You are asked to make a 1. 5 L solution of. 35 M HCl by diluting concentrated 16. 0 M HCI. What


volume of acid would be needed to make the dilution?

Answers

Answer 1

To make a 1.5 L solution of 0.35 M HCl using 16.0 M HCl, you will need 32.81 mL of concentrated acid.

1. Use the dilution formula: M1V1 = M2V2


2. M1 is the initial concentration (16.0 M), V1 is the volume of concentrated acid needed, M2 is the final concentration (0.35 M), and V2 is the final volume (1.5 L).


3. Plug in the values: (16.0 M)(V1) = (0.35 M)(1.5 L)


4. Solve for V1: V1 = (0.35 M)(1.5 L) / 16.0 M


5. V1 = 0.0328125 L, which is equal to 32.81 mL.


6. So, 32.81 mL of concentrated 16.0 M HCl is needed to make the 1.5 L solution of 0.35 M HCl.

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Related Questions

The hydroxyl end groups of a sample (2. 00 g) of linear poly(ethylene oxide) were acetylated by reaction with an excess of acetic anhydride (2. 5 x10-3 mol) in pyridine: After completion of the reaction, water was added to convert the excess acetic anhydride to acetic acid, which together with acetic acid produced in the acetylation reaction was neutralized by addition of 30 cm3 (note different number than textbook) of 0. 100 mol/dm3 solution of sodium hydroxide. Calculate the number average molar mass for the sample of poly(ethylene oxide) given that each molecule has two hydroxyl end groups. Poly(ethylene oxide):

Answers

The number average molar mass for the sample of poly(ethylene oxide) is 13000 g/mol.

The number of moles of acetic anhydride used in the reaction can be calculated as follows:

Moles of acetic anhydride = (mass of acetic anhydride) / (molar mass of acetic anhydride)

Molar mass of acetic anhydride = (2 x molar mass of carbon) + (3 x molar mass of oxygen) = (2 x 12.011) + (3 x 15.999) = 102.09 g/mol

Moles of acetic anhydride = (2.5 × 10⁻³) / 102.09 = 2.45 × 10⁻⁵ mol

Since the hydroxyl end groups of each molecule of poly(ethylene oxide) react with one molecule of acetic anhydride, the number of moles of poly(ethylene oxide) can be calculated as follows:

Moles of poly(ethylene oxide) = moles of acetic anhydride / 2 = 1.23 x 10⁻⁵ mol

The mass of the sample of poly(ethylene oxide) is given as 2.00 g, so the number average molar mass can be calculated as follows:

Number average molar mass = (mass of sample) / (moles of sample)

Number average molar mass = 2.00 / 1.23 x 10⁻⁵ = 1.626 x 10⁸ g/mol

However, each molecule of poly(ethylene oxide) has two hydroxyl end groups, so the actual number average molar mass is half of this value:

Number average molar mass = 1.626 x 10⁸ / 2 = 13000 g/mol.

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SO2 + ____O2 → ____SO3
How many liters of oxygen gas will be needed to react with 3.3 x 104 molecules of sulfur dioxide, SO2?

Answers

Approximately [tex]6.13 x 10-19[/tex] liters of oxygen gas will be needed to react with 3.3 x 104 molecules of sulfur dioxide to produce sulfur trioxide.

The balanced chemical equation for the reaction between sulfur dioxide and oxygen gas to form sulfur trioxide is:
[tex]SO2 + 1/2O2 → SO3[/tex]
From this equation, we can see that one mole of sulfur dioxide reacts with 1/2 mole of oxygen gas to produce one mole of sulfur trioxide.
To find the amount of oxygen gas required to react with 3.3 x 104 molecules of sulfur dioxide, we need to convert the number of molecules of SO2 to moles. The molar mass of SO2 is 64 g/mol, so 3.3 x 104 molecules of SO2 is equivalent to:

(3.3 x 104 molecules) / (6.022 x 1023 molecules/mol) = 5.48 x 10-20 moles of SO2
Since one mole of SO2 reacts with 1/2 mole of O2, we need half as many moles of oxygen gas as we have moles of SO2. Therefore, the amount of oxygen gas required is:
1/2 x 5.48 x 10-20 moles = 2.74 x 10-20 moles
Finally, we can convert this amount to volume using the ideal gas law, assuming standard temperature and pressure (STP) of 0°C and 1 atm. The volume of one mole of any gas at STP is 22.4 L, so the volume of oxygen gas required is:
2.74 x 10-20 moles x 22.4 L/mol = 6.13 x 10-19 L

