proportional to temperature. Thus low temperature results in cooling of the system.

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Answer:

The He₂ 2+ ion is more stable since it has a higher bond order (bond order = 1) than the He₂ + ion (bond order = 1/2).

Explanation:

Molecular orbital of He₂⁺

[tex]1\sigma_{1s}^21\sigma(star)_{1s}^1[/tex]

There are two electrons in bonding and 1 electron in antibonding orbital

Bond order = [tex]\frac{(2-1)}{2}[/tex]    

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

Molecular orbital of He₂⁺²

[tex]1\sigma_{1s}^21\sigma(star)_{1s}^0[/tex]

There are two electrons in bonding and 0 electron in antibonding orbital

Bond order = [tex]\frac{(2-0)}{2}[/tex]

= 1

So bond order of He₂⁺² is 1 which is more stable than He₂⁺ whose bond order is   [tex]\frac{1}{2}[/tex]   .

In comparing the He2 2+ and He2 + ions and viewing their stability through the lens of Molecular Orbital Theory, we see that the He2 + ion is more stable due to its higher bond order.

Using Molecular Orbital Theory, He2 2+ and He2 + ions can be compared based on their bond order. Bond order is a measure of the stability of a bond. A higher bond order signifies greater stability. The He2 + ion has a bond order of 1, while the He2 2+ ion has a bond order of 0.5. Therefore, the major difference between He2 2+ and He2+ comes from their bond orders, and the He2+ ion is considered more stable because its bond order is higher.

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Answer:

Part A

 The volume of the gaseous product  is  [tex]V = 787L[/tex]

Part B

The volume of the the engine’s gaseous exhaust is  [tex]V_e = 2178 \ L[/tex]

Explanation:

Part A

From the question we are told that

    The temperature is  [tex]T = 350^oC = 350 +273 =623K[/tex]

     The pressure is  [tex]P = 735 \ torr = \frac{735}{760} = 0.967\ atm[/tex]

     The of  [tex]C_8 H_{18} = 100.0g[/tex]

The chemical equation for this combustion is

               [tex]2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}[/tex]

 The number of moles of  [tex]C_8 H_{18}[/tex] that reacted is mathematically represented as

               [tex]n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }[/tex]

The molar mass of  [tex]C_8 H_{18}[/tex] is constant value which is

                  [tex]M = 114.23 \ g/mole[/tex]  

So          [tex]n = \frac{100 }{114.23} }[/tex]

             [tex]n = 0.8754 \ moles[/tex]

The gaseous product in the reaction is [tex]CO_2_{(g)}[/tex] and water vapour

Now from the reaction

    2 moles of [tex]C_8 H_{18}[/tex]  will react with 25 moles of [tex]O_2[/tex] to give (16 + 18) moles of [tex]CO_2_{(g)}[/tex] and  [tex]H_2 O_{(g)}[/tex]

So

    1 mole of [tex]C_8 H_{18}[/tex] will  react with 12.5 moles of  [tex]O_2[/tex] to give 17 moles of [tex]CO_2_{(g)}[/tex] and  [tex]H_2 O_{(g)}[/tex]

This implies that

    0.8754 moles of [tex]C_8 H_{18}[/tex] will react with (12.5 * 0.8754 ) moles of [tex]O_2[/tex] to give  (17 * 0.8754) of [tex]CO_2_{(g)}[/tex] and  [tex]H_2 O_{(g)}[/tex]

So the no of moles of gaseous product is

         [tex]N_g = 17 * 0.8754[/tex]

         [tex]N_g = 14.88 \ moles[/tex]

From the ideal gas law

       [tex]PV = N_gRT[/tex]

making V the subject

        [tex]V = \frac{N_gRT}{P}[/tex]

Where R is the gas constant with a value [tex]R = 0.08206 \ L\cdot atm /K \cdot mole[/tex]

Substituting values

          [tex]V = \frac{14.88* 0.08206 *623}{0.967}[/tex]

          [tex]V = 787L[/tex]

Part B

From the reaction the number of moles of oxygen that reacted is

         [tex]N_o = 0.8754 * 12.5[/tex]

