a 1.0 l solution of mgf2 was electrolyzed for 11.2 h to give 44.75 g of magnesium. assuming the minimum voltage needed was available, what amperage would be needed to complete the electrolysis in the given time?

Answers

Answer 1

The amperage would be needed to complete the electrolysis in the given time is I = 8.37 A.

The length of time needed to create the same quantity of hydrogen in an electrolysis cell would be cut in half if the amperage in the cell were increased by a factor of 2. This is so because the amount of current passing through the cell directly proportionately affects the pace of electrolysis.

MgF₂ solution was electrolyzed for 19.6 h to give 74.4 g of Magnesium.

So, atomic mass of Magnesium = 24.3 g/mole.

Mass of Magnesium on electrolysis = 74.4 g

no. of Moles of Magnesium = 74.40/24.3 = 3.06 moles.

For deposition of Magnesium on Cathode and ! mole of deposition of Magnesium; electricity required = 2F

= 2 x 96500 C = 193000C

For 3.06 mole deposition of Magnesium,

electricity required = 3.06 x 193000 C

= 590580 C

Time = 19.6 h = 19.6 x 60 x 60 = 70560 seconds

Uisng relation ,

Q = I x t

I = Q/t = 193000 x 3.06 / 70560 = 590580/70560 = 8.369 A

I = 8.369 ≈ 8.37 A.

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Related Questions

a sample of groundwater has 90 mg/l of ca2 and 45 mg/l of mg2 . find the total hardness expressed in meq/l and mg/l as caco 3 . how would this water be classified in terms of hardness?

Answers

To calculate the total hardness of the water sample, we need to first convert the concentrations of Ca2+ and Mg2+ from mg/L to mEq/L:

1 mEq/L = atomic weight (mg/L) / valence

The atomic weights and valences for Ca2+ and Mg2+ are:

Ca2+: atomic weight = 40.08 g/mol, valence = 2

Mg2+: atomic weight = 24.31 g/mol, valence = 2

Converting the concentrations to mEq/L:

Ca2+: (90 mg/L) / (40.08 g/mol / 2) = 2.24 mEq/L

Mg2+: (45 mg/L) / (24.31 g/mol / 2) = 1.85 mEq/L

The total hardness of the water sample is the sum of the molar concentrations of Ca2+ and Mg2+:

Total hardness = 2.24 mEq/L + 1.85 mEq/L = 4.09 mEq/L

To express the total hardness as mg/L as CaCO3, we can use the conversion factor of 1 mEq/L = 50 mg/L as CaCO3:

Total hardness = 4.09 mEq/L x 50 mg/L as CaCO3/mEq = 204.5 mg/L as CaCO3

Based on the total hardness value, this water sample would be classified as "hard water."

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consider a reaction between two gaseous reactants (4 mol of a and 4 mol of b) in the closed flasks shown below. assume that the two reactions are both at room temperature. which reaction will occur faster?

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Answer:

....................................................

Factors such as pressure, volume, and the presence of catalysts can affect the rate of the reaction.

Figure out the reaction between two gaseous reactants?

The two gaseous reactants (4 mol of A and 4 mol of B) in the closed flasks shown below will occur faster, I would need more information about the specific conditions in each flask. Factors such as pressure, volume, and the presence of catalysts can affect the rate of the reaction.

If you could provide more details about the flasks and the conditions, I would be happy to help you determine which reaction will occur faster.

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what is the ph of a solution prepared by mizing 100ml of 0.020m ba(oh)2 with 50ml of 0.400m of koh? assume that the volumes are addative

Answers

The pH of the solution is approximately 12.73.

First, we need to find the moles of each solution:

moles of Ba(OH)2 = 0.020 mol/L x 0.100 L = 0.002 mol

moles of KOH = 0.400 mol/L x 0.050 L = 0.020 mol

Next, we need to find the total volume of the solution:

Vtotal = 100 mL + 50 mL = 150 mL = 0.150 L

Now, we can find the total concentration of OH- ions:

[OH-] = moles of Ba(OH)2 + moles of KOH / Vtotal

[OH-] = (0.002 mol + 0.020 mol) / 0.150 L = 0.187 mol/L

Finally, we can find the pH of the solution using the following formula:

pH = 14 - log([OH-])

pH = 14 - log(0.187) = 12.73

Therefore, the pH of the solution is approximately 12.73.

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what do you suspect is the solid or oil that was not soluble in hexanes after synthesizing the adipoyl chloride?

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Without more information about the synthesis process and the specific substances used, it's difficult to say exactly what the solid or oil that was not soluble in hexanes might be. However, there are a few possibilities to consider.

One possibility is that the solid or oil is an impurity that was introduced during the synthesis process. For example, it could be a side product or a reactant that did not fully react with the adipoyl chloride. In this case, the substance may not be soluble in hexanes because it has different chemical properties than the desired product.

