For the following reaction, draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
When drawing hydrogen atoms on a carbon atom, either include all hydrogen atoms or none on that carbon atom, or your structure may be incorrect.
Select the correct IUPAC name for the organic reactant:
a) 2-methylbutene
b) 2-methyl-1-butene
c) 3-methyl-3-butened) 3-methylbutene

Answer:

Hydrogen peroxide is ans

Answer: 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]{\text{Moles of} H_2}=\frac{1.35g}{2.01g/mol}=0.672moles[/tex]

[tex]\text{Moles of} S_8=\frac{6.86g}{256.5g/mol}=0.0267moles[/tex]

[tex]8H_2+S_8\rightarrow 8H_2S[/tex]

According to stoichiometry :

1 mole of [tex]S_8[/tex] require = 8 moles of [tex]H_2[/tex]

Thus 0.0267 moles of [tex]S_8[/tex] will require=[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex]  of [tex]H_2[/tex]

Thus [tex]S_8[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.

As 1 mole of [tex]S_8[/tex] give = 8 moles of [tex]H_2S[/tex]

Thus 0.0267 moles of [tex]S_8[/tex] give =[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex]  of [tex]H_2S[/tex]

Mass of [tex]H_2S=moles\times {\text {Molar mass}}=0.214moles\times 34.08g/mol=7.29g[/tex]

Thus 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.

Answer:

1 particle of the gas would occupy a volume of 3.718*10⁻²³L

Explanation:

Hello,

1. The sample has a particle of 6.022×10²²particles

2. Volume of the sample = 22.4L

3. Temperature of the sample = 0°C = (0 +273.15)K = 273.15K

4. Pressure of the sample = 1.0atm

What volume would 1 particle of the gas occupy?

But we remember that 1 mole of any substance = 6.022×10²² molecules or particles or atoms

What would be the number of moles for 1 particule?

1 mole = 6.022×10²² particles

X moles = 1 particle

X = (1 × 1) / 6.022×10²² particles

X = 1.66×10⁻²⁴ moles

Therefore, 1 particle contains 1.66×10⁻²⁴ moles

Since we know our number of moles, we can proceed to use ideal gas equation,

Ideal gas equation holds for all ideal gas and is defined as

PV = nRT

P = pressure of the ideal gas

V = volume the gas occupies

n = number of moles of the gas

R = ideal gas constant = 0.082 L.atm / mol.K

T = temperature of the gas

PV = nRT

Solving for V,

V = nRT/ P

We can now plug in our values into the above

equation.

V = (1.66*10⁻²⁴ × 0.082 × 273.15) / 1

V = 3.718*10⁻²³L

Therefore, 1 particule of the gas would occupy a volume of 3.718*10⁻²³L.

Answer:

1. Acid-Base Reaction

Explanation:

Fe(OH)2(s) + HCl(aq) = FeCl2(aq) + H2O(l)

base               acid

This a reaction between base and acid.

Ferrous hydroxide is an inorganic alkaline compound whereas hydrochloric acid is an acid. The reaction between Fe(OH)₂and HCl is an acid-base reaction. Thus, option 1 is correct.

An acid-base reaction is a chemical change that occurs and takes place when the reactant constitutes an acid and a base. They are characterized by the exchange of protons that results in the formation of conjugate bases and acids or salt.

The acid-base chemical reaction is shown as,

Fe(OH)₂(s) + HCl(aq) ⇒ FeCl₂(aq) + H₂O(l)

Here, ferrous hydroxide is a base with hydroxide ions and hydrochloric acid is an acid with hydrogen ions. HCl donates its proton to form water molecules with hydroxide ions of ferrous hydroxide.

Therefore, in option 1. the reaction is an acid-base reaction.

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Answer:

The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)

Explanation:

The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)

Answer:

Explanation:

From the question, osmotic pressure exerted by a solution is equal to the MOLARITY multiplied by the absolute TEMPERATURE and the GAS CONSTANT r.

