Select the answer with the correct number of significant figures for each calculation. (6.022 × 1023) × 2.58 = 1.55 × 1024 1.554 × 1024 1.5537 × 1024

The length of the half-life is 1.8 seconds (rounded to two significant figures).

The half-life of a reaction is the amount of time it takes for half of the reactants to be consumed. The following is a solution to the problem:

If the reaction is 39% complete at the end of 23 seconds, we can assume that the remaining 61% of reactants will require another half-life to be consumed, which means that half of 61% (or 30.5%) will be consumed in the second half-life.

The percentage remaining after one half-life is 50%, and the percentage remaining after two half-lives is 50% of 50%, or 25%. Therefore, the reaction will be 61% complete after the first half-life, 30.5% complete after the second half-life, and 15.25% complete after the third half-life.

Since the reaction is 39% complete after the first half-life, we can use the following equation to find the length of the half-life: 39% = 100% × (1/2)^(t/h)where t/h represents the length of the half-life. In order to solve for t/h, we can divide both sides by 100% and take the logarithm of both sides:

ln(0.39) = ln(0.5) × (t/h). We can now solve for t/h by dividing both sides by ln(0.5):t/h = ln(0.39) / ln(0.5) = 1.79.

Therefore, the length of the half-life is 1.8 seconds (rounded to two significant figures).

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The season the northern hemisphere is experiencing is A, summer.

Summer: June solstice to September equinox. Summer is the season that follows spring and precedes fall. It typically begins around June 20th or 21st and lasts until around September 22nd or 23rd.

Fall (Autumn): September equinox to December solstice. Fall is the season that follows summer and precedes winter. It typically begins around September 22nd or 23rd and lasts until around December 20th or 21st.

Winter: December solstice to March equinox. Winter is the season that follows fall and precedes spring. It typically begins around December 20th or 21st and lasts until around March 20th or 21st.

Spring: March equinox to June solstice. Spring is the season that follows winter and precedes summer. It typically begins around March 20th or 21st and lasts until around June 20th or 21st.

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Image transcribed:

BAND HALL

Earth and Space

What season is the northern hemisphere experiencing?

A Summer

B. Spring

C. Winter

D. Fall

850 kiloliters are equal to 850,000,000 millilitres.

To convert 850,000,000 milliliters to kiloliters, we can use the conversion factor 1 kiloliter = 1,000 liters and 1 liter = 1,000 milliliters.

First, we need to convert the milliliters to liters by dividing 850,000,000 by 1,000:

850,000,000 milliliters / 1,000 = 850,000 liters

Then, we can convert the liters to kiloliters by dividing 850,000 by 1,000:

850,000 liters / 1,000 = 850 kiloliters

Therefore, 850,000,000 milliliters is equivalent to 850 kiloliters.

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Yes I would live in Aerobics environment because I need oxygen to thrive, and survive.

Humans require an aerobic environment to live. Aerobic means "with oxygen," and the human body needs oxygen to produce energy through a process called cellular respiration. Without oxygen, cells cannot produce enough energy to sustain life. In contrast, anaerobic environments lack oxygen and can be toxic to humans.

In general, humans have evolved to live in aerobic environments, and our bodies are well-equipped to handle normal levels of oxygen in the air. However, exposure to high levels of oxygen can also be harmful, as it can lead to oxidative stress and damage to cells. Similarly, exposure to low levels of oxygen, such as in high-altitude environments, can also be challenging for humans.

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The number of years it will take for the population of orcas to double, given the births and deaths is 5. 10 years .

