Choose the most appropriate indicator for the titration of a weak acid with NaOH, where the expected equivalence point of the titration is at pH 8.8.

a. methyl orange, pH range 3.2-4.4
b. methyl red, pH range 4.8 6.0
c. bromothymol blue, pH range 6.0-7.6
d. phenolphthalein, pH range 8.2-10.0
e. alizarin yellow R. pH range 10.1-12.0

Answers

Answer 1

Answer:

D phenolphthalein,pH range 8.2-10.0


Related Questions

Charcoal from the dwelling level of the Lascaux Cave in France gives an average count of 0.97 disintegrations of ^14 C per minute per gram of sample. Living wood gives 6.68 disintegrations per minute per gram. Estimate the date of occupation and hence the probable date of the wall painting in the Lascaux Cave. Hint: Disintegrations per minute per gram" has the same units as the time-derivative of concentration for a radioactive decay model. (You may use the fact that the half-life of ^14C is 5568 years.)

Answers

Answer:

Explanation:

count given by old sample = .97 disintegrations per minute per gram

count given by fresh sample = 6.68 disintegrations per minute per gram

Half life of radioactive carbon = 5568 years

rate of disintegration

dN / dt = λ N

In other words rate of disintegration is proportional to no of radioactive atoms present . As number reduces rate also reduces .

Let initial no of radioactive be N₀ and after time t , number reduces to N

N₀ / N = 6.68 / .97

Now

[tex]N=N_0e^{-\lambda t}[/tex]

[tex]\frac{N}{N_0} =e^{-\lambda t}[/tex]

[tex]\frac{6.68}{.97} = e^{\lambda t}[/tex]

λ is disintegration constant

λ = .693 / half life

= .693 / 5568

= .00012446 year⁻¹

Putting the values in the equation above

[tex]\frac{6.68}{.97} = e^{.00012446\times t}[/tex]

[tex]6.8866 = e^{.00012446\times t}[/tex]

1.929577 = .00012446 t

t = 15503.6 years .  

For a single substance at atmospheric pressure, classify the following as describing a spontaneous process, a nonspontaneous process, or an equilibrium system.
A) Solid melting below its melting point
B) Gas condensing below its condensation point
C) Liquid vaporizing above its boiling point
D) Liquid freezing below its freezing point
E) Liquid freezing above its freezing point
F) Solid melting above its melting point
G) Liquid and gas together at boiling point with no net condensation or vaporization
H) Gas condensing above its condensation point
I) Solid and liquid together at the melting point with no net freezing or melting

Answers

Answer:

Spontaneous process- This is the process that occurs on its own without the application of any external energy or other factor. They include

B) Gas condensing below its condensation point

C) Liquid vaporizing above its boiling point

D) Liquid freezing below its freezing point

F) Solid melting above its melting point

Non spontaneous - This is the process that doesn’t occurs on its own and requires the application of any external energy or factor. They include

A) Solid melting below its melting point

E) Liquid freezing above its freezing point

H) Gas condensing above its condensation point

Equilibrium system

G) Liquid and gas together at boiling point with no net condensation or vaporization

I) Solid and liquid together at the melting point with no net freezing or melting

The matching is as follows:

A) Solid melting below its melting point - nonspontaneous process

B) Gas condensing below its condensation point - spontaneous process

C) Liquid vaporizing above its boiling point - spontaneous process

D) Liquid freezing below its freezing point - spontaneous process

E) Liquid freezing above its freezing point - nonspontaneous process

F) Solid melting above its melting point - spontaneous process

G) Liquid and gas together at boiling point with no net condensation or vaporization - Equilibrium system

H) Gas condensing above its condensation point - nonspontaneous process

I) Solid and liquid together at the melting point with no net freezing or melting -  Equilibrium system

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When electrons are filling energy levels, the lowest energy sublevels are occupied first. This is ____________.

Answers

When electrons are filling energy levels, the lowest energy sublevels are occupied first. This is Hund's rule.

Hund's rules state that:

Every orbital in a sublevel has to be singularly occupied before any other orbital is able to be doubly occupied.

All of the electrons in single occupied orbitals have to have the same spin to maximize the total spin.