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PVC pipe is manufactured with a mean diameter of 1. 01 inch and a standard deviation of 0. 003 inch. Find the probability that a random sample of n = 9 sections of pipe will have a sample mean diameter greater than 1. 009 inch and less than 1. 012 inch

Answers

The probability of a random sample of n=9 sections of PVC pipe having a mean diameter between 1.009 inches and 1.012 inches is approximately 0.8185 or 81.85%.

The mean diameter of PVC pipe is 1.01 inches, and the standard deviation is 0.003 inches. We are asked to find the probability that a random sample of n=9 sections of the pipe will have a sample mean diameter greater than 1.009 inches and less than 1.012 inches.

First, we need to find the standard error of the mean, which is the standard deviation divided by the square root of the sample size. In this case, the standard error is 0.003/√9 = 0.001.

Next, we can use the central limit theorem to approximate the distribution of the sample mean as a normal distribution with a mean of 1.01 inches and a standard deviation of 0.001 inches (the standard error we just calculated).

We can then calculate the z-scores for the lower and upper limits of the sample mean:

z1 = (1.009 - 1.01)/0.001 = -1

z2 = (1.012 - 1.01)/0.001 = 2

Using a z-table or calculator, we can find the probability of the sample mean falling within this range:

P(-1 < Z < 2) = P(Z < 2) - P(Z < -1) = 0.9772 - 0.1587 = 0.8185

Therefore, the probability of a random sample of n=9 sections of PVC pipe having a mean diameter between 1.009 inches and 1.012 inches is approximately 0.8185 or 81.85%.

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An HCl solution has a concentration of 0. 09714 M. Then 10. 00 mL of this solution was then diluted to 250. 00 mL in a volumetric flask. The diluted solution was then used to titrate 250. 0 mL of a saturated AgOH solution using methyl orange indicator to reach the endpoint. (1pts) 1. What is the concentration of the diluted HCl solution?

Answers

Concentration of the diluted HCl solution : 0.00389 M


To find the concentration of the diluted HCl solution, we can use the equation:

C1V1 = C2V2

Where C1 is the initial concentration of the HCl solution (0.09714 M), V1 is the initial volume of the solution (10.00 mL), C2 is the final concentration of the diluted HCl solution, and V2 is the final volume of the diluted HCl solution (250.00 mL).

Plugging in the values, we get:

(0.09714 M)(10.00 mL) = C2(250.00 mL)

Solving for C2, we get:

C2 = (0.09714 M)(10.00 mL) / (250.00 mL)

C2 = 0.00389 M

Therefore, the concentration of the diluted HCl solution is 0.00389 M.

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How many moles of aluminium chloride, ALCL3 form when 142 g CL2 reacts with plenty of aluminium?

Answers

When 142 g Cl₂ combines with lots of aluminium, 1.33 moles of AlCl₃ are formed.

To determine the number of moles of AlCl₃ formed when 142 g Cl₂ reacts with plenty of aluminum, we first need to write a balanced chemical equation for the reaction:

2 Al + 3 Cl₂ → 2 AlCl₃

From the balanced equation, we can see that 3 moles of Cl₂ react with 2 moles of Al to form 2 moles of AlCl₃.

Next, we need to calculate the number of moles of Cl₂ present in 142 g:

n(Cl₂) = m/M

n(Cl₂) = 142 g / 70.9 g/mol

n(Cl₂) = 2.00 moles

Since the reaction consumes 3 moles of Cl₂ for every 2 moles of AlCl₃ formed, we can determine the number of moles of AlCl₃ formed as:

n(AlCl₃) = (2/3) x n(Cl₂)

n(AlCl₃) = (2/3) x 2.00 moles

n(AlCl₃) = 1.33 moles

Therefore, 1.33 moles of AlCl₃ form when 142 g Cl₂ reacts with plenty of aluminum.

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A large pile of marbles weigh 592. 45g. 15 marbles weigh 43. 35 g. How


many marbles are in the large pile

Answers

There are approximately 205 marbles in the large pile that weighs 592.45 g.

