Calculate the equilibrium constant K c for the following overall reaction: AgCl(s) + 2CN –(aq) Ag(CN) 2 –(aq) + Cl –(aq) For AgCl, K sp = 1.6 × 10 –10; for Ag(CN) 2 –, K f = 1.0 × 10 21.

Answers

Answer 1

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

1.6x10¹¹ = Kc

Related Questions

Rank the following amine derivatives from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
Highest acidity
anilinium ion
aniline
ammonium ion
secondary amine
amide
Lowest acidity

Answers

Answer:

anilinium ion > ammonium ion > amide > aniline > secondary amine

Explanation:

Acidity of amine derivatives can derived from their pKa values.

The rule of thumb for acidity with relation to pKa values is that:

As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.

Hence the stronger the acid , the  lower pKa value  and the weaker the acid , the stronger the pKa value.

So the pKa value for anilinium ion = 4.6

ammonium ion = 9.4

Amide = 15

Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.

The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.

In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).

the amine derivatives ranking is as follows:

anilinium ion > ammonium ion > amide > aniline > secondary amine

At what temperature in K will 0.750 moles of oxygen gas occupy 10.0 L and exert 2.50 atm of pressure

Answers

Answer:

406 K.

Explanation:

The following data were obtained from the question:

Number of mole (n) = 0.750 mole

Volume (V) = 10.0 L

Pressure (P) = 2.50 atm

Temperature (T) =.?

Note: Gas constant (R) = 0.0821 atm.L/Kmol

The temperature, T can be obtained by using the ideal gas equation as follow:

PV = nRT

2.5 x 10 = 0.75 x 0.0821 x T

Divide both side by 0.75 x 0.0821

T = (2.5 x 10) /(0.75 x 0.0821 )

T = 406 K.

Therefore, the temperature is 406 K.

Answer: 406 K

Explanation:

We can rewrite the ideal gas law to solve for T:

PV = nRT

T=PV / nR

We are given the following from the problem:

n=0.750 mol P=2.50 atm V=10.0 L

Plugging in our values and using R=0.08206 L⋅atm / K⋅mol we get:

T=(2.50 atm)(10.0 L) / (0.750 mole)(0.08206L ⋅ atm ⋅ mole K) = 406 K

As a reaction proceeds, the ratio between the rate of consumption of reactant and the rate of formation of product:

Answers

Answer:

Depends on the reaction.

Explanation:

Hello,

In this case, the answer is depends on the reaction since the ratios between the rates of both consumption and formation depend upon the stoichiometric coefficients in the chemical reaction. For instance, for the reaction:

A -> 2B

The relationship is:

[tex]\frac{1}{-1}r_A =\frac{1}{2} r_B[/tex]

Therefore, we can see that the rate of consumption of A half the rate of formation of B, but is we consider the following chemical reaction:

2A -> B

The relationship is:

[tex]\frac{1}{-2}r_A =\frac{1}{1} r_B[/tex]

Therefore we can see that the rate of consumption of A doubles the rate of consumption of B.

Best regards.

A student mixes 1.0 mL of aqueous silver nitrate, AgNO3 (aq), with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube. What will the student observe

Answers

Answer:

AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.

Explanation:

If a student mixes 1.0 mL of aqueous silver nitrate AgNO3 (aq)  with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube.

The sodium chloride is being acidified with dilute trioxonitrate (V) acid. Then a few drops of  silver trioxonitrate(V) is added afterwards. A  white precipitate of silver chloride, which dissolves readily in aqueous ammonia indicates the presence of sodium chloride.

The reaction proceeds as follows:

[tex]\mathtt{AgNO_{3(aq)} + NaCl _{(aq)} \to AgCl _{(s)} + NaNO_3_{(aq)}}[/tex]

From the reaction between AgNO3 (aq) and NaCl (aq), AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.