It is called Aufbau Principle

Determine the volume occupied by 10 mol of helium at 27 ° C and 82 atm

please.

Answers

Answer:

3.00 L

Explanation:

Convert the pressure to Pascals.

P = 82 atm × (101325 Pa/atm)

P = 8,308,650 Pa

Convert temperature to Kelvins.

T = 27°C + 273

T = 300 K

Use ideal gas law:

PV = nRT

(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)

V = 0.00300 m³

If desired, convert to liters.

V = (0.00300 m³) (1000 L/m³)

V = 3.00 L

Answer:

[tex]\large \boxed{\text{3.0 L}}[/tex]

Explanation:

[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]

How many molecules of CaCl2 are equivalent to 75.9g CaCl2 (Ca=40.08g/mol, CL=35.45g/mol)

Answers

Answer:

[tex]\large \boxed{4.12 \times 10^{23}\text{ formula unis of CaCl}_{2}}$}[/tex]

Explanation:

You must calculate the moles of CaCl₂, then convert to formula units of CaCl₂.

1. Molar mass of CaCl₂

CaCl₂ = 40.08 + 2×35.45 = 40.08 + 70.90 = 110.98 g/mol

2. Moles of CaCl₂ [tex]\text{Moles of CaCl}_{2} = \text{75.9 g CaCl}_{2} \times \dfrac{\text{1 mol CaCl}_{2}}{\text{110.98 g CaCl}_{2}} = \text{0.6839 mol CaCl}_{2}[/tex]

3. Formula units of CaCl₂

[tex]\text{No. of formula units} = \text{0.6839 mol CaCl}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules CaCl}_{2}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{4.12 \times 10^{23}}\textbf{ formula units CaCl}_{2}\\\text{There are $\large \boxed{\mathbf{4.12 \times 10^{23}}\textbf{ formula units of CaCl}_{2}}$}[/tex]

An atom with 19 protons and 18 neutrons is a(n)

A. Isotope of potassium(K)
B. Standard atom of argon(Ar)
C. Standard atom of (K)
D. Isotope of argon (Ar)

Answers

Answer:

A

Explanation:

The number of protons indicates the element so we know it's potassium. To get the number of neutrons you subtract the number of protons (19) from the mass number which for potassium is 39.

39-19=20 neutrons

Because you have 18 neutrons then yours would be an isotope.

Answer: A. Isotope of potassium(K)

Explanation: Founders Educere answer.

Determine whether each of the following salts will form a solution that is acidic, basic, or pH-neutral. Drag the appropriate items to their respective bins.
Al(NO3)3
C2H5NH3NO3
NaClO
RbI
CH3NH3CN

Answers

Answer:

Al(NO₃)₃: Acidic.

C₂H₅NH₃NO₃: Acidic.

NaClO: Basic

RbI: pH-neutral

CH₃NH₃CN: Solution basic

Explanation:

The general rules to determine if a solution is acidic, basic or neutral are:

If it is a salt of a strong acid and base, the solution will be pH-neutral. If it is a salt of a strong acid and a weak base, the solution will be acidic due to the hydrolysis of the weak base component (cation). If it is a salt of a strong base and a weak acid, the solution will be basic due to the hydrolysis of the weak acid component (anion).

For the salts:

Al(NO₃)₃. The repective acid is HNO₃ (Strong acid) and the base is Al(OH)₃ (Weak base). As the salt comes from strong acid and weak base. SOLUTION ACIDIC

C₂H₅NH₃NO₃. The acid is HNO₃ (Strong acid) and the base C₂H₅NH₃OH (Weak base). SOLUTION ACIDIC.

NaClO. Tha acid is HClO (weak acid), and the base NaOH (Strong base). SOLUTION BASIC.

RbI: The acid is HI (Strong acid) and the base RbOH (Strong base). pH-NEUTRAL

CH₃NH₃CN. The acid is HCN (weak acid; pKb = 4.79) and  the base CH₃NH₃OH (weak base; pKa = 10.64). Both weak acid and base will produce each hydrolisis. The lower pK will predominate. That is the weak acid. SOLUTION BASIC

Solution of Al(NO₃)₃ and C₂H₅NH₃NO₃ salts is acidic, NaClO is basic and of RbI & CH₃NH₃cyanide is neutral in nature.