         [tex]N_o = 10.94 \ moles[/tex]

The volume is

      [tex]V_o = \frac{10.94 * 0.08206 *623}{0.967}[/tex]

      [tex]V_o = 579 \ L[/tex]

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         [tex]V_e = V_o * \frac{0.79}{0.21}[/tex]

Substituting values

       [tex]V_e = 579 * \frac{0.79}{0.21}[/tex]

       [tex]V_e = 2178 \ L[/tex]

The total volume of gaseous products produced from the combustion of 100g C8H18 is approximately 382 L. By considering the composition of air and that N2 and Ar do not react, you can calculate the total volume of the gaseous exhaust.

Upon seeing this question, the first step is to write the balanced chemical equation for the combustion of C8H18, which is ubiquitous in gasoline:

C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O

This looks complex, but the Ideal Gas Law simplifies everything. According to the stoichiometric ratio, 1 mole of C8H18 will produce 8 moles of CO2 and 9 moles of H2O when burnt, totalling 17 moles of gas. The molar mass of C8H18 is around 114 g/mol, so 100g is approximately 0.877 moles.

Therefore, upon combustion, 0.877 * 17 = 14.9 moles of gas are produced. Applying the Ideal Gas Law (PV=nRT), keeping in mind that 350°C is 623K and 1 atm is 760 torr, we have: (nR(623K)/735torr) = V. Plugging the numbers and making sure to keep the units consistent, we get V = 14.9 x 0.0821 x 623 / 0.967 = approximately 382 L.

Coming to part (b), considering the composition of air and assuming no N2 or Ar reacts, you find that for every L of O2 burned, 1L of air is consumed. Adding the total volume of gas from part (a) and the extra nitrogen and argon, you find the total volume of the engine's gaseous exhaust.

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Answer:

Functional groups are group of atoms which are responsible for the reactivity of compound. In ethylene glycol, organic functional group is OH which is ALCOHOL.

Explanation:

Ethylene glycol, the starting material in the reaction, contains two reactive hydroxyl groups (-OH). These groups enable it to react with terephthalic acid, which has carboxylic acid functional groups, resulting in a polymerization reaction that forms a plastic named polyethylene terephthalate.

Ethylene glycol, also known as 1,2-ethanediol, is an organic compound with two hydroxyl groups (-OH). These hydroxyl groups are the reactive functional groups in ethylene glycol, making it a type of alcohol. In the context of the reaction mentioned, ethylene glycol can react with terephthalic acid, another organic molecule with carboxylic acid (-COOH) as the functional group, which will yield a polymer called polyethylene terephthalate (PET).

PET is a type of plastic used commonly for making bottles, films, and fibers for clothing. The formation of plastics by reaction of alcohols (like ethylene glycol) and carboxylic acids (like terephthalic acid) is known as a polymerization reaction. In these reactions, many small monomer units (the alcohol and the acid, in this case) join together to form a large polymer molecule.

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When two or more atoms loss or gain electrons to produce an ion there is a formation of  ionic bonds.

The bond is produced when an atom, specially a metal, loses an electron or electrons, and becomes a positive ion, or cation. Another atom, typically a non-metal, is able to get the electron(s) to become a negative ion, or anion.

Ionic bonds affect a cation and an anion.Ionic bonds are produce only between metals and nonmetals.

The ionic bond is produce through the transportation of electrons from the metal atoms to the non-metal atoms. The metal atoms lose their valence electrons to attain a stable noble gas electron arrangement.

Thus, The strong electrostatic forces of attraction between the oppositely-charged ions is known as ionic bond.

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Answer:

Fiber is "Not Digestible"

Explanation:

Carbohydrates that contain fiber cannot be completely digested. the indigestible components of fiber are measured in the calorimeter, but they are not accessible for energy in the human body.

While estimating the Caloric content of the chemical constituents present in the food item, the carbohydrate amount is discounted to some extent compared to the average value measured using bomb calorimetry. This is because the fiber content is not digestible.

