what is the molality of a solution

Answer:

[H₃O⁺] = 2.5 × 10⁻¹³ M

pH = 12.6

Explanation:

Step 1: Given data

Concentration of OH⁻: 0.04 M

Step 2: Calculate the concentration of H₃O⁺

Let's consider the self-ionization of water reaction.

2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

The ionic product of water is:

Kw = [OH⁻] × [H₃O⁺] = 10⁻¹⁴

[H₃O⁺] = 10⁻¹⁴ / [OH⁻]

[H₃O⁺] = 10⁻¹⁴ / 0.04

[H₃O⁺] = 2.5 × 10⁻¹³ M

Step 3: Calculate the pH

The pH is:

pH = -log [H₃O⁺] = -log 2.5 × 10⁻¹³ = 12.6

Answer:

atoms contain tiny negatively charged subatomic particles or electrons

Answer:

13.11 g.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below :

N2 + O2 —> 2NO

Next, we shall determine the mass of O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32 g.

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 2 x 30 = 60 g.

From the balanced equation above,

32 g of O2 reacted to produce 60 g of NO.

Finally, we shall determine the theoretical yield of NO as follow:

From the balanced equation above,

32 g of O2 reacted to produce 60 g of NO.

Therefore, 6.99 g of O2 will react to produce = (6.99 x 60)/32 = 13.11 g of NO.

Therefore, the theoretical yield of nitrogen monoxide, NO is 13.11 g.

Answer:158 days (D)

Explanation:

It will take 158 days for grass contaminated with Sr-90 to be safe for cattle to eat. Therefore, option (A) is correct.

The half-life of a radioactive material is defined as the time that is needed to reduce the initial quantity of a radioactive element to half after disintegration.

The half-life of a radioactive element can be described as the characteristic of the element and does not influence by the initial amount of the radioactive substance.

Given, the half-life of the Strontium-90 = 28 days

The rate constant of the decay can be determined as:

[tex]t_{\frac{1}{2} } =\frac{0.693}{k}[/tex]

[tex]k=\frac{0.693}{t_{\frac{1}{2} } }[/tex]

k = 0.693/28

k = 0.025 day⁻¹

The concentration of Strontium-90 reduced to less than 2% is safe. Therefore final concentration [A] = 2 % = 0.02

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

[tex]t = \frac{2.303}{0.02475} log \frac{1}{[0.02]}[/tex]

t = 158 days

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Answer:

C;  ΔSuniv>0

Explanation:

In this question, we want to select which of the options must be true.

What we should understand is that for a process to be spontaneous, the change in entropy must be greater than 0 i.e the change in entropy must be positive.

Looking at the options we have; option C is the correct answer.

Option B looks correct but it is wrong. This is because if change in universal entropy is greater than zero, then change in Gibbs free energy must be less than zero for spontaneity to occur

It can be deduced that for a spontaneous process, B. ΔG>0.

It should be noted that a spontaneous process simply means a process that occurs without input of matter or electrical energy.

In this case, for a spontaneous process, it's true that ΔG>0, it should be noted that a spontaneous process related to the second law in thermodynamics. This is characterized by an increase in entropy.

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Answer:

half life=0.693/rate constant =188.3

The half-life of this element is 188.32 years

The formula for calculating the half-life of an element is expressed according to the equation:

[tex]t_{1/2}=\frac{ln 2}{\lambda}[/tex]

Given the following parameters:

The rate constant for the decay = 3.68 × 10⁻³ day⁻¹.

Substitute into the formula for calculating the half-life as shown:

[tex]3.68\times 10^{-3}=\frac{0.693}{\lambda}\\ 0.00368=\frac{0.693}{\lambda}\\\lambda=\frac{0.693}{0.00368}\\\lambda = 188.32 years[/tex]

Hence the half-life of this element is 188.32 years

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Answer:

Heat of vaporization refers to condensation and heat of fusion refers to melting.

Explanation:

Heat of vaporization or heat of evaporation, is defined as the energy required to convert liquid substance into a gas which creates condensation. As a gas condenses to a liquid, heat is released.

Heat of fusion refers to melting because heat of fusion is defined as the energy required to change any amount of substance when it melts.

Hence, the correct answer is "Heat of vaporization refers to condensation and heat of fusion refers to melting.".