What is pH?

pH of any solution tells about the acidity or basicity of the solution, pH of any solution ranges from 0 to 14 and from acidity to basicity.

Al(NO₃)₃ is a salt which is formed by the mixing of strong acid HNO₃ (Nitric acid) and weak base Al(OH)₃, so the resultant solution of the salt is acidic in nature.C₂H₅NH₃NO₃ salt is formed by the mixing of strong acid HNO₃ (Nitric acid) and weak base C₂H₅NH₃OH, so the resultant solution of the salt is acidic in nature.NaClO is a salt of weak acid is HClO and strong base NaOH, so the resultant solution of the salt is basic in nature.RbI salt is formed by the combination of strong acid HI and strong base RbOH, so the resultant solution of the salt is neutral in nature.CH₃NH₃Cyanide is a salt of weak acid hydrogen cyanide and weak base CH₃NH₃OH,  so the resultant solution of the salt is neutral in nature.

Hence, appropriate differentiation was done above.

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Which of the following is an alkaline earth metal?
A. Silicon (Si)
B. Magnesium (Mg)
C. Carbon (C)
D. Aluminum (AI)

Answers

Answer:

B

Explanation:

The alkaline earth metals are the elements located in Group 2. The only element out of our choices that is in Group 2 is Magnesium.

Magnesium (Mg) is an alkaline earth metal. So, option (B) is the correct answer.

What are Alkaline Earth Metals?

Alkaline earth metals are metals with two valence electrons in their outermost shell.

Group 2A (or IIA) elements:
- Beryllium (Be)
- Magnesium (Mg)
- Calcium (Ca)
- Strontium (Sr)
- Barium (Ba)
- Radium (Ra)


The six alkaline earth metals are barium, radium, strontium, magnesium, and calcium. By giving their outermost electrons, which boosts their stability, they acquire the electron configuration of noble gases. The electrons are transported to an electronegative atom, giving the alkaline earth metals a positive charge.

The second column of the periodic table contains alkaline earth metals, which are very reactive metals. Every single object in the world is made using these metals. Sulfates of these metals may be found all throughout nature.

Minerals come in a variety of forms, some of which include gypsum, calcium sulfate, epsomite, magnesium sulfate, barite, and barium sulfate.

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Calculate the molality of a solution prepared by dissolved 19.9 g of kcl in 750ml of water

Answers

First, we find the molar mass of KCl which is AK+ACl=39+35.5=74.5g/mole

now we find out the number of moles by dividing the given mass to the molar mass n=m/M=19.9/74.5=0.26 moles. The molarity of a solution is equal to the number of moles divided by the volume of the solution. =0.26moles/0.75liters=0.346M

How long should you hold the iron on the hair to heat the strand and set the base ?


A) 5 seconds
B) 15 seconds
C) 30 seconds
D) 1 minute

Answers

A) 5 seconds
That’s what I did
A) 5 seconds
That is what I use and what most people in my life use

Select the correct answer from each drop-down menu.
Pictures that we receive from space are of the
✓ because it takes time for
to reach Earth.

Answers

Answer:

Pictures that we receive from space are of the past because it takes time for light to reach Earth.

Explanation:

For example, Mars is so far away that, depending on its position in orbit, a picture from Mars takes between 4 min and 24 min to reach Earth.

Answer: Pictures that we receive from space are of the

past

because it takes time for

light

to reach Earth.

Explanation:

BEARINGS
Question 12 (SSCE 1994 May/June)
(a) A village P is 10km from a lorry station, Q, on
a bearing 065º. Another village R, is 8km from
Q on a bearing 155º. Calculate
(i) the distance of R from P to the nearest
kilometer
(ii) the bearing of R from P to the nearest degree
(b) M is a village on PR such that QM is
perpendicular to PR. Find the distance of M
from P to the nearest kilometer.​

Answers

Answer:

a. (i) the distance of R from P is 13 Km to the nearest  kilometer

(ii) the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree

b. the distance of m from P is 11 Km to the nearest kilometre

Explanation:

a) A triangle PQR is formed. Q = 90°,  p = 8 km; r = 10 km; distance of R from P is q is to be found.