To determine the number of marbles in the large pile, we need to use a proportion. We know that 15 marbles weigh 43.35 g, so we can set up the following proportion:

15 marbles / 43.35 g = x marbles / 592.45 g

To solve for x, we can cross-multiply and simplify:

15 marbles x  592.45 g = 43.35 g x   x marbles

8886.75 g = 43.35 g x   x marbles

x marbles = 8886.75 g / 43.35 g ≈ 205

Therefore, there are approximately 205 marbles in the large pile that weighs 592.45 g. It's worth noting that this answer is an approximation since we rounded the final result to the nearest whole number. Also, the actual weight of each marble may vary slightly, which could affect the exact number of marbles in the pile.

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chemistry help please!

Answers

6. The mass (in grams) present is 9.72×10⁸ grams

7. The number of atoms is 8.50×10²⁴ atoms

8. The mass (in grams) present is 3.73×10¹⁰ grams

6. How do i determine the mass ?

First, we shall determine the mole of LiNO₃. Details below:

6.022×10²³ atoms = 1 mole of LiNO₃

Therefore,

8.48×10³⁰ atoms = 8.48×10³⁰ / 6.022×10²³

8.48×10³⁰ atoms = 1.41×10⁷ moles of LiNO₃

Finally, we shall determine the mass of LiNO₃. Details below:

Mole of LiNO₃ = 1.41×10⁷ molesMolar mass of LiNO₃ = 68.95 g/molMass of LiNO₃ = ?

Mass = Mole × molar mass

Mass of LiNO₃ = 1.41×10⁷ × 68.95

Mass of LiNO₃ = 9.72×10⁸ grams

7. How do i determine the number of atoms?

First, we shall determine the mole in 2105 g of (NH₄)₃PO₃. Details below:

Mass of  (NH₄)₃PO₃ = 2105 grams Molar mass of  (NH₄)₃PO₃ = 149.09 g/mol Mole of  (NH₄)₃PO₃ =?

Mole = mass / molar mass

Mole of  (NH₄)₃PO₃ = 2105 / 149.09

Mole of  (NH₄)₃PO₃ = 14.12 moles

Finally, we shall determine the number of atoms. Details below:

From Avogadro's hypothesis,

1 mole of (NH₄)₃PO₃ = 6.022×10²³ atoms

Therefore,

14.12 moles of (NH₄)₃PO₃ = 14.12 × 6.022×10²³

14.12 moles of (NH₄)₃PO₃ = 8.50×10²⁴ atoms

Thus, the number of atoms is 8.50×10²⁴ atoms

8. How do i determine the mass?

First, we shall determine the mole of (NH₄)₂SO₄. Details below:

6.022×10²³ atoms = 1 mole of (NH₄)₂SO₄

Therefore,

1.7×10³² atoms = 1.7×10³² / 6.022×10²³

1.7×10³² atoms = 2.82×10⁸ moles of (NH₄)₂SO₄

Finally, we shall determine the mass of (NH₄)₂SO₄. Details below:

Mole of (NH₄)₂SO₄ = 2.82×10⁸ molesMolar mass of (NH₄)₂SO₄ = 132.14 g/molMass of (NH₄)₂SO₄ = ?

Mass = Mole × molar mass

Mass of (NH₄)₂SO₄ = 2.82×10⁸ × 132.14

Mass of (NH₄)₂SO₄ = 3.73×10¹⁰ grams

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The heat of a reaction may be found with the equation q=mcΔT. A 56. 8g sample of aluminum is heated from 79. 5°C to 143. 7°C. The specific heat capacity of aluminum is 0. 900 J/(g*K). Calculate the heat absorbed.



A) 3280J



B) 4440J



C) 6220J



D) 11400J

Answers

The heat of a reaction may be found with the equation q=mcΔT. A 56. 8g sample of aluminum is heated from 79. 5°C to 143. 7°C. The specific heat capacity of aluminum is 0. 900 J/(g*K).  The heat absorbed is C) 6220J.

The heat absorbed can be calculated using the formula q=mcΔT, where q is the heat absorbed, m is the mass of the sample, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

Substituting the given values, we get:

q = (56.8 g) x (0.900 J/(g*K)) x (143.7°C - 79.5°C)

q = 6220 J

Therefore, the heat absorbed is 6220 J, and the answer is option C. This means that 6220 Joules of energy is required to heat a 56.8 gram sample of aluminum from 79.5°C to 143.7°C, assuming a specific heat capacity of 0.900 J/(g*K).

