The nitric acid solution used in a lab had a hydronium ion concentration of 0.53 M.
a. Calculate the pH of the solution
b. Calculate the pOH of the solution.
c. Calculate the hydroxide ion concentration.

Answers

Answer 1

(a) The pH of the nitric acid solution is approximately 0.28.

(b) The pOH of the nitric acid solution is approximately 13.72

(c) The hydroxide ion concentration of the nitric acid solution is approximately 1.89 x 10^-14 M.

What is the pH of the solution?

a. To calculate the pH of the solution, we can use the formula:

pH = -log[H3O+]

where;

[H3O+] is the hydronium ion concentration.

Substituting the given value:

pH = -log(0.53) ≈ 0.28

b. The pOH of the solution can be calculated using the formula:

pOH = -log[OH-]

where;

[OH-] is the hydroxide ion concentration.

To find the pOH, we need to first calculate the [OH-]. We know that:

Kw = [H3O+][OH-] = 1.0 x 10^-14

where;

Kw is the ion product constant for water.

Rearranging the equation, we can solve for [OH-]:

[OH-] = Kw / [H3O+]

[OH-] = 1.0 x 10^-14 / 0.53

[OH-] ≈ 1.89 x 10^-14

Now, we can calculate the pOH:

pOH = -log(1.89 x 10^-14) ≈ 13.72

c. We can use the [OH-] concentration calculated in part (b) to find the hydroxide ion concentration:

[OH-] = 1.89 x 10^-14

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Related Questions

A sample of iodine-131 has an activity of 200 mCi. If the half-life of iodine-131 is 8. 0 days, what activity is observed after 16 days?


0. 5 half lives


Step [1]: Determine the number of half lives.


x 128


5. 88 x10-37


mci


16 days


128 half lives


2 half lives


80 half lives


1 day


Step [2]: Find the final activity.


50. 0 mci


200. Mci


200. Mci


(initial activity)

Answers

Iodine-131 is a radioactive isotope of iodine that has a half-life of 8.0 days. This means that after 8.0 days, half of the original amount of iodine-131 will have decayed, and after another 8.0 days (a total of 16 days), half of the remaining iodine-131 will have decayed again.

The activity of a radioactive sample is a measure of the number of radioactive decays that occur in a given time period. It is measured in units of becquerels (Bq) or curies (Ci). One curie is equal to 3.7 x 10^10 becquerels.

In this case, we are given that the initial activity of the sample is 200 mCi (milliCuries). To find the activity after 16 days, we can use the following equation:

Activity = Initial activity x (1/2)^(t/half-life)
where t is the time elapsed and half-life is the half-life of the isotope.
Substituting the given values, we get:

Activity = 200 mCi x (1/2)^(16/8)
Activity = 200 mCi x (1/2)^2
Activity = 200 mCi x 0.25
Activity = 50 mCi

Therefore, the activity observed after 16 days is 50 mCi. This means that half of the original iodine-131 has decayed in that time period. It is important to note that the actual number of atoms remaining in the sample will also be halved after 16 days, but the activity will be reduced by a factor of four (since activity is proportional to the number of decays per unit time).

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4. if 10.0 moles of naoh are dissolved in water to make 250.0 l of solution, what is the molarity of the
solution?
5. if 80.0 moles of naoh are dissolved in water to make 1.00 liter of solution, what is the molarity of the
solution?
6. if you have 1.00 liter of a 1.0 m solution of nacl, how many moles of nacl were dissolved in the water to
make that solution?
7. if you have 1.0 liter of a 1.00 m solution of nacl, how many moles of nacl were dissolved in the water to
make that solution?
write complete sentences.
8. how would you make 100.0 l of 1.0 m naoh?

Answers

If 10.0 moles of NaOH are dissolved in water to make 250.0 l of solution, the molarity of the solution is 0.04 moles

Molarity is defined as the number of moles of solute present in 1 litre of a solution. It is denoted by M and the formula is represented as

Molarity = number of moles of solute/ volume of the solution in L

According to given data

Number of moles of solute = 10 moles

volume of the solution = 250 L

Therefore, molarity = 10 moles/250 L

molarity = 0.04 moles.

Thus, molarity of the solution is 0.04 moles.

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∆E = −33 kJ/mol Ea = 20 kJ/mol What is E a′ ?
Answer in units of kJ/mol.

Answers

The activation energy (Ea) is the minimum energy required for a reaction to occur, and it is defined as the energy difference between the reactants and the activated complex or transition state. In an exothermic reaction, the products have lower energy than the reactants, so the change in energy (∆E) is negative.

The activation energy of the forward reaction is given as 20 kJ/mol. This means that 20 kJ/mol of energy must be provided to the reactants to reach the activated complex and initiate the forward reaction.

