In general ethers have a much lower boiling point than their isomeric alcohols. why?

Point defects are possible for al2o3 as an impurity in mgo. then 2 Al3+ must be added to form 1 Mg2+  must be added to form each of these defects

Point defects are zero-dimensional lattice defects, meaning they lack lattice structure in any dimension. Impurity atoms in a pure metal, vacancies, and self-interstitials are examples of common point defects.

Point defects are zero-dimensional lattice defects, which means they lack any kind of lattice structure. Point defects include, but are not limited to, atoms of impurity in a pure metal, vacancies, and self-interstitials.

The term "point defect" refers to the flaw or imperfection that results from the placement of a point-like constituent particle, such as an atom, an ion, or a molecule, in a crystalline structure.

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The final answer is 11.71

The concentration of a given solution (H2NCH2CH2NH2)  is 0.314 M

pa1 = 6.848

a1 = 1.42 X [tex]10^{-7}[/tex]

Pa2 = 9.928

a2 = 1.18 X [tex]10^{-10}[/tex]

The equation is given as;

NH2CH2CH2NH2 + H2O ---> NH2CH2CH2NH3+ + OH-

Initial 0.314 0 0

Change -x +x +x

Equilibrium 0.314 -x x x

b1 = Kw / Ka2 = 10^-14 / 1.18 X 10^-10 = 8.47 X 10^-5 = [NH2CH2CH2NH3+][OH-] / [NH2CH2CH2NH2]

8.47 X 10^-5 = x^2 / 0.314 -x

8.47 X 10^-5 X 0.314 - 8.47 X 10^-5x = x^2

x^2 - 2.66 X 10^-5 + 8.47 X 10^-5x = 0

on solving

x = 0.0051 M = [NH2CH2CH2NH3+] = [OH-]

[NH2CH2CH2NH2] = 0.314 - 0.0051 = 0.3089 M

For the second dissociation,

NH2CH2CH2NH3+ + H2O ---> +NH3CH2CH2NH3+ + OH-

Initial 0.0051 0 0.0051

Change -x +x +x

Equilibrium 0.0051 -x x 0.0051 +x

b2 = Kw / Ka1 = 10^-14 / 1.42 X 10^-7 = 7.04 X 10^-8 = [+NH3CH2CH2NH3+] [OH-] / [+NH3CH2CH2NH2]

7.04 X 10^-8 = x (0.0051+x) / (0.0051-x)

We may ignore x in the denominator as b2 is very low

7.04 X 10^-8 = x (0.0051+x) / (0.0051)

3.59 X 10^-10 = 0.0051x + x^2

x = 7.039 X 10^-8 M

[OH-] = 0.0051 + x = 0.0051 approx

[+NH3CH2CH2NH3+] = x = 7.039 X 10^-8 M

NH2CH2CH2NH3+ = 0.0051 +x = 0.0051 M approx

pOH = --log [OH-] = -log 0.0051 = 2.29

pH = 14 - pOH = 11.71

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There are 1157.14 milliliters of a 9. 0 M [tex]H_{2} SO_4}[/tex] solution are needed to make 0. 45 L of a 3. 5 M solution.

Calculation,

The formula for the dilution of a solution is given as:

[tex]M_{1} V_{1}[/tex] = [tex]M_{2} V_{2}[/tex]

Where M is molarity and V is the volume of the solution in liters ( L ).

Given data,

[tex]M_{1}[/tex] =  9. 0 M

[tex]V_{1}[/tex] = 0. 45 L

[tex]M_{2}[/tex] = 3. 5 M

[tex]V_{2}[/tex] = ?

9. 0 M ×0. 45 =  3. 5 M × [tex]V_{2}[/tex]

[tex]V_{2}[/tex] in Lit = 9. 0 M ×0. 45/ 3. 5 M = 1.15 L

[tex]V_{2}[/tex] in the ml = 1.15 × 1000ml = 1157.14 ml

Volume in milliliters is 1157.14 .

