Determine the pH of a buffer which is a 0.20 M solution of trimethylamine (N(CH3)3) and a 0.40 M solution of trimethylammonium chloride (NH(CH3)3Cl). The Kb of trimethylamine at 25°C is 6.3x10-5.

Answer:

D) in chemical changes and physical changes.

Explanation:

Mass and energy are conserved in chemical changes and physical changes. Therefore, option D is correct.

Physical modifications are those that affect a chemical substance's form but not its chemical content. Physical changes may normally be used to separate compounds into chemical elements or simpler compounds, but they cannot be used to separate mixtures into their component compounds.

A change in physical attributes accompanies a physical change. Melting, turning into a gas, changing strength or durability, changing crystal structure or texture, and changing shape, size, color, volume, or density are a few examples of physical qualities.

In a physical change, the substance's shape or appearance changes, but the type of matter it contains stays the same. In contrast, when matter undergoes a chemical transformation, at least one new substance with novel features is created.

Thus, option D is correct.

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Answer:

milligram, centigram, decigram

Hope this answer correct :)

Answer:

C

Explanation:

hope it helps.

Answer:

a) 7.94 x 10²³ molecules, b) 5.62 g SO2, c) 1.97 L

Explanation:

a) We need to first convert Cu2S to moles. Since molar mass of Cu2S is 159.14 g/mol, 14.0 g = 0.0880 mol. Using molar ratios (3 mol O2/2 mol Cu2S, 0.0880 mol of Cu2S = 0.132 mol O2. Since 1 mol contains 6.02 x 10²³ molecules, 0.132 mol O2 = 7.94 x 10²² molecules O2.

b) Since the molar ratios of Cu2S to SO2 is 1:1, 0.0880 mol of Cu2S produces 0.0880 mol SO2. To convert mol to grams, we use the molar mass of SO2 (64.06 g/mol) to figure out that 0.0880 mol SO2 = 5.63 g SO2.

c) At STP, 1 mol occupies 22.4 liters. 0.0880 mol SO2 x 22.4 L/1 mol = 1.97 L

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Answer:

1.34x10⁻⁵ mol / L is the solubility of AgCl

Explanation:

Ksp of AgCl is defined as:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

If 0.010M of AgCl is added, some Ag⁺ and Cl⁻ will be produced until:

Ksp = [Ag⁺] [Cl⁻]

Ksp for AgCl = 1.8x10⁻¹⁰ (Taken from ALEKS Data tab):

Some Ag⁺ and Cl⁻ are produced, you can take this "some" as X:

[Ag⁺] = X

[Cl⁻] = X

Where X is the amount of AgCl that dissolvesin water. X = solubility:

Ksp = 1.8x10⁻¹⁰ = [Ag⁺] [Cl⁻]

1.8x10⁻¹⁰ = [X] [X]

1.8x10⁻¹⁰ = X²

X =

Answer:

Groups

Explanation:

1 A, 3 B, and 7 A are examples of  group number on the periodic table

Group 1A are  the alkali metals includes lithium sodium and potassium.

Group 3B in most periodic tables, lanthanum and actinium are considered to be a part of Group3B.

Group 7A are the halogens includes chlorine ,bromine and iodine.

They are all soft, silver metals. Due to their low ionization energy, these metals have low melting points and are highly reactive. The group 1A elements with their ns1 valence electron configurations are very active metals. They lose their valence electrons very readily.

All the Group 3B elements are rather soft, silvery-white metals, although their hardness increases with atomic number. They have higher ionization energies than the Group 1A and 2A elements, and are ionized to form a 3+ charges.

They have very high electronegativity. They have seven valence electrons . They are highly reactive, especially with alkali metals and alkaline earths metals.