To find the activation energy of the reverse reaction (Ea′), we can use the equation:

Ea′ = Ea + ∆E

where Ea is the activation energy of the forward reaction and ∆E is the change in energy of the reaction. Since we are given ∆E as -33 kJ/mol, which represents the change in energy for the forward reaction, we can substitute the values and solve for Ea′.

Plugging in the given values, we get:

Ea′ = 20 kJ/mol + (-33 kJ/mol)

Ea′ = -13 kJ/mol

Therefore, the activation energy of the reverse reaction (Ea′) is -13 kJ/mol. This negative value means that the reverse reaction has a lower activation energy than the forward reaction, which is consistent with the fact that the reaction is exothermic. A lower activation energy for the reverse reaction means that it is easier for the products to convert back to the reactants, which is why exothermic reactions tend to be more favorable in the forward direction.

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If an alveolus with an initial volume of 3 ml of air with a total pressure of 760 mmhg decreases in volume to 2 ml, what would the new pressure be and in which direction would air flow? assume you are at sea level.

Answers

The new pressure be and in which direction would air flow is 1140 mmHg.

Using Boyle's law, we know that the pressure and volume of a gas are inversely proportional. Therefore, if the volume of the alveolus decreases from 3 ml to 2 ml, the pressure inside the alveolus will increase by a factor of 3/2 or 1.5 times. The new pressure inside the alveolus will be 760 mmHg x 1.5 = 1140 mmHg.

According to the principles of gas flow, air moves from an area of higher pressure to an area of lower pressure. Therefore, in this scenario, air would flow out of the alveolus since the pressure inside the alveolus (1140 mmHg) is now higher than the atmospheric pressure outside the body (760 mmHg).

It's important to note that this scenario assumes that all other factors affecting the pressure inside the alveolus, such as temperature and the number of gas molecules, remain constant.

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How many grams of chlorine gas must be reacted with excess sodium iodide if 10 grams of sodium chloride are needed?

Answers

Answer:

sorry xouldnt answer all

Explanation:

thier is ¹² equations ln tour answer

Need help can u tell how to answer questions like this

Answers

The volume/concentration of the above questions are as follows:

5000mL3M1M

How to solve dilution questions?

The amount of volume or concentration of a substance can be calculated using the following expression;

CaVa = CbVb

Where;

Ca and Va are initial and final concentrations respectivelyCb and Vb are initial and final volume respectively

1. 10 × 250 = 0.5 × Vb

2500 = 0.5Vb

Vb = 5000mL

2. 0.400 × 15 = 2 × Cb

6 = 2Cb

Cb = 3M

3. 50 × 20 = 1000 × Cb

1000 = 1000Cb

Cb = 1M

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2. A student in the group next to you is not following the safety rules. He manages to spill a large amount of solution on his clothes and THEN he catches himself on fire! His burning clothes give off a beautiful bright red color. What chemical compound did he spill on himself? How do you know?​

Answers

Based on the scenario described, it is likely that the student spilled a solution containing a flammable compound such as ethanol or methanol. These compounds are commonly used in chemistry labs and can easily catch fire if not handled properly. The bright red color of the flames is likely due to the presence of a metal salt in the solution, which can produce colored flames when heated. It is important to always follow safety rules in a lab setting to prevent accidents like this from happening.

How does burn ethanol?

Ethanol can be burned in the presence of oxygen to produce carbon dioxide (CO2) and water (H2O) in a process known as combustion. The chemical formula for ethanol combustion is:

C2H5OH + 3O2 → 2CO2 + 3H2O

In this reaction, the ethanol (C2H5OH) reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The reaction releases heat, which can be used as a source of energy.

To burn ethanol, it is typically mixed with air or oxygen and then ignited. The combustion process can be controlled by adjusting the amount of ethanol and oxygen that is mixed together, as well as the temperature and pressure of the reaction.

In some cases, ethanol is burned in internal combustion engines, such as those used in cars and other vehicles. In these engines, the combustion of ethanol is used to power the engine and generate mechanical energy.

It's important to note that the combustion of ethanol releases carbon dioxide, a greenhouse gas that contributes to climate change. As such, efforts are being made to reduce the amount of greenhouse gas emissions from burning ethanol and other fuels, through the use of renewable energy sources and more efficient combustion processes.

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12. what is the weight/volume percent concentration of 100. ml of a 30.0% (w/v) solution of
vitamin c after diluting to 200. ml?

Answers

A 30% (w/v) solution of vitamin C was diluted to 200 ml. The weight/volume percent concentration of the resulting solution is 15%.