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Answer:

102 g Al₂O₃

Explanation:

The theoretical yield is the mass of product calculated via the molar masses and balanced chemical equation.

The limiting reagent is the reactant that is completely reacted before the other reactant(s) are used up. Since Al produces the smaller amount of product, it appears that Al is the limiting reactant. You can only make as much product as the limiting reactant allows. As such, the theoretical yield is 102 grams Al₂O₃.

Using the gas laws, the volume of nitrogen produced is 6.6 L.

We know that the product of a reaction is obtained from the balanced reaction equation. The reactants combine together to yield the products hence we have; N2 + 3H2 ----> 2NH3

We know that 1 mole of hydrogen occupies 22.4 L

x moles of hydrogen occupies 25500 L

x = 1 mole * 25500 L/ 22.4 L

x = 1138.4 moles

If 3 mole of hydrogen reacts with 1 mole of nitrogen

1138.4 moles  reacts with 1138.4 moles * 1 mole/ 3 moles = 379.5 moles

Mass of nitrogen = 379.5 moles * 28 g/mol = 10626 g

Now 3 moles of hydrogen produces 44.8 L of NH3

1138.4 moles  produces 1138.4 moles * 44.8 L/ 3 moles = 17000 L

Now;

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 1 * 17000 * 848/5.5 * 273

V2 =9601 L

2) 2NH4NNO3(s) ---> 4H2O(g) + 2N2(g) + O2(g)

Number of moles = 12g/ 80 g/mol = 0.15 moles

Now;

2 moles of NH4NNO3 produced 2 moles of nitrogen hence 0.15 moles of N2 was produced.

1 mole of N2 occupies 22.4 L

0.15 moles of N2 occupies 0.15 moles  *  22.4 L/1 mole = 3.36 L

P1 = 760 torr

V1 = 3.36 L

T1 = 273 K

P2 = 745 torr

T2 = 527 degrees or 800 K

V2 = ?

V2 =  P1V1T2/P2T1

V2 =  760 *  3.36 * 527/745 * 273

V2 = 6.6 L

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Percent ionization = 3.17%

We utilize the supplied acid equilibrium constant (Ka) to calculate the percentage of the acid that is ionized. It is the ratio of the acid and dissociated ion equilibrium concentrations. The HF acid would dissociate in the manner described below:

HF = H+ + F-

The following is how the acid equilibrium constant might be written:

Ka = [H+][F-] / [HF] = 3.5 x 10-4

We utilize the ICE table to determine the equilibrium concentrations.

        HF             H+              F-

I      0.337           0                 0

C      -x              +x               +x

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

E    0.337-x        x                   x

3.5 x 10-4 = [H+][F-] / [HF]

3.5 x 10-4 = [x][x] / [0.337-x]

Solving for x,

x = 0.01069 = [H+] = [F-]

percent ionization = 0.01069 / 0.337 x 100  = 3.17%

percent ionization = 3.17%

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Answer:

This is the answer

Explanation:

charges passed = current x time = 10 x 3 x 60

= 1800 C

mole of charge = 1800 / 96500

= .01865 moles

Au+3 contains 3 positive charges

3 mole of charge will deposit 1 mole of Au .01865 moles will deposit .01865 / 3 mole

= 6.2167 x 10-3 moles.

The name for aqueous [tex]H_{2} CO_{3}[/tex] is carbonic acid.

So, option C is correct one.

Binary hydrogen compounds with non metals may form H+ (proton) and an anion dissolved in water. The acidic solutions are named as if they were molecular acids , using the usual name for the compound itself , replacing hydrogen with hydro- and suffix -ide with ic. The word acid is then use.

The inorganic acid is an acid drive from one or more inorganic compounds. All inorganic acids form hydrogen ions and the conjugate base ions when dissolved in water.

Example: carbonic acid ( [tex]H_{2} CO_{3}[/tex] )

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From the addition of 1 equivalent of HBr to 1,3-butadiene, 3-bromo-1-butene is obtained as a product.