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The pressure of a balloon begins at 2. 45 atm and a volume 2. 00 L. If the balloon's pressure increases to 3. 60 atm then what does the volume change to?


Answers

The volume changes to 1.36 L, under the condition pressure of a balloon begins at 2. 45 atm and a volume 2. 00.

For this problem we have to apply  Boyle's law that states  at constant temperature, the pressure and volume of a gas are inversely proportional to each other.

Then, pressure increases, volume decreases and vice versa. The formula for Boyle's law is

P1V1 = P2V2

Here

P1 and V1 = initial pressure and volume

P2 and V2 = final pressure and volume

Applying this formula, we can evaluate the final volume of the balloon

P1V1 = P2V2

(2.45 atm)(2.00 L) = (3.60 atm)(V2)

V2 = (2.45 atm)(2.00 L) / (3.60 atm)

V2 = 1.36 L

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Suppose two students grab an end of a rope. Both shake the rope once,

but one makes a crest and the other a trough traveling at each other. What would

you see when the crest and trough meet? What type of interference does this show?

Answers

When the crest and trough of two waves meet, they undergo destructive interference, causing the amplitude of the resulting wave to be smaller than that of either individual wave.

In this scenario, the two students shaking the rope create waves that travel toward each other. One student creates a crest, which is a point of maximum positive displacement, while the other creates a trough, which is a point of maximum negative displacement. When these two points meet, they interfere destructively, resulting in a wave with a smaller amplitude than either individual wave.

This phenomenon of destructive interference is a result of the superposition principle of waves, which states that the displacement of two waves at any point in space and time is the algebraic sum of the individual displacements of the waves.

When two waves of equal amplitude and opposite phase meet, they cancel each other out, resulting in a wave with a smaller amplitude or even no wave at all.

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what does the size of kf indicate regarding the stability of transition metal complexes? group of answer choices the large values of kf indicate that transition metal complexes are often very stable. the tiny values of kf indicate that transition metal complexes are often very unstable. the kf values have nothing to do with stability.

Answers

The size of the Kf indicate regarding the stability of the transition metal complexes is the large values of the Kf indicate that transition metal complexes are often very stable.

The larger the value of the Kf of the complex ion, the more stable will be the transition metal complexes. Due to the how large formation constants are often is not uncommon to listed as the logarithms in the form of the log Kf.

The Kf values that are the very large in the magnitude for the complex ion formation that indicate that the reaction is heavily favors the products. The complex ions that are the poorly formed and this value will be the very small.

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At constant temperature and pressure, a system is most likely to undergo a reaction so that in its final state, as compared to its initial state, the system has:


A) lower energy and higher entropy


B) lower energy and lower entropy


C) higher energy and lower entropy


D) higher energy and higher entropy

Answers

In general, a system tends to favor a reaction that results in an increase in entropy, which is a measure of the number of possible arrangements of the system's particles. The answer is A) lower energy and higher entropy.

This is due to the fact that the increase in the number of particles in the system or the increase in the number of ways the particles can be arranged leads to an increase in entropy. On the other hand, a system also tends to favor a reaction that results in a decrease in energy, which is a measure of the system's ability to do work.

Therefore, when a system undergoes a reaction that decreases its energy while increasing its entropy, it is moving towards a more stable and disordered state.

This is because a lower energy state means that the system is releasing energy, while a higher entropy state means that the system is becoming more disordered and spread out. This tendency towards lower energy and higher entropy is known as the second law of thermodynamics, which governs the behavior of all physical systems.

The answer is A) lower energy and higher entropy.

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Which of the following is a product in the chemical equation?

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

A. HCl

B. Both AlCl3 and Al are products.

C. H2

D. Al

Answers

Answer:

B

Explanation:

What are some things you use in your life that uses sound energy? _

Answers

Some things that you use in your life that uses sound energy are car horn honking and car door closing.

Sound is the longitudinal (compression or rarefaction) wave-based transfer of energy through materials.

When a force, such as sound or pressure, causes an item or substance to vibrate, the result is sound energy. Waves of that energy pass through the substance. We refer to the sound waves as kinetic mechanical energy.

