A 27g pancake (61 Calories) is burned in a calorimeter that contains 2.500 kg of water. How much does the temperature of the water increase if all of the energy from the pancake is transferred into the water?

Answer: S. Auresus

Explanation:I just took it

A microbiologist studies the microbes and the disease they cause. The bacterium that will be more responsive to antibiotic treatment will be Staphylococcus aureus.

Antibiotics are said to be the medication and drug class that works against bacterial infections. They are used to kill and slow the rate of growth or reproduction of microbes like bacteria to stop diseases or infections like flu, cold, cough, etc.

The bacteria can be resistant or susceptible to the antibiotic and can be tested by exposing them to antibiotics. As Staphylococcus aureus species are vulnerable they will not be able to grow under the antibiotic treatment. Whereas, Escherichia coli being resistant will show some growth.

Therefore, Staphylococcus aureus will be more responsive to antibiotics.

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Answer:

The answer to the question above is explained below

Explanation:

There are so many differences between the two ecosystem (terrestrial ecosystems and aquatic ecosystems). The most important ones will be highlighted:

1. Aquatic environments are so rich in nutrients they support more live than equivalent terrestrial ecosystems as aquatic ecosystems have availability of water more than the terrestrial ecosystems. Presenting the consequent importance of water as a limiting factor in the terrestrial ecosystems.

2. Aquatic ecosystems are much more stable than terrestrial ecosystems, with smaller fluctuations in temperature and other variables.

3. In terrestrial ecosystems, there is hardly ever a shortage of light, while it can be a limiting factor in some aquatic ecosystems.

4. Gravity has so much influence on terrestrial animals. That is not the case for their aquatic counterparts where water supports aquatic organisms.

An ecosystem is a large community of living organisms (plants, animals and microbes) in a particular area in conjunction with the non-living components of their environment, interacting as a system.

Terrestrial Ecosystem includes: Grasslands, Forests, Desert.

Aquatic Ecosystem includes: Freshwater ecosystem, Marine Ecosystem.

Answer:

He added 992 milimoles of zinc nitrate (Zn(NO3)2)

Explanation:

Step 1: Data given

Volume of zinc nitrate = 435.0 mL =0.435 L

Molarity of zinc nitrate = 2.28 M

Step 2: Calculate moles zinc nitrate

Moles = molarity*  volume

Moles Zn(NO3)2 = 2.28 M * 0.435 L

Moles Zn(NO3)2 = 0.9918 moles

Step 3: Convert moles to milimoles

Moles Zn(NO3)2 = 0.9918 moles

Moles Zn(NO3)2 = 0.9918 * 10^3 milimoles

Moles Zn(NO3)2 = 991.8 milimoles ≈ 992 milimoles

He added 992 milimoles of zinc nitrate (Zn(NO3)2)

Final answer:

To determine the millimoles of zinc nitrate added, convert 435.0 mL to liters, use the molarity of 2.28 M to find moles, and then multiply by 1000 to get 992.4 millimoles.

Explanation:

To calculate the millimoles of zinc nitrate a chemist has added to the reaction flask, you first need to know the concentration of the solution and the volume of the solution used. Here, we have a 2.28 M zinc nitrate solution, and the volume used is 435.0 mL. To find the millimoles, first convert the volume from milliliters to liters (since molarity is moles per liter), then use the molarity to find the moles of zinc nitrate, and finally convert that to millimoles.

Convert volume to liters: 435.0 mL × (1 L / 1000 mL) = 0.435 L

Calculate moles of Zn(NO3)2: 2.28 moles/L × 0.435 L = 0.9924 moles

Convert moles to millimoles: 0.9924 moles × 1000 mmol/mole = 992.4 mmol

Therefore, the chemist has added 992.4 millimoles of zinc nitrate to the flask.

Answer:

the concentration of sulfuric acid in the flask is  0.375 M

Explanation:

H2SO4 (aq) + 2 NaOH (aq) → 2 H2O (l) + Na2SO4 (aq)  

Moles of NaOH = 15 x 0.25 /1000

                        = 0.00375

Moles of H2SO4 needed to neutralize = 0.00375 /2

                       = 0.001875

Molarity of H2SO4

                      = 0.001875 x 1000 /20

                      = 0.09375 M

concentration of sulfuric acid in the flask 0.09375 M

Moles

V2 = 20 ml

N2 =?