At its most fundamental, a calorie is a measure of electricity. One Calorie (same as one kilocalorie, or 1,000 energy) is the quantity of power that is required to warm one kilogram of water 1 degree Celsius at sea stage. The strength content of meals become traditionally measured by the usage of bomb calorimetry.

As an alternative, the entire caloric price is calculated by using including the calories supplied by using the strength-containing nutrients: protein, carbohydrate, fats, and alcohol.

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Answer:

The limiting reactant is Carbon dioxide, [tex]CO_{2}[/tex]

Explanation:

The balanced reaction equation is:

[tex]2NH_{3} + CO_{2}[/tex] → [tex]CH_{4} N_{2} O + H_{2} O[/tex]

The mole ratio of ammonia to carbon dioxide is 2:1

142100/17g = 8358.8 mol of NH3

211400/44g = 4, 804.5 mole of CO2

Now:

4,804.5 mol of CO2 × [tex]\frac{2 mol NH_{3} }{1 molCO_{2} }[/tex] = 9,609 mol of NH3  present

8,358.8 mole of NH3 × [tex]\frac{1 moleCO_{2} }{2moles NH_{3} }[/tex] = 4,179.4 mol of CO2 present

NH3 needs 8, 358.8 moles but had 9, 609 moles⇒ excess reactant

CO2 needs 4, 804,5 mol but had 4, 179.4 moles⇒ limiting reactant (used up completely)

The limiting reactant is carbon dioxide, CO2.

Ammonia is the limiting reactant in the synthesis of urea from ammonia and carbon dioxide.

The reaction between ammonia (NH3) and carbon dioxide (CO2) produces urea (CH4N2O) and water (H2O) according to the balanced equation: 2NH3(aq) + CO2(aq) → CH4N2O(aq) + H2O(l).

To determine the limiting reactant, we need to compare the amount of each reactant used to the amount of urea produced. From the given information, 142.1 kg of ammonia and 211.4 kg of carbon dioxide react to produce 171.4 kg of urea.

We can use the stoichiometry of the balanced equation to find the theoretical yield of urea from both reactants:

Since the actual yield of urea is 171.4 kg, which is less than both the theoretical yields from ammonia and carbon dioxide, the limiting reactant is ammonia (NH3).

Therefore, the limiting reactant is ammonia (NH3).

Answer:

Explanation:

In ion exchange softener , the minerals  of hard water are in the form of ions of Mg⁺² and Ca⁺² . They are replaced by Na⁺ ion . The Na⁺ that resides on the bed of resin used in ion exchange softener make all heavy ions like Mg⁺² and Ca⁺² . In this way all the hard ions of hard water are removed and they are replaced by Na⁺ ion . hard ions are removed because they form insoluble compounds so they are precipitated out.

The ion that remains in soft water is Na⁺ ion.

The negative aspect of consuming this ion is that it is harmful for those suffering from heart ailment like heart pressure.

Answer: Please See answers below

Explanation:

A. Ions in mobile phase are attracted to counterions covalently attached to stationary phase  -----Ion Exchange Chromatography

B. Solute in mobile phase is attracted to specific groups covalently attached to stationary phase -----Affinity Chromatography

C. Solute equilibrates between mobile phase and surface of stationary phase  -----Adsorption Chromatography

D. Solute equilibrates between mobile phase and film of liquid attached to stationary phase  --- Partition Chromatography

E. Different-sized solutes penetrate pores in stationary phase to different extents. Large solutes are eluted first. ---Molecular Exclusion Chromatography

Answer:

SiH4 is nonpolar and BBr3 is nonpolar and SiF4 is nonpolar.

Explanation:

SiH4 is a non-polar compound. Though the Si–H bonds are polar, as a result of different electronegativities of Si and H. However, as there are 4 electron repulsions around the central Si atom, the polar bonds are arranged symmetrically around the central atom having a tetrahedral shape hence they cancel out making the compound nonpolar.

SiF4 is a nonpolar molecule because the fluorine atoms are arranged symetrically around the central silicon atom in a tetrahedral molecule with all of the regions of negative charge cancelling each other out just like in SiH4.