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Answer:

the wavelength of radiation emitted  is [tex]\mathbf{\lambda= 2169.62 \ nm}[/tex]

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

[tex]\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}[/tex]

For n = 7:

[tex]\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}[/tex]

[tex]\mathtt{E_7 =-0.27755 \ eV}[/tex]

For n = 4

[tex]\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}[/tex]

[tex]\mathtt{E_4 =- 0.85\ eV}[/tex]

The  electron goes from the n = 7 to the n = 4, then :

[tex]\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}[/tex]

[tex]\mathtt{= 0.57245\ eV}[/tex]

Wavelength of the radiation emitted:

[tex]\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}[/tex]

where;

hc  = 1242 eV.nm

[tex]\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}[/tex]

[tex]\mathbf{\lambda= 2169.62 \ nm}[/tex]

Answer:

The amount of dissolved gases in the body. Have a good day! =)

Explanation:

Answer:

Explanation:

Amidation is a reaction with the formation of an amide. It involves a process where series of reaction between amine and carboxylic acid takes place.

The main objective of this question is to draw with the aid of a diagram.

the structure of the amine and carboxylic acid reactants required to form the following amide in an amidation reaction.

But before that,

we are to draw the starting amine and to be sure to include the nonbonding electron pairs.

Also, to draw  starting carboxylic acid and to be sure to include the nonbonding electron pairs.

The whole series of the diagrammatic expression for the amine formation can be found in the diagram attached below.

Answer:

70

Explanation:

100/20 =5

5 x each day over 2 weeks (14 days) = 70

The vials are necessary for the entire treatment is 70 vials

We have to first calculate the total number of antibiotics taken by the patient in two weeks, then find the total vials neccessary for the entire treatment.

The total antibiotics for a period of two weeks = (100 mg/day)(2 weeks)

Total = 100 mg/day×14 days

Total = 1400 mg for two weeks.

If the antibiotics contain vials in 20 mg/vials of the drugs,

Then the vials necessary for the entire treatment is

vial for the entire treatment  = 1400/20

total vials = 70 vials

Hence the vials are necessary for the entire treatment is 70 vials

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Answer:

1.pp

2.cp

3.pp

4.pp

5.pp

6.pp

Answer:

3.

Explanation:

Hello,

In this case, considering the given number:

[tex]2.30x10^5[/tex]

We can see that it does not have zeros to left of the first nonzero digit which is 2, therefore, we just have to consider the 2, 3 and 0 as significant figures because the right-handed zeros are actually considered. Hence, such measurement has 3 significant figures.

Answer:

a. minimum voltage that must be supplied for a redox reaction to occur

c. always equal to Eanode - Ecathode

Explanation:

In an electrolytic cell;  The electromotive force(the maximum standard potential difference) of the cell formed by the system is defined as the standard electrode potential of the right handed electrode minus the standard electrode potential of the left hand electrode. (i.e [tex]\mathbf{E^{\theta}_{cell}=E_{anode} - E_{cathode}}[/tex] )

As we all known that the process by which chemical energy is being converted to electrical energy is called the Electrochemical cell. It consists of two half cells , an oxidation half cell reaction and a reduction half-cell reaction.The overall redox reaction results in a flow of electrons in an electric current which is produced by a minimum voltage.

Therefore, option a and c are both correct.

Answer:

b. H-3

Explanation:

The half life of the following isotopes is given as;

a. 0-19

26.88 seconds

b. H-3

12 years

c. Zn-71

2.4 minutes

d. Rb-85

This is a stable isotope. It does not decomposes.

Comparing the half lives, the isotope with the longest half life is Hydrogen - 3. With half life of 12 years.

Answer:

459.6kj

Explanation:

the chemical equation =

3mn(s) + 2CO2(g) --- Mn3O4(s)

we have change in H to be -1378.83

we are required to find the amount of heat that is liberated for a mole of manganese.

3 moles of manganese = -1378.83

1 mole of manganese = ?

when we cross multiply, we will have a ratio that we will use to find this amount

1378.83/3 = 459.6kj

Answer:

1.86 m

Explanation:

Step 1: Calculate the mass of solute and solvent in 100 grams of solution

We have a 15.0% by mass solution of MgCl₂ (solute) in H₂O (water). In 100 g of solution, there are 15.0 g of solute and 100.0 g - 15.0 g = 85.0 g of solvent.

Step 2: Calculate the moles corresponding to 15.0 g of MgCl₂

The molar mass of MgCl₂ is 95.21 g/mol.