To find the population doubling time, we first need to find the rate at which the population of orcas grew in the current year:

= ( 15 births - 8 deaths ) / 51 orcas

= 7 / 51 x 100 %
= 13. 7 %

Then, we can use the Rule of 70 to find the doubling time. The Rule of 70 shows the periods till doubling as :

= 70 / growth rate

= 70 / 13.7

= 5. 10 years

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The percent by volume of a solution formed by mixing 349 mL of isopropanol with 380mL of water. The mass of a solute in a solution with 65% (m/m) of a solute and a mass of the solution is 327g. The molarity of 171g of KBr dissolved in 829. 0 mL pure water

1. The percent by volume of the solution formed by mixing 349 mL of isopropanol and 380 mL of water is 52.2% and 47.8%.

To find the percent by volume of the solution, we need to add the volumes of the two components and then calculate the percentage of each component in the total volume:

Total volume = [tex]349 mL + 380 mL = 729 mL[/tex]

Percent by volume of isopropanol = [tex](349 mL / 729 mL) * 100percent = 47.8 percent[/tex]

% by volume of water = [tex](380 mL / 729 mL) * 100 percent = 52.2percent[/tex]

Therefore, the solution contains 47.8% (v/v) of isopropanol and 52.2% (v/v) of water.

2. The mass of the solute in a solution with 65% (m/m) of a solute and a mass of the solution of 327.0 g is 212.55 g.

We are given the mass percent (m/m) of the solute and the total mass of the solution. Therefore, we can calculate the mass of the solute using the following formula:

Mass of solute = Mass of solution x Mass percent of solute

Mass of solute = [tex]327.0 g * 0.65 = 212.55 g[/tex]

Therefore, the mass of the solute is 212.55 g.
3. The molarity of 171g of KBr dissolved in 829.0 mL of pure water is 1.74 M.

To calculate the molarity of KBr in the solution, we need to first calculate the number of moles of KBr present in the solution using the following formula:

Number of moles = Mass of solute / Molar mass of KBr

The molar mass of KBr is 119 g/mol (39 g/mol for K and 80 g/mol for Br).

Number of moles = 171 g / 119 g/mol = 1.44 mol

The volume of the solution is given in mL, so we need to convert it to liters:

Volume of solution = 829.0 mL = 0.8290 L

Now, we can calculate the molarity using the following formula:

Molarity = Number of moles / Volume of solution

Molarity = 1.44 mol / 0.8290 L = 1.74 M

Therefore, the molarity of the KBr solution is 1.74 M.

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Answer:

A. Ethyl acetate (ethyl alcohol + acetic acid)

H

|

H--C--O--C--H

|

CH3

B. Butyl formate (butyl alcohol + formic acid)

H

|

H--C--O--CH3

|

CH3CH2CH2CH2

C. Methyl benzoate (methyl alcohol + benzoic acid)

H

|

H--C--O--C6H5

|

CH3

D. Ethyl butyrate (ethyl alcohol + butyric acid)

H

|

H--C--O--C3H7

|

CH2CH3

E. Propyl propionate (propyl alcohol + propionic acid)

H

|

H--C--O--C--CH3

| |

CH3 CH2CH3

F. Methyl propanoate (methyl alcohol + propanoic acid)

H

|

H--C--O--C2H5

|

CH3

G. Butyl benzoate (butyl alcohol + benzoic acid)

H

|

H--C--O--C6H5

|

CH3CH2CH2CH2

H. Ethyl hexanoate (ethyl alcohol + hexanoic acid)

H

|

H--C--O--C5H11

|

CH2CH3

I. Butyl pentanoate (butyl alcohol + pentanoic acid)

H

|

H--C--O--C4H9

|

CH3(CH2)2

J. Methyl pentanoate (methyl alcohol + pentanoic acid)

H

|

H--C--O--C4H9

|

CH3

(Please could you kindly mark my answer as brainliest you could also follow me so that you could easily reach out to me for any other questions)

It takes approximately 9,510.77 hours for the concentration of 11-trans-retinal to reach [tex]3.14*10^{-3}M[/tex]

The concentration of 11-trans-retinal can be calculated using the following equation:
[tex]ln(At/A0) = kt[/tex]
Where At is the concentration of 11-trans-retinal at time t, A0 is the initial concentration of 11-trans-retinal, and k is the rate constant.
Since we know the rate constant is [tex]1.02*10^{-5} /s[/tex] and the concentration of 11-trans-retinal is [tex]3.14*10^{-3}M[/tex], we can solve for t.
ln([tex]3.14*10^{-3}M/A0[/tex]) = [tex](1.02*10^{-5})t[/tex]
Solving for t, we get:
[tex]t = ln(3.14*10^{-3} /A0)/(1.02*10^{-5} )[/tex]
Therefore, it takes approximately 9,510.77 hours for the concentration of 11-trans-retinal to reach [tex]3.14*10^{-3}M[/tex].