Answer:

315 g

Explanation:

Step 1: Write the thermochemical equation

2 H₂O(l) → 2 H₂(g) + O₂(g)      ΔH = +572 kJ

Step 2: Calculate the molar of water decomposed by 5.00 × 10³ kJ of energy

According to the thermochemical equation, 572 kJ are required to decompose 2 moles of water.

5.00 × 10³ kJ × (2 mol/572 kJ) = 17.5 mol

Step 3: Calculate the mass corresponding to 17.5 moles of water

The molar mass of water is 18.02 g/mol.

17.5 mol × 18.02 g/mol = 315 g

Answer:

1. 0.00352 M

2. 2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)

3. 0.00534 M

Explanation:

1.

Mass of strontium hydroxide= 10.45 g

Volume of solution = 41.00 ml

Number of moles = mass of Sr(OH)2/molar mass of Sr(OH)2 = 10.45g/121.63 g/mol= 0.0859 moles

Molarity= number of moles × volume = 0.0859 ×41/1000 = 0.00352 M

2.

2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)

3.

Concentration of acid CA= the unknown

Volume of acid VA= 31.5 ml

Concentration of base CB= 0.00352 M

Volume of base VB= 23.9 ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB = NA/NB

CAVANB= CBVBNA

CA= CBVBNA/VANB

CA= 0.00352 × 23.9 ×2/31.5 ×1

CA= 0.00534 M

A. The molarity of the Sr(OH)₂ solution is 2.09 M

B. The balanced equation for the reaction is

2HNO₃ + Sr(OH)₂ —> Sr(NO₃)₂ + 2H₂O

C. The molarity of the acid, HNO₃ is 3.17 M

A. Determination of the molarity of the Sr(OH)₂ solution

Mass of Sr(OH)₂ = 10.45 g

Molar mass of Sr(OH)₂ = 88 + 2(16 + 1) = 122 g/mol

Mole = mass / molar mass

Mole of Sr(OH)₂ = 10.45 / 122

Mole of Sr(OH)₂ = 0.0857 mole

Volume = 41 mL = 41 / 1000 = 0.041 L

Molarity = mole / Volume

Molarity of Sr(OH)₂ = 0.0857 / 0.041

B. The balanced equation for the reaction.

C. Determination of the molarity of the acid, HNO₃.

From the balanced equation above,

The mole ratio of the acid, HNO₃ (nA) = 2

The mole ratio of the base, Sr(OH)₂ (nB) = 1

From the question given above,

Volume of base, Sr(OH)₂ (Vb) = 23.9 mL

Molarity of base, Sr(OH)₂ (Mb) = 2.09 M

Volume of acid, HNO₃ (Va) = 31.5 mL

MaVa / MbVb = nA/nB

(Ma × 31.5) / (2.09 × 23.9) = 2

(Ma × 31.5) / 49.951 = 2

Cross multiply

Ma × 31.5 = 49.951 × 2

Ma × 31.5 = 99.902

Divide both side by 31.5

Ma = 99.902 / 31.5

Thus, molarity of the acid, HNO₃ is 3.17 M

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Answer:

[tex]Kp=5.14[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

Thus, the equilibrium expression is written as:

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

And in terms of the reaction extent:

[tex]Kp=\frac{(2x)^2}{(10-x)(10-3x)^3}[/tex]

Thus, from the equilibrium pressure of ammonia we can compute the reaction extent:

[tex]p_{NH_3}=2x=6.0 atm\\\\x=3.0atm[/tex]

Therefore, the equilibrium constant turns out:

[tex]Kp=\frac{(2*3.0)^2}{(10.0-3.0)(10.0-3*3.0)^3}\\\\Kp=5.14[/tex]

Regards.

Answer:

Underneath the frozen upper layer, the water remains in its liquid form and does not freeze. Also, oxygen is trapped beneath the layer of ice. As a result, fish and other aquatic animals find it possible to live comfortably in the frozen lakes and ponds. ... This irregular expansion of water is called anomalous expansion.

Explanation:

so pretty much its because there is water under the top frozen layer and air is trapped underneath the ice

Answer:

glue that would make the strongest craft stick tower.

Explanation:

Independent variable: In statistics and research methods, the term "independent variable" is determined as a variable that is being changed, controlled, or altered in an experiment or research by the researcher or the experimenter to see its effect on DV or dependent variable. However, it is said that independent variable directly effect the dependent variable.