(i) Using the cosine rule: q² = p² + r² - 2prCosQ

q² = 8² + 10² - 2 * 8 * 10 * Cos90

q² = 64 + 100 + 0

q² = 164

q = 13 Km  to the nearest kilometre

Therefore, the distance of R from P is 13 Km to the nearest  kilometer

(ii) the bearing of R from P

The angle at P is found using the formula Cos P = (q² + r² - p²)/2qr

Cos P = 13² + 10² - 8²/2 * 13 *10

Cos P = 0.7884

P = Cos⁻¹ 0.7884

P = 38°

Therefore, the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree

Note : 25° is alternate (Northwest) to 65°at P

b) A right-angled triangle  QMP is formed

using the trigonometrical ratios; cos Θ = adjacent/hypotenuse

where the hypotenuse side = 10 km, adjacent side = distance of M from P, x

cos P = x/10

x = cos 38 * 10

x = 11 Km to the nearest kilometre

Therefore, the distance of m from P is 11 Km to the nearest kilometre

Which is one characteristic of producers?
They recycle nutrients.
They do not eat other organisms.
They break down waste for energy.
They use other organisms for energy.

Answers

Answer: They use other organisms for energy

Explanation:

Answer:

they use other organisms for energy

Explanation: d

Amanda is doing a report for her Earth Science class about the four seasons. Which of the following would be an effective scientific model to incorporate in her project? a. A calendar indicating the first days of autumn, winter, spring, and summer b. Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season c. A poster board with pictures of weather characteristic of the four seasons d. A bar graph indicating average high and low temperatures for Amanda’s area in the autumn, winter, spring, and summer Please select the best answer from the choices provided

Answers

Answer:

(B)

Explanation:

edg 2020

The four seasons in earth is originating from the earth's revolution around the sun. Therefore, the most suitable model for Amanda is  Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season.

What are seasons?

Seasons in earth is originating from the difference in the distance from the sun over each time period. Hence, revolution of earth around sun make these seasons.

The time period at which earth comes closer to sun more hot will be earth's surface and we experience summer season. When we far from sun winter season occurs.

Therefore, different season are coming based on the distance of earth from sun at each revolution point. This is also affected by the tilt of earth's in its own axis.

Hence, different earth models with different distance from sun is most suitable model here for Amanda. Thus option B is correct.

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Which of the following sets of quantum numbers is NOT allowed? a. n = 5, l= 4, ml= –2, ms = +1/2 b. n = 2, l = 1, ml= 0, ms = +1/2 c. n = 4, l= 4, ml= 0, ms = –1/2 d. n = 3, l= 2, ml= –1, ms = +1/2 e. n = 2, l= 0, ml= 0, ms = -1/2

Answers

Answer:

C

Explanation:

the n value must always be greater than the l value

Out of the following set of quantum numbers ,set C is not allowed as the azimuthal and principal quantum numbers are same.

What are quantum numbers?

Quantum numbers are the numbers which describe the values of conserved quantities  with respect to the dynamics of a quantum system.They correspond to the Eigen values of operators  which commute with the Hamiltonian quantities.

The Hamiltonian quantities can be known with precision simultaneously with the system's energy.Quantum numbers can take values of discrete sets of integers or even half-integers  even though they can approach infinity in some cases.

They can specifically describe energy levels of electrons, and can also explain angular momentum,spin,etc.These are used to describe the path of an electron in an atom ,when the quantum numbers of all atoms are combined they must comply with the Schrodinger equation.