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The type of reaction is represented by this graph is D. Exothermic Reaction progress

Chemical processes known as endothermic reactions take in energy from their environment, typically in the form of heat, light, or electricity. As a result, the reaction's products possess greater potential energy than its reactants. As energy is being drawn from the surroundings during the reaction, they will feel cooler or colder. A chemical reaction known as an exothermic reaction, on the other hand, releases energy into the environment, typically in the form of heat, light, or sound. Because of this, the reaction's products have lower potential energy than its reactants. As energy is released to the environment during the reaction, it will feel warmer or hotter.

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The number of mole of the sample containing 8.45×10²³ molecules CH₄ is 1.4 mole

Avogadro's hypothesis gives a well defined relationship between number of mole and number of molecules. This is given below:

6.022×10²³ molecules = 1 mole of substance

Thus, we can say that 1 mole of methane, CH₄ will be equivalent to 6.022×10²³ molecules as shown below:

6.022×10²³ molecules = 1 mole of CH₄

With the above information, we can determine the number of mole containing  8.45×10²³ molecules. Details below:

6.022×10²³ molecules = 1 mole of CH₄

8.45×10²³ molecules = 8.45×10²³ / 6.022×10²³

8.45×10²³ molecules = 1.4 mole of CH₄

Thus, the number of mole  is 1.4 mole

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Complete question:

What is the number of mole of a sample containing 8.45×10²³ molecules CH₄

According to the question the mass of 12 needed to react exactly with 35 g Al is 420 g.

React is an open-source JavaScript library created by Face book for building user interfaces. It is a declarative, efficient, and flexible JavaScript library for building user interfaces. React allows developers to create large web applications that use data and can change over time without reloading the page. It is easy to use and requires minimal coding. React is used for developing complex and interactive UIs for web and mobile applications, as well as for creating single-page applications.

The balanced equation for the reaction is:
12 mol A l + 12 mol O2 -> 13 mol Al2O3
We can calculate the mass of 12 needed to react with 35 g A l using the following equation:
Mass (12) = (35 g Al / (12 mol A l/mol 12)) x (12 mol 12/1 mol Al)
= 420 g 12
Therefore, the mass of 12 needed to react exactly with 35 g A l is 420 g.

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Therefore, the rate at which the ethanol was being pumped into the tank is 1.21 liters per minute.

To solve this problem, we need to use the fact that the concentration of ethanol in the mixture is 15% and that the flow rate is constant.

Let's assume that the rate at which the ethanol is being pumped into the tank is x liters per minute. Then, after t minutes, the total volume of the mixture in the tank will be (40.5 + xt) liters, and the amount of ethanol in the mixture will be 0.15xt liters.

We also know that after 25 minutes, the amount of water in the tank was 84.2 liters. Therefore, the amount of ethanol in the mixture after 25 minutes was (40.5 + 25x - 84.2) * 0.15 liters.

Setting these two expressions equal to each other, we get:

0.15xt = (40.5 + 25x - 84.2) * 0.15

Simplifying and solving for x, we get:

x = 1.21 liters per minute

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Note: The complete question would be as bellow,

Ethanol was pumped into a plastic tank containing 40.5 liters of water initially, with a constant flow rate. If the concentration of ethanol in the mixture is 15%, what is the rate at which the ethanol was being pumped into the tank?

Answer:

3, 2, 1, 6

Explanation:

Let's do some algebra lol

Let's call coefficient for Cu(NO3)2 "a"

Let's call coefficient for K3PO4 "b"

Let's call coefficient for Cu3(PO4)2 "c"

Let's call coefficient for KNO3 "d"

Cu3(PO4)2 has 3x as many moles of Cu compared to Cu(NO3)2, so we know that 3c = a

Cu(NO3)2 has 2x as many moles of NO3 compared to KNO3, so we know that 2a = d

Repeat this process for K and PO4 --> you get equations 3b = d and 2c = b respectively

2a = d = 3b = d so 2a = 3b, let's see if a = 3, b = 2 works

plug a and b into other two equations --> c = 1, d = 6

these are all whole numbers so it works! (if they're not whole numbers than multiply every coefficient by their LCM to make it whole)

so your coefficients for each of them are 3, 2, 1, 6

The grams of KCIO3 are 4.05 g.