To find the weight/volume percent concentration after diluting, we need to use the formula:

C1V1 = C2V2

Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

Given:

C1 = 30% (w/v)

V1 = 100 mL

V2 = 200 mL

Using the formula, we can solve for C2:

C1V1 = C2V2

(30%)(100 mL) = C2(200 mL)

C2 = (30%)(100 mL) / (200 mL)

C2 = 15%

Therefore, the weight/volume percent concentration of the 100 mL of 30.0% (w/v) solution of vitamin C after diluting to 200 mL is 15%.

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If re glazing do you cover what was forgotten(in the 2nd firing) or you fàreglaze everywhere

Answers

When reglazing, you cover the areas that were forgotten in the 2nd firing as well as reglaze everywhere for a uniform appearance.

Reglazing involves applying a new layer of glaze to a previously fired ceramic piece to improve its appearance, fix any issues from previous firings, or to achieve a specific effect.

In your case, if some areas were missed or improperly glazed during the 2nd firing, you would want to apply glaze to those forgotten areas to ensure a consistent finish.

However, it's important to reglaze the entire piece, not just the missed areas, to maintain a uniform appearance and avoid any inconsistencies in the final result. Before reglazing, make sure the ceramic piece is clean and free of dust or debris.

Apply the glaze evenly, using an appropriate technique such as brushing or dipping, and then fire the piece again according to the glaze's specific firing temperature and instructions. This should result in a well-glazed and visually appealing ceramic piece.

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why does the pinacol rearrangement more often use sulfuric acid, h 2 s o 4 , as the acid catalyst rather than hydrochloric acid, h c l ?

Answers

The pinacol rearrangement more often use the sulfuric acid, H₂SO₄ , as the acid catalyst rather than the hydrochloric acid, HCl is because H₂SO₄ have the more proton than that of the HCl.

The pinacol rearrangement process will takes place through 1,2-rearrangement. This rearrangement will involves the shift of the two adjacent atoms. Pinacol is the compound which has the two hydroxyl groups, each of the attached to the vicinal carbon atom. It is the solid organic compound which is the white.

The H₂SO₄ have the more proton than that of the HCl. This will makes the pinacol rearrangement more often use the sulfuric acid H₂SO₄ , as the acid catalyst and rather than the hydrochloric acid, HCl.

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H2 + Br2 → 2HBr. How many liters of hydrogen gas are needed to react with 9.0 g of bromine?

Answers

We need 2.77 liters of hydrogen gas to react with 9.0 g of bromine.



From this equation, we can see that 1 mole of hydrogen gas (H2) reacts with 1 mole of bromine (Br2) to produce 2 moles of hydrogen bromide (HBr).
we need to use the molar mass of bromine, which is 79.9 g/mol.
Number of moles of Br2 = mass / molar mass = 9.0 g / 79.9 g/mol = 0.113 moles
Since 1 mole of H2 reacts with 1 mole of Br2, we need 0.113 moles of H2 to react with the given amount of Br2.
To find out how many liters of H2 are needed, we need to use the ideal gas law, which relates the number of moles of a gas to its volume:

PV = nRT

where P is the pressure of the gas (in atm), V is the volume (in liters), n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature (in Kelvin).

Assuming standard conditions of temperature and pressure (STP), which are 0°C (273 K) and 1 atm, respectively, we can simplify the equation to:
V = nRT/P = (0.113 mol)(0.0821 L·atm/mol·K)(273 K)/(1 atm) = 2.77 L

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A student determined that the mass of the carbon product was less than the mass of the sulfuric acid and sugar that were combined. how would you account for this loss of mass? use evidence and scientific reasoning to support your answer.

Answers

The loss of mass could be accounted for by the release of gases such as water vapor, carbon dioxide, and sulfur dioxide during the reaction between sulfuric acid and sugar.

This is due to the fact that both sugar and sulfuric acid are organic compounds, and when heated, they undergo dehydration and decomposition reactions respectively, producing gases that escape the system. The carbon product is also likely to be less dense than the reactants, resulting in a further loss of mass.

Additionally, some of the sugar may have not fully reacted due to incomplete mixing, resulting in residual sugar that was not accounted for in the mass measurement. Overall, the loss of mass is expected in any chemical reaction, and in this case, it can be attributed to the production of gases and incomplete reaction.

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Cause and Effect. Unbalanced forces acting on a Nebula result in___


a. A constant linear motion


b. Equilibrium of the nebula


c. A change in its motion

Answers

Unbalanced forces acting on a Nebula result in a change in its motion.(C)

When unbalanced forces act on a nebula, they disrupt its equilibrium and cause a change in its motion. This is due to Newton's First Law of Motion, which states that an object at rest or in constant linear motion will continue in that state unless acted upon by an unbalanced force.