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2
22. A sodium chloride solution is 15.0% m/m%. Calculate mass of sodium chloride in 219 g solution.
14.2g
80.38
11.2 g
32.9 g

Answers

Answer: The mass of sodium chloride in 219 g solution is 32.9 g

Explanation:

To calculate the mass percent of element in a given compound, we use the formula:

[tex]\text{Mass percent of A}=\frac{\text{Mass of A}}{\text{mass of A +mass of B}}\times 100[/tex]

To find mass of sodium chloride in solution:

[tex]\text{Mass percent of sodium chloride}=\frac{\text{Mass of sodium chloride}}{\text{mass of solution}}\times 100[/tex]

Mass percent of sodium chloride= 15.0 %

Mass of solution = 219g

[tex]15=\frac{\text{Mass of sodium chloride}}{219}\times 100[/tex]

[tex]{\text{Mass of sodium chloride}=32.9g[/tex]

Thus mass of sodium chloride in 219 g solution is 32.9 g

A calibration curve constructed from absorbance values of solutions containing a known concentration of permanganate ions has the following best-fit line:
y = (3.62× 10^3 L/mol) x
where y is the absorbance of the solution at 525 nm and x is the concentration of MnO4- (aq) in mol/L. The path length of the cuvettes used in the experiment is 1 cm. Based on this information, what is the molar absorptivity of MnO4- (aq) at 525 nm?

Answers

Answer:

3.62×10³ L/mol

Explanation:

Beer-Lambert law relates the absorbance of a sample and its concentration. Its formula is:

A = ε×C×l

Where A is absorbance of the sample, ε is molar absorptivity (A constant f each sample), C its concentration and l is path length

Now, the formula obtained was:

y = (3.62×10³ L/mol) x

Where Y ia absorbance = A, x its concentration = C and 1cm is path length.

You can write:

A = (3.62×10³ L/mol)×C×l

That means, molar absorptivity of your sample under the meaured conditions is:

3.62×10³ L/mol

A glass cylinder contains 2 gases at a pressure of 106 kPa. If one gas is at 7 kPa, what is the pressure of attributed to the other gas? a) 9 kPa b) 99 kPa c) 113 kPa d) 7 kPa e) 2 kPa (URGENT)

Answers

Answer:

b) 99 kPa

Explanation:

According to Daltons law of partial pressure, the total pressure of a mixture of two or more non reactive gases is the sum of their individual pressures. Let the total pressure of a mixture of n number of gases be [tex]P_{total}[/tex] and their individual pressure be [tex]P_1,P_2,P_3,\ .\ .\ .\ ,\ P_n[/tex], According to Daltons partial pressure law:

[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n[/tex]

Since A glass cylinder contains 2 gases at a pressure of 106 kPa, therefore n = 2. Also one gas ([tex]P_1[/tex]) is at 7 kPa. Using Daltons partial pressure law:

[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n\\P_{total}=P_1+P_2\\106\ kPa=7\ kPa+P_2\\P_2=106\ kPa-7\ kPa\\P_2=99\ kPa[/tex]

Identify the energy transformations in the following actions.
a. Turning on a space heater
b. Dropping an apple core into the garbage
c. Climbing up a rope ladder
d. Starting a car
e. Turning on a flashlight
Think about each action as a before and after. For instance, before the space heater is turned on, what kind of energy is there? After the space heater is turned on, what happens? This change from before to after will help you identify the two types of energy involved.
Identify five more types of energy transformations that you see at home, at school, or outdoors. Make sure to name the action, such as turning on a light, as well as the two types of energy involved. Remember, for energy to be transformed, the type of energy before and after a task must be different.

Answers

Answer:

to. Turn on a heater (electrical energy that is transformed into heat energy)

yes. Throwing an apple core in the trash (chemical energy that is transformed into nutritional energy for decomposers)

C. Climbing a rope ladder (chemical energy from our food into mechanical energy that allows us to climb the ladder)

re. Starting a car (electrochemical energy in mechanical energy)

me. Turning on a flashlight (chemical energy from the battery into light energy)

Five daily actions that exemplify the transformation of energy:

Do physical activity (change of metabolic energy in mechanics)

Turn on a heater (electrical energy that is transformed into heat energy)

Lighting a stove (Chemical energy product of combustion that results in the transformation of that energy into heat)

Turn on a cell phone (chemical energy characteristic of the battery that is transformed into sound and light energy)

Riding a bicycle (nutritional energy or energy of the metabolism that is transformed into mechanical energy)

Explanation:

The energy is never lost, it is always transformed.

It is one of the great principles of physics and is the reason why the universe is infinite.