Answer:

OPTION C is correct

C) 1.07 g

Explanation:

CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)3(aq)

But we know molarity molarity= number of moles of solute/ volume of the solution

M= n/V

From the equation above

number of moles of Cacl2 = (37.5 ×0 .100 × 10^-3) = 0.00375 moles.

Then

1 mole of Cacl2 yields 2 moles of Agcl2

0.00375 moles of Cacl2 will produce let say Y.

Y= (0.00375 ×2)/1

= 0.0075 moles.

Number of moles of Agcl2 = mass /molar mass of Agcl

Molar mass of Agcl = 178

Then mass = 178 ×0.0075 = 1.047

Therefore, the mass of agcl precipitate is

1.07 g

The given question is incomplete. The complete question is:

Compound X has a molar mass of 153.05 g/mol and the following composition:

element mass %

carbon 47.09%

hydrogen 6.59%

chlorine 46.33%

Write the molecular formula of X.

Answer: The molecular formula of X is [tex]C_6H_{10}Cl_2[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 47.09 g

Mass of H = 6.59 g

Mass of Cl = 46.33 g

Step 1 : convert given masses into moles.

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{47.09g}{12g/mole}=3.92moles[/tex]

Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.59g}{1g/mole}=6.59moles[/tex]

Moles of Cl =[tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{46.33g}{35.5g/mole}=1.30moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{3.92}{1.30}=3[/tex]

For H = [tex]\frac{6.59}{1.30}=5[/tex]

For Cl =[tex]\frac{1.30}{1.30}=1[/tex]

The ratio of C : H: Cl= 3: 5 :1

Hence the empirical formula is [tex]C_3H_5Cl[/tex]

The empirical weight of [tex]C_3H_5Cl[/tex] = 3(12)+5(1)+1(35.5)= 76.5g.

The molecular weight = 153.05 g/mole

Now we have to calculate the molecular formula.

[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{153.05}{76.5}=2[/tex]

The molecular formula will be=[tex]2\times C_3H_5Cl=C_6H_{10}Cl_2[/tex]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can realize that water has a very simple atomic structure which consists of two hydrogen atoms bonded to one oxygen atom. The nature of the atomic structure of water causes its molecules to have unique electrochemical properties. The hydrogen side of the water molecule has a slight positive charge whereas at the other side of the molecule a negative charge exists. This molecular polarity causes water to be a powerful solvent and is responsible for its strong surface tension.

Moreover, water is involved in several both inorganic and organic chemical reactions leading to hydration, for example, the conversion of alkenes to alcohols, the hydrolysis of acyl halides, anhydrides, esters and amides to carboxylic acids and the hydration of a raft of inorganic salts that exist as hydrates only, such as copper (II) sulfate pentahydrate and so.

Best regards.

Answer:

The answer is

Explanation:

1 mole of acid.

Hope this helps....

Have a nice day!!!!

A buffer that is 1 M in acid and base will have the greatest capacity of buffer, and therefore the greatest buffer capacity.

A weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base, are combined to form the buffer solution, a water-based solvent solution.

In a biological system, a buffer's keep intracellular and extracellular pH levels within a relatively small range and to withstand pH fluctuations brought on by both internal and external factors.

A buffer is a substance that can withstand a pH shift when acidic or basic substances are added. It may balance out little quantities of additional acid or base, keeping the pH stable.

Thus, 1 M in acid and base solution has the greatest buffer capacity.

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Answer:

804 J

Explanation:

Step 1: Given data

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.0235J·g⁻¹K⁻¹ × (1.50 × 10³g) × [15.0°C-(-7.8°C)]

Q = 804 J

Answer:

CH2Cl2 is polar

Explanation:



Answer:

The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.  

Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams

The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,  

Moles of water = weight of water/molecular weight

= 42.1 grams / 18 = 2.3

The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,  

= 214.2 * 0.2190 = 46.91 grams

The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,

= 214.2 * 0.4314 = 92.40 grams

The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,  

= 214.2 * 0.3497 = 74.91 grams

The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17

The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,  

92.40/78.96 = 1.17

The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,  

74.91/15.999 = 4.68.  

Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O

Answer:

1.6x10¹¹ = Kc

Explanation:

For the reaction:

AgCl(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + Cl⁻(aq)

Kc is defined as:

Kc = [Ag(CN)₂⁻] [Cl⁻] / [CN⁻]²

Ksp of AgCl is:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

Where Ksp is:

Ksp = [Ag⁺] [Cl⁻] = 1.6x10⁻¹⁰

In the same way, Kf of Ag(CN)₂⁻ is:

Ag⁺(aq) + 2CN⁻ ⇄ Ag(CN)₂⁻

Kf = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = 1.0x10²¹

The multiplication of Kf with Ksp gives:

[Ag⁺] [Cl⁻] *  [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = Ksp*Kf

[Ag(CN)₂⁻] [Cl⁻] / [CN⁻]² = Ksp*Kf

Obtaining the same expression of the first reaction

That means Ksp*Kf = Kc

1.6x10⁻¹⁰*1.0x10²¹ = Kc

Answer:

15. Critical angle of glass and water combination, θ = 62.45°

16. Critical angle for the interface between Mystery A and glass, θ = 37.93°

Note; The question is incomplete. The complete question is as follows:

Medium Air Water Glass Mystery A Mystery B Table-2 Speed (m/s) 1.00 C 0.75 c 0.67 0.41 c 0.71 c n 1.00 1.33 1.50 Index of Refraction n of a given medium is defined as the ratio of speed of light in vacuum, c to the speed of light in a medium, v. n = c/v

Table-4: Incident Angle (degrees) Reflected Angle Refracted angle (degrees) (degrees) % Intensity of reflected ray 0 10 20 30 40 50 N/A N/A N/A 30 40 50 0 11.3 22.7 34.2 46.3 59.5 N/A N/A N/A 0.67 1.22 3.08 % Intensity of refracted ray 100 100 100 99.33 98.78 96.92

When rays travel from a denser medium to a less dense medium, we can define a critical angle of incidence θ such that refracted angle θ₂ = 90°. Applying Snell's law: Critical angle θ = sin-1(n₂/n₁).

When the angle of incidence is greater than the critical angle, 100% of the light intensity is reflected. This is called total internal reflection because all the light is reflected.

15. Calculate the critical angle of glass and water combination. Show your calculation.

16. What is the critical angle for the interface between Mystery A and glass?

Explanation:

15.  Applying Snell's law; Critical angle θ = sin-1(n₂/n₁).

where n₂,refractive index of water = 1.33, n₁, refractive index of glass = 1.50 since glass is denser than water

θ = sin-1(1.33/1.50)

θ = 62.45°

Critical angle of glass and water combination, θ = 62.45°

16.  Refractive index of mystery A , n = c/v

where v = 0.41 c

therefore, n = c / 0.41 c = 2.44

Critical angle for the interface between Mystery A and glass, θ = sin-1(n₂/n₁).

where n₂,refractive index of glass = 1.50, n₁, refractive index of mystery A = 2.44 since mystery A is denser than glass as seen from its refractive index

θ = sin-1(1.50/2.44)

θ = 37.93°

Critical angle for the interface between Mystery A and glass, θ = 37.93°

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

Moles of KHP =

According to reaction, 2 moles of KHP  with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

Volume of the calcium hydroxide solution used is 0.0235 mL.

Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.

Molality can be defined as the mass of solute present in 1000 grams of solvent.

Molality = [tex]\rm \dfrac{moles\;of\;solute}{mass\;of\;solvent\;(kg)}[/tex]

The moles of NaCl has been calculated as;

3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{mass\;of\;water\;(kg)}[/tex]

3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{2.692\;kg}[/tex]

Moles of NaCl = 3.140 [tex]\times[/tex] 2.692

Moles of NaCl = 8.45288 mol

The moles can be expressed as;

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Molecular weight of NaCl = 58.5 g/mol

The mass of NaCl can be calculated as:

8.45288 mol = [tex]\rm \dfrac{mass\;of\;NaCl}{58.5\;g/mol}[/tex]

Mass of NaCl = 58.5 g/mol [tex]\times[/tex] 8.45288 mol

Mass of NaCl = 494.493 grams.

The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.

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Answer:

Yes

Explanation:

The balanced equation of the reaction is;

FeBr2 (aq) + Cl2 (g) → FeCl2 (aq) + Br2 (aq)

This reaction is possible because chlorine is more electronegative than bromine and can displace it from its salt.