Substitute in the equation and find N2

N2 = 2 x 15 x 0.25 / 20

     = 0.375 M

Thus, the concentration of sulfuric acid in the flask is  0.375 M

Answer:

The correct name for the following compound is

2 - methylbicyclo[4.3.0]nonane

Explanation:

Rule 1

Numbering of bicyclic compound is done from bridgehead carbon

Rule 2

The system is numbered with one of the bridgeheads, numbering proceeding by the longest possible path to the second bridgehead.

Name of the given bicyclic compound is 2 - methylbicyclo[4.3.0]nonane

Answer:

the reported molar concentration of the sodium hydroxide solution will be too high.

Explanation:

The term "standardization" in science or in analytical Chemistry simply means the process involved in the determination of a standard or the concentration of a particular substance.

One of the advantages and the most important of the advantages of Standardization is that it Helps in making sure that we get a result with the least error.

So, let me explain the answer. The reported molar concentration of the sodium hydroxide solution will be TOO HIGH because the potassium hydrogen phthalate is NOT completely dry. The weight of the moisture will create an additional weight which will increase the weight of potassium hydrogen phthalate in the solution

If potassium hydrogen phthalate is not completely dry, it will make the reported molar concentration of the sodium hydroxide solution too low. This is due to the added water from the moisture which causes a dilution effect.

In the situation where the potassium hydrogen phthalate is not completely dry, the reported molar concentration of the sodium hydroxide solution will be too low. This is because the water from the phthalate's moisture can dilute the sodium hydroxide solution, which leads to a lower than actual molar concentration. It's essential to have a completely dry potassium hydrogen phthalate when standardizing sodium hydroxide solutions to get an accurate molar concentration.

For instance, when potassium hydroxide, a highly soluble ionic compound, is dissolved in a dilute solution, it completely dissociates, giving a certain concentration of [OH-] ions. Suppose this reaction occurs in an environment where excess water is introduced due to the moisture in potassium hydrogen phthalate. In that case, this equilibrium will be affected, leading to a lower concentration of hydroxide ions and thus, a lower reported molar concentration of the sodium hydroxide solution.

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Answer:

see below

Explanation:

The rate constant is missing in question, but use C(final) = C(initial)e^-kt = 0.200M(e^-k·10). Fill in k and compute => remaining concentration of reactant

Answer:

C(initial)e^-kt = 0.200M(e^-k·10).

Explanation:

The rate constant is missing in question, but use C(final) = C(initial)e^-kt = 0.200M(e^-k·10).

Fill in k and compute => remaining concentration of reactant

Answer:

Answer: (1R,2S) / (1S, 2R) , (1R,2R) / (1S, 2S)  

Explanation:

Sodium borohydride reduction of benzoin will give four possible stereo isomers out of which are (1R,2S) - (1S, 2R) isomers and (1R,2R) - (1S, 2S) isomers which are known as enantiomers.

In general enantiomers show single spot in the TLC as they do not show any difference in Rf value (i.e) (1R,2S) - (1S, 2R) isomers show only one spot although they are two compounds and also (1R,2R) - (1S, 2S) isomers also show one spot. That is the reason why you are observing two spots in the TLC ( of reaction mixture) other than starting materilal.

Answer:

Explanation:

find the solution below

Answer:

D)

Explanation:

I just answered the question and got it right.

To find the final molarity of bromide ions in a solution, we need the mass of the dissolved potassium bromide and the volume of the solution. Then we do a simple molar calculation. The molarity of KBr and Br- are equivalent as KBr disassociates fully in solution.

In order to determine the final molarity of the bromide anion, we first need to know the amount of potassium bromide (KBr) dissolved and the volume of the solution it's dissolved in. However, these key data are missing in the question.

Regardless, the steps to calculate molarity (M) are as follows:

In this particular scenario, we also know that one mole of KBr results in one mole of bromide ions (Br-), as it disassociates fully upon dissolution. So, the molarity of KBr would be equal to the molarity of Br-.

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Answer:

c. Z is a catalyst, and ZA2 and A are reaction intermediates.