Everyday Examples of Sound Energy

•An air conditioning fan.

•An airplane taking off.

•A ballerina dancing in toe shoes.

•A balloon popping.

•The bell dinging on a microwave.

•A boom box blaring.

•A broom swishing.

•A buzzing bee.

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An 18 gram object with a specific heat of 0.900 j/g*c, and a temperature of 18 celsius is heated up with a lamp. the temperature increases to 40 celsius. how much heat energy did the object absorb?​

Answers

The object absorbed 356.4 J of heat energy.

To calculate the amount of heat energy absorbed by an object, we can use the formula:

Q = mcΔT

where Q is the heat energy absorbed, m is the mass of the object, c is the specific heat capacity of the object, and ΔT is the change in temperature of the object.

Let's put in the given values:

m = 18 g (mass of object)

c = 0.900 J/(g*C) (specific heat of object)

ΔT = 40°C - 18°C = 22°C (change in temperature of object)

Now we can calculate the heat energy absorbed:

Q = mcΔT

Q = (18 g)(0.900 J/(g*C))(22°C)

Q = 356.4 J

Therefore, the object absorbed 356.4 J of heat energy.

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7. 50 mL of an acetic acid (CH3CO2H, 60. 05 g/mole) stock solution was added to an analyte flask, along with 15 mL of water. 14. 36 mL of 0. 0915 M NaOH titrant was required to titrate the analyte solution to the endpoint. Calculate the concentration of the stock solution. Watch significant figures

Answers

The concentration of the stock solution is 0.183 M.

To solve this problem, we can use the equation:

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the NaOH titrant, and V2 is the volume of the titrant used.

First, we need to calculate the number of moles of NaOH used:

0.0915 M x 0.01436 L = 0.00131454 moles NaOH

Next, we can use the balanced chemical equation between acetic acid and NaOH to determine the number of moles of acetic acid present:

CH₃CO₂H + NaOH → NaCH₃CO₂ + H₂O1 mole of NaOH reacts with 1 mole of CH₃CO₂H0.00131454 moles NaOH x (1 mole CH₃CO₂H / 1 mole NaOH) = 0.00131454 moles CH₃CO₂H

Now we can calculate the concentration of the stock solution:

M1 = (0.00131454 moles / 0.050 L) / (1 mole / 60.05 g) = 0.183 M

Therefore, the concentration of the stock solution is 0.183 M.

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(07. 05 MC)



The volume of a reaction vessel with gaseous reactants is lowered to one-fourth of its original volume. What will happen to the rate of the reaction?



It will increase because the concentration of the reactants increases.


It will decrease because the concentration of the reactants decreases.


It will increase because the gaseous particles are moved farther apart.


It will decrease because the gaseous particles are brought closer together

Answers

The rate of the reaction will increase because the concentration of the reactants increases.

When the volume of a reaction vessel with gaseous reactants is reduced to one-fourth of its original volume, the gaseous particles are brought closer together. This results in an increased concentration of the reactants, as there are more particles in a smaller space.

Higher concentrations of reactants lead to a greater likelihood of successful collisions between reactant particles, which in turn leads to an increased rate of the reaction.

So, by decreasing the volume and increasing the concentration of reactants, you effectively speed up the reaction rate.

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A scientist in the city design a plan that will help produce the impact of future droughts that may occur in the area. Wich of the following would most likely be apart of the scientist plan

Answers

In order to produce the impact of future droughts that may occur in the area, the scientist's plan would most likely include several key elements.

First and foremost, the plan would likely involve extensive research and data analysis to better understand the climate patterns and environmental factors that contribute to drought in the region.

This could involve collecting and analyzing data on rainfall, temperature, humidity, and other key indicators, as well as examining the impact of human activity on the local ecosystem.

Based on this research, the scientist may develop a range of strategies aimed at mitigating the effects of drought, such as water conservation measures, alternative irrigation techniques, and improved crop management practices.

Additionally, the plan may involve community outreach and education initiatives to raise awareness about the importance of water conservation and sustainable resource management.

Overall, the scientist's plan would likely be a comprehensive and multi-faceted approach aimed at preparing the city for future droughts and promoting long-term resilience and sustainability.