1. For the following reaction, 4.86 g of magnesium nitride are mixed with excess water. The reaction yields 7.18 g of magnesium hydroxide.
magnesium nitride(s) + water(1) –> magnesium hydroxide (aq) + ammonia (aq)
What is the ideal yield of magnesium hydroxide?
What is the percent yield for this reaction?
2. For the following reaction, 6.41 g of hydrogen gas are mixed with excess nitrogen gas. The reaction yields 26.2 g of ammonia.
nitrogen(g) + hydrogen(g) –> ammonia(g)
What is the ideal yield of ammonia?
What is the percent yield for this reaction?
3. For the following reaction, 3.79 g of water are mixed with excess chlorine gas. The reaction yields 8.70 g of hydrochloric acid.
chlorine(g) + water(1) –> hydrochloric acid(aq) + chloric acid (HCIO3)(aq)
What is the ideal yield of hydrochloric acid?
What is the percent yield for this reaction?

Answers

Answer:

See explanation

Explanation:

1)

Mg3N2(s) + 6H2O(l) ------------> 3Mg(OH)2 + 2NH3(g)

Number of moles of magnesium nitride= mass/molar mass= 4.86g/100.9494 g/mol = 0.048 moles

1 mole of magnesium nitride yields 3 moles of magnesium hydroxide

0.048 moles of magnesium nitride yields 0.048 moles × 3= 0.144 moles of magnesium hydroxide

Theoretical yield of magnesium hydroxide = 0.144 moles × 58.3197 g/mol = 8.398 g

Percent yield= actual yield/ theoretical yield × 100

Percent yield= 7.18/8.398 × 100/1 = 85.5%

2)

N2(g) + 3H2(g) -------> 2NH3(g)

Number of moles of hydrogen gas = mass/ molar mass = 6.41g/ 2gmol-1 = 3.205 moles of hydrogen gas.

From the balanced reaction equation;

3 moles of hydrogen gas yields 2 moles of ammonia

3.205 moles of hydrogen gas yields 3.205 × 2/3 = 2.1367 moles of ammonia

Theoretical yield of ammonia = 2.1367 moles × 17 gmol-1 = 36.3 g

Percent yield = actual yield/ theoretical yield ×100

Percent yield = 26.2/36.3 ×100

Percent yield = 72.2%

3)

3Cl2(g) + 3H2O(l) ------> HOCl3(aq) + 5HCl(aq)

Number of moles of water= mass/ molar mass = 3.79g/18 gmol-1 = 0.21 moles

Since

3 mole of water yields 5 mole of HCl

0.21 moles of water yields 0.21 × 5/3 = 0.35 moles of HCl

Theoretical yield of HCl = 0.35 moles × 36.5 gmol-1 = 12.775 g

Percent yield = actual yield/ theoretical yield × 100/1

Percent yield = 8.70/12.775 ×100

Percent yield = 68.1%

When the nuclide carbon-14 undergoes beta decay: The name of the product nuclide is . The symbol for the product nuclide is

Answers

Answer:

Explanation:

The equation for the decay is given as;

¹⁴₆C  --> X + ⁰₋₁e

For conservation of matter, the mass number and atomic number has to be the same in both the reactant and product side f he equation;

Mass number;

14 = x + 0

x = 14

Atomic Number;

6 = x + (-1)

x = 6 + 1 =7

¹⁴₆C  --> ¹⁴₇N + ⁰₋₁e

The name of the product nuclide with atomic number of 7 is Nitrogen. The symbol is; ¹⁴₇N

Identify four general properties that make an NSAID unique as compared to the NSAID aspirin. List specific properties that make aspirin, naproxen, and ibuprofen unique from one another

Answers

Answer:

NSAIDs are steroidal anti-inflammatories, their action is on the phospholipase A2 enzyme, this enzyme is responsible for breaking down the phospholipids of the membrane to trigger an inflammatory response. This is how steroidal anti-inflammatory drugs inhibit ALL inflammatory pathways (not like NSAIDs that they only inhibit the COX pathway).

These corticosteroid drugs cannot exceed the systemic mineralocorticoid value 1 in the body, since this corticosteroid hormone is also released by the adrenal cortex.