Dimensional analysis is used to cancel units and convert between different quantities (moles, grams, etc.) in a systematic and logical way. By using conversion factors, we can ensure that our calculations are accurate and that we arrive at the correct units for the final answer.

We need to convert the moles of O2 to moles of KCIO3 because we are ultimately interested in finding the mass of KCIO3. By using the conversion factor of 2 moles KCIO3/3 moles O2, we can relate the two quantities and determine the number of moles of KCIO3 required to react with the given mass of O2.

To solve the problem, we need to use the given conversion factors and dimensional analysis to cancel units and find the grams of KCIO3.

Step 1: Write down the given conversion factors:

1 mole KCIO3 = 122.55 g KCIO3

1 mole O2 = 32.00 g O2

2 moles KCIO3 = 3 moles O2

2 moles KCIO3 = 2 moles KCI

Step 2: Write down the given mass of O2 and use it to find the moles of KCIO3:

0.8 g O2 × (1 mole O2/32.00 g O2) × (2 moles KCIO3/3 moles O2) = 0.0333 moles KCIO3

Step 3: Convert the moles of KCIO3 to grams:

0.0333 moles KCIO3 × (122.55 g KCIO3/1 mole KCIO3) = 4.05 g KCIO3

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Answer:

Neither of these changes are caused by cooling, but both are chemical changes.

The initial volume is equal to the value of nRT/P when the pressure is 800.0 mm Hg, so the initial volume is 0.50 l.

We can use the ideal gas law to solve this problem: PV = nRT, where P is the pressure, V is the volume, n is the amount of substance, R is the ideal gas constant, and T is the temperature. Since the temperature is constant, we can assume that the value of nRT is also constant.

We can rearrange the equation to find the volume: V = (nRT/P).

First, we can calculate the value of nRT/P for the initial volume, 740.0 mm Hg:
(nRT/740.0) = 0.50.

Now, we can solve for nRT/P when the pressure is 800.0 mm Hg:
(nRT/800.0) = 0.50.

The initial volume is equal to the value of nRT/P when the pressure is 800.0 mm Hg, so the initial volume is 0.50 l.

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The correct order of the reactions for the decomposition reaction of ozone to oxygen is as follows:

This series of radical reactions occurs in the stratosphere, where ozone O3 and an oxygen atom (O). The oxygen atom then reacts with another ozone molecule in the second reaction, forming two oxygen molecules. The third reaction is similar to the first, with an ozone molecule reacting with an oxygen atom to form two oxygen molecules. These reactions continue until all of the ozone has been converted to oxygen.

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The equilibrium concentrations are 0.247 M for CH4 and CCl4, and 0.254 M for CH2Cl2.

The equilibrium constant, Kc, is given by the expression:

Kc = [CH2Cl2]² / ([CH4] [CCl4])

We are given the initial concentrations of CH4 and CCl4:

[CH4] = 0.374 M
[CCl4] = 0.374 M

Let x be the change in concentration at equilibrium. The equilibrium concentrations can be expressed as:

[CH4] = 0.374 - x
[CCl4] = 0.374 - x
[CH2Cl2] = 2x

Substituting these values into the expression for Kc, we get:

9.52×10-2 = (2x)² / ((0.374 - x) (0.374 - x))

Solving for x, we get:

x = 0.127 M

Therefore, the equilibrium concentrations are:

[CH4] = 0.374 - 0.127 = 0.247 M
[CCl4] = 0.374 - 0.127 = 0.247 M
[CH2Cl2] = 2(0.127) = 0.254 M

Answer: The equilibrium concentrations are 0.247 M for CH4 and CCl4, and 0.254 M for CH2Cl2.

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We need to add 100 mL (0.1 liter) of the 10 M solution to 900 mL (0.9 liter) of solvent to make 1 liter of a 1 M solution. Option C is correct.