Explanation:

Overall reaction is given as;

A2 + 2 B → 2 BA

Mechanism:

Step 1: A2 + Z → ZA2

Step 2: ZA2 + B → BA + Z + A

Step 3: A + B → BA

The options given are centered upon catalysts and reaction intermediates. So before proceeding, we have to understand the difference between the two and how to identify them.

A catalyst is basically a reaction booster to speed up the rate of the reaction and the reaction intermediate is an unstable, temporl species formed from the reactants before getting to the products.

The difference is given as;

Catalysts are present as reactants in the very beginning and products at the end of the reaction.

Intermediates, on the other hand, are not present in the initial reaction but are produced within one of the steps and then consumed within another step.

Following the above, we can deduce that;

Z is a catalyst because it is present as a reactant in the beginning.

ZA2 and A are a reaction intermediates because they are not present in the overall reaction but are produced and consumed in one of the steps.

Correct option is given as;

c. Z is a catalyst, and ZA2 and A are reaction intermediates.

Answer:

2) ΔS°rxn is positive, and the process is spontaneous at high temperatures only.

Explanation:

ΔS = ΔH / T , ΔS is change in entropy , ΔH is change in enthalpy

Since ΔH is positive , ΔS is positive .

ΔG = ΔH - TΔS

For spontaneous reaction . ΔG should be negative .

As ΔS is positive , at high temperature the value of TΔS will be more and hence the value of TΔS will be higher than Δ H . Hence ΔG will be negative.

Hence at higher temperature , the reaction will be spontaneous.

The predicted sign of ΔS°rxn and the conditions under which this reaction would be spontaneous should be option  2) ΔS°rxn is positive, and the process is spontaneous at high temperatures only.

We know that

ΔS = ΔH / T

Here ΔS means a change in entropy

And, ΔH means a change in enthalpy

Since ΔH is positive so  ΔS is positive.

Now

ΔG = ΔH - TΔS

However, For spontaneous reaction. ΔG should be negative .

Since ΔS is positive, at high temperature the value of TΔS should be more and due to this the value of TΔS will be more than Δ H . Thus,  ΔG will be negative.

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The molar mass or gram formula mass is the conversion factor used to convert between grams and moles in chemistry.

The conversion factor used to convert between grams and moles in chemistry is called the molar mass or the gram formula mass. It is defined as the mass of one mole of a substance. To convert grams to moles, you divide the given mass by the molar mass of the substance. To convert moles to grams, you multiply the given moles by the molar mass.

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Answer:Neptunium; Protactinium; Thorium; Uranium

Explanation:

Answer:

a. 0.8125 moles of sucrose

b. 277.8 g of sucrose

Explanation:

Consider these relation's value:

Molarity = Mol / Volume(L)

Mol = Molarity . Volume(L)

Volume(L) = Mol / Molarity

So, Molarity = 3.25M

Volume(L) = 0.25L

Molarity . Volume(L) = Mol → 3.25mol/L . 0.25L = 0.8125 mol

Let's convert the moles to mass → 0.8125 mol . 342 g /1mol = 277.8 g of sucrose

Answer:

We have 0.8125 moles sucrose in this solution. This is 277.9 grams of sucrose

Explanation:

Step 1: Data given

Molarity of a sucose solution = 3.25 M

Molar mass of sucrose = 342 g/mol

Step 2:  Calculate moles sucrose

Moles sucrose = molarity sucrose solution * volume solution

Moles sucrose = 3.25 M * 0.25 L

Moles sucrose =  0.8125 moles sucrose

Step 3: Calculate mass of sucrose

Mass sucrose = moles sucrose * molar mass sucrose

Mass sucrose = 0.8125 moles * 342 g/mol

Mass sucrose = 277.9 grams

We have 0.8125 moles sucrose in this solution. This is 277.9 grams of sucrose

Answer:

Low molecular weight carboxylic acids are volatile.

One should use caution when handling low molecular weight carboxylic acids.

Final answer:

Low molecular weight carboxylic acids are volatile and can have strong odors. They are not completely safe for handling without protection due to possible irritation from the vapors. Therefore, caution and appropriate safety measures are necessary when handling these substances.