The NSAIDs generate: sporadic peaks in blood glucose, hypertension, fluid retention, increase in body fat mass, possible suppression of the adrenal cortex over time, inhibiting endogenous synthesis of corticosteroids.

On the other hand, naproxen and ibuprofen are NSAIDs, that is, non-steroidal anti-inflammatory drugs that do not influence both routes of inflammation, but only COX, this enzyme is abbreviated as COX but is called cyclooxygenase, and is responsible for a single route of inflammation.

NSAIDs such as naproxen and ibuprofen can cause gastric disorders such as ulcers or gastritis if they are consumed in a very repetitive manner.

In addition, both drugs are anti-inflammatory, analgesic and antipyretic. Although its two main functions are the first two, it was shown to have an effect in lowering body temperature.

That they are anti-inflammatory means that they inhibit the path of inflammation and analgesics the path of pain.

Explanation:

Both types of drugs generate the same effect but by different mechanisms.

Some are steroids and others are not, although steroids are considered to have a greater risk of benefit that is why they are administered against more systematically compromised instances such as anaphylactic shock.

NSAIDs such as naproxen and ibuprofen are the most prescribed today, since they have few risks and very good benefits, meaning that their adverse effects are not lethal or highly relevant and have a good effect on symptoms.

Both must be administered with care when treating a diabetic patient since corticosteroids generate glycemic peaks or increase in blood glucose, and NSAIDs compete for plasma protein with oral hypoglycemic agents, thus generating that these are in higher free concentrations. high diffusing better through the tissues and increases the potency of the adverse effects of these.

For each of the processes, determine whether the entropy of the system is increasing or decreasing.

a. A snowman melts on a spring day.
b. A document goes through a paper shredder.
c. A water bottle cools down in a refrigerator.
d. Silver tarnishes
e. Dissolved sugar precipitates out of water to form rock candy.

Answers

Explanation:

Entropy refers to the degree of disorderliness of a system.

a. A snowman melts on a spring day.

Entropy is increasing because there is a change in state of matter from solid to liquid. Liquid particles have more freedom f movement compared to solids.

b. A document goes through a paper shredder.

Entropy increases because random, disorganized bits of paper are left.

c. A water bottle cools down in a refrigerator.

Entropy  decreases because temperature is directly proportional to entropy.

d. Silver tarnishes

Entropy increases because random bits of the sliver particles are formed.

e. Dissolved sugar precipitates out of water to form rock candy.

Entropy decreases because the random dissolved sugar precipitates are ordered into a rock candy.

Which of the following processes is spontaneous? Select the correct answer below:
a. carbon graphite turning to carbon diamond at normal pressure
b. carbon diamond turning to carbon graphite at normal pressure
c. carbon diamond turning to carbon graphite at high pressure
d. none of the above

Answers

Answer:

b. carbon diamond turning to carbon graphite at normal pressure.

Explanation:

Graphite is the stable form of the carbon element under normal ambient pressure.  The process of conversion of carbon diamond to carbon graphite is a spontaneous process under typical ambient conditions, since diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions, and the process of conversion can take place without a continuous input of energy from an external source.

How many moles of bromine will react with 0.0500 mole of C 2H 2 in the reaction C 2H 4 + Br 2 → C 2H 4Br 2?

Answers

Answer:

0.05 mole of Br2.

Explanation:

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

C2H4 + Br2 —› C2H4Br2

From the balanced equation above,

1 mole of C2H4 reacted with 1 mole of Br2 to produce 1 mole of C2H4 Br2.

Finally, we shall determine the number of moles bromine that will react with 0.05 mole of C2H2.

The number of mole of Br2 needed for the reaction can be obtained as follow:

From the balanced equation above,

1 mole of C2H4 reacted with 1 mole of Br2.

Therefore, 0.05 mole of C2H4 will also react with 0.05 mole of Br2.

Therefore, 0.05 mole of Br2 is needed for the reaction.

PLEASE HELP WILL GIVE BRAINLIEST!!!
Refining aluminum takes place in steel vats lined with graphite. What function does the graphite lining serve?
A. It is the cathode.
B. It is the anode.
C. It soaks up the waste oxygen released from bauxite (Al2O3)
D. It serves as a grounding layer to prevent electrical surges.