[tex]15.0g \times \frac{1mol}{95.21 g} = 0.158mol[/tex]

Step 3: Convert the mass of water to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]85.0g \times \frac{1kg}{1,000g} = 0.0850kg[/tex]

Step 4: Calculate the molality of the solution

[tex]m = \frac{0.158mol}{0.0850kg} = 1.86m[/tex]

The molality is 1.86 m.

Given:

Mass solution of MgCl₂ in H₂O= 15%

Density= 1.127g/mL

To find:

Molality=?

Let's solve this question step by step:

Step 1: Calculate the mass of solute and solvent in 100 grams of solution

As it is given that 15.0% by mass solution of MgCl₂ (solute) in H₂O (water).

In 100 g of solution, there are 15.0 g of solute and 100.0 g - 15.0 g = 85.0 g of solvent.

Step 2: Calculate the moles corresponding to 15.0 g of MgCl₂

The molar mass of MgCl₂ is 95.21 g/mol.

Number of moles is given mass divided by the molar mass.

[tex]\text{Number of moles}= \frac{15g}{95.21g/mol} =0.158mol[/tex]

Step 3: Convert the mass of water to kilograms.

As we know, 1 kg = 1,000 g.

[tex]=\frac{85}{1000} =0.085kg[/tex]

Step 4: Calculate the molality of the solution.

Molality is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.

[tex]\text{Molality}=\frac{0.158\text{mol}}{0.085\tetx{kg}} =1.86 m[/tex]

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Explanation:

Moles of 0.042 L of Pure water = 0.042 / 22.4

[tex]Moles \: of \: 0.042 L \: of \: Pure \: water \: = 1.87 \times {10}^{ - 3} [/tex]

[tex]Molecules \: in \: 0.042 L \: of \: Pure \: water \: = moles \times avogadro \: number[/tex]

Molecules in 0.042 L of Pure Water

= 1.126 * 10^21

Answer:

See explanation

Explanation:

Anode;

Sn(s) ------> Sn^2+(aq) + 2e

Cathode;

Mn^2+(aq) + 2e ------> Mn(s)

The minimum voltage required to drive the reaction is the cell voltage. The cell voltage is obtained from;

E°cell= E°cathode - E°anode

E°cell= -1.19 - (-0.14)

E°cell= -1.05 V

Answer:

\large \boxed{\text{150 g TiCl}_{4}}  

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:     189.68                  79.87

           TiCl₄ + 2H₂O ⟶ TiO₂ + 4HCl

m/g:                                 50.0

To solve this stoichiometry problem, you must

1. Theoretical yield of TiO₂

[tex]\text{Theoretical yield} = \text{50.0 g actual} \times \dfrac{\text{100 g theoretical}}{\text{78.9 g actual}} = \text{63.37 g theoretical}[/tex]

2.  Moles of TiO₂

[tex]\text{Mass of TiO}_{2} = \text{63.37 g TiO}_{2} \times \dfrac{\text{1 mol TiO}_{2}}{\text{79.87 g TiO}_{2} } = \text{0.7934 mol TiO}_{2}[/tex]

3,  Moles of TiCl₄

The molar ratio is 1 mol TiO₂:1 mol TiCl₄.

[tex]\text{Moles of TiCl}_{4} = \text{0.7934 mol TiO}_{2} \times \dfrac{\text{1 mol TiCl}_{4}}{\text{1 mol TiO}_{2}} = \text{0.7934 mol TiCl}_{4}[/tex]

4.  Mass of TiCl₄

[tex]\text{Mass of TiCl}_{4} = \text{0.7934 mol TiCl}_{4} \times \dfrac{\text{189.98 g TiCl}_{4}}{\text{1 mol TiCl}_{4}} =\textbf{150 g TiCl}_{\mathbf{4}} \\\\\text{You must use $\large \boxed{\textbf{150 g TiCl}_{\mathbf{4}}}$}[/tex]

Answer:

(CH3)3CCHO

Explanation:

Carbonyl compounds that contain alpha hydrogen atoms are the only ones that can undergo aldol condensation reaction.