Explanation:

Low molecular weight carboxylic acids tend to be volatile, implying they can easily vaporize at room temperature, and often have strong, sharp odors. One commonly known example is ethanoic acid (acetic acid), which is found in household vinegar. While they are usually colorless and can be found in everyday products such as foods and household items, it is not accurate to say that they are completely safe and can be handled without protection. The vapors of low molecular weight carboxylic acids can be irritating and, in some cases, harmful if inhaled in large amounts. Therefore, caution should be used when handling these substances.

Carboxylic acids are weak acids that do not completely ionize in water. This characteristic makes them less dangerous than strong acids; however, they can still cause irritation and should be treated with respect in a laboratory setting. Carboxylic acids such as propionic acid, while not highly toxic, can still pose safety risks if not handled properly, reinforcing the need for appropriate safety measures such as gloves and eye protection when handling these compounds.

Answer:

The value of Ka [tex]= 1.1*10^{-2}[/tex]

It is a weak  acid

Explanation:

   From the question we are told that

             The concentration of [tex][HClO_2]=0.24M[/tex]

             The concentration of  [tex][H^+]=0.051M[/tex]

             The concentration of  [tex][ClO_2^-]=0.051M[/tex]

Generally the equation for the ionic dissociation of [tex]HClO_2[/tex] is

                [tex]HClO_2_(aq) -------> H^{+}_{(aq)} + ClO_2^{-}_{(aq)}[/tex]

The equilibrium constant is mathematically represented as

                         [tex]Ka = \frac{concentration \ of \ product }{concentration \ of \ reactant }[/tex]

                               [tex]= \frac{[H^+][ClO_2^-]}{[HClO_2]}[/tex]

Substituting values since all value of concentration are at equilibrium

                    [tex]Ka = \frac{0.051 * 0.051}{0.24}[/tex]

                          [tex]= 1.1*10^{-2}[/tex]

Since the value of  is less than 1 it show that in water it dose not completely

disassociated  so it an acid that is weak

Answer:

A mixture is made when two or more substances are combined, but they are not combined chemically. A chemical reaction is a transformation from one set of chemicals into another set.

Explanation:

A mixture can be homogeneous or heterogeneous, when homo, you cannot see each variable but hey are there. Hetero, means you can see each variable in the mixture. It is considered a mixture and not a chemical reaction because it can be reversed. It will always be able to go back to its separate forms and keep its original composition. When there is a chemical reaction, it is hard to separate and go back to the original variables because there is a molecular bond.

The given question is incomplete. The complete question is:

Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H2O). If 1.60 g of sodium chloride is produced from the reaction of 1.8 g of hydrochloric acid and 1.4 g of sodium hydroxide, calculate the percent yield of sodium chloride. Be sure your answer has the correct number of significant digits in it.

Answer: Thus the percent yield of sodium chloride is 78.0%

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} HCl=\frac{1.8g}{36.5g/mol}=0.049moles[/tex]

[tex]\text{Moles of} NaOH=\frac{1.4g}{40g/mol}=0.035moles[/tex]

[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]

According to stoichiometry :

1 mole of [tex]NaOH[/tex] require = 1 mole of [tex]HCl[/tex]

Thus 0.035 moles of [tex]NaOH[/tex] will require=[tex]\frac{1}{1}\times 0.035=0.035moles[/tex] of [tex]HCl[/tex]

Thus [tex]NaOH[/tex] is the limiting reagent as it limits the formation of product and [tex]HCl[/tex] is the excess reagent.

As 1 mole of [tex]NaOH[/tex] give = 1 mole of [tex]NaCl[/tex]

Thus 0.035 moles of [tex]NaOH[/tex] give =[tex]\frac{1}{1}\times 0.035=0.035moles[/tex]  of [tex]NaCl[/tex]

Mass of [tex]NaCl=moles\times {\text {Molar mass}}=0.035moles\times 58.5g/mol=2.05g[/tex]

[tex]{\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%[/tex]

[tex]{\text {percentage yield}}=\frac{1.60g}{2.05g}\times 100\%=78.0\%[/tex]

Thus the percent yield of sodium chloride is 78.0%

Explanation:

1. When sodium phosphate is added to copper (II) sulfate it gives solid precipitate of copper(II) phosphate and solution of sodium sulfate.