Answers

Answer: It's the anode broski (B)

Explanation: I'm taking the Chem summer course too broski, this was the correct answer. Cheers broski

Given the data: Ag2O(s), = ‑31.1 kJ mol-1, S° = +121.3 J mol-1 K-1 Ag(s), = 0.00 kJ mol-1, S° = +42.55 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = +205.0 J mol-1 K-1 Calculate the temperature at which = 0 for the reaction, Ag2O(s) → 2 Ag(s) + ½ O2(g). Assume that, since the physical states do not change, and are independent of tempera­ture between ‑50.0 °C and 950.0 °C.

Answers

Answer:

[tex]T=469.1K\\\\T=195.9\°C[/tex]

Explanation:

Hello,

In this case, for the given decomposition reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):

[tex]\Delta _rH=2\Delta _fH_{Ag}+\frac{1}{2} \Delta _fH_{O_2}-\Delta _fH_{Ag_2O}\\\\\Delta _rH=2*0.00+\frac{1}{2} *0.00-(-31.1)=31.1kJ/mol[/tex]

Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):

[tex]\Delta _rS=2S_{Ag}+\frac{1}{2} S_{O_2}-S_{Ag_2O}\\\\\Delta _rS=2(42.55)+\frac{1}{2} (205.0)-(121.3)=66.3J/(mol*K)[/tex]

Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the unknown temperature, for such Gibbs energy equaling 0, the temperature (in K and °C) turns out:

[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\0=31.1kJ/mol-T(66.3\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\T=\frac{31.1kJ/mol}{0.0663kJ/(mol*K)} =469.1K\\\\T=195.9\°C[/tex]

Which is within the given rank.

Best regards.

If 1 mol of a pure triglyceride is hydrolyzed to give 2 mol of RCOOH, 1 mol of R'COOH, and 1 mol of glycerol, which of the following compounds might be the triglyceride?
CHOC(O)R
A. CHOC(O)R
CHOC(O)R
CH,OC(O)R
B. CHOC(O)R
CH2OC(O)R
CHOC(O)R
C. CHOC(O)R
CHOC(O)R
CHOC(O)R
D. CHOC(O)R
CHOC(O)R

Answers

Answer:

The correct option is C.

Note the full question and structure of the moleculesis found in the attachment below.

Explanation:

Triglycerides or triacylglycerols are non-polar, hydrophobic lipid molecules composed of three fatty acids linked by ester bonds to a molecule of glycerol.

The fatty acids linked to the glycerol molecule are denoted by R and may be of the same kind or different. when the R group is the same, the R is attached in all the three positions for ester bonding in the glycerol molecule but when they are different are denoted by R, R' and R'' respectively.

During the hydrolysis of triglycerides, the three fatty acids molecules are obtained as well as a glycerol molecule.

From the question, when 1 mole of the triglyceride is hydrolysed, 2 moles of RCOOH, 1 mole of R'COOH and 1 mole of glycerol is obtained. The triglyceride must then be composed of two fatty acids which are the same denoted by R, and a different fatty acid molecule denoted by R'.

The correct option therefore, is C

The respiration rate of a goldfish is measured. The goldfish is then placed in cold water and the respiration rate is measured again. What is the INDEPENDENT variable?

Answers

Answer:

Temperature of the water

Explanation:

In every study, there must be independent and dependent variables. An independent variable is the variable that is changed in order to obtain a response. In this case, the temperature of the water is being changed, the response in this experiment is the respiration rate of the goldfish.

Thus the respiration rate of the goldfish is the dependent variable because it is controlled by the temperature of the water and changes accordingly.

Summarily, the independent variable is the temperature of the water while the dependent variable is the respiration rate of the goldfish.

Consider the compound hydrazine N2H4 (MW = 32.0 amu). It can react with I2 (MW = 253.8 amu) by the following reaction 2 I2 + N2H4 ------------- 4 HI + N2 (a) How many grams of I2 are needed to react with 36.7 g of N2H4? (b) How many grams of HI (MW = 127.9 amu) are produced from the reaction of 115.7 g of N2H4 with excess iodine?