The 3 bromine atoms all lie in the same plane thus the geometry of the compound will be trigonal planar. The BBr3 will be non polar because the three B-Br bonds will cancel out each others' dipole moment given that they are in the same plane.

A non-polar chemical is SiH4. Despite the fact that the Si-H bonds are polar, this is because Si and H have differing electronegativities.

Thus, The polar bonds are symmetrically organized around the center Si atom, which has a tetrahedral form, and thus cancel out because there are four electron repulsions around it, rendering the combination nonpolar.

Because the fluorine atoms are symmetrically positioned around the center silicon atom in a tetrahedral molecule with all of the negative charge regions canceling each other out, SiF4 is a nonpolar molecule and electronegativities.

The geometry of the compound will be trigonal planar since all three bromine atoms are located in the same plane. The three B-Br bonds will cancel, making the BBr3 non-polar.

Thus, A non-polar chemical is SiH4. Despite the fact that the Si-H bonds are polar, this is because Si and H have differing electronegativities.

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Answer:

1.822 g of magnesium hydroxide would be produced.

Explanation:

Balanced reaction: [tex]2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl[/tex]

     Compound                                 Molar mass (g/mol)

         NaOH                                           39.997

         [tex]MgCl_{2}[/tex]                                           95.211

        [tex]Mg(OH)_{2}[/tex]                                        58.3197

So, 2.50 g of NaOH = [tex]\frac{2.50}{39.997}[/tex] mol of NaOH = 0.0625 mol of NaOH

      4.30 g of [tex]MgCl_{2}[/tex]  = [tex]\frac{4.30}{95.211}[/tex] mol of [tex]MgCl_{2}[/tex] = 0.0452 mol of [tex]MgCl_{2}[/tex]

According to balanced equation-

2 mol of NaOH produce 1 mol of [tex]Mg(OH)_{2}[/tex]    

So, 0.0625 mol of NaOH produce [tex](\frac{0.0625}{2})[/tex] mol of NaOH or 0.03125 mol of NaOH

1 mol of [tex]MgCl_{2}[/tex] produces 1 mol of [tex]Mg(OH)_{2}[/tex]

So, 0.0452 mol of [tex]MgCl_{2}[/tex] produce 0.0452 mol of [tex]Mg(OH)_{2}[/tex]

As least number of moles of [tex]Mg(OH)_{2}[/tex] are produced from NaOH therefore NaOH is the limiting reagent.

So, amount of [tex]Mg(OH)_{2}[/tex] would be produced = 0.03125 mol

                                                                           = [tex](0.03125\times 58.3197)[/tex] g

                                                                           = 1.822 g

To find the grams of magnesium hydroxide produced, you need to balance the chemical equation and calculate stoichiometry.

To find the grams of magnesium hydroxide produced, we need to balance the chemical equation and calculate the stoichiometry.

The balanced chemical equation for the reaction is:
2NaOH + MgCl2 → Mg(OH)2 + 2NaCl

Using the molar masses of sodium hydroxide (NaOH), magnesium chloride (MgCl2), and magnesium hydroxide (Mg(OH)2), we can calculate the grams of magnesium hydroxide produced:

After performing these calculations, we find that 4.59g of magnesium hydroxide would be produced.

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Answer:

OH OH ONa NaOH OH

Explanation:

See attached image

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = [tex]\mathbf{10^{-14.1963} }[/tex] × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = [tex]\mathbf{10^{-17.5804} }[/tex] × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

The cathode is where reduction occurs, it's regarded as the 'positive' electrode, and in electrochemical cells, it's where metal is plated out. Negative ions do not necessarily flow toward the cathode.

In an electrochemical cell, some statements are always true for the cathode. The cathode is considered the "positive" electrode in the cell, it is where the reduction reaction (gain of electrons) occurs, and it is the location where metal is plated out (in electrolytic cells). Negative ions do not flow toward the cathode; ions flow based on their charge and the charge of the electrode. Positive ions (cations) move toward the cathode.

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Answer:

the mesopelagic, dysphotic, or twilight zone

Explanation:

Marine zones are the divisions of the ocean. The ocean is divided into two basic parts; the pelagic or open ocean, and the benthic or sea floor.