Another possibility is that the substance is a byproduct of the reaction between the adipoyl chloride and another substance, such as a solvent or a catalyst. In this case, the substance may not be soluble in hexanes because it has a different chemical structure than the desired product and is not compatible with hexanes.

Alternatively, it's possible that the solid or oil is a form of the adipoyl chloride itself. For example, if the adipoyl chloride was not fully purified or if it was synthesized using impure starting materials, it could contain other compounds that are not soluble in hexanes.

Overall, without more information about the synthesis process and the specific substances used, it's difficult to determine the exact nature of the solid or oil that was not soluble in hexanes. Further analysis, such as chromatography or spectroscopy, may be necessary to identify the substance and determine its origin.

discussion and conclusion on how to determine the reaction enthalpy of sodium hydroxide and hydrochloric acid​

Answers

Discussion:

You can describe the reaction that took place in terms of enthalpy, by writing a fully balanced equation (and net ionic equation) for the reaction, as well as drawing an energy change diagram for the reaction, clearly indicating the measured quantity of heat energy change.Is the reaction exothermic or endothermic? Explain this in terms of bonds breaking and formingDiscuss the method and set up of the experimentWas the experiment accurate (calculate percentage error). [tex]|\frac{theoretical-experimental}{theoretical} |[/tex] × 100%. The theoretical value is -55.84 kJ/mol. error from 0% to ≈30% is accurate.Was the experiment reliable? (are results of each trial close to each other?) (only if applicable)Was the experiment valid? (is it both reliable and accurate)What can be done to improve the experiment? to improve reliability, validity, accuracy?

Conclusion: You can use this basic outline, to structure your conclusion, and expand it from there.

By investigating/measuring/using a....... it was determined that........ This is consistent/not consistent with the expected result/theory of...... due to/because of...........

how many moles of naf must be dissolved in 1.00 liter of a saturated solution of pbf2 at 25˚c to reduce the [pb2 ] to 1 x 10–6 molar? (ksp pbf2 at 25˚c = 4.0 x 10–8)

Answers

The moles of NaF that must be dissolved in 1.00 liter of a saturated solution of PbF₂ at 25˚C to reduce the [Pb²⁺] to 1 x 10⁻⁶ molar is 2.0 x 10⁻⁵.

The solubility product expression for PbF₂ is given by:

Ksp = [Pb²⁻][F-]²

At equilibrium, the product of the ion concentrations must be equal to the solubility product constant. We are given that the [Pb²⁺] in the saturated solution is 1 x 10⁻⁶ M. Therefore, we can use the Ksp expression to calculate the concentration of F- in the solution:

Ksp = [Pb²⁺][F⁻]²4.0 x 10⁻⁸ = (1 x 10⁻⁶)([F⁻]²)[F⁻]² = 4.0 x 10⁻²[F⁻] = 2.0 x 10⁻¹

Now, we can calculate the amount of NaF needed to reduce the [F⁻] concentration to 2.0 x 10⁻¹ M. Since NaF is a 1:1 electrolyte, the concentration of F- will be equal to the concentration of NaF added.

Number of moles of NaF = (2.0 x 10⁻¹) mol/L x 1.00 L = 2.0 x 10⁻¹ moles

However, we need to dissolve this amount of NaF in a saturated solution of PbF₂. Therefore, we need to check that the amount of NaF we added will not exceed the maximum amount that can dissolve in the solution at 25˚C.

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why do you think scientists chose the top of mauna loa, hawaii, as the best place to measure atmospheric co2 concentrations?

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The scientists chose the top of Mauna Loa, Hawaii, is the best place to measure the atmospheric CO₂ concentrations is because to measure the CO₂ in the air masses which could be representative the Northern Hemisphere, and the globe.

To measure the CO₂ in the air masses which could be representative the Northern Hemisphere, and the globe. The rise in level of the atmospheric CO₂ concentrations and this resulted in the global warming and the climate change.

The climate change is the serious consequences, it also including the rising sea levels, it will be more frequent and the severe weather events, it will increased the risk of the droughts and the wildfires.

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Write the expression for the equilibrium constant for each of the following reaction:
2Fe2O3(s)+3C(s)⇌4Fe(s)+3CO2(g)
A) Kc=[CO2]3
B) Kc=[Fe]4[CO2]3[Fe2O3]2[C]3
C) Kc=[Fe2O3]2[C]3[Fe]4[CO2]3
D) Kc=2[Fe2O3]3[C]4[Fe]3[CO2]

Answers

The correct expression for the equilibrium constant (Kc) for the reaction:
[tex]2Fe2O3(s) + 3C(s) ⇌ 4Fe(s) + 3CO2(g)[/tex] is:  [tex]Kc=[Fe]4[CO2]3/[Fe2O3]2[C]3[/tex]



The equilibrium constant expression for the given reaction, [tex]2Fe2O3(s) + 3C(s) ⇌ 4Fe(s) + 3CO2(g)[/tex]  is written as the ratio of the product concentrations raised to their respective coefficients divided by the reactant concentrations raised to their respective coefficients.

The ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the equilibrium equation at equilibrium is K, according to the law of mass action. The equilibrium constant expression is known as the ratio, a condition where there is a balance between opposing and static forces.

In this case, it would be:
[tex]Kc = ([Fe]^4[CO2]^3)/([Fe2O3]^2[C]^3)[/tex]

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The correct expression for the equilibrium constant for the given reaction is:
C) Kc=[Fe2O3]2[C]3[Fe]4[CO2]3

How to write the equilibrium constant of a reaction?

The equilibrium constant (Kc) for a chemical reaction is written using the concentrations of the species involved in the reaction. Here's the general format for writing the equilibrium constant expression:

For the generic reaction:

aA + bB ⇌ cC + dD

The equilibrium constant (Kc) expression would be: Kc = [C]^c [D]^d / [A]^a [B]^b

where [A], [B], [C], and [D] represent the concentrations of the respective species at equilibrium, and a, b, c, and d are the stoichiometric coefficients of the species in the balanced chemical equation.

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Lab: Relative and Absolute Dating Lab Report What is the purpose of the lab?

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The goal of a Relative and Absolute Dating Lab Report is to discover and utilize the concepts of relative and absolute dating methods for determining the age of geological materials like rocks and fossils.

What is the point of absolute dating?

Geologists frequently need to know the age of the material they find. They use absolute dating methods, also known as numerical dating, to give rocks an exact date, or date range, in years. This is distinct from relative dating, which only places geological events in chronological order.

What exactly is the concept of relative dating?

Relative dating is the process of determining whether one rock or geologic event is older or younger than another without knowing their exact ages that is, how many years ago the object was formed.

Where can the relative dating method be used?

Relative dating is used to order geological events and the rocks they leave behind. Stratigraphy is the process of reading the order. Relative dating does not yield precise numerical dates for the rocks.

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You want to use Le Chatelier's Principle to help push the reaction to the right, so you know that one reagent needs to be added in excess. You know acetic acid is cheap, but you do not want to have to neutralize excess acid at the end of the reaction. You choose to add an excess of isoamyl alcohol. You look in the research lab, and all the isoamyl alcohol (d = 0.810 g/mL) you could find was 55 mL. You decide to use it all.
If you use all 55 mL of isoamyl alcohol, and you want to add it a five fold excess, how much volume (in mL) of of glacial acetic acid (17 M) should you add?

Answers

We need to add 100.59 mL of glacial acetic acid to achieve a 5-fold excess of isoamyl alcohol.

To calculate the volume of glacial acetic acid needed to add, we need to determine the number of moles of isoamyl alcohol and the number of moles of acetic acid required to react with it in a 5:1 ratio.

First, let's calculate the number of moles of isoamyl alcohol:

55 mL x 0.810 g/mL = 44.55 g

44.55 g / 130.23 g/mol = 0.342 moles

For the reaction, the ratio of isoamyl alcohol to acetic acid is 5:1, so we need 5 times the amount of moles of acetic acid as isoamyl alcohol:

0.342 moles isoamyl alcohol x 5 = 1.710 moles acetic acid

Now, we can calculate the volume of 17 M glacial acetic acid needed:

1.710 moles x (1 L / 17 mol) x (1000 mL / 1 L) = 100.59 mL

Therefore, we need to add 100.59 mL of glacial acetic acid to achieve a 5-fold excess of isoamyl alcohol.

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You should add 149 mL of glacial acetic acid (17 M) to react with the excess isoamyl alcohol and push the reaction to the right.

Based on Le Chatelier's Principle, adding an excess of isoamyl alcohol will push the reaction to the right. To achieve a five-fold excess, you will need to add 5 times the amount of isoamyl alcohol you have.

First, let's calculate the mass of 55 mL of isoamyl alcohol:
55 mL x 0.810 g/mL = 44.55 g

To get a five-fold excess, you will need to add 5 x 44.55 g = 222.75 g of isoamyl alcohol.

Next, let's calculate the amount of acetic acid needed to react with this excess of isoamyl alcohol. The balanced chemical equation for the reaction between isoamyl alcohol and acetic acid is:

isoamyl alcohol + acetic acid ⇌ isoamyl acetate + water

Since the reaction is in equilibrium, we can use Le Chatelier's Principle to predict the effect of adding excess isoamyl alcohol. The system will shift to the right to use up the excess alcohol and produce more isoamyl acetate and water. Therefore, we need to add enough acetic acid to react with all the excess alcohol, plus some extra to ensure the reaction goes to completion.

The molar ratio of isoamyl alcohol to acetic acid in the reaction is 1:1. This means that for every mole of isoamyl alcohol, we need one mole of acetic acid to react with it. The molecular weight of isoamyl alcohol is 88.15 g/mol, so we can calculate the number of moles of excess alcohol we have:

222.75 g / 88.15 g/mol = 2.528 mol

Therefore, we need to add at least 2.528 mol of acetic acid to react with all the excess alcohol.