An amphoteric salt is one that contains an anion that can act as either an acid or a base in water. Baking soda, NaHCO3, is an example. By combining the ionization and hydrolysis reactions of the anion, you get the principle reaction that occurs when this salt is dissolved in water.
2HCO3-(aq) CO32-(aq) + H2CO3(aq)
The pH for such, a solution is given by
pH = pKa1 + pKa2/2
where Ka1 and Ka2 are the ionization constants of the acid (in this case, H2CO3). Note that the pH of the solution is independent of the salt concentration. Calculate the pH of a NaHCO3 solution.

Answers

Answer:

pH = 8.34

Explanation:

The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:

H₂CO₃ ⇄ HCO₃⁻ + H⁺ Ka1 -Here, HCO₃⁻ is acting as a base-

HCO₃⁻⇄ CO₃²⁻ + H⁺ Ka2 -Here, is acting as an acid-

Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:

pKa1 = 6.37; pKa2 = 10.32

As the pH of amphoteric salts is:

pH = (pKa1 + pKa2) / 2

pH = 8.34

The volume of ammonia gas at 1.14 atm of pressure is gradually decreased from 90.3 mL to 43.4 mL. What is the final pressure of ammonia if there is no change in temperature

Answers

Answer:

THE FINAL PRESSURE OF AMMONIA IF THERE IS NO CHANGE IN TEMPERATURE AND A DECREASE IN VOLUME FROM 90.3 mL TO 43.4 mL IS 2.91 atm.

Explanation:

At constant temperature, the pressure of a given mass of gas is inversely proportional to the volume. This question follows Boyle's law of gas laws.

Mathematically written as:

P1V1 = P2V2

Re-arranging the formula by making P2 the subject of the formula;

P2 = P1V1 / T2

P1 = 1.4 atm

V1 = 90.3 mL

V2 = 43.4 mL

P2 = unknown

So therefore, we have:

P2 = 1.4 * 90.3 / 43.4

P2 = 2.91 atm

The final pressure of ammonia is therefore 2.91 atm.

Answer:

2.37 atm

Explanation:

Step 1: Given data

Initial pressure of ammonia (P₁): 1.14 atmInitial volume of ammonia (V₁): 90.3 mLFinal pressure of ammonia (P₂): ?Final volume of ammonia (V₂): 43.4 mLConstant temperature

Step 2: Calculate the final pressure of ammonia

Since the temperature is kept constant, we can calculate the final pressure of ammonia using Boyle's law.

[tex]P_1 \times V_1 = P_2 \times V_2\\P_2 = \frac{P_1 \times V_1}{V_2} = \frac{1.14atm \times 90.3mL}{43.4mL} = 2.37 atm[/tex]

According to the collision theory, all collisions do not lead to reaction. Which choice gives both reasons why not all collisions between reactant molecules lead to reaction? 1. The total energy of two colliding molecules is less than some minimum amount of energy. 2. Molecules cannot react with each other unless a catalyst is present. 3. Molecules that are improperly oriented during collision will not react. 4. Solids cannot react with gases.

Answers

Answer: 3. Molecules that are improperly oriented during collision will not react.

Explanation:

According to the collision theory , the number of collisions that take place per unit volume of the reaction mixture is called collision frequency. The effective collisions are ones which result into the formation of products.

It depends on two factors:-

1. Energy factor:  For collision to be effective,  the colliding molecules must have energy more than a particular value called as threshold energy.

2. Orientation factor: Also the colliding molecules must have proper orientation at the time of collision to result into formation of products.

Thus not all collisions between reactant molecules lead to reaction because molecules that are improperly oriented during collision will not react.

A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and measures the initial reaction rates (The data from the three experiments is in the table). 1. Write the rate law 2. Solve for k.

Answers

Answer:

1. [tex]Rate =k [NO]^{2}[Cl_{2}][/tex]

2. [tex]k= 0.42 \frac{L^{2}}{mol^{2}*s}[/tex]

Explanation:

[tex]Rate =k [NO]^{m}[Cl_{2}]^{n}[/tex]

[tex]Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2[/tex]

[tex]Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1[/tex]

[tex]Rate =k [NO]^{2}[Cl_{2}]^{1}[/tex]

[tex]Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}[/tex]

3. There are many different primary standards that could be used in a standardization titration. What are the criteria for a primary standard

Answers

Answer:

High purity.