To make a 1 liter solution of 1 M concentration, we need to dilute the 10 M solution by a factor of 10.

The dilution factor is the ratio of the final volume to the initial volume, which is 1 liter / 0.1 liter = 10.

So, we need to add 1 part of the 10 M solution to 9 parts of solvent (usually water) to make a total of 10 parts, which will result in a 1 M solution.

Therefore, we need to add 100 mL (0.1 liter) of the 10 M solution to 900 mL (0.9 liter) of solvent to make 1 liter of a 1 M solution.

Hence, C. 100 mL is the correct option.

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--The given question is incomplete, the complete question is

"How much of a 10 M solution is needed to make 1 liter of a 1 M solution? A) 1 mL B) 10 mL C) 100 mL D) 1000 mL​."--

The correct answer is None of the other answers.  According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This means that a gas with a lower molar mass will effuse faster than a gas with a higher molar mass.

The equation for Graham's Law of Effusion is:
Rate of effusion of gas 1/Rate of effusion of gas 2 = √(Molar mass of gas 2/Molar mass of gas 1).

In this case, we are given the rate of effusion of oxygen gas (12.7 minutes) and asked to find the rate of effusion of argon gas.
The molar mass of oxygen gas is 32 g/mol and the molar mass of argon gas is 40 g/mol.

Plugging in the given values into the equation, we get:
Rate of effusion of argon/12.7 minutes = √(32 g/mol/40 g/mol)
Cross-multiplying and solving for the rate of effusion of argon, we get:
Rate of effusion of argon = 12.7 minutes × √(32 g/mol/40 g/mol) = 11.3 minutes.

Therefore, the approximate time it would take for 1.0 L of argon gas to effuse is 11.3 minutes. This is not one of the answer choices, so the correct answer is None of the other answers.

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The procedure for the amino acid chromatography states that you are to be careful not to touch the paper with your fingers, except along the edges, to prevent contamination of the paper or the sample. Chromatography is a technique used in the separation of different molecules or components of a mixture. It involves the movement of the components of a mixture through a stationary phase, which is usually a solid or liquid, in a mobile phase or a gas or liquid.

Chromatography is used in the separation and identification of amino acids, which are the building blocks of proteins. In amino acid chromatography, the stationary phase is a special paper or a silica gel-coated plate, and the mobile phase is a solvent, usually a mixture of water and an organic solvent.

When performing amino acid chromatography, it is crucial to avoid contamination of the paper or the sample with any foreign substances, such as dust, oil, or bacteria, which can interfere with the separation and identification of the amino acids. Therefore, it is essential to handle the paper with care and avoid touching it with bare hands.

The oils and sweat present on the hands can leave behind residues on the paper that can interfere with the separation process. Therefore, one should avoid touching the paper except along the edges, where there is less chance of contaminating the sample. To handle the paper, it is best to use forceps or gloves that do not leave behind any residues.

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model of the particles in solid carbon dioxide (dry ice):

   __         __

  /    \       /    \

 |  o  |      |  o  |

\____/  \____/

model of the particles in gaseous carbon dioxide:

  o       o

     o

o       o

The arrows showing the flow of thermal energy into the solid carbon dioxide could be represented as:

  __      __

/    \  /    \

|  →  |  |  o  |

\____/  \____/

The arrow depicts how heat energy is transferred into the solid carbon dioxide, causing it to sublime and become gaseous carbon dioxide.

Carbon dioxide exists in a solid form as dry ice. It directly transforms into a gas when brought to room temperature, a process known as sublimation.

Thermal energy is introduced into the solid carbon dioxide during this process, causing its particles to separate and turn into a gas.

The discharge of the gas is caused by the gaseous carbon dioxide particles' increased freedom of movement and spreading out.

In conclusion, dry ice is an intriguing substance that transforms through a special process known as sublimation from a solid to a gas. This process happens as a result of thermal energy entering the solid and forcing its particles to disperse and turn into a gas.

Science-related fields like cryogenics, food preservation, and even special effects can benefit from understanding the behavior of dry ice and the underlying concepts of sublimation.