The pelagic zone is further divided into five broad zones according to how far down sunlight penetrates and they are:

1) the epipelagic, euphotic, or sunlit zone: the top layer of the ocean where enough sunlight penetrates for plants to carry on photosynthesis.

2) the mesopelagic, dysphotic, or twilight zone: a dim zone where some light penetrates, but not enough for plants to grow.

3) the bathypelagic, aphotic, or midnight zone: the deep ocean layer where no light penetrates.

4) the abyssal zone: the pitch-black bottom layer of the ocean; the water here is almost freezing and its pressure is immense.

5) the hadal zone: the waters found in the ocean's deepest trenches.

Answer:

The concentration of NaOH is 2.47 M

Explanation:

The equation of the reaction is

HCl (aq) + NaOH(aq) ------> NaCl(aq) + H2O(l)

Given;

Concentration of acid CA= 1.25 M

Volume of acid VA= 5.0L

Concentration of base CB= ????

Volume of base VB= 2.53 L

Number of moles of acid NA= 1

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CB= CAVANB/VBNA

CB= 1.25 × 5.0 ×1 / 2.53×1

CB= 2.47 M

Therefore the concentration of NaOH is 2.47 M

Answer:

2 unpaired Electrons

Explanation:

Answer:

I was told 2 I'm not so sure, But if it's wrong I'm Sorry :( I'm having a sorta bad day...I'm trying to help everyone to make my troubles go away god bless you and have a wonderful day, ~~ Night.

Answer:

the concentration of Pb2+ ions is  [tex][Pb^{2+}] } = 0.15 mol/L[/tex]

Explanation:

From the question we are told that

     The concentration of [tex]Cl^{-}[/tex] is [tex]C_s = 0.0400 M[/tex]

The reaction is  

           [tex]Pb^{2+} + 2Cl^- ----> PbCl_2[/tex]

The solubility constant for this reaction is mathematically represented as

        [tex]K_{sp} = [Pb^{2+}][Cl^{-}]^2[/tex]

Substituting values

      [tex]2.4 0 * 10^{-4} = 0.0400^2 * [Pb^{2+}][/tex]

      [tex][Pb^{2+}] } = \frac{2.4 0 * 10^{-4} }{0.0400^2 }[/tex]

       [tex][Pb^{2+}] } = 0.15 mol/L[/tex]

Answer:

8.0*10^2

Explanation:

What volume (in milliliters) of a 15% (m/v) NaOH solution contains 120 g NaOH?

For 15% (m/v), the conversion factor can be 15%/100 mL or 100 mL/15%. In this case, 100 mL/15% is used because we need mL as our final expression. 100 mL/15% * 120 g NaOH= 800 mL NaOH. In scientific notation: 8.0*10² or 8.0e² mL NaOH.

To find the volume of a 15% (m/v) NaOH solution that contains 120 grams of NaOH, use the formula V = (mass of solute x 100) / %m/v, which results in 800 mL. Therefore, the correct answer is E) 800 mL.

To find the volume of a 15% (m/v) NaOH solution that contains 120 grams of NaOH, we will use the formula for mass/volume percentage concentration:

%m/v = (mass of solute/volume of solution) x 100%

Given:

Mass of NaOH (solute) = 120 g

Concentration of NaOH solution = 15% (m/v)

We need to find the volume of the solution (V).

Rearranging the formula to solve for V:

V = (mass of solute x 100%) / %m/v

Substituting the given values:

V = (120 g x 100%) / 15%

Converting the percentage:

V = (120 g x 100) / 15 = 12,000 / 15 = 800 mL

Therefore, 800 mL of a 15% NaOH solution contains 120 g NaOH. The correct answer is E) 800 mL.

What volume (mL) of a 15% (m/v) NaOH solution contains 120 g NaOH?

A) 18 mL

B) 0.13 mL

C) 13 mL

D) 120 mL

E) 800 mL

Answer:

CO HAS A TRIPLE BOND WHILE C-O BOND IN CO2 IS A DOUBLE BOND

CO HAS A LONE PAIR ON CARBON WHILE CO2 DOES NOT

Explanation:

Bond dissociation bond enthalpy or energy is the energy needed to break 1 mole of a divalent molecule into separate atoms mostly in the gaseous state.