The concentration of the acetic acid is given as 17 M, which means it contains 17 moles of acetic acid per liter of solution. To calculate the volume of acetic acid needed, we can use the following equation:

moles of acetic acid = concentration * volume (in liters)


We can rearrange this equation to solve for the volume:
volume (in liters) = moles of acetic acid / concentration

Plugging in our values, we get:
volume (in liters) = 2.528 mol / 17 M = 0.149 L

Finally, we need to convert liters to milliliters:
volume (in mL) = 0.149 L x 1000 mL/L = 149 mL

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The presence of an alcohol group (-OH), __________ the ΔT value of a molecule compared to the presence of a methyl group (-CH3).
A. increases
B. decreases
C. stays the same

Answers

The presence of an alcohol group (-OH) in a molecule, compared to the presence of a methyl group (-CH3), increases the ΔT value of a molecule.


The presence of an alcohol group (-OH) leads to the formation of hydrogen bonds, which are stronger than the van der Waals forces present in molecules with a methyl group (-CH3). As a result, more energy is required to break these hydrogen bonds, leading to a higher ΔT value (a greater change in temperature during phase transitions).

Therefore the correct answer is A. increases.

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does this suggest that your reaction worked? use three key signals to justify your answer 1-methoxy-2-chloro-4-nitrobenzene

Answers

Yes, the reaction worked. Three key signals that suggest the reaction worked include the appearance of the product, the presence of the expected starting material, and the absence of any other byproducts.

The product, 1-methoxy-2-chloro-4-nitrobenzene, can be identified by its distinct color, smell, and boiling point. Additionally, if the expected starting material is present, then it shows that the reaction has taken place.

Lastly, the absence of any other byproducts such as unreacted starting material implies that the reaction was successful. All together, all three signals indicate that the reaction worked.

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If you could change the volume and keep the number of particles the same, what law(s) could you demonstrate? Explain.

Answers

Answer:

Avogadro's Law:

Explanation:

Volume and Amount. Avogadro's law states that at the same temperature and pressure, equal volumes of different gases contain an equal number of particles.

An old Magi cube camera flash bulb (1960s) used Mg metal sealed in bulb with oxygen. Calculate ∆G for its reaction Mg + 1/2 O2= MgO. Where S° Mg= 32. 7, 1/2 O2= 205. 0, MgO= 26. 9 J/mol/K, ΔΗf° -601. 2 kJ/mol

Answers

The value of ∆G for the reaction Mg + 1/2 O₂ = MgO is -557.7 kJ/mol.

To determine ∆G for the reaction, we can use the Gibbs free energy equation;  ∆G = ∆H - T∆S

where; ∆H will be the enthalpy change

T will be the temperature in Kelvin

∆S will bethe entropy change

First, we need to find the values of ∆H and ∆S for the reaction. We can use the enthalpy of formation (∆Hf°) values to calculate ∆H;

∆Hf°(Mg) = 0 kJ/mol

∆Hf°(O₂) = 0 kJ/mol

∆Hf°(MgO) = -601.2 kJ/mol

∆H = ∆Hf°(MgO) - ∆Hf°(Mg) - (1/2)∆Hf°(O₂)

∆H = -601.2 kJ/mol - 0 kJ/mol - (1/2)(0 kJ/mol)

∆H = -601.2 kJ/mol

Next, we need to calculate the entropy change (∆S) for the reaction;

∆S = S°(MgO) - S°(Mg) - (1/2)S°(O₂)

∆S = 26.9 J/mol/K - 32.7 J/mol/K - (1/2)(205.0 J/mol/K)

∆S = -147.2 J/mol/K

Now we can calculate ∆G for the reaction at room temperature (298 K);

∆G = ∆H - T∆S

∆G = -601.2 kJ/mol - (298 K)(-147.2 J/mol/K)

∆G = -601.2 kJ/mol + 43.5 kJ/mol

∆G = -557.7 kJ/mol

Negative sign, indicates that the reaction is spontaneous and will proceed in the forward direction.

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2. calculate the ph of a solution prepared by mixing 25.0 ml of 0.60 m hc2h3o2 and 15.0 ml of 0.60 m naoh?

Answers

The Ph of a solution is 8.46

The reaction is:

[tex]HC_2H_3O+2 + NaOH - > NaC_2H_3O_2 + H_2O[/tex]
This is a neutralization reaction, where the acid HC2H3O2 reacts with the base NaOH to form the salt NaC2H3O2 and water.