In the case of a nebula, the unbalanced forces may come from nearby stellar explosions, passing stars, or gravitational interactions.

These forces can cause parts of the nebula to compress and collapse, initiating the formation of new stars and planetary systems. As a result, the motion of the nebula changes over time as it evolves and develops under the influence of these forces.(C)

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I take 50.0 ml of 0.50 m hcl and add it to 150.0 ml of 0.10 m hno3. what is the ph of the resulting solution?

Answers

The pH of the resulting solution is calculated to be 1.40.

To determine the pH of the resulting solution, we need to first calculate the moles of each acid present.

Moles of HCl = (0.50 mol/L) x (0.050 L) = 0.025 mol
Moles of HNO3 = (0.10 mol/L) x (0.150 L) = 0.015 mol

Since the two acids are both strong acids, they will completely dissociate in solution. This means that the resulting solution will contain 0.025 mol of H+ ions from HCl and 0.015 mol of H+ ions from HNO3.

To calculate the pH of this solution, we can use the equation:

pH = -log[H+]

[H+] = (0.025 mol + 0.015 mol) / (0.050 L + 0.150 L) = 0.040 mol/L

pH = -log(0.040) = 1.40

Therefore, the pH of the resulting solution is 1.40.

In summary, when 50.0 ml of 0.50 M HCl is added to 150.0 ml of 0.10 M HNO3, the resulting solution contains 0.025 mol of H+ ions from HCl and 0.015 mol of H+ ions from HNO3. The pH of the resulting solution is calculated to be 1.40.

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Describe the bonding you would expect to find in a coin made out of copper?

Answers

The bonding you would expect to find in a coin made out of copper is metallic bonding.

Metallic bonding is the strong attraction between positively charged metal ions and the surrounding delocalized electrons in a metallic lattice. Copper is a metal and its atoms are closely packed together, forming a lattice structure. In the case of a copper coin, copper atoms lose their outermost electrons to form positively charged copper ions (Cu⁺), which are then held together by the delocalized electrons that move freely throughout the metal lattice. This type of bonding gives copper its characteristic properties such as electrical conductivity, malleability, and ductility making it an ideal material for coinage.

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Question 1 (3 points)
Fe +
Cl₂ -->
FeCl3

Answers

Answer:

2Fe + 3Cl_2 → 2FeCl 3

Explanation:

To balance the equation, we need to make sure that the number of atoms of each element is the same on both sides of the equation. In this case, we have two iron atoms and six chlorine atoms on the left-hand side, and two iron atoms and six chlorine atoms on the right-hand side. To balance the equation, we can add a coefficient of 2 in front of FeCl3 to get:

2Fe + 3Cl_2 → 2FeCl 3

Now we have two iron atoms and six chlorine atoms on both sides of the equation, and the equation is balanced.

Lead can be prepared from galena (lead II sulfide) by roasting the galena in the presence of oxygen to form lead II oxide and sulfur dioxide. Heating the metal oxide with more galena creates the molten metal and more sulfur dioxide. If we start with 25 mol of PbS, how many moles of SO2 do we create from both steps of the reaction? How many moles of lead do we form?



PbS + O2 -> PbO + SO2


PbO + PbS -> Pb + SO2

Answers

Here, 50 mol of SO2 will be created, and 25 mol of lead will be formed from both steps of the reaction.

To determine the moles of SO2 created and moles of lead formed in both steps of the reaction, we'll first need to examine each step individually.

Step 1: PbS + O2 -> PbO + SO2
Starting with 25 mol of PbS, this reaction occurs in a 1:1 molar ratio with SO2. Thus, 25 mol of SO2 will be created in this step.

Step 2: PbO + PbS -> Pb + SO2
Since 25 mol of PbO is created in step 1, the same amount of PbS is available to react in step 2. This reaction also occurs in a 1:1 molar ratio with SO2, meaning that another 25 mol of SO2 will be created in this step.

The total amount of SO2 created in both steps is the sum of the moles produced in each step:
25 mol (from step 1) + 25 mol (from step 2) = 50 mol of SO2

Additionally, since the second step forms lead (Pb) in a 1:1 molar ratio with PbS, we will have 25 mol of lead formed.

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Identify and explain one alternate view to the neoclassical view. What does this view consider that the neoclassical view does



not?

Answers

Alternate view to the neoclassical view is the Post-Keynesian view is Post-Keynesians believe that the neoclassical view does not adequately account for the role of uncertainty in economic decision-making, the importance of historical and institutional factors, and the potential for instability in markets.

Post-Keynesians argue that economic agents do not have perfect information and face uncertain future outcomes, which can lead to irrational decision-making and result in market failures. They also stress the importance of historical and institutional factors, such as power relations and social norms, in shaping economic outcomes.