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Enolate ion of an aromatic ketone reacts faster with an aldehyde group (producing a crossed-aldol reaction) than with the carbonyl group of another molecule of ketone because of the electronic effect of the substituent.

An enolate ion of an aromatic ketone reacts faster with an aldehyde group to produce a crossed aldol reaction due to electronic effects of the substituent. In case of a ketone, the alpha-proton (C-H bond) is less acidic as compared to that of an aldehyde. The difference in the acidities of alpha-proton atoms is caused by the electron-withdrawing nature of the ketone carbonyl group. This is due to the electronic effects of the substituent.The cross aldol reaction is the reaction between an aldehyde and a ketone to produce a β-hydroxy ketone or aldol. The enolate of a ketone reacts with an aldehyde (it's carbonyl carbon) to form the β-hydroxy ketone or aldol.Crossed aldol reactions occur more frequently and are of greater interest than simple aldol reactions. The main reason for this is the possibility of forming different products by using different aldehydes and ketones.

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Answer:

The molar mass of oxygen is 32 g. Hence the number of moles of oxygen are: 78.2 g / (32 g/mole) = 78.2/32 moles Since 1 mole of oxygen produces 2 moles of copper oxide, the number of moles of copper oxide generated are: (78.2/32) x 2 moles = 4.89 moles of copper oxide.

Explanation:

the chemist can report that the molar heat capacity of  [tex]C_{3} H_{9}N[/tex]  is 134.0 J/mol·K (rounded to three significant digits).

To calculate the molar heat capacity of  [tex]C_{3} H_{9}N[/tex] , we need to know the number of moles of  [tex]C_{3} H_{9}N[/tex]  in the sample and the amount of heat absorbed by the sample. We can use the following formula to calculate the number of moles of  [tex]C_{3} H_{9}N[/tex]

n = m/M

where:

n = number of moles

m = mass of [tex]C_{3} H_{9}N[/tex] (809.0 mg)

M = molar mass of  [tex]C_{3} H_{9}N[/tex]

The molar mass of [tex]C_{3} H_{9}N[/tex]  can be calculated as follows:

M = (3 x M(C)) + (9 x M(H)) + M(N)

Using the atomic masses of the elements from the periodic table, we can calculate the molar mass of  [tex]C_{3} H_{9}N[/tex]  as follows:

M(C) = 12.01 g/mol

M(H) = 1.008 g/mol

M(N) = 14.01 g/mol

M = (3 x 12.01) + (9 x 1.008) + 14.01 = 59.11 g/mol

Now we can calculate the number of moles of  [tex]C_{3} H_{9}N[/tex]

n = 809.0 mg / 59.11 g/mol = 0.01368 mol

Next, we can use the following formula to calculate the molar heat capacity of  [tex]C_{3} H_{9}N[/tex]

Cp = q/nΔT

We are given that q = 1834 J and we need to assume a value for ΔT. Let's assume that the temperature of the sample increased by 10.0°C (which is equivalent to 10.0 K). Then we can calculate the molar heat capacity of  [tex]C_{3} H_{9}N[/tex] as follows:

Cp = 1834 J / (0.01368 mol x 10.0 K) = 134.0 J/mol·K

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2Al + 6HCl → 3H2 + 2AlCl3, It is a redox reaction ; Cu(OH)2 → H2O + CuO and this is a decomposition reaction.

Balanced chemical equation is that equation where number of atoms of each type in the reaction is same on both the reactants and product sides.

Balanced chemical equation for the reaction between aluminum (Al) and hydrochloric acid (HCl) is: 2Al + 6HCl → 3H2 + 2AlCl3

This is a redox reaction, where aluminum is oxidized to Al3+ and hydrogen ions (H+) are reduced to hydrogen gas (H2).

Balanced chemical equation for the given reaction is: Cu(OH)2 → H2O + CuO

To balance this equation, we need to put a coefficient of 1 in front of Cu(OH)2, 1 in front of H2O, and 1 in front of CuO.

This is a decomposition reaction where Copper (II) hydroxide (Cu(OH)2) breaks down into water (H2O) and Copper (II) oxide (CuO).