The carbon and oxygen in carbon monoxide form a triple bond as carbon monoxide has 10 electrons in their outermost shell which results into six shared electrons in 3 bonding orbitals as against the double bond formed by other carbon compounds. Four electrons come from oxygen and the remaining two from carbon and due to this, two electrons from oxygen will occupy one orbital and this forms a dative bond. Also because of the triple bond, carbon monoxide is often regarded as a more stable compound than carbon dioxide with a double bond. This gives it its higher bond dissociation enthalpy value and more energy is needed to break it into its separate atoms. This is in conjunction with a larger bond length similar to the bong length in a triple bond. This makes it more stronger than the bond dissociation enthalpy of carbon dioxide having a double bond.

Answer: provided in the answer segment

Explanation:

Below is a step by step process to analyzing this problem

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From 1H-NMR singlet at 5.80 ppm show N-H peak as shown in structure.

Here, H(A) hydrogen has no neighbor hydrogen so it appears integrated singlet at 7.38 ppm.      

H(C) hydrogen has the next 2 neighbor hydrogen H(B) and H(D) so it appears as a triplet at region 7.23-7.29 ppm.

H(B) hydrogen has next to one neighbor hydrogen H(C) show doublet at 6.92-6.98 ppm.

H(D) hydrogen has next to one neighbor hydrogen H(C) show doublet at 7.28-7.32 ppm.

(b). From our basic chemistry knowledge, we know that benzene molecule is planer so H(A) is more deshielding because of two substituent groups than H(C), which makes the delta value of H(A) is greater than H(C).

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The 1H NMR spectrum of 3-nitroaniline compounds has four different protons which can be differentiated by their chemical shift and coupling pattern. HA is usually the most deshielded due to its closeness to the nitro group. The coupling pattern helps especially in differentiating protons like HA and HC.

In 3-nitroaniline compound, the protons, designated as HA, HB, HC, and HD all show up distinctly in the 1H NMR spectrum. The key to differentiating them comes down to their chemical shift and their coupling pattern.

The most deshielded proton (highest chemical shift) is usually HA due to its proximity to the nitro group. HB and HC are the aromatic protons on the benzene ring, and their positions relative to the aniline (amino) group aids in their differentiation. Finally, HD is usually the most shielded proton (lowest chemical shift), as it is the amino proton.

Coupling helps to differentiate between otherwise closely overlapping protons. Due to the nature of the coupling constants involved, HA and HC will generally have distinct coupling patterns that allow them to be differentiated.

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Answer:

Explanation:

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Final answer:

The number of particles in a mole is far greater than the number of grains of sand on Earth's beaches, showcasing the mole's significance in representing vast quantities of atoms or molecules in chemistry.

Explanation:

When comparing the number of grains of sand on Earth’s beaches to the number of particles in a mole, it becomes evident that a mole represents a vastly larger quantity. The number of particles in a mole is 6.022x10²³, which dwarfs the estimated 7.5x10¹¸ grains of sand on Earth's beaches. This expansive difference illustrates that the mole is an incredibly large unit, used to easily convey numbers of particles or atoms that would otherwise be incomprehensible in scale. A mole's enormity is further illustrated by examples such as a mole of paper sheets extending more than a million times the distance from the Earth to the Sun, or a mole of sand filling a cube about 32 kilometers on a side. Therefore, the number of particles in a mole is far greater than the number of grains of sand on Earth’s beaches, highlighting the mole's utility in chemistry for working with the huge numbers of atoms and molecules involved in even small chemical samples.

The number of particles in a mole is about  [tex]\(7.5 \times 10^{18}\),[/tex] (blank) the number of grains of sand on Earth’s beaches.

The number of particles in a mole, also known as Avogadro's number, is approximately [tex]\(6.022 \times 10^{23}\).[/tex]

Comparing this to the estimated number of sand grains on Earth's beaches, which is on the order of [tex]\(7.5 \times 10^{18}\),[/tex]  we can conclude that the number of particles in a mole is significantly larger than the number of sand grains on Earth's beaches.