Additionally, Post-Keynesians believe that markets are not inherently stable and can experience periods of instability and crisis, contrary to the neoclassical view that markets naturally tend toward equilibrium. The Post-Keynesian view emphasizes the role of uncertainty, history, and institutional factors in shaping economic outcomes, as well as the potential for instability in markets, which are not fully accounted for in the neoclassical view.

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The Keynesian view is an alternate view to the neoclassical view. It considers that there is a role for government in managing the economy through fiscal and monetary policy.

What is neoclassical?

Neoclassical is an art and design style that emerged in the mid-18th century and is based on the classical styles of ancient Greece and Rome. Neoclassical art and design sought to revive the aesthetic principles of antiquity and emphasized the use of symmetry, order, and balance in its works. This style was seen in art, architecture, and furniture, and often included motifs from classical mythology.

It assumes that markets are not always efficient and that people may not always act rationally. This view considers that the economy may not always be in equilibrium and that there may be periods of recession or depression. It also considers that individuals and companies may not always respond to economic changes in the same way, and that government intervention may be necessary to ensure economic stability. This view does not assume that the market is self-regulating and that it will always reach equilibrium.

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If you have a 6.2 l container with a pressure of 1.5 atm, how many moles are present if the temperature is 38 o c? (0.0821 l atm/mol k)

a
2.28
b
0.28
c
0.31
d
0.36

Answers

If there is a container with a volume of 6.2 liters and a pressure of 1.5 atmospheres, the number of moles present in the container is approximately 0.28 moles. Therefore, the correct answer is option b) 0.28.

To calculate the number of moles present in the container, we can use the ideal gas law equation:

PV = nRT

Where:

P = pressure in atm

V = volume in liters

n = number of moles

R = ideal gas constant (0.0821 L atm / (mol K))

T = temperature in Kelvin

First, we need to convert the temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15

T(K) = 38 °C + 273.15 = 311.15 K

Now we can rearrange the equation to solve for the number of moles (n):

n = PV / RT

Substituting the given values:

P = 1.5 atm

V = 6.2 L

R = 0.0821 L atm / (mol K)

T = 311.15 K

n = [tex](1.5 \text{ atm} \times 6.2 \text{ L}) / (0.0821 \text{ L atm/(mol K)} \times 311.15 \text{ K})[/tex]

n ≈ 0.28 moles

Therefore, the number of moles present in the container is approximately 0.28 moles.

The correct answer is option b) 0.28.

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What set of coefficients will balance the chemical equation below:

___KNO3 (aq) + ___PbO (s) ___Pb(NO3)2 (aq) + ___K2O (s)

A. 2,1,1,1

B. 1,3,1,3

C. 2,2,2,1

D. 1,2,1,2

Answers

Answer:

Explanation:

The correct answer is A.2,1,1,1 ;

As Our balancing equation is totally a Mathematics calculation In which We have to make coefficients in a manner to have all the atoms got equal on both side of the reactants.

We do balancing for Conservation of Mass.

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If a 2000g block of metal lost 3120 j of heat energy is cooled from 212 c to 200 c, what is the specific heat of the metal

Answers

Explanation:

3120 j / (2000 g * (212-200 C) )  = .13 j /( g C)  

A 29. 7-gram piece of aluminum is sitting on a hot plate. A student accidentally left the hot plate on. The aluminum now is very hot and has to be cooled. You fill a beaker with 250 grams of water. The aluminum is placed in the water. You are curious so you place a thermometer in the beaker. The water warms from 22. 3 C to 30. 8 C. The C (aluminum) is 0. 900 J/gC, and the C (water) is 4. 18 J/gC Do you have enough information to calculate the amount of energy transferred in this situation? Explain in 2-3 complete sentences

Answers

Yes, we have enough information to calculate the amount of energy transferred in this situation. We can use the equation Q = mCΔT.

Q is the amount of energy transferred, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature. We know the mass and specific heat capacity of the aluminum and water, as well as the change in temperature of the water.

Using this information, we can calculate the amount of energy transferred from the aluminum to the water.

To be specific, we can use the equation Q(aluminum) = m(aluminum) x C(aluminum) x ΔT(water) to find the amount of energy transferred from the aluminum to the water.

Since the aluminum starts at a higher temperature than the water, it will lose energy and transfer it to the water until both reach thermal equilibrium.

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A) Determine the [OH−] of a solution that is 0. 145 M in CO32− ( Kb=1. 8×10−4 ).


Express your answer using two significant figures.



B) Determine the pH of a solution that is 0. 145 M in CO32−.


Express your answer to two decimal places.