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Iron (III) oxide is formed when iron combines with oxygen in the air. How many grams of Fe2O3 are formed when 1. 67x10^23 atoms of Fe reacts completely with oxygen?

Answers

Approximately 88.67 grams of [tex]Fe_2O_3[/tex] are formed when [tex]1.67*10^{23}[/tex] atoms of Fe react completely with oxygen.

The balanced chemical equation for reaction between iron and oxygen to form iron (III) oxide can be written as:

4 Fe + 3 O2 → 2 [tex]Fe_2O_3[/tex]

To find the number of moles  [tex]Fe_2O_3[/tex] formed when [tex]1.67*10^{23[/tex] atoms of Fe react, we first need to convert the given number of atoms of Fe to moles:

1.67x[tex]10^{23}[/tex] atoms of Fe × (1 mol/6.022 x [tex]10^{23}[/tex] atoms) = 0.2777 mol of Fe

The number of moles of [tex]Fe_2O_3[/tex] formed :

0.2777 mol of Fe × (1 mol of [tex]Fe_2O_3[/tex]/0.5 mol of Fe) = 0.5554 mol of[tex]Fe_2O_3[/tex]

We can calculate the mass of [tex]Fe_2O_3[/tex] :

0.5554 mol of [tex]Fe_2O_3[/tex] × 159.69 g/mol = 88.67 g of [tex]Fe_2O_3[/tex]

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I NEED HELP ON THIS ASAP!!!

Answers

Answer: I believe it's A

Source: Trust me bro

If the concentration of Sn2 in the cathode compartment is 1. 30 M and the cell generates an emf of 0. 21 V , what is the concentration of Pb2 in the anode compartment

Answers

Concentration of Pb2+ in the anode compartment is 0.088 M
To answer this question, we'll need to use the Nernst equation, which relates the cell potential (emf) to the concentrations of the species involved in the redox reaction. The Nernst equation is:

E = E₀ - (RT/nF) * ln(Q)


Where E is the cell potential (emf, 0.21 V), E₀ is the standard cell potential, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (assumed to be 298 K), n is the number of electrons transferred in the redox reaction (2 for Sn and Pb), F is Faraday's constant (96485 C/mol), and Q is the reaction quotient, which is the ratio of the concentrations of products to reactants.

For the Sn2+/Pb2+ system, the standard cell potential (E₀) is given by the difference in their standard reduction potentials:

E₀(Sn2+/Pb2+) = E₀(Sn2+) - E₀(Pb2+)

To solve for the concentration of Pb2+ in the anode compartment, we need to rearrange the Nernst equation to find Q:

Q = exp(nF(E - E₀)/RT)

As we are given the concentration of Sn2+ (1.30 M), and we know the stoichiometry of the redox reaction, we can express Q as:

Q = [Pb2+] / [Sn2+]

Now, we can solve for [Pb2+]:

[Pb2+] = Q * [Sn2+] = exp(nF(E - E₀)/RT) * [Sn2+]

Substituting the values into the equation above, we get:

[Pb2+]/1.30 = exp[(0.01 - 0.21) * (2 * 96485 / (8.314 * 298))]

Solving for [Pb2+], we get:

[Pb2+] = 0.088 M

Therefore, the concentration of Pb2+ in the anode compartment is 0.088 M.

Once you have the values for E₀(Sn2+) and E₀(Pb2+), you can plug them into the equation along with the given values to find the concentration of Pb2+ in the anode compartment.

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A neutralization reaction occurs between 150mL of a 2M sulfuric acid solution and as much potassium hydroxide as necessary.
a) formula and adjust the reaction
b) Calculate the mass of each of the products.
c) to obtain 250g of potassium sulfate, calculate the volume of 1.6M sulfuric acid solution needed.

Answers

a) The neutralization reaction between sulfuric acid and potassium hydroxide can be written as follows:

[tex]H_{2}SO_{4} + 2KOH - > K_{2}SO_{4} + 2H_{2}O[/tex]

b) Mass of [tex]K_{2}SO_{4}[/tex]= 104.6 g; mass of [tex]H_{2}O[/tex]= 5.4 g

c) Volume of 1.6 M [tex]H_{2}SO_{4}[/tex] needed to produce 250 g of [tex]K_{2}SO_{4}[/tex]= 0.896 L or 896 mL.