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Explanation:

To calculate the amount of water needed to dilute the solution of potassium hydroxide and change its pH from 12.0 to 11.0, we need to use the formula for calculating the pH of a diluted solution.

The formula is:

pH = -log[H+]

where [H+] is the concentration of hydrogen ions in moles per liter.

Since we are diluting the solution by adding water, the concentration of [OH-] (hydroxide ions) will decrease proportionally to the volume of water added. This means that we can use the following equation to calculate the new concentration of [OH-]:

[OH-]1V1 = [OH-]2V2

where V1 is the initial volume of the solution, [OH-]1 is the initial concentration of hydroxide ions, V2 is the final volume of the solution after dilution, and [OH-]2 is the final concentration of hydroxide ions.

We know that the initial pH is 12.0, which means that [OH-]1 = 10^-2.0 M = 0.01 M.

We want to change the pH to 11.0, which means that [OH-]2 = 10^-11.0 M = 1 x 10^-11 M.

We also know that we are adding water to dilute the solution, but we don't know how much water we need to add yet. Let's call this volume of water "Vw".

Using the equation above, we can solve for V2:

[OH-]1V1 = [OH-]2V2

(0.01 M)(100 ml) = (1 x 10^-11 M)(100 ml + Vw)

V2 = (0.01 M)(100 ml)/(1 x 10^-11 M) - Vw

V2 = 10^12 ml - Vw

Now we can use this value for V2 in the pH formula to calculate the new pH:

pH = -log([H+])

[H+] = Kw/[OH-]

Kw is the ion product constant for water, which is equal to 1 x 10^-14 at room temperature.

[H+] = (1 x 10^-14)/(1 x 10^-11)

[H+] = 1 x 10^-3 M

pH = -log(1 x 10^-3)

pH = 3

We want to achieve a pH of 11.0, so we need to add enough water to bring down the pH from 12.0 to 11.0. This means that we need to add enough water so that V2 becomes:

V2 = (0.01 M)(100 ml)/(1 x 10^-11 M) - Vw = 10^11 ml

Therefore, we need to add:

Vw = V2 - initial volume of solution

Vw = (10^11 ml) - (100 ml)

Vw = 99999900 ml or approximately 100 million ml or 100 cubic meters of water.

So, in order to change the pH of a solution of potassium hydroxide with a pH of 12.0 to a pH of 11.0 by adding water only, you would need to add approximately 100 million milliliters or about 100 cubic meters of water.

Part A: No, 250 mg NaOH would not exceed the capacity of the buffer to neutralize it.

Part B: No, 350 mg KOH would not exceed the capacity of the buffer to neutralize it.

Part C: Yes, 1. 25 g HBr would exceed the capacity of the buffer to neutralize it.

Part D: Yes, 1. 35 g HI would exceed the capacity of the buffer to neutralize it.

Part A:

We first need to calculate the pH of the buffer solution using the Henderson-Hasselbalch equation to see if 250 mg NaOH would surpass the buffer's ability to neutralise it:

pH = pKa + log([[tex]A^-[/tex]]/[HA])

where

pKa is the acid dissociation constant of [tex]HNO_2[/tex],

[[tex]A^-[/tex]] is the conjugate base concentration ([tex]NO_2^-[/tex]),

[HA] is the acid concentratio ([tex]HNO_2[/tex]).

The pKa of [tex]HNO_2[/tex] is 3.15, so:

pH = 3.15 + log([[tex]NO_2^-[/tex]]/[[tex]HNO_2[/tex]])

pH = 3.15 + log(0.150/0.100)

pH = 3.40

The buffer is a basic buffer since its pH is higher than 7.