Let P = osmotic pressure,

C = molarity, then

T = absolute temperature

r=gas constant

The Osmotic pressure Equation exerted by a solution [tex]P=C*T*r[/tex]

[tex]P=CTr[/tex]

Then it was required in the question to write an equation that will let you calculate the molarity c of this solution, and this equation should contain ONLY symbols

C= molarity of the solution

P=osmotic pressure

r = gas constant

T= absolute temperature

[tex]C=P/(rT)[/tex]

The equation that will let us calculate the molarity c of this solution = [tex]C=P/(rT)[/tex]

Answer:

Explanation:

In cobalt - 65 ,

no of protons is 27 ( p )

no of neutron = 65 - 27 ( n )

= 38

n / p ratio

=  38 / 27

= 1.41

If case of emission of alpha particle

no of proton p = 27 - 2 = 25

no of neutrons = 38 - 2 = 36

n / p ratio = 36 / 25

= 1.44

So it increases

In case of emission of beta particle

No of neutron n = 38 - 1 = 37

No of proton = 27 + 1 = 28

n / p ratio = 37 / 28

= 1.32

Hence ratio decreases.

Hence beta ray decay will result in decrease in n / p ratio.

Answer:

21 mL of NaOH is required.

Explanation:

Balanced reaction: [tex]HBr+NaOH\rightarrow NaBr+H_{2}O[/tex]

Number of moles of HBr in 11.0 mL of 0.30 M HBr solution

= [tex](\frac{0.30}{1000}\times 11.0)[/tex] moles = 0.0033 moles

Let's say V mL of 0.16 M NaOH solution is required to reach equivalence point.

So, number of moles of NaOH in V mL of 0.16 M NaOH solution

= [tex](\frac{0.16}{1000}\times V)[/tex] moles = 0.00016V moles

According to balanced equation-

1 mol of HBr is neutralized by 1 mol of NaOH

So, 0.0033 moles of HBr are neutralized by 0.0033 moles of NaOH

Hence, [tex]0.00016V=0.0033[/tex]

           [tex]\Rightarrow V=\frac{0.0033}{0.00016}=21[/tex]

So, 21 mL of NaOH is required.

Answer: they blow up

Explanation: add them together and they will blow up

Answer:

Magnesium nitrate Reactions

Magnesium nitrate has a high affinity towards water. Therefore, heating it results to decompose into magnesium oxide, nitrogen oxides, and oxygen. 2 Mg(NO3)2 → 2 MgO + 4 NO2 + O2.

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid

Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

The molarity of acetic acid is 0.11M and the percent by weight is 0.666%.

Molarity of any solution is used to define their concentration and it will be calculated as:

M = n/V, where

n = moles

V = volume

Molarity of acetic acid will be calculated as:

M₁V₁ = M₂V₂, where

M₁ = molarity of acetic acid = ?

V₁ = volume of acetic acid = 37.54mL = 0.037L

M₂ = molarity of NaOH = 0.1062M

V₂ = volume of NaOH = 39.26mL = 0.039L

On putting all these values on the above equation we can calculate the molarity as:

M₁ = (0.1062)(39.26) / (37.54) = 0.11M

Now we calculate the moles of acetic acid by using the molarity formula as:

n = 0.11M × 0.037L = 0.00407 moles

Molar mass of acetic acid = 60g/mole

Mass of 0.00407 moles of acetic acid = 4.1x10⁻³ moles×(60g / mol) = 0.2501 g

Density of solution = 1.00 g/mL

So, 37.54mL in 1g/mL = 37.54g/mL

Percent by weight will be calculated as:

%w/w = 0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

Hence, molarity and %(w/w) of acetic acid is 0.11M and 0.666% respectively.