Aldol condensation is one of the condensation reactions in organic chemistry that occurs when an enol or an enolate ion reacts with another carbonyl compound to yield a β-hydroxyaldehyde or β-hydroxyketone, this is immediately followed by the dehydration of the product to yield a conjugated enone.

(CH3)3CCHO lacks alpha hydrogen atoms hence it does not undergo the aldol condensation.

Answer:

[tex]\%m=0.15\%[/tex]

Explanation:

Hello,

In this case, we are asked to compute the by mass percent representing the toxicity of ethylene glycol in the body mass. In such a way, since the by mass percent is computed with the shown below formula:

[tex]\%m=\frac{m_{ethylene \ glycol}}{m_{ethylene \ glycol}+m_{body\ mass}}*100\%[/tex]

We can use the given masses to obtain:

[tex]\%m=\frac{1.5g}{1.5g+1000g}*100\%\\ \\\%m=0.15\%[/tex]

Best regards.

Answer: The mole fraction of benzene will be 0.34 and mole fraction of toluene is 0.66

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

[tex]p_1=x_1p_1^0[/tex] and [tex]p_2=x_2P_2^0[/tex]

where, x = mole fraction in solution

[tex]p^0[/tex] = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

[tex]p_{total}=p_1+p_2\\[/tex][tex]p_{total}=x_{benzene}p_{benzene}^0+x_{toluene}P_{toluene}^0[/tex]

[tex]x_{benzene}=x[/tex],

[tex]x_{toluene}=1-x_{benzene}=1-x[/tex],

[tex]p_{benzene}^0=75torr[/tex]

[tex]p_{toluene}^0=22torr[/tex]

[tex]p_{total}=40torr[/tex]

[tex]40=x\times 75+(1-x)\times 22[/tex]

[tex]x=0.34[/tex]

Thus (1-x0 = (1-0.34)=0.66

Thus the mole fraction of benzene will be 0.34 and that of toluene is 0.66

Answer:

diffraction

Explanation:

Louis de Broglie in 1924 proposed the idea of wave particle duality. In his proposition, matter could exhibit wavelike properties.

Electrons are generally regarded as particles. Electrons may also display wavelike properties such as diffraction patterns. Generally, diffraction is regarded as a property of waves.

Hence, electron diffraction effect owes to the wavelike nature of of electrons when they are passed near matter.

The volume of KOH required for the neutralization of acid has been 38.9 ml. Thus, option B is correct.

Neutralization reaction has resulted in the formation of salt and water with the reaction of acid and base.

In the neutralization reaction, the strength of the acid and base can be given by:

Molarity of acid [tex]\times[/tex] Volume of acid = Molarity of base [tex]\times[/tex] Volume of base

Given, the molarity of KOH base = 0.124 M

The volume of acid (HCl) = 23.4 ml

Molarity of acid (HCl) = 0.206 M.

Substituting the values:

0.206 [tex]\times[/tex] 23.4 = 0.124 × Volume of base (KOH)

Volume of base (KOH) = 38.874 ml.

Volume of KOH = 38.9 ml

The volume of KOH required for the neutralization of acid has been 38.9 ml. Thus, option B is correct.

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Answer:

Traces of sodium impurity

Explanation:

Metal ions are identified by a characteristic colour imparted to flame by the metal ion solution. Various metals have various colours which they impart to a flame.

The energy of a flame is not sufficient for the excitation of electrons of Mg to a higher energy level. As a result of this, Mg do not give any color in Bunsen flame.

However, a few tinges of yellow-orange colour which is characteristic of sodium metal do appear when magnesium ions are exposed to a flame as a consequence of traces of sodium impurity in the magnesium ion solution.