The balanced equation is given as:

[tex]2Na_3PO_4(aq)+3CuSO_4(aq)\rightarrow Cu_3(PO_4)_2(s)+3Na_2SO_4(aq)[/tex]

2. When sodium phosphate is added to iron(III) chloride it gives solid precipitate of iron (III) phosphate and solution of sodium chloride.

The balanced equation is given as:

[tex]Na_3PO_4(aq)+Fe(Cl)_3(aq)\rightarrow FePO_4(s)+3NaCl(aq)[/tex]

Answer:

Order of increasing ionic radius starting from the smallest is

Mg2+ < Na+ < F- < O2- < N3-

Explanation:

Ionic radius is the distance between the nucleus and the electrons in the outermost shell of an ion. In the ions given about, it can be deducted that they all have the same number of electrons and are said to be isoelectronic. This shows us that they all have the same electrons and we can only seperate them by checking the respective charges on the ions. A more positive charge will have a smaller radius while a more negative will have a larger radius. This is because when an atom loses an electron and form a cation (+), the remaining electrons move closer to the nucleus and hence a smaller radius of the atom occurs. But when an atom gains n electron,it forms anion (-), the new electron leads to the shielding of the remaining electrons from the nucleus and a large radius of the atom occurs.

So therefore, the above ions all have 10 electrons in their shells and magnesium with atomic number 12 and having lost 2 electrons to become positively charge has the smallest radius compared to others.

Mg2+ = 12 protons, 10 electrons

Na+ = 11 protons, 10 electrons

F- = 9 protons, 10 electrons

O2- = 8 protons, 10 electrons

N3- = 7 protons, 10 electrons

N3- has the largest ionic radius and Mg2+ has the lowest ionic radius.

The order is thus, Mg2+ , Na+, F-, O2-, N3-

The order of increasing ionic radius is N3-, O2-, F-, Na, Mg2+

The ionic radius is a measure of the size of an ion. In general, ions with more electrons have larger radii. Let's arrange the ions in order of increasing ionic radius:

The larger the negative charge on an ion, the larger its ionic radius. Therefore, N3- has the smallest radius, followed by O2-, F-, Na, and finally Mg2+, which has the largest radius among the given ions.

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Answer:

d. I and II

Explanation:

In chemistry, it is an old analytical technique to determine the amount of free chloride ions by silver chloride precipitation. This precipitation method could both be used to determine the initial chloride ion concentration in different analyte solutions or to determine the number of uncoordinated chloride ions in a complex. The number of moles of chloride ions in the silver chloride, shows the number of uncoordinated ions in a chloride ion containing complex.

The concentration of chloride ions in an analyte could be gravimetrically determined by silver chloride precipitation.

The Ksp of the solution is [tex]1.607*10^-^1^4[/tex]

Data;

This is the point or temperature in which a solute is completely soluble in a solvent.

The solubility of iron(ii) hydroxide can be calculated by using the equation of reaction.

[tex]Fe(OH)_2 \to Fe^2^+ + 2OH^-\\[/tex]

But the solubility of the solution is given as 1.59*10^-5 mol/L

[tex][Fe^2^+] = s\\\\[/tex]

[tex][OH^-] = 2s\\K_s_p = s * (2s)^2 = 4s^3[/tex]

Let's substitute the values and solve

[tex]K_s_p = 4 *( 1.59*10^-^5)^3 = 1.607*10^-^1^4[/tex]

The Ksp of the solution is [tex]1.607*10^-^1^4[/tex]

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Final answer:

The Ksp of iron(II) hydroxide at 25 °C is calculated to be 1.61 x 10−14, based on its given solubility and the stoichiometry of its dissociation in water.

Explanation:

The solubility product constant, Ksp, for iron(II) hydroxide, Fe(OH)2, at 25 °C can be calculated using the mol/L solubility given and the stoichiometry of the dissolution reaction:

Fe(OH)2 (s) → Fe2+ (aq) + 2OH− (aq)

For iron(II) hydroxide, the solubility is 1.59 x 10−5 mol/L. This is the concentration of Fe2+. Since the ratio of Fe2+ to OH− is 1:2, this means the concentration of OH− is twice this value, thus 3.18 x 10−5 mol/L. The Ksp is the product of these ion concentrations:

Ksp = [Fe2+][OH−]2 = (1.59 x 10−5)(3.18 x 10−5)2

Calculating the above expression:

Ksp = 1.59 x 10−5 x (1.01 x 10−9)

Ksp = 1.61 x 10−14

This is the Ksp of iron(II) hydroxide at 25 °C, expressed in scientific notation to two significant figures.