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Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is [tex]K_c = 14.39[/tex]

Explanation:

The chemical equation for this decomposition of ammonia is

                [tex]2 NH_3[/tex]  ↔   [tex]N_2 + 3 H_2[/tex]

The initial concentration of ammonia is mathematically represented a

          [tex][NH_3] = \frac{n_1}{V_1} = \frac{29}{75}[/tex]

          [tex][NH_3] = 0.387 \ M[/tex]

The initial concentration of nitrogen gas  is mathematically represented a

         [tex][N_2] = \frac{n_2}{V_2}[/tex]

         [tex][N_2] = 0.173 \ M[/tex]

So  looking at the equation

   Initially (Before reaction)

      [tex]NH_3 = 0.387 \ M[/tex]

      [tex]N_2 = 0 \ M[/tex]

      [tex]H_2 = 0 \ M[/tex]

During reaction(this is gotten from the reaction equation )

        [tex]NH_3 = -2 x[/tex](this implies that it losses two moles of concentration )

         [tex]N_2 = + x[/tex]  (this implies that it gains 1 moles)

         [tex]H_2 = +3 x[/tex](this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

        [tex]NH_3 = 0.387 -2x[/tex]

       [tex]N_2 = x[/tex]

        [tex]H_2 = 3 x[/tex]

Now since

     [tex][NH_3] = 0.387 \ M[/tex]

     [tex]x= 0.387 \ M[/tex]    

[tex]H_2 = 3 * 0.173[/tex]    

[tex]H_2 = 0.519 \ M[/tex]    

[tex]NH_3 = 0.387 -2(0.173)[/tex]

[tex]NH_3 = 0.041 \ M[/tex]

Now the equilibrium constant is

           [tex]K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]

substituting values

           [tex]K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}[/tex]

           [tex]K_c = 14.39[/tex]

Answer:

N2H4

Explanation:

A hydride is a binary compound of hydrogen and another element. Binary compounds contain only two atoms. We have to x-ray the hydrides of nitrogen given in the question in order to make our choice. Remember that we were told that that the hydride contains its own volume of nitrogen and twice its volume of hydrogen.

Now consider the hydride N2H4.

N2H4(g) -----> N2(g) + 2H2(g)

The volume ration of nitrogen gas to hydrogen gas in N2H4 is 1:2.

The molecular mass of the compound is;

N2H4= 2(14) + 4(1)= 28+4= 32

Since

molecular mass= 2 vapour density

Vapour density= molecular mass/2

Vapour density= 32/2

Vapour density = 16

Therefore the hydride of nitrogen referred to in the question is N2H4

Answer : The net ionic equation will be,

[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

[tex]Ba(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(aq)+BaSO_4(s)[/tex]

The ionic equation in separated aqueous solution will be,

[tex]Ba^{2+}(aq)+2OH^-(aq)+2H^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2H^+(aq)+2OH^{-}(aq)[/tex]

In this equation, [tex]H^+\text{ and }OH^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]

Answer:  a.CCl4 aka carbon tetrachloride

Explanation:

ionic compounds and polar molecules can be dissolved in water which is a polar solvent.

choice d (KF) is a salt (an ionic compound) and can be dissolved in water /(K+ and F- ions would be formed in water).

choice c (NH3 or ammonia) is a very polar molecule and thus can be dissolved in water(Hydrogen bonding).

choice b (CH3Cl) is slightly polar because the atoms surrounding the central carbon atom are different(3 H atoms and 1 chlorine atom) and can be dissolved in water(Dipole-dipole interaction).

choice a is nonpolar and cannot be dissolved in water.

Answer:

b) sp^2

Explanation:

Hybridization refers to the concept that atomic orbitals fuse to form newly hybridized orbitals, which in turn, influences molecular geometry and bonding properties. In chemistry, orbital hybridisation (or hybridization) is the implies the mixing of atomic orbitals to form hybrid orbitals (with different energies, shapes, etc., different from that of the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds according to the principles of the valence bond theory.