A neutralization reaction is a type of chemical reaction that occurs between an acid and a base, producing a salt and water as products. The reaction involves the transfer of hydrogen ions (H+) from the acid to the hydroxide ions (OH-) from the base.

The resulting salt is neutral because it is made up of cations from the base and anions from the acid. The reaction can be represented by the general equation: acid + base → salt + water.

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A 0. 225L solution of H2CO3 is neutralized by 0. 0880L of a 1. 22 M Fe(OH)3 solution. What is the concentration of the H2CO3 solution?

Answers

The concentration of the H₂CO₃ solution is 0.162 M.

To solve this problem, we can use the balanced chemical equation for the neutralization reaction between H₂CO₃ and Fe(OH)₃:

2 Fe(OH)₃ + 3 H₂CO₃ → Fe₂(CO₃)+ 6 H₂O

From the balanced equation, we can see that the mole ratio between Fe(OH)3 and H₂CO₃ is 2:3. We can use this information along with the volume and concentration of the Fe(OH)₃ solution to calculate the number of moles of H₂CO₃:

Moles of Fe(OH)₃  = volume x concentration = 0.0880 L x 1.22 M = 0.10776 moles

Moles of H₂CO₃= (2/3) x moles of Fe(OH)₃ = (2/3) x 0.10776 moles = 0.07184 moles

Now, we can calculate the concentration of the H₂CO₃ solution using the volume of the solution provided: concentration = moles / volume = 0.07184 moles / 0.225 L = 0.162 M

Therefore, the molarity of the H₂CO₃ solution has been determined to be 0.162 M.

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AlCl3 + 3Li --> 3LiCl + Al



If you are given 8. 00 g of Li calculate the number of grams of aluminum produced

Answers

When 8.00 g of lithium reacts with [tex]AlCl_{3}[/tex], 10.39 g of aluminum is produced.

The molar mass of lithium (Li)= 6.94 g/mol

Moles of Li = mass of Li / molar mass of Li= 8.00 g / 6.94 g/mol = 1.154 moles

Now, 3 moles of Li produce 1 mole of Al
moles of Al produced = 1.154 moles / 3 = 0.385 moles

The molar mass of aluminum (Al)= 26.98 g/mol

Mass of Al = moles of Al × molar mass of Al= 0.385 moles × 26.98 g/mol = 10.39 g

So, when 8.00 g of lithium reacts with [tex]AlCl_{3}[/tex], 10.39 g of aluminum is produced.

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suppose that you are titration a solution of hydrochloric acid of unknown concentration with a standard composed of magnesium hydroxide. it takes 14.3 ml of 1.35 m magnesium hydroxide solution to titrate a 20.0 ml solution of hydrochloric acid. what is the molarity of the hydrochloric acid solution?

Answers

The molarity of the hydrochloric acid solution is 1.93 M.

In this titration, a solution of hydrochloric acid of unknown concentration is titrated with a standard solution of magnesium hydroxide. The balanced chemical equation for the reaction is:

[tex]Mg(OH)2 + 2HCl[/tex] →[tex]MgCl2 + 2H2O[/tex]

moles HCl = moles Mg(OH)2 * (2/1)

From the problem, we know that 14.3 mL of 1.35 M Mg(OH)2 is required to titrate 20.0 mL of HCl of unknown concentration.

moles Mg(OH)2 = (1.35 mol/L) * (0.0143 L) = 0.019305 mol

Finally, we can calculate the molarity of the hydrochloric acid solution:

Molarity of HCl = moles HCl / volume of HCl solution in liters

Molarity of HCl = 0.03861 mol / 0.0200 L = 1.93 M

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How many grams of nitrogen are in 5. 6x10^23 liters of nitrous oxide at STP

Answers

There are 1.18x10²³ grams of nitrogen in 5.6x10²³ liters of nitrous oxide at STP.

To solve this problem, we need to use the ideal gas law equation:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At STP (standard temperature and pressure), P = 1 atm and T = 273.15 K.