Answer:

Combustion

Explanation:

Combustion reactions are the one where the oxygen is a reactant and the products are always water and carbon dioxide.

It is also a redox type because the oxygen is reduced and the carbon from the C₄H₁₀O oxidized to CO₂

C₄H₁₀O + 6O₂ → 4CO₂ + 5H₂O

1 mol of butyl alcohol burns in prescence of 6 moles of oxygen in order to produce 4 moles of carbon dioxide and 5 moles of water.

Answer:

This is a combustion reaction

Explanation:

Step 1: Data given

A single-replacement reaction is a reaction where we will replace one element with a similar element in the compound.

A single-replacement or displacement reaction has the following form:

A+BC→AC+B

Element  A  is usually a metal and will replace element  B in the compound , this should also be a metal.

If the elemnent isn't a metal ( so a nonmetal), the replacing element will either be a metal.

In a double-replacement reaction we will exchange positive and negative ions of two ionic compounds, what will cause the formation of two new compounds.

A double-replacement reaction has the following form:

AB+CD→AD+BC

IIn this example,  A  and  C  are positively-charged ions (cations), and   B  and  D  are negatively-charged ions (anions).

In a decomposition reaction a compound will break down into two or more substances ( those are smaller, more simple).

A decomposition reaction uses the following form.

AB→A+B

In a combustion reaction a substance will react with oxygen gas (O2). Oxygen gas is needed to start the reaction.CO2 gas and water vapor (H2O) will be produced.

In a combination reaction we will combine two or more substances to form a single new substance. This is also known as a synthesis reaction, and uses the general form:

A+B→AB

Step 2: The equation

C4H10(g) + 6O2(g) → 4CO2(g) + 5H2O(g)

Step 3: What kind of reaction is this

Since the reaction requires oxygen and will produce carbon dioxie and water vapor. This means this reaction is a combustion reaction.

(Combustion of butane).

Answer:

the pH of HCOOH solution is 2.33

Explanation:

The ionization equation for the given acid is written as:

[tex]HCOOH\leftrightarrow H^++HCOO^-[/tex]

Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

and the equilibrium concentration for each of the product would be x

Equilibrium expression for the above equation would be:

[tex]\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}[/tex]

[tex]1.8*10^-^4=\frac{x^2}{c-x}[/tex]

From given info, equilibrium concentration of the acid is 0.12

So, (c-x) = 0.12

hence,

[tex]1.8*10^-^4=\frac{x^2}{0.12}[/tex]

Let's solve this for x. Multiply both sides by 0.12

[tex]2.16*10^-^5=x^2[/tex]

taking square root to both sides:

[tex]x=0.00465[/tex]

Now, we have got the concentration of [tex][H^+] .[/tex]

[tex][H^+] = 0.00465 M[/tex]

We know that, [tex]pH=-log[H^+][/tex]

pH = -log(0.00465)

pH = 2.33

Hence, the pH of HCOOH solution is 2.33.

Answer:

The correct answer is 2.34

Explanation:

HCOOH is formic acid. It is a weak acid so it does not dissociates completely in water. At the beggining (I) the initial concentration is 0.12 M. In water it will dissociate in a certain grade x as follows:

          HCOOH    →    H⁺ + HCOO⁻

I             0.12 M           0          0                    

C           - x                   x          x

E          (0.12 M - x)       x          x

The mathematical expression for the equilibrium constant (Ka) is the following:

[tex]K_{a} = \frac{[H^{+} ][HCOO^{-} ]}{[HCOOH]}[/tex]

[tex]1.8 x 10^{-4} = \frac{(x x)}{(0.12 M -x)}[/tex]

As the value of Ka is too small in comparison with the initial concentration 0.12 M, we can approximate: 0.12 M - X ≅ 0.12 M. Then, we calculate x:

1.8 x 10⁻⁴ = x²/0.12 M

⇒ x= [tex]\sqrt{0.12 x 1.8 x 10^{-4} }[/tex]= 4.65 x 10⁻³

Since x = 4.65 x 10⁻³ , from the equilibrium we have:

[H⁺] = x = 4.65 x 10⁻³

From the definition of pH, we have:

pH = -log [H⁺] = -log (4.65 x 10⁻³)= 2.34

Answer:

see below

Explanation:

for A + 2B => Products ...