Answer:

71.3 g

Explanation:

molar mass of C₇H₆O₃ = 138.13 g

molar mass of [tex]C_{9}H_{8}O_{4}[/tex] = 180.17 g

find the moles of reactant: C₇H₆O₃

C₇H₆O₃ that reacted = mass/molar mass = 57.6 g C₇H₆O₃ / 138.13 g C₇H₆O₃

= 0.417 mol [tex]C_{7}H_{6}O_{3}[/tex]

From the reaction equation, 1 mole of C₇H₆O₃ yields one mole of aspirin

find theoretical yield of aspirin:

0.417 mol C7H6O3 x 180.17 g C9H8O4 / 1 mol C9H8O4[tex]\frac{0.417 mol C_{7}H_{6}O_{3}}{} x \frac{180.17 g C_{9}H_{8}O_{4}}{1 mol C_{9}H_{8}O_{4}}[/tex]

= 75.1 g C9H8O4

Actual yield= % yield × theoretical yield/100

Actual yield = 95.0 × 75.1/100

Actual yield = 71.3 g

Taking into account the reaction stoichiometry and actual yield, the mass of C₉H₈O₄ produced from 57.6 g of C₇H₆O₃ assuming 95.0% yield is 57.0285 grams.

The balanced reaction is:

22 C₇H₆O₃ (s) + 32 C₄H₆O₃ (s) → 24 C₉H₈O₄ (s) + 33 HC₂H₃O₂ (aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The mass molar of each compound is:

Then by reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:

Then you can apply the following rule of three: if by stoichiometry 3036 grams of C₇H₆O₃ produce 3264 grams of C₉H₈O₄, 57.6 grams of C₇H₆O₃ will produce how many mass of C₉H₈O₄?

[tex]mass of C_{9} H_{8} O_{4} =\frac{57.6 grams of C_{7} H_{6} O_{3} x3264 gramsof C_{9} H_{8} O_{4} }{3036 grams of C_{7} H_{6} O_{3}}[/tex]

mass of C₉H₈O₄= 60.03 grams

On the other side, actual yield is the amount of product actually obtained from a reaction. Assuming 95.0% yield  and considering the previously calculated mass the maximum amount of product that can be produced in the reaction, you can apply the following rule of three:  If 100% equals 60.03 grams of the compound produced, 95% equals how much mass of the compound?

[tex]mass of C_{9} H_{8} O_{4} =\frac{95 percentx60.03gramsof C_{9} H_{8} O_{4} }{100 percent}[/tex]

mass of C₉H₈O₄= 57.0285 grams

Finally, the mass of C₉H₈O₄ produced from 57.6 g of C₇H₆O₃ assuming 95.0% yield is 57.0285 grams.

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Answer:

d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2

Explanation:

Hello,

In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:

[tex]pH=pKa+log(\frac{[Base]}{[Acid]} )[/tex]

We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:

a. [tex]\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36[/tex]

b. [tex]\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417[/tex]

c. [tex]\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868[/tex]

d. [tex]\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959[/tex]

Therefore, the d. solution has the best buffering capacity.

Regards.

Answer:

7.89 g

Explanation:

Step 1: Write the balanced equation

S₈ + 16 F₂(g) → 8 SF₄

Step 2: Calculate the moles corresponding to 2.34 g of S₈

The molar mass of S₈ is 256.52 g/mol.

[tex]2.34g \times \frac{1mol}{256.52g} = 9.12 \times 10^{-3} mol[/tex]

Step 3: Calculate the moles of SF₄ produced from 9.12 × 10⁻³ mol of S₈

The molar ratio of S₈ to SF₄ is 1:8. The moles of SF₄ produced are 8/1 × 9.12 × 10⁻³ mol = 0.0730 mol

Step 4: Calculate the mass corresponding to 0.0730 moles of SF₄

The molar mass of SF₄ is 108.07 g/mol.

[tex]0.0730 mol \times \frac{108.07 g}{mol} = 7.89 g[/tex]

Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

0.085 moles of Al are required to form 23.6 g of AlBr₃.

Let's consider the following balanced equation for the synthesis reaction of AlBr₃.

2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)

First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).

[tex]23.6 g \times \frac{1mol}{266.69g} = 0.0885 mol[/tex]

The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885  moles of AlBr₃ are:

[tex]0.0885molAlBr_3 \times \frac{2molAl}{2molAlBr_3} = 0.0885molAl[/tex]

0.085 moles of Al are required to form 23.6 g of AlBr₃.

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Volume ..........is the answer