C) Determine the pOH of a solution that is 0. 145 M in CO32−.


Express your answer to two decimal places

Answers

A)  The value of concentration [OH⁻] = √(Kb*[CO₃²⁻]) = √(1.8×10⁻⁴*0.145) = 0.0034 M.

B)  The pH of the solution is pH = -log[H⁺] = -log(2.24×10⁻¹²) = 11.65.

C)  The pOH of the solution is pOH = -log(0.0034) = 2.47.

A) To determine the [OH⁻] of a solution that is 0.145 M in CO₃²⁻ (Kb=1.8×10⁻⁴), we can use the Kb expression of CO₃²⁻ and the fact that Kw (the ion product constant) is equal to [H⁺][OH⁻] to solve for [OH⁻].

The Kb expression for CO₃²⁻ is: Kb = [HCO₃⁻][OH⁻]/[CO₃²⁻]. Since the concentration of HCO₃⁻ is negligible compared to the concentration of CO₃²⁻, we can assume that [HCO₃⁻] is equal to 0.

B) To determine the pH of a solution that is 0.145 M in CO₃²⁻, we need to find the concentration of H⁺ in the solution. Since CO₃²⁻ can act as a base, it can react with water to form HCO₃⁻ and OH⁻.

The Kb expression for CO₃²⁻ can be rewritten as: Kw/Kb = [H⁺][OH⁻]/[CO₃²⁻] = [H⁺][OH⁻]/0.145. Solving for [H⁺], we get [H⁺] = Kw/[OH⁻][CO₃²⁻] = 1.0×10⁻¹⁴/(0.0034*0.145) = 2.24×10⁻¹² M.

C) To determine the pOH of a solution that is 0.145 M in CO₃²⁻, we can use the fact that pOH = -log[OH⁻]. From part A, we know that the [OH⁻] of the solution is 0.0034 M.

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0. 18 g of a
divalent metal was completely dissolved in 250 cc of acid
solution containing 4. 9 g H2SO4 per liter. 50 cc of the
residual acid solution required 20 cc of N/10 alkali for
complete neutralization. Calculate the atomic weight of
metal.
39.
Ans: 36​

Answers

The atomic weight of the metal is 36 g/mol.

To solve this problem, we need to use the concept of equivalent weight. The equivalent weight of a divalent metal is equal to its atomic weight divided by its valency, which in this case is 2.

First, let's calculate the number of equivalents of H2SO4 present in the solution.

4.9 g of H2SO4 per liter of solution means that there are 4.9/98 = 0.05 moles of H2SO4 per liter.

So in 250 cc (or 0.25 liters) of solution, there are 0.05 x 0.25 = 0.0125 moles of H2SO4.

Since H2SO4 is a diprotic acid, each mole of H2SO4 can donate 2 equivalents of H+. Therefore, the total number of equivalents of H+ present in the solution is 2 x 0.0125 = 0.025.

Now let's calculate the number of equivalents of alkali (which we know is N/10 or 0.1 N) required to neutralize 50 cc of the solution.

20 cc of N/10 alkali is equal to 0.002 equivalents of alkali (since N/10 alkali has a normality of 0.1, which means it can donate 0.1 equivalents of OH- per liter of solution).

Since the acid and alkali react in a 1:1 ratio, this means that there are also 0.002 equivalents of H+ in 50 cc of the solution.

Therefore, the initial number of equivalents of H+ in the solution must have been 0.025 + 0.002 = 0.027.

Now we can use this information to calculate the number of equivalents of metal present in the solution.

Since the metal is divalent, it will donate 2 equivalents of metal ions for every 1 equivalent of H+ that it reacts with.

Therefore, the number of equivalents of metal present in the solution is 0.027/2 = 0.0135.

Finally, we can calculate the atomic weight of the metal using the formula:

Atomic weight = Equivalent weight x Valency

In this case, the equivalent weight is equal to the atomic weight divided by 2 (since the metal is divalent).

So:

Atomic weight = Equivalent weight x 2

Atomic weight = (0.018 g / 0.0135 equivalents) x 2

Atomic weight = 36 g/mol

Therefore, the atomic weight of the metal is 36 g/mol.

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What valume of 0.1mol /dm hydrochloric acid will be required to neutralized 20cm of 2.0mol/dm sodium hydroxide?

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0.21 dm³ of 0.1 mol/dm³ hydrochloric acid is required to neutralize 20 cm³ of 2.0 mol/dm³ sodium hydroxide.