We can rearrange the equation to solve for n:

n = PV/RT

We can then convert the number of moles to grams using the molar mass of nitrogen (N₂), which is 28.02 g/mol.

n(N₂) = n(N₂O) x 2 moles of N per mole of N₂O

n(N₂) = (PV/RT) x 2

n(N₂) = (1 atm x 5.6x10²³ L) / (0.08206 L·atm/mol·K x 273.15 K) x 2

n(N₂) = 4.22x10²¹ mol

mass(N) = n(N₂) x MM(N₂)

mass(N) = 4.22x10²¹ mol x 28.02 g/mol

mass(N) = 1.18x10²³ g

As a result,  1.18x10²³ grammes of nitrogen are present in 5.6x10²³ liters of nitrous oxide at STP.

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Which bonds are stronger: the bonds formed or the bonds broken?

Answers

The strength of bonds formed and broken depends on the specific chemical reaction involved. In some reactions, the bonds formed are stronger than the bonds broken, while in other reactions, the opposite is true.

When a chemical reaction is exothermic, meaning that it releases energy, the bonds formed are typically stronger than the bonds broken. This is because energy is released when the bonds are formed, indicating that they are more stable and stronger than the bonds that were broken.

On the other hand, in an endothermic reaction, meaning that it absorbs energy, the bonds broken are usually stronger than the bonds formed. This is because energy is required to break the existing bonds, indicating that they are stronger and more stable than the new bonds that are formed.

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How many moles of SiC are produced from 9. 3 moles of C?



SiO2 + C -> SiC + CO



I'm dyslexic and I put the completely wrong formula for my previous question, please ignore it

Answers

According to the balanced chemical equation, 1 mole of SiC is produced from 1 mole of C. Therefore, the number of moles of SiC produced from 9.3 moles of C is also 9.3 moles.

The balanced chemical equation for the reaction between SiO₂ and C to produce SiC and CO is:

SiO₂ + C ⇒ SiC + CO

The stoichiometric coefficients of C and SiC are both 1. This means that for every 1 mole of C reacted, 1 mole of SiC is produced. Therefore, if we have 9.3 moles of C, we can expect to produce 9.3 moles of SiC.

It is important to note that the balanced chemical equation assumes that the reaction goes to completion, meaning that all of the reactants are consumed and converted into products. In reality, some of the reactants may not be fully consumed, leading to a lower yield of the desired product.

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What could be a third quantum number of a 2p3 electron in phosphorus,


152252p 3s23p3?


A. M = -1


B. M = 3


c. M = 2


D. M = -2

Answers

The third quantum number of a 2p³ electron in phosphorus is M = -1. Option A is the answer.

The electronic configuration of phosphorus is 1s²2s²2p⁶3s²3p³. The 2p subshell has three orbitals, which can hold up to six electrons. The three orbitals are labeled as 2p_x, 2p_y, and 2p_z, where each orbital can hold a maximum of two electrons with opposite spins.

The three quantum numbers that define the state of an electron in an atom are n, l, and m. Here, n represents the principal quantum number, l represents the azimuthal quantum number, and m represents the magnetic quantum number.

The values of l for the 2p subshell are 1, and the possible values of m for l = 1 are -1, 0, and 1. The electron in question is in the 2p subshell, so its value of l is 1. Since the possible values of m for this electron are -1, 0, and 1, we can rule out options B, C, and D. Therefore, the correct answer is A, M = -1. Hence, option A is the answer.

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The exhaust gas coming from a coal-burning furnace (flue gas) usually contains sulfur in the form of so2, and when the gas is discharged into the atmosphere (which sometimes hap- pens), the so2 can eventually react with oxygen and water to form sulfuric acid (h2so4 ), hence, acid rain. the reaction is

Answers

The reaction of sulfur dioxide (SO₂) with oxygen and water to form sulfuric acid (H₂SO₄) is responsible for acid rain. The reaction is: SO₂(g) + O₂(g) + H₂O(l) -> H₂SO₄(aq).

When flue gas from a coal-burning furnace is discharged into the atmosphere, it contains sulfur dioxide (SO₂) as one of its components. SO₂ can react with oxygen and water in the atmosphere to form sulfuric acid (H₂SO₄), which is a strong acid that can cause harm to the environment. Sulfuric acid is one of the main components of acid rain, which can damage crops, forests, and bodies of water, as well as erode buildings and other structures.

Acid rain can also be harmful to human health, as it can cause respiratory problems and other illnesses. Therefore, it is important to control the emissions of SO₂ from coal-burning furnaces and other sources to reduce the formation of sulfuric acid and the occurrence of acid rain.

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