Next, we need to calculate the amount of each reagent used in the reaction. To do this, we use the equation:

Molarity (M) = moles (mol) / volume (L)

For [tex]HC_2H_3O_2[/tex]:

M = 0.60 M

Volume = 25.0 ml = 0.025 L

moles = M x volume = 0.60 M x 0.025 L = 0.015 mol

For NaOH:

M = 0.60 M

Volume = 15.0 ml = 0.015 L

moles = M x volume = 0.60 M x 0.015 L = 0.009 mol

Since the reaction is a 1:1 stoichiometry, we can see that 0.009 mol of NaOH is enough to react with all the HC2H3O2 in the solution, leaving some excess NaOH. Therefore, we need to calculate the concentration of the remaining NaOH in the solution:

moles of NaOH remaining = moles of NaOH added - moles of HC2H3O2 reacted

= 0.009 mol - 0.015 mol = -0.006 mol (negative sign indicates there is no excess NaOH remaining)

To calculate the concentration of the NaOH that reacted, we need to subtract the moles of NaOH remaining from the total moles of NaOH added:

moles of NaOH reacted = moles of NaOH added - moles of NaOH remaining

= 0.009 mol - (-0.006 mol) = 0.015 mol

The volume of the final solution is:

Total volume = volume of HC2H3O2 + volume of NaOH

= 25.0 ml + 15.0 ml = 0.040 L

The concentration of NaC2H3O2 in the final solution is:

Molarity (M) = moles / volume

M = 0.015 mol / 0.040 L = 0.375 M

Now, we need to calculate the pH of the solution. NaC2H3O2 is the conjugate base of HC2H3O2, which means it will hydrolyze in water to form OH- ions:

NaC2H3O2 + H2O ⇌ NaOH + HC2H3O2

The equilibrium constant for this reaction is called the base dissociation constant (Kb) and is given by:

Kb = [NaOH] [HC2H3O2] / [NaC2H3O2]

We can use the relationship:

Kw = Ka x Kb

Where Kw is the ion product constant for water, which is 1.0 x 10^-14 at 25°C, and Ka is the acid dissociation constant for HC2H3O2, which is 1.8 x 10^-5 at 25°C.

Rearranging the equation, we get:

Kb = Kw / Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.6 x 10^-10

Next, we need to calculate the concentration of HC2H3O2 and NaOH that are present in the solution after hydrolysis. Since NaC2H3O2 is a strong electrolyte,

it will completely dissociate in water to form Na+ and C2H3O2- ions. Therefore, the concentration of Na+ ions will be equal to the concentration of NaC2H3O2, which is 0.375 M.

The concentration of OH- ions can be calculated from the Kb expression:

Kb = [OH-]^2 / [HC_2H_3O_2]

[OH-]^2 = Kb x [[tex]HC_2H_3O_2[/tex]] = 5.6 x 10^-10 x 0.015 M = 8.4 x 10^-12

[OH-] = 2.9 x 10^-6 M

The pH of the solution can be calculated from the relationship:

pH + pOH = 14

pOH = -log [OH-] = -log (2.9 x 10^-6) = 5.54

pH = 14 - pOH = 14 - 5.54 = 8.46

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__________________ is the application of pulling force to hold a bone in alignment.

Answers

Answer:

Traction

Explanation:

Traction is a set of mechanisms for straightening broken bones or relieving pressure on the spine and skeletal system

A balloon is rubbed against a wall. The picture on the left shows the balloon and the wall before rubbing. The picture on the right shows the balloon and the wall after rubbing.

What happened when the balloon was rubbed against the wall? (5.b)


2. A balloon is rubbed against a wall. The picture on the left shows the balloon and the wall before rubbing. The picture on the right shows the balloon and the wall after rubbing.




What happened when the balloon was rubbed against the wall?


A. Electrons were transferred from the wall to the balloon.

B. Protons were transferred from the wall to the balloon.

C. Electrons were transferred from the balloon to the wall.

D. Protons were transferred from the balloon to the wall.

Answers

Answer: The answer should be A

Explanation:

each of the following can act as both an brönsted acid and a brönsted base except:
(A) HCO3
(B) NH4+
(C) HS
(D) H2PO4

Answers

The answer is (C) HS.

Each of the other options can donate a proton (act as a Brönsted acid) in certain conditions and accept a proton (act as a Brönsted base) in other conditions. However, HS is only capable of acting as a Brönsted base and accepting a proton, but it cannot donate a proton and act as a Brönsted acid.

Out of the given options, the one that cannot act as both an acid and a base is (C) HS. This is because HS can only act as a brönsted acid by donating a proton to a brönsted base, but it cannot act as a brönsted base by accepting a proton from a brönsted acid. This is because it lacks a lone pair of electrons on the sulfur atom, which is necessary for accepting a proton.

On the other hand, [tex]HCO_{3}[/tex] ,[tex]NH_{4}[/tex]+, and [tex]H_{2}[/tex][tex]O_{4}[/tex]P can all act as both brönsted acids and bases depending on the reaction conditions.

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(B) NH4⁺,  cannot act as both a Brønsted acid and a Brønsted base.

What is Bronsted Acid-Base pairs?


A Brønsted acid is a species that can donate a proton (H⁺), while a Brønsted base is a species that can accept a proton (H⁺).