The volume of 0.1 mol/dm³ hydrochloric acid required to neutralize 20 cm³ of 2.0 mol/dm³ sodium hydroxide can be calculated using the formula:

Volume of acid = (Volume of alkali x Concentration of alkali x Molar mass of acid) / (Molar mass of alkali x Concentration of acid)

Firstly, we need to convert the volume of alkali from cm³ to dm³, which gives us 0.02 dm³. The molar mass of hydrochloric acid (HCl) is 36.5 g/mol, and the molar mass of sodium hydroxide (NaOH) is 40 g/mol.

Substituting these values and the given concentrations into the formula, we get:

Volume of acid = (0.02 x 2.0 x 40) / (36.5 x 0.1) = 0.21 dm³

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write balanced equations for each of the processes described below. (use the lowest possible coefficients. omit states-of-matter.)

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1.  Balanced equation for the combustion of propane: [tex]C_3H_8 + 5O_2\ - > 3CO_2 + 4H_2O.[/tex]

2. Balanced equation for the reaction between hydrochloric acid and sodium hydroxide:[tex]HCl + NaOH\ - > NaCl + H_2O.[/tex]

3. 3. Balanced equation for the decomposition of calcium carbonate upon heating: [tex]CaCO_3\ - > CaO + CO_2.[/tex]

1. [tex]C_3H_8 + 5O_2\ - > 3CO_2 + 4H_2O.[/tex]

This reaction shows that propane[tex](C_3H_8)[/tex] reacts with oxygen[tex](O_2)[/tex] from the air to produce carbon dioxide[tex](CO_2)[/tex] and water[tex](H_2O)[/tex] in a balanced chemical equation.

2. [tex]HCl + NaOH\ - > NaCl + H_2O.[/tex]

This reaction demonstrates that hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to produce sodium chloride (NaCl) and water [tex](H_2O)[/tex] in a balanced chemical equation.

3. [tex]CaCO_3\ - > CaO + CO_2[/tex].

This reaction illustrates that when calcium carbonate[tex](CaCO_3)[/tex] is heated, it decomposes to produce calcium oxide (CaO) and carbon dioxide [tex](CO_2)[/tex] in a balanced chemical equation.

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--The complete Question is, Write balanced equations for each of the processes described below:

1. Combustion of propane (C3H8) in air to produce carbon dioxide and water.

2. Reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) to produce sodium chloride (NaCl) and water (H2O).

3. Decomposition of calcium carbonate (CaCO3) upon heating to produce calcium oxide (CaO) and carbon dioxide (CO2). --

Calculate E0, E, and ΔG for the following cell reactions (a) Mg(s) + Sn2+(aq) ⇌ Mg2+(aq) + Sn(s) where [Mg2+] = 0. 055 M and [Sn2+] = 0. 055 M

Answers

The E° for cell reaction is - 2.37 V and  2.23 V and E for cell reaction  = 2.22V and ΔG = - 428.39kJ/mol.

The formula for solving the equation for given cell is as follows :

                      E°cell , Ecell and Δ[tex]G_{rnx}[/tex]

The standard cell potential is the potential of cell at standard condition of 1MConcentration and pressure 1 atm E°cell

calculation :

                E°cell = E° cathode - E° anode          it is calculated using the Nernst equation which is discussed below :

                  Ecell = E°cell -- [tex]\frac{RT}{nF}[/tex] 1n K  = E°cell -- [tex]\frac{0.0591}{n}[/tex]log [tex]\frac{Products}{Reactants}[/tex]

Here, F is the Faraday's constant, R is the gas constant, T is the temperature, and n is the number of transferred electrons. K is the equilibrium constant.

The Gibbs free energy is the greatest work that is finished by a framework . The standard cell potential is without like energy by the recipe as follows;and F is Faraday's steady.

A system's maximum amount of work is referred to as its Gibbs free energy. The standard cell potential is connected with the free energy by the recipe as follows:    Δ G = -n F Ecell

Here, E cell is cell potential

Δ G is the free energy n is the quantity of electrons moved and F is Faraday's steady.

The given net cell equation is as follows: Mg + Sn²⁺⇒ Mg²⁺ + Sn

 Oxidation :                                        

Mg ⇒ Mg ²⁺ + 2e⁻ E⁰anode = - 2.37 V

Reduction:Sn²⁺ + 2e⁻⇒ Sn E⁰

So,  cathode = - 0.14V

The standard cell potential is calculated as follows:E⁰ cell = - 0.14 V- (- 2.37 V ) =  2.23 V

The half reaction potentials for the oxidation and reduction are determined. They are subbed in the equation and the standard cell potential is determined.

Number of electrons transferred ,      n = 2   ,[Mg²⁺]   = 0.055M   ,  [ Sn²⁺ ]  = 0.030 M               The Nernst equation for reaction :

Ecell = E °cell = [tex]\frac{0.0591}{n}[/tex]log Mg ²⁺ / Sn²⁺

The cell potential for reaction is :

                       Ecell = 2.23V - [tex]\frac{0.0591}{2}[/tex]log[tex]\frac{0.055M}{0.030M}[/tex]= 2.22V

 The values are substituted for the reaction calculated here in the Nernst equation and cell potential.     