What is moles?

Moles are defined as the smallest base unit that is equivalent to [tex]6.02214 X 10^{23}[/tex] units.

Moles are chemical entities, which are used to measure the mass or molecular mass of the compound, molecule, or atom. The moles are, for example can be explained as equal to the mass of the atom. For example, there is one mole in 12 grams of carboncarbon.

Thus the kinetic product formed is 3-bromo-1-butene.

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The type of reaction that is represented by these diagrams (attached) are given as:  "condensation polymerization" (Option D)

Any type of polymer in polymer science that undergoes a condensation reaction during the polymerization process is referred to as a condensation polymer (i.e. a little molecules, methanol or water is derived as a metabolite).

The provided reaction is an example of condensation polymerization because it involves the combination of two monomers, which produces a big polymer and water as a byproduct.

Condensation polymers make significant contributions to the packaging, insulation, and textile sectors as far as real-world application of same is concerned.

Also examples of times that are condensation polymers and that have silicon rather than carbon as part of their molecular or structural make up are;

In conclusion, it is to be noted that The type of reaction that is represented by these diagrams (attached) are given as:  "condensation polymerization"

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Answer:

condensation polymerization

Explanation:

The enthalpy of reaction (ΔH) of the above chemical reaction is equal to -143 kJ/mol.

A chemical reaction can be defined as a chemical process that involves the continuous transformation (rearrangement) of the ionic, atomic or molecular structure of a chemical element by breaking down and forming chemical bonds, in order to produce a new chemical compound while new bonds are formed.

A balanced chemical equation can be defined as a chemical equation wherein the number of atoms on the reactant (left) side is equal to the number of atoms on the product (right) side.

Next, we would write the properly balanced chemical equation for this chemical reaction:

                                    3/2O₂ -----> 2/2O₃

Therefore, the enthalpy of reaction (ΔH) of the above chemical reaction is half the enthalpy of reaction (ΔH) of the initial chemical reaction:

Enthalpy of reaction (ΔH) = ½ × -286 kJ/mol

Enthalpy of reaction (ΔH) = -143 kJ/mol.

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There are 2  peaks are in the proton spin decoupled 13C NMR spectrum of 1,3,5-trinitrobenzene .

The 13C NMR spectrum give the peak which is directly about the carbon skeleton not just the proton attached to it  . The number of signals tell us how many different carbons or set of equivalent carbons . The splitting of a signal tells us how many hydrogens  are attached to each carbon.

In 1,3,5-trinitrobenzene molecule , there are 2  peaks are in the proton spin decoupled 13C NMR spectrum three carbon give one signal and another three carbon give another one signal .

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A chemist reacted 0. 2 moles sodium benzoate with 0. 25 moles of hydrochloric acid. if she generated 22 g benzoic acid, the percent yield is 90%.

Reactant + HCl ------ Product

0.2 mol 0.25 22g

mol

Number of moles of reactant = 0.2mol

Molar mass of product = 122.12g

Number of moles of product = given mass/ molar mass

= 22g/ 122.12 g

= 0.180mol

Percent yield = (number of moles of product/ number of moles of reactant) × 100

= (0.180/0.2) × 100

= 90%

Thus we concluded that the percent yield of the given solution is 90%.

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The potential of the hydrogen in water gives the pH of the acidity and basicity of the solution. The pH of the solution is 9.06. Thus, option iii is correct.

The pH of a substance is given by the subtraction of the pOH from 14 which is the range of the pH scale.

The dissociation reaction for ammonia is given as,

NH₃ (aq) + H₂O (l)  ⇄  NH₄⁺ (aq) + OH⁻ (aq)

Here, the concentration of ammonia is [NH₃] - x, ammonium ion is [NH₄⁺] + x, and hydroxide ion is x.