In 1931 Linus Pauling proposed the idea of “mixing” the orbitals or “hybridizing” them to account for certain observed bonding patterns. Pauling proposed a sort of a combination of the orbitals giving you an orbital that has partial characters.

Hybridization is merely a mathematical construct. It is never an actual “process” that occurs within orbitals . Hybridization is a mathematical model that describes how the atomic orbitals would’ve looked like based on the observable molecular orbitals.

sp2 hybridization leads to the formation of a double bond. sigma bonds may also be formed depending on the valency of the central atom. In alkenes, an sp2 hybridized carbon atom forms a double bond in addition to two sigma bonds to other atoms.

The predicted hybridization is:

b) [tex]sp^2[/tex]

It is the integration of atomic orbitals to shape new orbitals with exclusive energies and shapes than the unique orbitals.

Given: An atom that's surrounded with the aid of using a double bond and unmarried bonds.

[tex]sp^2[/tex]  hybridization ends in the formation of a double bond. sigma bonds can also be shaped relying at the valency alkenes, an   [tex]sp^2[/tex]sigma bonds to different atoms.

Thus, correct option is b.

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Answer:

THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K

Explanation:

Mass of first sample of water = 106 g

Initial temp of first sample = 21.4  °C = 21.4 + 273 K = 294.4 K

Mass of second sample = 64.3 g

Final temp of theresulting mixture = 46.8  °C = 46.8 + 273 K = 319.8 K

Specific heat capacity of water = 4.184 J/g K

It is worthy to note that;

Heat gained by the first sample = Heat lost by the second sample

Since heat = mass * specific heat capacity * change in temperature, we have

Mass * specific heat * change in temp of the first sample  = Mass * specific heat * change in temp. of the second sample

MC (T2 - T1) = MC (T2-T1)

106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)

106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)

11 265.0016 = 269.0312 (319.8 - T1)

Since the change in temperature = 319.8 -T1

Change in temperature =11265.0016 / 269.0312

Change in temperature =  41.87

Change in temperature = 319.8 -T1

41.87 = 319.8 - T1

T1 = 319.8 - 41.87

T1 = 277.93 K

T1 = 4.93  °C

So therefore, the initial temperature of the sacond sample is 4.73  °C or 277.93 K

Answer:

Explanation:

a )

m = m₀ [tex]e^{-\lambda t[/tex]

m is mass after time t . original mass is m₀ , λ is disintegration constant

λ = .693 / half life

= .693 / 1590

= .0004358

m = m₀ [tex]e^{- 0.0004358 t}[/tex]

b )

m = 50 x [tex]e^{-.0004358\times 500}[/tex]

= 40.21 mg .

c )

40 = 50 [tex]e^{-.0004358t[/tex]

.8 = [tex]e^{-.0004358t[/tex]

[tex]e^{.0004358t[/tex] = 1.25

.0004358 t = .22314

t = 512 years .

Answer:

A. ethane > butane > decane > dodecane > heptadecane

Explanation:

In fractionating column, crude oil is separated by means of fractional distillation due to the wide range of boiling point of the crude products such as ethane, propane, butane pentane etc.

The product with the least weight rises to top height while the product with highest weight will move down.

For the given hydrocarbon products, the ranking according to their molecular weight, starting with the lighter product to heavier product is

ethane (C2), butane (C4), decane(C10), dodecane (C12), heptadecane(C17).

Thus, the correct ranking, starting with the product that will rise highest is ethane > butane > decane > dodecane > heptadecane

Answer:

B and D

Explanation:

If we use the info given we have a band a 3300 cm-1 and 2200 cm-1 this indicates that we have an alkyne functional group. Additionally, the hydrogenation of the unknown molecule will consume two moles of hydrogens this fits with the 2 pi bonds in the alkyne functional group. So, we can discard "a" and "e". The product of this hydrogenation is 2-methylhexane therefore we can discard c because the methyl group is placed on carbon 3. Structures b and d can work.

See figure 1

I hope it helps!