Rate Law => Rate =k[A][B]ˣ

As shown in expression, A & B are included, C is not.

Answer:

Exponent of a is 1, exponent of b is 2 and exponent of c = 0

Explanation:

or the rate equation is the expression which relates the rate of the reaction with the concentration of pressure of the reactants. The rate law is expressed in terms of the molar concentration of the reactants with each term raised to power its stoichiometric coefficient.

The new volume of the Cl2 gas under the conditions at the bottom of Hellas Planitia on Mars would be approximately 4.2 litres, as calculated using the ideal gas law.

The question is asking about the adjustment of gas volume under different conditions of pressure and temperature, so the ideal gas law is applicable here which states that: PV = nRT, where P is pressure, V is volume, T is temperature, n is number of moles of the gas and R is gas constant.

At Standard Temperature and Pressure (STP), the conditions are 1 atm pressure and 0°C (273 K). Please note that the volume given is already at STP, so to find the new volume, we should rearrange the ideal gas law to V2 = V1 * P1/P2 * T2/T1.

However, note that all the measurements need to be in the same standard units. Here, the initial pressure (P1) is 1 atm, converted to kPa becomes approximately 101.3 kPa, the final pressure (P2) is given as 1.16 kPa. The initial temperature (T1) is 273K and the final temperature (T2) needs to be converted from Celsius to Kelvin, making it 268 K (-5°C + 273). Thus, applying the equation, we get:

V2 = 0.175L * 101.3 kPa / 1.16 kPa * 268 K / 273 K = 4.2 L approximately

Therefore, the new volume of the Cl2 gas, if it is transported to the conditions at the bottom of Hellas Planitia on Mars, would be approximately 4.2 litres.

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Using the combined gas law, the new volume of Cl₂ gas at 1.16 kPa pressure and -5.0°C temperature at the bottom of Hellas Planitia on Mars is calculated to be approximately 15.06 liters.

To determine the new volume of Cl₂ gas at the conditions provided, we need to use the combined gas law which is given by the equation: [tex]\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}[/tex], where P is pressure, V is volume and T is temperature in Kelvin.

At STP (Standard Temperature and Pressure), we have a volume of 175 mL (or 0.175 liters), a pressure of 101.3 kPa, and a temperature of 273.15 K. The conditions at the bottom of the Hellas Planitia are 1.16 kPa and -5.0°C or 268.15 K.

First, we convert the initial volume into liters: 175 mL = 0.175 L.
Next, we rearrange the combined gas law to solve for V₂:

Therefore, the new volume of Cl₂ gas at the bottom of Hellas Planitia will be approximately 15.06 liters.

Final answer:

In a galvanic cell, half-reactions occur at separate electrodes: reduction at the cathode and oxidation at the anode. The cathode is the positive electrode, while the anode is the negative electrode. Standard cell potential can be calculated but requires the standard reduction potentials for the specific half-reactions.

Explanation:

Galvanic Cell Half-Reactions, Electrodes, and Potentials

In a galvanic cell, a spontaneous redox reaction occurs that drives the flow of electrons from the anode to the cathode through an external circuit. The original question does not provide a complete redox reaction, but assuming it involves bromine (Br2), hydrogen (H2), and hydroxide ions (OH-), the following can be considered:

For the provided redox reaction, the half-reaction that occurs at the cathode (reduction) might be:

2H+ + 2e- → H2

And the half-reaction at the anode (oxidation) could be:

2Br- → Br2 + 2e-

The cathode is the positive electrode, as it undergoes reduction, and the anode is the negative electrode, as it undergoes oxidation. Without actual potential values given, the cell voltage under standard conditions cannot be calculated here. However, the standard cell potential can be found by using standard reduction potential tables and subtracting the anode potential from the cathode potential.