The molar concentration of ammonia is,

M = (0.12 M × 0.2 L) ÷ 0.45 L = 0.053 M

The molar concentration of ammonium ion is,

M = (0.15 M × 0.25 L) ÷ 0.45 L = 0.083 M

From the base dissociation constant and previous concentration from the reaction, the value of x or hydroxide ion is calculated as,

Kb = [NH₄⁺][OH⁻] ÷ [NH₃]

1.8 × 10⁻⁵ (0.053 - x) - (0.083 + x) × x = 0

x = [OH⁻] = 1.15 x 10⁻⁵

The pH from the hydroxide ion is calculated as,

pOH = - log [OH⁻]

= - log (1.15 x 10⁻⁵)

= 4.94

Further,

pH = 14 - pOH

= 14 - 4.94

= 9.06

Therefore, option iii. 9.06 is the pH of the solution.

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Answer:

Explanation:

Take a look at the balanced chemical equation for this synthesis reaction

N

2(g]

+

3

H

2(g]

→

2

NH

3(g]

Notice that you have a

1

:

3

mole ratio between nitrogen gas and hydrogen gas. This means that, regardless of how many moles of nitrogen gas you have, the reaction will always consume twice as many moles of hydrogen gas.

So, if you have

2

moles of nitrogen taking part in the reaction, you will need

2

moles N

2

⋅

3

moles H

2

1

mole N

2

=

6 moles H

2

The pH is 13.31.

The smaller the pH value the more acidic, the larger the pH value the more basic.

The range is 0 to 14, with 7 representing neutrality. Acidity is indicated by pH values below 7, whereas baseness is shown by pH values above 7.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per liters of a solution.

Molarity of NaOH = 0.40 M

Volume of NaOH = 10.0 mL

Molarity of HCL = 0.50 M

Volume of HCL = 20.0 ml

[tex][OH^-] = (M_2V_2 -M_1V_1)/(V_1+V_2)\\[/tex]

         [tex]= (0.50*20.0 - 0.40*10.0) / (20+10)[/tex]

         [tex]= (10 -4)/ 30\\[/tex]

         [tex]= 6/30[/tex]

         [tex]=0.2[/tex]

[tex]pOH = - log (OH ^-)[/tex]

[tex]pOH = - log (0.2)\\[/tex]

[tex]pOH = - ( - 0.698)[/tex]

[tex]pOH = 0.69[/tex]

[tex]pH + pOH = 14[/tex]

[tex]pH = 14 -0.69[/tex]

[tex]pH = 13.31[/tex]

Therefore, the pH after 10.0 mL of 0.40 M NaOH is 13.31.

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Mass of silver can be plated onto an object in 33. 5 minutes at 8. 70 a of current is 19.55 g

Given, I=8.70A, t=33.5min=33.5*60=2010 sec, m of Ag = 19600g M=107.87 g/mol

From  faradays law we can write q=I*t= 8.70*2010=17487 C.

moles of Ag =17487/ 19600=0.1812 moles

Ag deposited = 0.1812*107.87 = 19.55 g

Having the atomic number 47 and the Latin word argentum, which means "bright" or "white," silver is a chemical element. Its symbol is Ag. The highest electrical, thermal, and reflectivity of any metal are displayed by this transition metal, which is soft, white, and lustrous. The metal can be found in the Earth's crust in three different forms: as an alloy with other metals like gold and in minerals like argentite and chlorargyrite. It can also be found in its pure, free elemental form, also known as "native silver." During the refining of copper, gold, lead, and zinc, the majority of silver is created as a byproduct. As a valuable metal, silver has long been treasured.