As a result, we must determine how many moles of [tex]NO_2^-[/tex] there are in 500.0 mL of the buffer solution:

moles of [tex]NO_2^-[/tex] = concentration x volume

moles of [tex]NO_2^-[/tex] = 0.150 mol/L x 0.500 L

moles of [tex]NO_2^-[/tex] = 0.075 mol

It is necessary to convert 250 mg of NaOH into moles in order to assess whether the buffer can neutralise it:

moles of NaOH = mass / molar mass

moles of NaOH = 0.250 g / 40.00 g/mol

moles of NaOH = 0.00625 mol

Since

[tex]NaOH + HNO_2[/tex] → [tex]NaNO_2 + H_2O[/tex]

The amount of [tex]HNO_2[/tex] consumed by 0.00625 mol of NaOH is:

moles of [tex]HNO_2[/tex] consumed = 0.00625 mol

Since

the buffer initially contained 0.100 mol/L of [tex]HNO_2[/tex], the number of moles of [tex]HNO_2[/tex] in 500.0 mL of the buffer solution is:

moles of [tex]HNO_2[/tex] = concentration x volume

moles of [tex]HNO_2[/tex] = 0.100 mol/L x 0.500 L

moles of [tex]HNO_2[/tex] = 0.050 mol

Consequently, 0.050 mol of  [tex]HNO_2[/tex] can be neutralised by the buffer, while 0.00625 mol of  [tex]HNO_2[/tex] is actually consumed by 0.00625 mol of NaOH. The buffer can neutralise 250 mg of NaOH because the amount of  [tex]HNO_2[/tex] used by the NaOH is less than the amount of  [tex]HNO_2[/tex] present initially.

Part B:

Evaluate if 350 mg KOH would be too much for the buffer to neutralise.

We must first determine the buffer solution's pH:

pH = pKa + log([[tex]A^-[/tex]]/[HA])

pH = 3.15 + log([tex][NO_2^-]/[HNO_2][/tex])

pH = 3.15 + log(0.150/0.100)

pH = 3.40

Since

The buffer is a basic buffer since its pH is higher than 7.

The concentration of the conjugate base in the buffer solution determines a basic buffer's ability to neutralize a base (like KOH). As a result, we must determine how many moles of [tex]NO_2^-[/tex] there are in 500.0 mL of the buffer solution:

moles of [tex]NO_2^-[/tex] = concentration x volume

moles of [tex]NO_2^-[/tex]  = 0.150 mol/L x 0.500 L

moles of [tex]NO_2^-[/tex]  = 0.075 mol

To find whether the buffer can neutralize 350 mg KOH, we need to convert 350 mg to moles:

moles of KOH = mass / molar mass

moles of KOH = 0.350 g / 56.11 g/mol

moles of KOH = 0.00624 mol

Since

KOH is a strong base, it will react completely with the [tex]HNO_2[/tex] in the buffer to form [tex]KNO_2[/tex] and water:

[tex]KOH + HNO_2[/tex] → [tex]KNO_2 + H_2O[/tex]

The amount of [tex]HNO_2[/tex] consumed by 0.00624 mol of KOH is:

moles of [tex]HNO_2[/tex]  consumed = 0.00624 mol

Since

[tex]HNO_2[/tex] was initially present in the buffer at a concentration of 0.100 mol/L; hence, there are 500.0 mmol of  [tex]HNO_2[/tex]   in the buffer solution.

moles of  [tex]HNO_2[/tex]   = concentration x volume

moles of  [tex]HNO_2[/tex]   = 0.100 mol/L x 0.500 L

moles of  [tex]HNO_2[/tex]   = 0.050 mol

As a result, 0.050 mol of  [tex]HNO_2[/tex] can be neutralised by the buffer, while 0.00624 mol of  [tex]HNO_2[/tex] is actually consumed by 0.00624 mol of KOH. The buffer can neutralise 350 mg of KOH because the amount of  [tex]HNO_2[/tex]used by the KOH is smaller than the amount of  [tex]HNO_2[/tex] present at first in the buffer.

Part C:

We must first decide if 1.25 g of HBr is an acid or a basic in order to assess whether it would be too much for the buffer to neutralise.

As HBr is an acid and the problem's buffer is a basic buffer, an acid cannot be neutralised.

Consequently, we may deduce that the buffer is unable to neutralise 1.25 g HBr without having to conduct any computations.

Part D:

We must first decide if 1.35 g of HI is an acid or a basic in order to assess whether it would be too much for the buffer to neutralise.

As HI is an acid and the problem's buffer is a basic buffer, an acid cannot be neutralised by it.

Consequently, we may deduce that the buffer is unable to neutralise 1.35 g of HI without having to conduct any computations.

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