In fact, the number of particles in a mole is approximately 80 million times greater than the number of sand grains on Earth's beaches.

This stark difference highlights the immense size of a mole, indicating that it represents an enormous quantity of particles.

Therefore, a mole encompasses an incredibly large number of particles, underscoring its importance in chemistry and the sciences.

Answer:

Explanation:

Diiodine pentoxide is I₂O₅

Reaction with water

I₂O₅ + H₂O = 2HIO₃

2 )   HIO₃  is called iodic acid.

Final answer:

Nonmetal oxides react with water to form acids. One example is the reaction of diiodine pentaoxide with water to produce iodic acid.

Explanation:

Nonmetal oxides react with water to form acids. One example is the reaction of diiodine pentaoxide with water to produce iodic acid.

(1) Equation:

Diiodine pentaoxide + Water → 2HI + HIO3

(2) Acid Name:

Iodic acid

Answer:

[H+] = 0.0562 M

Explanation:

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H⁺ concentration in a solution with a pH of 1.25 is 0.0562.

pH may be defined as a measure of  acidity of a substance.

More precisely, pH is defined as the negative logarithm of the hydrogen ion concentration.

                                               pH = - log [H⁺]

The range of pH extends from zero to 14.

A pH value of 7 is neutral, because pure water has a pH value of exactly 7. Values lower than 7 are acidic; values greater than 7 are basic or alkaline.

pH measurements can be made using litmus paper or  pH paper known to change colors around a certain pH value.

A universal indicator is a mixture of indicator solutions intended to provide a color change over a pH range of 2 to 10.

pH measurements can also be done with pH meters that give the exact value of pH.

Given,

pH = 1.25

We know that, pH = - log [H⁺]

                        [H⁺] = antilog [ -pH]

                          [H⁺] = antilog [ -1.25]

                           = 0.0562 M

Therefore, H⁺ concentration in a solution with a pH of 1.25 is 0.0562.

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Answer:

Explanation:

conjugate acid, based on Brønsted–Lowry acid–base theory, is a chemical compound that is formed by the reception of a proton by a base

a. CH₃COOH + H₂O ⇌ H₃0⁺ + CH₃C00-

Acid <> CH₃COOH

Base <> H₂O

Conjugate acid <> H₃0 +

Conjugate base <>CH₃C00-

b. HCO₃ + H₂O ⇌ H₂CO₃⁻ + OH⁻

Acid <> H₂O

Base <> HCO₃

Conjugate acid <> H₂CO₃⁻

Conjugate base <>OH⁻

C. HNO₃ + SO₄²⁻ ⇌ HSO₄⁻ + NO₃⁻

Acid <>HNO₃

Base <>SO₄²⁻

Conjugate acid <>HSO₄⁻

Conjugate base <>NO₃⁻

A Bronsted acid is reffered to as a proton donor while a Bronsted base is a proton acceptor

In the reaction CH3COOH + H2O -> H3O+ + CH3C00-, CH3COOH is the proton donor and H2O is the proton acceptor. In the reaction HCO3- + H2O -> H2CO3 + OH-, HCO3- is the proton donor and H2O is the proton acceptor. In the reaction HNO3 + SO42- -> HSO4- + NO3-, HNO3 is the proton donor and SO42- is the proton acceptor.

a. In the reaction CH3COOH + H2O -> H30+ + CH3C00-, CH3COOH is the proton donor (acid) and H2O is the proton acceptor (base). The conjugate acid-base pairs are CH3COOH/CH3COO- and H3O+/H2O.

b. In the reaction HCO3- + H2O -> H2CO3 + OH-, HCO3- is the proton donor (acid) and H2O is the proton acceptor (base). The conjugate acid-base pairs are HCO3-/H2CO3 and H2O/OH-

c. In the reaction HNO3 + SO42- -> HSO4- + NO3-, HNO3 is the proton donor (acid) and SO42- is the proton acceptor (base). The conjugate acid-base pairs are HNO3/NO3- and HSO4-/SO42-