What mass of silver can be plated onto an object in 33. 5 minutes at 8. 70 a of current? ag (aq) e- → ag(s) 19. 6 g 9. 78 g 0. 326 g 3. 07 g 0. 102 g

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When 0.34 of HNO₃ is titrated to equivalence using 0.14 l of 0.1 m NaOH then the concentration of HNO₃ is 0.041 M

The reaction of neutralization of HNO₃ with NaOH is

HNO₃ + NaOH → H₂O + NaNo₃

When 1 mole of HNO₃ react with 1 mole of NaOH, based on chemical rection the moles of NaOH at equivalence point are equal to moles of HNO₃ present in solution: -

With the mole and volumes, we can find molarity as follows:

            Moles of NaOH = moles HNO₃

            ⁼ 0.14 L X (0.1 mol NaOH/L) = 0.014 mole NaOH

            =0.014 mol HNO₃

Molarity: -

              [tex]\frac{(Mole of HNO₃)}{(volume of HNO₃)}= \frac{0.014}{0.34}[/tex]

               = 0.041 M

Thus, from above solution we concluded that the concentration of HNO₃ solution is 0.041 M.

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Taking into account the scientific notation,  the result of the subtraction is 6.5×10⁵.

First, remember that scientific notation is a quick way to represent a number using powers of base ten.

The numbers are written as a product:

a×10ⁿ

where:

You want to subtract two numbers in scientific notation. It should be noted that when the numbers to be added do not have the same base 10 exponent, the base 10 power with the highest exponent must be found. In this case, the highest exponent is 5.

Then all the values ​​are expressed as a function of the base 10 exponent with the highest exponent. In this case: 5.00×10⁴=0.500×10⁵

Taking the quantities to the same exponent, all you have to do is subtract what was previously called the number "a". In this case:

7.00×10⁵ - 0.500×10⁵= (7.00- 0.500)×10⁵= 6.5×10⁵

Finally, the result of the subtraction is 6.5×10⁵.

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Now , we know that one mole of SO² Contains 2 Moles ( two ) of atoms of oxygen . Hence , 16 Grams of SO² Contains 0.5 Mole(s) of Oxygen

0.5

hope this helps you. :)

The piece of plastic has   1.256× 10¹³  protons  more than  that of electrons.

A proton is a subatomic particle with a defined mass of 1 and a charge of +1 (positive charge). The symbol for the proton is either p or p+.

An electron is a subatomic particle that has a negative electric charge and that actively orbits the nucleus of the atom .

To calculate ,

given ,

magnitude of charge , q=+2μC=+2×10⁻⁶

Since, protons carry positive charge. Then, we know that standard number of proton in 1 Coulomb of charge is,

1C=6.28×10¹⁸ protons

Number of protons in 1 microcoulomb charge is,

=6.28×10¹⁸×10⁻⁶

=6.28×10¹²

Then for the net charge of q, the number of protons is given as,

np= q×6.28×10¹²

np= 2×6.28×10¹²

np=1.256×10¹³ protons

Thus, the piece of plastic has 1.256×10¹³   more protons than electrons.

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There are  1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

Calculation ,

Given ; Current ( I ) = 5. 68 A

In [tex]Fe(NO_{3} )_{2}[/tex] , the valancy of Fe is +2 .

2 moles of [tex]e^{-}[/tex] are required for the decomposition of 1 mole of Fe .

7. 20 g  of Fe in moles  = 7. 20 g /55.845 g/mol =0.12 mole

x moles of  [tex]e^{-}[/tex] are required for the decomposition of 0.128 mole of Fe .

moles of   [tex]e^{-}[/tex] are required = 0.256 moles

Charge on 1 mole  of   [tex]e^{-}[/tex] = 96500 C

Charge on 0.256  mole  of   [tex]e^{-}[/tex] = 24704 C

Current ( I )= Q/t  

t =Q / I = 24704 C/5. 68 A = 4349 sec = 1.2 hr

Therefore , there are  1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

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193 grams of NaCl are needed to make 1.50 L of a 2.20-M NaCl solution.

Calculation of the mass of NaCl required:

The only thing we need to know is that a solution's molarity informs us of the exact amount of moles of solute that are contained in 1 L of a solution.

In this situation, a NaCl (sodium chloride) solution with a 2.20-M concentration will have 2.20 moles of the solute (sodium chloride) in every 1 L of the solution.