In group seventeen, electro negativity decreases down the group. Hence as we move down the group, elements become less electronegative and can be displaced from their salt by more electronegative elements found earlier in the group.

Hence chlorine can displace bromine in FeBr2 to form FeCl2.

Answer:

Yes, because Cl2 has higher activity than Br2

Explanation:

Answer:

[tex]1.25~mol~H_2O[/tex] and [tex]0.627~mol~N_2[/tex]

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ([tex]N_2H_4[/tex]) and oxygen ([tex]O_2[/tex]). So, we can start with the reaction between these compounds:

[tex]N_2H_4~+~O_2~->~N_2~+~H_2O[/tex]

Now we can balance the reaction:

[tex]N_2H_4~+~O_2~->~N_2~+~2H_2O[/tex]

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for [tex]N_2H_4[/tex] and 31.99 g/mol for [tex]O_2[/tex]):

[tex]20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4[/tex]

[tex]20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2[/tex]

In the balanced reaction we have 1 mol for each reagent (the numbers in front of [tex]O_2[/tex] and [tex]N_2H_4[/tex] are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is [tex]N_2H_4[/tex].

With this in mind, we can calculate the number of moles for each product. In the case of [tex]N_2[/tex] we have a 1:1 molar ratio (1 mol of [tex]N_2[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2[/tex]

We can follow the same logic for the other compound. In the case of [tex]H_2O[/tex] we have a 1:2 molar ratio (2 mol of [tex]H_2O[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O[/tex]

I hope it helps!

Answer:

e is the suitable answer for that. I think it is correct.

Answer: The correct option is E ( the reactant that is completely used up by a reaction).

Explanation:

A LIMITING REACTANT can be defined as the reagent or the substance that is involved in a chemical reaction which determines when the reaction will stop. This is because it is COMPLETELY used up in the reaction. Reactants that are called limiting reactants is because the quantity of these reagents are capable of limiting the amount of products formed. And by doing so, the chemical reaction cannot proceed further with the absence of this reactant. Using an attached diagram below to illustrate further:

The reagents D and E reacts to form F as the product. In this reaction, reactant E is the limiting reagent because there is still some left over D in the products. Therefore, D was in excess when E was all USED UP.

Therefore the CORRECT option is E which states that the reactant that is completely used up by a reaction, best describes a limiting reactant.

Option A is WRONG because it's the concentration of both reactants in chemical equation can limit the speed of that reaction.

Option B is WRONG because it's when the concentration of a particular reactant is either increased or decreased can affect the position of equilibrium.

Option C and D are wrong because the reactant that remains in the end of a reaction and can produce the greatest amount of product is the one in EXCESS.

Answer:

a. Methanol remains the same

b. Methanol decreases

c. Methanol increases

d. Methanol remains the same

e. Methanol increases

Explanation:

Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.

a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.

b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.

c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.

d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.

e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.

Explanation:

The equation is given as;

2 A + B + 3 C → Products

The order of the reaction refers to the extent at which the rate depends n the concentration of the reactant.

The order of reaction is experimentally obtained. It can also be obtained from the rate law of the reaction.

If the rate law is given as;

rate law = k [A]²[B][C]³

Then the order is second order with respect to A.

The order is second order with respect to A.

Given that;

2A + B + 3C → Products at 470 K

Find:

Order of reaction with respect to A

Computation:

The reaction that takes place refers to how much the rate is influenced by the reactant concentration.

The reaction order is determined empirically. This can also be derived from the reaction's rate law.

Rate law = k[A]²[B][C]³

So, The order is second order with respect to A.

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Answer:

When balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.

Explanation:

Oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The oxidation and reduction half-reactions, in a basic medium, adjust the oxygens and hydrogens as follows:

In the member of the half-reaction that presents excess oxygen, you add as many water molecules as there are too many oxygen. Then, in the opposite member, the necessary hydroxyl ions are added to fully adjust the half-reaction. Normally, twice as many hydroxyl ions, OH-, are required as water molecules have previously been added.

In short, you first adjust the oxygens with OH-, then you adjust the H with H₂O, and finally you adjust the charge with e-

So, when balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.

Answer:

c. balance the reaction as though under acidic conditions

Explanation:

When balancing redox reactions under basic conditions, a good technique is to first balance the reaction as though under acidic conditions. We then adjust the result to reflect the basic conditions.

Answer:

Explanation:

Answer in attached file .