Calculation for the free energy for reaction ,

                                             ] Δ[tex]G_{rxn}[/tex] = -nFE cell

       = - 2 × 96485 C/ mol ×2.22 V

                          = --428393J/mol × [tex]\frac{1KJ}{1000J}[/tex]  = - 428.39kJ/mol

                                 

The cell potential for the response is subbed in the recipe and free energy for the response is determined

Nernst equation :

The standard electrode potential, absolute temperature, the number of electrons involved in the redox reaction, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation, respectively, can all be used to calculate the reduction potential of a half-cell or full cell reaction using the Nernst equation, a chemical thermodynamic relationship.

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What is the molar solubility of ag2cr04 in water? (ksp of ag2cro4 is 8.0 x 10-12)

Answers

The molar solubility of Ag₂CrO₄ in water is approximately 1.24 x 10^-4 mol/L.

The solubility of a salt in water can be calculated using its solubility product constant (Ksp) value. The Ksp expression for Ag₂CrO₄ is:

[tex]Ag_2CrO_4[/tex](s) ⇌ [tex]2Ag^+(aq)[/tex] + [tex]CrO_4^{2-}(aq)[/tex]

The Ksp expression for this equilibrium is:

Ksp = [Ag+]^2[[tex]CrO_4^{2-[/tex]]

where [Ag+] and [CrO₄²-] are the concentrations of Ag+ and CrO₄²- ions in the equilibrium, respectively.

Let's assume that the molar solubility of [tex]Ag_2CrO_4[/tex] in water is x mol/L. Since the Ag₂CrO₄ dissociates into 2 Ag+ ions and 1 [tex]CrO__4^2-[/tex] ion, the concentration of Ag+ ions and [tex]CrO_4^{2-}[/tex] ions in the equilibrium will be 2x and x, respectively. Substituting these values into the Ksp expression, we get:

Ksp = (2x)^2(x) = 4x^3

Now, we can solve for x:

Ksp = [tex]4x^3[/tex]

8.0 x [tex]10^-12[/tex] = [tex]4x^3[/tex]

[tex]x^3[/tex] = (8.0 x [tex]10^-12[/tex])/4

[tex]x^3[/tex] = 2.0 x [tex]10^{-12}[/tex]

x = (2.0 x [tex]10^{-12}[/tex])^(1/3)

x = 1.24 x [tex]10^{-4[/tex] mol/L

Therefore, the molar solubility of Ag₂CrO₄ in water is approximately 1.24 x [tex]10^{-4[/tex] mol/L.

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A jewel thief has two fish tanks in his house, neither of which have fish in them. Supposedly the thief hide his jewels in one of the tanks. As you look, you notice that both of the tanks have little treasure chests at the bottom. Just before you each in you notice electric wires laying in the water, so you quickly pull back. Upon closer inspection you see that the right tank has residue on the sides, which turns out to be salt. The left tank has no salt in it. Which tank probably has the jewels in it and why?

Answers

It is likely that the jewels are hidden in the tank with salt residue on their sides.

Using salt to set up an electric system

The presence of the electric wires in both tanks suggests that the thief has set up a security system to protect the treasure chests.

The purpose of the salt in the right tank is likely to act as a conductor, completing an electric circuit if someone were to touch the chest or the wires. This would trigger an electric shock and serve as a deterrent to potential thieves.

Since the thief is unlikely to have set up the security system in the tank without the jewels, the lack of salt in the left tank suggests that it is a decoy, intended to mislead potential thieves.

Therefore, the tank with the salt residue is the more likely hiding place for the jewels.

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Questlon 25 of 25
henry lifted a box that weighed 500 n to a height of 0.75 meters. it took him
1.5 seconds. how much work did henry do?
o a. 667 j
b. 750 j
c. 500 j
d. 375 j

Answers

The work done by Henry can be calculated by multiplying the weight of the box (500 N) with the distance it was lifted (0.75 m). Thus, the work done is 375 J (Joules).(D)

In physics, work is defined as the energy transferred when a force is applied to move an object through a distance. The unit of work is Joule, which is the same as Newton-meter. In this question, Henry lifted the box with a force equal to its weight, and the box was lifted through a distance of 0.75 m.

Therefore, Henry did work on the box by transferring 375 J of energy to it. This work is equal to the potential energy gained by the box due to its vertical displacement. The time taken (1.5 seconds) is not relevant to the calculation of work.

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