The solution's molarity of 2.20 M requires that every 1 L of this solution include 2.20 moles of sodium chloride, which implies that every 1.50 L of this solution needs to contain

[tex]1.5 L solution .\frac{2.20 moles NaCl}{1L solution} = 3.30 moles NaCl[/tex]

Utilize the compound's molar mass to translate the amount of NaCl (sodium chloride) in moles to grams.

[tex]3.30 moles NaCl . \frac{58.4 gm}{1 mole NaCl} = 192.7 gm[/tex] ≈ 193 gm

Therefore it is concluded that the final answer is 193 gm.

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The pH of a  a 0. 10 m solution of NaCN at 25°C is 11.15

Calculation

The reaction in the solution is given below

                   [tex]CN^{-} + H_{2} O[/tex] → HCN + [tex]OH^{-}[/tex]

initial                    0.1

change           ( -x)                (+x)

equilibrium     (  0.1 - x )         x

Kb = [HCN] [[tex]CN^{-}[/tex] ]/[[tex]CN^{-}[/tex] ]

Kb × Ka = Kw = 1.0 × [tex]10^{-14}[/tex]

Kb =  1.0 × [tex]10^{-14}[/tex] / 4.9  × [tex]10^{-10}[/tex] =  [HCN] [[tex]CN^{-}[/tex] ]/[[tex]CN^{-}[/tex] ] = [tex]x^{2}[/tex]/ (  0.1 - x )

Kb =  2.04× [tex]10^{-5}[/tex] = [tex]x^{2}[/tex]/ (  0.1 - x )

Since , [NaCN] /Kb > 100 , we can simplify the above equation to

=  2.04× [tex]10^{-5}[/tex] = [tex]x^{2}[/tex]/ (  0.1 )

x = 1.43 × [tex]10^{-3}[/tex] M = [HCN] = [ [tex]OH^{-}[/tex]]

Then pOH = 2.84

pH + pOH = 14

pH = 14- pOH = 14 - 2.84 = 11.15

Therefore , the pH is 11.15

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The atoms become larger, the intermolecular forces between the atoms become stronger and more energy is needed to overcome these forces. Hence, the boiling point of noble gases increases down the group.

What are noble gases?

Trends in Physical Properties of Noble Gases:

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Ester is the major product formed in the reaction involving an acid anhydride dissolved in alcohol.

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The same amount of energy is released when they recombine.

The decomposition of water:

H₂O → H₂ + O₂                                          Δ = +285.8 kJ/mol

The formation of water:

H₂ + O₂ → H₂O                                          Δ = -285.8 kJ/mol

Endothermic water decomposition requires 285.8 kJ of energy input per mole of degraded water. One mole of water is created from hydrogen and oxygen in the opposite reaction, which is exothermic and produces 285.8 kJ of energy. The energy level is the same, but the energy change sign is different.

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The mass, in grams, of the chlorine gas will be 0.84 grams

At standard temperature and pressure, 22.4 L of gas has 1 mole of the gas in it.

This time, what we have is 265 mL of the gas at STP. 265 ml is equivalent to: 265/1000 = 0.265 L

If.

22.4 L = 1 mole

Then,

0.265 L = 0.265 x 1/22.4 = 0.01183 moles

This means that 0.01183 moles of chlorine is present in the gas at STP.

Mass of 0.01183 moles of chlorine = moles x molar mass.

The molar mass of chlorine gas is 71

Mass of the chlorine gas = 0.01183 x 71 = 0.84 grams

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Answer:

2,2 - dibromo - 4,5,5 - trimethylhexane

Explanation:

When 2 equivalents of HBr are added to 4,5,5‑trimethyl‑1‑hexyne, the alkane (at the top right of the image) is formed. The bromine branches are added to the second carbon because the intermediate carbocations are the most stable on the secondary carbon. Below I have drawn the mechanism.