Answers

Answer:Cobb

Explanation:What y'all

What is the energy of a photon of electromagnetic radiation with a wavelength of 963.5 nm?​ (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J · s

Answers

Answer:

[tex]E=2.06\times 10^{-19}\ J[/tex]

Explanation:

Given that,

The wavelength of electromagnetic radiation is 963.5 nm.

We need to find the energy of a photon with this wavelength.

The formula used to find the energy of a photon is given by :

[tex]E=\dfrac{hc}{\lambda}\\\\E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{963.5\times 10^{-9}}\\\\E=2.06\times 10^{-19}\ J[/tex]

So, the energy of a photon is [tex]2.06\times 10^{-19}\ J[/tex].

Why can long chain fatty acids can form micelles in solutions with pH > 7 but are insoluble in pH < 5

Answers

Answer:

In basic conditions that is ( pH > 7 ), the equilibrium shifts towards right and produces a lot of (-ve) negatively charged fatty acids which are polar, In water, since they have long hydrophobic hydrocarbon part, this form micelles where the hydrocarbon part  remain inside the  sphere and -coo- group remain outside the sphere due to H-bonding interaction with water.

At ( pH < 5 ) I.e acidic conditions, the equilibrium shift to the left giving neutral molecules which can not have stronger H-bonding interaction .

So micelles cant form as they become insoluble.

A solution of LiCl in water has XLiCl = 0.0800. What is the molality? A solution of LiCl in water has XLiCl = 0.0800. What is the molality? 4.44 m LiCl 8.70 m LiCl 4.83 m LiCl 4.01 m LiCl

Answers

Answer:

mol LiCl = 4.83 m

Explanation:

GIven:

Solution of LiCl in water XLiCl = 0.0800

Mol of water in kg = 55.55 mole

Find:

Molality

Computation:

mole fraction = mol LiCl / (mol water + mol LiCl)

0.0800 = mol LiCl / (55.55 mol + mol LiCl)

0.0800 mol LiCl + 4.444 mol = mol LiCl

mol LiCl - 0.0800 mol LiCl = 4.444 mol

0.92 mol LiCl = 4.444 mol

mol LiCl = 4.83 m

Find the pH of these buffer solutions using the information provided: 1L solution containing 80g of lactic acid (MW

Answers

Answer:

pH of the solution is 2.0

Explanation:

The lactic acid is a weak acid that is in equilibrium with water as follows:

Lactic acid + H2O ⇄ Lactate + H₃O⁺

And Ka for lactic acid: 1.38x10⁻⁴

Ka = 1.38x10⁻⁴ = [Lactate] [H₃O⁺] / [Lactic acid]

Initial concentration of lactic acid is (MW: 112.06g/mol):

80g * (1mol / 112.06g) / 1L = 0.714M

The equilibrium concentration of the species in the equilibrium are:

[Lactate] = X

[H₃O⁺] = X

[Lactic acid] = 0.714-X

Replacing in Ka expression:

1.38x10⁻⁴ = [X] [X] / [0.714-X]

9.8532x10⁻⁵ -  1.38x10⁻⁴X = X²

9.8532x10⁻⁵ -  1.38x10⁻⁴X - X² = 0

Solving for X:

X = -1.0x10⁻². False solution, there is no negative concentrations

X = 9.86x10⁻³M. Right solution.

As [H₃O⁺] = X

[H₃O⁺] = 9.86x10⁻³M

and pH = -log [H₃O⁺] = -log 9.86x10⁻³M

pH = 2.0

pH of the solution is 2.0

What are some geographic features that could be found in the hydrosphere?

Answers

Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

Answer:

Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!

Explanation:

The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)

Check 0/1 ptRetries 5 Element R has three isotopes. The isotopes are present in 0.0825, 0.2671, and 0.6504 relative abundance. If their masses are 97.62, 109.3, and 138.3 respectively, calculate the atomic mass of element R.