Stability (low reactivity)

Low hygroscopicity (to minimize weight changes due to humidity)

Explanation:

There are different primary standards that could be used in a standardization titration in order to achieve the best and accurate result possible. These standards include high purity,stability and low hygoscropicity .

A high purity means the reactants lack impurities which could affect the result. Stability also ensures that there is non reactivity with elements/compounds in the atmosphere while low hygroscopicity ensures weight changes are minimized due to humidity.

g Air contains nitrogen, oxygen, argon, and trace gases. If the partial pressure of nitrogen is 592 mm Hg, oxygen is 160 mm Hg, argon is 7 mm Hg, and trace gas is 1 mm Hg, what is the atmospheric pressure

Answers

Answer:

Explanation:

Atmospheric pressure = partial pressure of nitrogen + partial pressure of oxygen + partial pressure of argon + partial pressure of trace element

putting the given values

Atmospheric pressure = 592 + 160 + 7 + 1

= 760 mm of Hg .

Generally, systems move spontaneously in the direction of increasing entropy. TRUE FALSE

Answers

Answer:

true

Explanation:

Convert the following measurement

Answers

Answer:

6.9 Kg/mol•dL

Explanation:

To convert 6.9×10⁴ g/mol•L to kg/mol•dL,

First, we shall convert to kg/mol•L.

This can be achieved by doing the following:

Recall: 1 g = 1×10¯³ Kg

1 g/mol•L = 1×10¯³ Kg/mol•L.

Therefore,

6.9×10⁴ g/mol•L = 6.9×10⁴× 1×10¯³

6.9×10⁴ g/mol•L = 69 Kg/mol•L

Finally, we shall convert 69 Kg/mol•L to Kg/mol•dL.

This is illustrated below:

Recall: 1 L = 10 dL

1 Kg/mol•L = 1×10¯¹ Kg/mol•dL

Therefore,

69 Kg/mol•L = 69 × 1×10¯¹

69 Kg/mol•L = 6.9 Kg/mol•dL

Therefore, 6.9×10⁴ g/mol•L is equivalent to 6.9 Kg/mol•dL.

The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed. is halved. has its sign changed. is unchanged.

Answers

Answer:

the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Hence the overall equation is now;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

Answer:

A.

Explanation:

Did it on Edge.

A reaction has the rate law, rate = k[A]3[B]. Which change will cause the great­est in­crease in the reaction rate?

Answers

Answer:

Increasing the concentration of A will cause the greatest change over the rate.

Explanation:

Hello,

In this case, considering the given rate law, which is third-order with respect to A, changing its concentration, the rate will be significantly modified. For instance, suppose a concentration of A and B of 1M and a symbolic rate constant (k),  this causes the rate to be:

[tex]r=k[1M]^3[1M]=1k\frac{M^4}{s}[/tex]

Then, if we change the concentration of A to 2 M holding the concentration of B in 1 M, the new rate constant will be:

[tex]r=k[2M]^3[1M]=8k\frac{M^4}{s}[/tex]

Nevertheless, if we hold the concentration of A in 1 M and the concentration of B is now 2 M (same change), the new rate constant is:

[tex]r=k[1M]^3[2M]=2k\frac{M^4}{s}[/tex]

It means that increasing the concentration of A will cause the greatest change over the rate.

Best regards.

The change in concentration of A will cause the great­est in­crease in the reaction rate.

Rate of reaction :

Given reaction has rate law ,

          [tex]rate=k[A]^{3} [B][/tex]

which is third-order with respect to concentration of A

From rate law equation, It is observed that the degree of concentration of A is three.

It means that small change in concentration of A result large change in rate of reaction.Thus, the change in concentration of A  will cause the great­est in­crease in the reaction rate.

Learn more about the rate of reaction here :

https://brainly.com/question/23637409

write the IUPAC name OF THE FOLLOWING COMPOUNDS

Answers

Answer:

Explanation:

a) 2 chloro butane

b) 2-3 dimethyl butane

c) 2 bromo 3 nitro pentane

d) 2-3 trimethyl pentane

e) 2-bromo,3-methyl,4-nitro hexane

f) 2-methyl cyclo butane

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