Answer:

7.5 atm

Explanation:

Initial pressure P1 = 1.0 ATM

Initial volume V1= 196 L

Final pressure P2= the unknown

Final volume V2= 26000ml or 26 L

From Boyle's law we have;

P1V1= P2V2

P2= P1V1/V2

P2= 1.0 × 196/26

P2 = 7.5 atm

Therefore, as the air is compressed, the pressure increases to 7.5 atm.

Answer:

C) Steam condensing on a mirror

Explanation:

This was just a change in the physical state.

Answer:

The correct answer is 5.30 * 10^-4 mol per L.

Explanation:

Based on Henry's law, in a solution solubility of the gas is directly proportional to the pressure, that is, C is directly proportional to P. Here P is the pressure and C is the concentration of the dissolved gases.  

Therefore, it can be written as,  

C2/C1 = P2/P1

Here, C1 is 6.8 * 10^-4 mol/L, P1 is 1 atm and P2 is 0.78 atm, then the value of C2 obtained by putting the values in the equation,  

C2/(6.8*10^-4) = 0.78/1

C2 = 0.78 * 6.8*10^-4

C2 = 5.30 * 10^-4 mol per L.  

Hence, the concentration of dissolved nitrogen at 0.78 atm is 5.30*10^-4 mol/L.  

Answer:

Metallic

Explanation:

"Metallic bonding is a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons and positively charged metal ions. It may be described as the sharing of free electrons among a structure of positively charged ions." -Wikipedia

Answer:

5.73 g Cu

Explanation:

M(CuSO4) = 159.6 g/mol

M(Al) = 27.0 g/mol

M(Cu) = 63.5 g/mol

14.4 g Al * 1 mol/27.0 g = 0.5333 mol Al

14.4 g CuSO4 * 1 mol/159.6 g = 0.0902 mol CuSO4

                                2 Al         +     3CuSO4 ------->  3Cu + Al2(SO4)3

from reaction           2 mol               3 mol

given                     (0.5333 mol )x       0.0902 mol

needed                   0.0601 mol

x= 2*0.0902/3 = 0.0601 mol   Al

Al is excess, CuSO4 is a limiting reactant.

                               2 Al         +     3CuSO4 ------->  3Cu + Al2(SO4)3

from reaction                               3 mol                  3 mol

given                                            0.0902 mol         x mol

x = 0.0902 mol Cu

0.0902 mol Cu * 1 mol/63.5 g Cu = 5.73 g Cu

Answer:

The percentage of the R-enantiomer is 26.18% while the percentage of the S-enantiomer is 73.82%

Explanation:

If the specific rotation of R enantiomer = -28.4, then the specific rotation of S = +28.4

Now, let us have x = % R, thus

% S = 100-x =y

Hence;

{- 28.4x + 28.4( 100 -x)}/100= 14.1

Thus;

-28.4x + 2840 -28.4x = 1410

-56.8x + 2840 = 1410

-56.8x = 1410-2840

-56.8x = -1430

x = 1430/56.8

x = 26.18%

y = 100-26.18% = 73.82%

Answer: Fossil fuels

Explanation:

Fossil fuels such as petroleum, oil,  and natural gas, are non-renewable energy resources which are formed from the remains of  prehistoric ancient  plants and animals beneath layers of rock of the earth surface.

By analyzing and studying fossil fuels using Radiocarbon analyses by archaeologists, earth scientists  etc, Information about the history of the earth can be obtained from the decomposition of dead organisms present in fossil fuels.

Answer:

[tex]_1^{1} {P}[/tex] is symbol for proton emission in the nucleus.

[tex]_2^{4} {He}[/tex] symbolises alpha emission, equivalent to helium atom emission of a radioactive particle

[tex]+_1^{0} {e}[/tex] is the radiation symbol for positiron particle. which occurs when beta + radioactive decay occurs

the second statement is the correct one quarks are needed to balance charges in all subatomic particles such as neutrons, protons and electrons