Answers

Answer:

Atomic mass = 127.198 amu

Explanation:

The average atomic mass is obtained by summing the masses of the isotopes each multiplied by its abundance.

Atomic mass = (97.62 * 0.0825) + (109.3 * 0.2671) + (138.3 * 0.6504)

Atomic mass = 8.05365 +  29.19403 + 89.95032

Atomic mass = 127.198 amu

The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.350 atm? (∆Hvap = 28.5 kJ/mol)

Answers

Answer:

The temperature at which the vapor pressure would be 0.350 atm is 201.37°C

Explanation:

The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.

Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.

The equation for Clausius- Clapeyron Equation can be expressed as:

[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]

where ;

[tex]P_1[/tex] is the vapor pressure at temperature 1

[tex]P_ 2[/tex] is the vapor pressure  at temperature 2

∆Hvap = enthalpy of vaporization

R = universal gas constant

Given that:

[tex]P_1[/tex] = 1 atm

[tex]P_ 2[/tex]  = 0.350 atm

∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol

[tex]T_1[/tex] = 282 °C  = (282 + 273) K = 555 K

R = 8.314 J/mol/k

Substituting the above values  into the Clausius - Clapeyron equation, we have:

[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]

[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }[/tex]

[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }[/tex]

[tex]\dfrac{1}{T_2} = 0.002108048708[/tex]

[tex]T_2 = \dfrac{1}{0.002108048708}[/tex]

[tex]\mathbf{T_2 }[/tex] =  474.37 K

To °C ; we have [tex]\mathbf{T_2 }[/tex] =   (474.37 - 273)°C

[tex]\mathbf{T_2 }[/tex] =  201.37 °C

Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C

The temperature of the liquid at the given vapor pressure is 201.5 ⁰C.

The given parameters;

boiling point temperature, = 282 ⁰Cvapor pressure, P₂ = 0.35 atmenthalpy of vaporization, ∆Hvap = 28.5 kJ/mol

The temperature of the liquid will be determined by applying Clausius- Clapeyron Equation;

[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )[/tex]

where;

R is ideal gas constant = 8.314 J/mol.kT₁ is the initial temperature in Kelvin = 282 + 273 = 555 KP₁ is the initial pressure = 1 atm

[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )\\\\ln(\frac{0.35}{1} ) = \frac{28.5 \times 10^3}{8.314} (\frac{T_2 - 555}{555T_2} )\\\\-1.049 = 6.176- \frac{3427.95}{T_2} \\\\\frac{3427.95}{T_2} = 6.176 + 1.049\\\\\frac{3427.95}{T_2} = 7.225\\\\T_2 = \frac{3427.95}{7.225} \\\\T_2 = 474.5 \ K\\\\T_2 = 474.5 - 273 = 201.5 \ ^0C[/tex]

Thus, the temperature of the liquid at the given vapor pressure is 201.5 ⁰C.

Learn more here:https://brainly.com/question/1077674

Which two layers are part of the thermosphere?

O exosphere and ionosphere

O ionosphere and mesosphere

mesosphere and stratosphere

O stratosphere and troposphere

Answers

The two layers are part of the thermosphere are  exosphere and ionosphere.

What is the exosphere layer?

The Exosphere is the topmost layer of the Earth's atmosphere.

and its gradually disappear into the vacuum of space.

It consist two parts that are:

exosphere and ionosphere.

Thus, option "A" is correct, the rest of the option is not a part of thermosphere.

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Answer:

hi hope your doing great the answer is A

Explanation:

its on Edge 2020

hope i helped :)

If you had a cup full of methanol and a pool full of methanol, would the mass change?

Answers

Answer:

the mass does not change

Explanation:

What creation would you make if you had access to any of the chemical elements? can someone answer this for me please.​

Answers

Answer:

Depends on the chemicals.

Explanation:

The creation of something depends on the chemical we have. If I have chemicals such as acid and base so I can produce salt and water by mixing acid and base. If I have ethanol, vegetable oil and sodium hydroxide which is  a catalyst, I can produce biodiesel which can be used in vehicles instead of fossil fuel. If I have sodium element and chlorine gas, I can produce sodium chloride also known as table salt.