(A) HCO3⁻ can act as an acid by donating a proton to form CO3²⁻ or as a base by accepting a proton to form [tex]H_{2}CO_{3}[/tex].
(C) HS⁻ can act as an acid by donating a proton to form S²⁻ or as a base by accepting a proton to form [tex]H_{2}S[/tex].
(D) H2PO4⁻ can act as an acid by donating a proton to form HPO4²⁻ or as a base by accepting a proton to form [tex]H_{3}PO_{4}[/tex].

However,
(B) NH4⁺ can only act as a Brønsted acid by donating a proton to form [tex]NH_{3}[/tex] but cannot act as a Brønsted base since it has no lone pair of electrons to accept a proton.

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How many moles are in 670 g of gold (|||) chloride

Answers

There are 2.208 moles of gold (III) chloride in 670 g.

To determine the number of moles in 670 g of gold (III) chloride, we need to first calculate the molar mass of gold (III) chloride, which is AuCl3.

The atomic mass of gold is 196.97 g/mol and the atomic mass of chlorine is 35.45 g/mol. Since there are three chlorine atoms in each molecule of gold (III) chloride, we multiply the atomic mass of chlorine by 3:

35.45 g/mol x 3 = 106.35 g/mol

Adding the atomic masses of gold and chlorine together gives us the molar mass of gold (III) chloride:

196.97 g/mol + 106.35 g/mol = 303.32 g/mol

Now, we can use this molar mass to convert 670 g of gold (III) chloride into moles:

670 g / 303.32 g/mol = 2.208 moles

Therefore, there are 2.208 moles of gold (III) chloride in 670 g.


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If I have an unknown quantity of gas at a pressure of 1.35 atm, a volume of 25 liters, and a temperature of 300. K, how many moles of gas do I have?

Answers

Answer:

We can use the ideal gas law to solve for the number of moles of gas:

PV = nRT

where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin.

Plugging in the given values:

(1.35 atm)(25 L) = n(0.0821 L·atm/mol·K)(300 K)

n = (1.35 atm)(25 L) / (0.0821 L·atm/mol·K)(300 K)

n = 1.29 mol

Therefore, there are 1.29 moles of gas in the container.

Help what's the answers?

Answers

The number of moles of bromine trifluoride needed to produce 23.2 L of fluorine gas according to the reaction would be 0.339 moles.

Stoichiometric problems

The balanced equation for the reaction is:

BrF3 → Br + 3F2

From the equation, we can see that 1 mole of BrF3 produces 3 moles of F2. Therefore, to calculate the number of moles of BrF3 needed to produce 23.2 L of F2 at 0°C and 1 atm, we need to use the ideal gas law:

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.

We can rearrange the ideal gas law to solve for n:

n = PV/RT

At 0°C (273 K) and 1 atm, the value of R is 0.08206 L·atm/mol·K. Substituting the values given, we get:

n = (1 atm) × (23.2 L) / (0.08206 L·atm/mol·K × 273 K)

n = 1.017 mol F2

Since 1 mole of BrF3 produces 3 moles of F2, we need 1/3 as many moles of BrF3:

n(BrF3) = 1.017 mol F2 × (1 mol BrF3 / 3 mol F2)

n(BrF3) = 0.339 mol BrF3

Therefore, 0.339 moles of BrF3 are needed to produce 23.2 L of F2 at 0°C and 1 atm.

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Question:
The Volume (V) of gas varies
directly as the temperature (T) and
inversely as the pressure (P). If the
volume is 225 cm³ when the
temperature is 300 K and the
pressure is 100 N/cm², what is the
volume when the temperature
drops to 270 K and the pressure is
150 N/cm²?

Answers

The volume of the gas when the temperature drops to 270 K and the pressure is 150 N/cm², is 135 cm³

How do I determine the volume of the gas?

The following data were obtained from the question.

Initial volume of gas (V₁) = 225 cm³Initial temperature of gas (T₁) = 300 KInitial pressure of gas (P₁) = 100 N/cm²New temperature (T₂) = 270 KNew pressure (P₂) = 150 N/cm²New volume of gas (V₂) = ?

The new volume of the gas can be obtained by using the combined gas equation as illustrated below:

P₁V₁ / T₁ = P₂V₂ / T₂

(100 × 225) / 300  = (150 × V₂) / 270

Cross multiply

300 × 150 × V₂ = 100 × 225 × 270

Divide both side by (300 × 150)

V₂ = (100 × 225 × 270) / (300 × 150)

V₂ = 135 cm³

Thus, the volume of the gas is 135 cm³

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How many molecules of carbon dioxide gas, CO2, are found in 0.125 moles

Answers

There are 7.52 x 10^22 molecules of carbon dioxide gas, CO2, in 0.125 moles.

        The number of molecules in a given number of moles can be calculated using Avogadro’s number, which is approximately 6.022 x 10^23. This number represents the number of particles (atoms or molecules) in one mole of a substance.

         To calculate the number of molecules in 0.125 moles of CO2, we can multiply the number of moles by Avogadro’s number: 0.125 moles x (6.022 x 10^23 molecules/mole) = 7.52 x 10^22 molecules.

         Avogadro’s number is a fundamental constant in chemistry and is used in many calculations involving moles and molar mass.  

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the most common constituent of gas in the disk of the milky way galaxy is ________.

Answers

The most common constituent of gas in the disk of the Milky Way galaxy is hydrogen gas.

Hydrogen gas is the most abundant element in the Milky Way galaxy, making up around 75% of its elemental mass. This is why hydrogen is often used as a tracer for studying the structure and dynamics of galaxies. The gas in the disk of the Milky Way is mostly composed of atomic hydrogen (H I) and molecular hydrogen (H2), with smaller amounts of other elements like helium and carbon. Studying the distribution and properties of this gas can provide insight into the formation and evolution of the Milky Way.

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The most common constituent of gas in the disk of the Milky Way galaxy is hydrogen gas.

Which gas is present in abundance in Milky Way?


The most common constituent of gas in the disk of the Milky Way galaxy is hydrogen. Hydrogen is the most abundant element in the universe and makes up the majority of the gas in the disk of the Milky Way galaxy, with its presence primarily in the form of atomic and molecular hydrogen.  It is often found in the form of molecular hydrogen ([tex]H_{2}[/tex]) in interstellar clouds, which are regions of gas and dust where stars are formed. Other common constituents of gas in the Milky Way galaxy's disk include helium (He), carbon (C), oxygen (O), nitrogen (N), and trace amounts of other elements.

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does the melting point obtained for your product indicate that your sample is indeed phenacetin? what additional evidence do you have that your product is phenacetin?

Answers

The melting point obtained for a product is an important indicator of its identity. The reported melting point of pure phenacetin is 133-136°C. If the melting point of the sample matches this range, then it is a good indication that the sample is indeed phenacetin.

Steps to find out if the product obtained is phenacetin:

Step 1: Measure the melting point of your sample using a melting point apparatus.

Step 2: Compare your obtained melting point with the known melting point of phenacetin (134-137°C).

Step 3: Assess if your sample's melting point is within the range of phenacetin's known melting point. If your sample's melting point falls within the range of 134-137°C, it could be an indication that your product is phenacetin.

However, the melting point alone cannot confirm the identity of the sample, as there may be other compounds with similar melting points. Additional evidence that can confirm the identity of the sample includes spectroscopic techniques such as IR or NMR spectroscopy, which can provide information about the chemical structure of the compound. Other tests such as chemical spot tests or thin-layer chromatography can also be used to confirm the identity of the compound.

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The melting point obtained for a product can provide an indication that the sample is indeed phenacetin, but it is not definitive proof.

Phenacetin has a melting point range of 134-137 °C, so if the melting point of the product falls within this range, it can suggest that the product is phenacetin. However, other compounds could have similar melting points, so further analysis is necessary to confirm the identity of the compound.

Additional evidence that the product is phenacetin can be obtained through techniques such as infrared spectroscopy, nuclear magnetic resonance (NMR) spectroscopy, or mass spectrometry (MS). These methods can provide information about the functional groups and molecular structure of the compound, allowing for comparison to known data for phenacetin. For example, infrared spectroscopy can show the presence of characteristic functional groups, such as the amide group in phenacetin. NMR spectroscopy can provide information about the number and arrangement of protons in the molecule, which can be compared to the known data for phenacetin. MS can also provide information about the molecular weight and fragmentation pattern of the compound, which can be compared to known data for phenacetin.

Overall, while the melting point can provide an initial indication of the identity of the compound, additional evidence from other analytical techniques is necessary to confirm the identity of phenacetin.

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What volume of chlorine gas at 46.0◦C and
1.60 atm is needed to react completely with
5.20 g of sodium to form NaCl?

Answers

The volume of chlorine gas at 46.0°C and 1.60 atm that is needed to react completely with 5.20 g of sodium to form NaCl is 1.85 L

How do i determine the volume of chlorine gas needed?

We'll begin by obtaining the mole of 5.20 g of sodium. Details below:

Mass of Na = 5.20 gMolar mass of Na = 23 g/mol Mole of Na =?

Mole = mass / molar mass

Mole of Na = 5.20 / 23

Mole of Na = 0.226 mole

Next, we shall determine the mole of chlorine gas needed. Details below:

2Na + Cl₂ -> 2NaCl

From the balanced equation above,

2 moles of Na reacted with 1 mole of Cl₂

Therefore,

0.226 mole of Na will react with = (0.226 × 1) / 2 = 0.113 mole of Cl₂

Finally, we shall determine the volume of chlorine gas, Cl₂ needed. This is shown below:

Temperature (T) = = 46 °C = 46 + 273 = 319 KPressure (P) = 1.60 atmGas constant (R) = 0.0821 atm.L/molKNumber of mole (n) = 0.113 moleVolume of chlorine gas, Cl₂ (V) =?