Why will the conjugate base of a weak acid affect pH? Select the correct answer below: it will react with hydroxide

Answers

Answer:

It will react with water

Explanation:

I know this is an older question, but I just wanted to provide the correct answer.

Since we are dealing with a weak base, and the acid is somewhat stronger, it will react with the water molecules to produce hydronium, which will affect the pH of the solution.

The conjugate base of a weak acid affect pH because it will react with hydronium ion.

A weak acid is an acid that does not dissociate completely in water. On the other hand, a strong acid achieves almost 100% dissociation in water.

Acids dissociate in water to yield the hydronium ion and a conjugate base. For instance, the weak acid, acetic acid is dissociated as follows;

CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)

We can see that the conjugate base( CH3COO-) could react with the hydronium ions thereby moving the equilibrium position to the left hand side and affecting the pH by decreasing the hydronium ion concentration.

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Missing parts;

Why will the conjugate base of a weak acid affect pH? Select the correct answer below: O it will react with hydroxide O it will react with water O it will react with hydronium O none of the above

Which of the following forms a molecular solid? Which of the following forms a molecular solid? C10H22 CaO gold C, graphite

Answers

Answer:

C10H22

Explanation:

Graphite is known as an allotrope of carbon. Its characteristics include being black and slippery and as used as lubricants.

Gold (Au) is an element on the periodic table with atomic number 79 and a mass number 197 which exists as a metal due to its hydrogen bonds.

C10H22 which is also known as decane belongs to the Alkane family.The General forces of attraction between the alkane family are weak but in the case of decade there is Van der waal force which makes Decane C10H22 a Molecular Solid.

Which of the following solutions is acidic? [H3O+] = 1.0 x 10-10 M [H3O+] < 1.0 x 10-7 M [OH-] = 1.0 x 10-10 M [OH-] = 1.0 x 10-7 M [OH-] > 1.0 x 10-7 M

Answers

Answer:

[OH-] = 1.0 x 10-10 M

Explanation:

The acidity of a solution can be determined directly from the concentration of the hydrogen ions and indirectly from the concentrations of the hydroxide ions.

Generally, for a neutral solution we have;

[H3O+] = [OH-] = 1.0 x 10-7 M

For an acidic solution;

[H3O+] > 1.0 x 10-7 M

[OH-] < 1.0 x 10-7 M

Comparing the options the correct option is;

[OH-] = 1.0 x 10-10 M

g The electronic structure of which ONE of the following species cannot be adequately described by a single Lewis formula? (In other words, the electronic structure of which one can only be described by drawing two or more resonance structures?) A) C2H4 B) SO3 2– C) SO3 D) C3H8 E) HCN

Answers

Answer:

C) SO3

Explanation:

Lewis formula shows the bonding between atoms of a molecule and expresses the lone pair present in the atoms.

SO3 or Sulfur trioxide cannot be adequately described by a single Lewis formula because it has majorly 3 resonance structures because Sulfur does not follow the octet rule and can expand electrons in its outer shell.

Hence, the correct answer is C) SO3

To refine aluminum from its ore, aluminum oxide is electrolyzed to form aluminum and oxygen. At which electrode does oxygen form? options: A) Both the anode and the cathode B) Cathode C) Neither electrode D) Anode

Answers

Answer:

im pretty sure its the anode

Explanation:

To solve such, we must know the concept of electrolysis reaction. The correct option is option D among all given options. At anode electrode oxygen forms.

What is chemical reaction?

Chemical reaction is a process in which two or more than two molecules collide in right orientation and energy to form a new chemical compound. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, double displacement reaction.

Electrolysis is the process of passing an electric current through a material to cause a chemical change. A chemical change occurs when a material loses or acquires the electron. To refine aluminum from its ore, aluminum oxide is electrolyzed to form aluminum and oxygen. At anode electrode oxygen forms.

Therefore, the correct option is option D among all given options. At anode electrode oxygen forms.

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