PV = nRT

1.6 × V = 0.113 × 0.0821 × 319

Divide both sides by 1.6

V = (0.113 × 0.0821 × 319) / 1.6

V = 1.85 L

Thus, the volume of chlorine gas, Cl₂ needed is 1.85 L

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why would it be necessary to slowly add the sulfuric acid to the p-cresol/acetic acid mixture in the test tube? simply to be sure the correct volumes are used. the reaction is exothermic which may boil and splatter the acidic solution out of the test tube. since the density of sulfuric acid is less than that for acetic acid, it requires a slower reaction time. the reaction is endothermic and the solution may solidify if the sulfuric acid is added too quickly.

Answers

The correct answer is option D. All of the above. It is necessary to slowly add the sulfuric acid to the p-cresol/acetic acid mixture in the test tube to prevent any accidents or injuries.

If sulfuric acid is added too soon, the solution may boil and the acid will spew out of the test tube, perhaps resulting in burns.

Sulfuric acid is also an endothermic reaction, which means it takes energy from its surroundings and has the potential to crystallise or cause the solution to harden.

Last but not least, adding the sulfuric acid gradually enables more precise measurement of the supplied sulfuric acid volume.

It is crucial to gradually add the sulfuric acid to the test tube mixture of p-cresol and acetic acid as a result of all these considerations.

Complete Question:

Why would it be necessary to slowly add the sulfuric acid to the p-cresol/acetic acid mixture in the test tube?

Options:

A.  To ensure accurate measurement of the volume of sulfuric acid added.

B. To prevent the solution from boiling and splattering the acidic solution out of the test tube.

C. To prevent the endothermic reaction from solidifying the solution.

D. All of the above.

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154.42g of oxygen gas (O2) react with an excess of ethane (C2H6) produces how many moles of water vapor (H2O)?

Answers

2.77 moles of water vapour (H2O) are created when 154.42 g of oxygen gas (O2) reacts with an excess of ethane (C2H6).

Calculation-

In order to create water vapour [tex](H_2O)[/tex], ethane [tex](C_2H_6)[/tex]and oxygen gas (O2) must be burned. The chemical equation for this reaction is:

[tex]C_2H_6 + 7O_2 -- > 4H_2O + 6CO_2[/tex]

We may deduce from the equation that when 1 mole of ethane (C2H6) interacts with 7 moles of oxygen gas (O2), 4 moles of water vapour (H2O) are created.

We must utilise its molar mass to translate the 154.42 g of oxygen gas (O2) consumed into moles. 32 g/mol (16 g/mol for each oxygen atom multiplied by two for O2) is the molar mass of oxygen gas.

Moles of oxygen gas (O2) = Mass of oxygen gas (O2) / Molar mass of oxygen gas (O2)

Moles of oxygen gas (O2) = 154.42 g / 32 g/mol

Moles of oxygen gas (O2) = 4.83 mol (rounded to two decimal places)

The balanced equation's stoichiometry predicts that 7 moles of oxygen gas [tex](O_2)[/tex]and 4 moles of water vapour [tex](H_2O)[/tex] will react. We can thus calculate the moles of water vapour [tex](H_2O)[/tex] created using the stoichiometric principle.

Moles of water vapor [tex](H_2O)[/tex] = Moles of oxygen gas [tex](O_2)[/tex] × (4 moles of [tex]H_2O[/tex] / 7 moles of O2)

Moles of water vapor [tex](H_2O)[/tex] = 4.83 mol × (4/7)

Moles of water vapour[tex](H_2O)[/tex] = 2.77 mol (rounded to two decimal places)

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2-thiosubstituted chlorocyclohexanes can undergo an sn2 reaction with intramolecular catalysis. which stereoisomer is the most reactive?

Answers

The axial stereoisomer is the most reactive in this type of reaction.

In an SN2 reaction with intramolecular catalysis, the most reactive stereoisomer is the one with an axial thioether group.

This is because in the axial position, the thioether group is closer to the leaving group (chlorine), allowing for more efficient overlap of their orbitals and a lower energy transition state.

The equatorial thioether group is farther away from the leaving group, resulting in a higher energy transition state and a slower reaction. Therefore, the axial stereoisomer is the most reactive in this type of reaction.

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the chemical composition of the interstellar medium is basically similar to that of (a) the Sun; (b) Earth; (e) Venus; (d) Mars

Answers

The chemical composition of the interstellar medium is not exactly the same as any of the listed options, but it is most similar to the composition of the Sun.



The interstellar medium is the matter that fills the space between stars in a galaxy, and it consists of gas (mostly hydrogen and helium) and dust particles. The gas in the interstellar medium is similar in composition to the gas in the Sun, with hydrogen being the most abundant element and helium being the second most abundant. Other elements are present in smaller amounts, but their relative abundances are similar to those in the Sun.

On the other hand, the chemical composition of Earth, Venus, and Mars is different from that of the interstellar medium and the Sun. These planets are composed of heavier elements, such as carbon, nitrogen, oxygen, and iron, which are not as abundant in the interstellar medium or the Sun. Additionally, the planets have undergone differentiation and have distinct layers with different compositions, while the interstellar medium is more homogeneous.

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