To prepare a solution of BF4-(aq), HF(g) is bubbled into a solution containing 50.0 g of H3BO3 in a 1 L reaction vessel. Calculate the maximum number of moles of BF4-(aq) that can be produced.

Answer: All of them are right there’s no wrong answer

Explanation:

Masses for the metal and the water in the calorimeter, temperature changes for the water and the metal, and the known specific heat of the water.

First, the masses of both the metal and the water within the calorimeter are crucial, as they determine the amount of substance undergoing the temperature change. Second, the temperature changes (Tw, final​ −Tw, initial​  and T metal, final −T metal, initial​ ) are necessary to quantify the heat exchange during the experiment. Lastly, the known specific heat of water (C water​ ) plays a role in the calculation, as it is a fundamental constant representing the amount of heat needed to raise the temperature of water.

The first law of thermodynamics, which states that the heat lost by the metal equals the heat gained by the water, and the ability of heat to flow from a hot object to a cooler one are overarching principles guiding the experimental setup and the interpretation of results. However, these principles are not directly inputted into the formula but are fundamental concepts in calorimetry and thermodynamics that inform the experimental design and the interpretation of the calculated specific heat values.

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Answer : The limiting reagent in this reaction is, [tex]Al_2(SO_4)_3[/tex] and number of moles of excess reagent is, 1.69 moles

Explanation : Given,

Mass of [tex]Al_2(SO_4)_3[/tex] = 500.0 g

Mass of [tex]Ca(OH)_2[/tex] = 450.0 g

Molar mass of [tex]Al_2(SO_4)_3[/tex] = 342.15 g/mol

Molar mass of [tex]Ca(OH)_2[/tex] = 74.1 g/mol

First we have to calculate the moles of [tex]Al_2(SO_4)_3[/tex] and [tex]Ca(OH)_2[/tex].

[tex]\text{Moles of }Al_2(SO_4)_3=\frac{\text{Given mass }Al_2(SO_4)_3}{\text{Molar mass }Al_2(SO_4)_3}[/tex]

[tex]\text{Moles of }Al_2(SO_4)_3=\frac{500.0g}{342.15g/mol}=1.461mol[/tex]

and,

[tex]\text{Moles of }Ca(OH)_2=\frac{\text{Given mass }Ca(OH)_2}{\text{Molar mass }Ca(OH)_2}[/tex]

[tex]\text{Moles of }Ca(OH)_2=\frac{450.0g}{74.1g/mol}=6.073mol[/tex]

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

[tex]Al_2(SO_4)_3(aq)+3Ca(OH)_2(aq)\rightarrow 2Al(OH)_3(s)+3CaSO_4(s)[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]Al_2(SO_4)_3[/tex] react with 3 mole of [tex]Ca(OH)_2[/tex]

So, 1.461 moles of [tex]Al_2(SO_4)_3[/tex] react with [tex]1.461\times 3=4.383[/tex] moles of [tex]Ca(OH)_2[/tex]

From this we conclude that, [tex]Ca(OH)_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Al_2(SO_4)_3[/tex] is a limiting reagent and it limits the formation of product.

Number of moles of excess reagent = 6.073 - 4.383 = 1.69 moles

Therefore, the limiting reagent in this reaction is, [tex]Al_2(SO_4)_3[/tex] and number of moles of excess reagent is, 1.69 moles

Answer:

15.33

Explanation:

Given parameters:

Mass of KOH = 121g

Volume of solution  = 100mL   = 0.1dm³

Molar mass of KOH  = 56.11g/mol

Unknown:

pH  of the solution = ?

Solution:

To find the pH, we must first know the concentration of the solution obtained by mixing KOH up to this volume.

This can be done by finding the molarity of the solution.

  Molarity  = [tex]\frac{number of moles }{volume of solution}[/tex]

  Number of moles of KOH  = [tex]\frac{mass}{molar mass}[/tex]    = [tex]\frac{121}{56.11}[/tex]   = 2.16mole

Input parameters;

 Molarity of solution = [tex]\frac{2.16}{0.1}[/tex]   = 21.6moldm⁻³

       KOH    →    OH⁻   +          K⁺

       21.6           21.6              21.6

In the solution we have 21.6moldm⁻³ of OH⁻ which is need to find the pH;

          pOH  = -log₁₀(OH⁻)

         pOH  =  -log₁₀21.6

          pOH  = -1.33

Since pH + pOH   = 14

          pH  = 14 - (-1.33)  = 15.33

3.41 is  the pH value of a sample that has ten times fewer hydronium ions than an equal volume of a vinegar sample with a pH value of 2.4.

Explanation:

pH of the first sample of acetic acid = 2.4

to know the [H+] concentration in the acetic acid solution, the equation used is:

pH = -log [H+]

[H+] = [tex]10^{-pH}[/tex]

[H+] = [tex]10^{-2.4}[/tex]

[H+]  = 3.98 X [tex]10^{-3}[/tex] M

The concentration of H+ ion in first case is 3.98 X [tex]10^{-3}[/tex] M, the second sample has ten times less hydronium ion so concentration of first case is divided by 10.

3.98 x [tex]10^{-4}[/tex] M

Now pH of the sample having 10 times fewer ions of acetic acid:

pH = -log [H+]

putting the values in the above equation:

pH = -log [3.98 x [tex]10^{-4}[/tex] M]

pH = 3.41

If solution has ten times less hydronium ion the pH will change to 3.41.

3.41 is the pH value of a sample that has ten times fewer hydronium ions. pH gives the concentration of Hydronium ion.

What information we have:

pH of the first sample of acetic acid = 2.4

To know the [H+] concentration in the acetic acid solution, the equation used is:

[tex]pH = -log [H^+]\\\\H^+ = 10^{-pH}\\\\H^+ = 10^{-2.4}\\\\H^+ = 3.98 *10^{-4} M[/tex]

The concentration of H+ ion in first case is [tex]3.98 *10^{-4} M[/tex], the second sample has ten times less hydronium ion so the concentration of first case is divided by 10.

[tex]3.98 *10^{-4} M[/tex],

Now pH of the sample having 10 times fewer ions of acetic acid:

pH = -log [H+]

pH = -log [ [tex]3.98 *10^{-4} M[/tex],]

pH = 3.41

If the solution has ten times less hydronium ion the pH will change to 3.41.

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Answer:

All these statements are true.

Explanation:

Many describe it as an "educated guess," based on prior knowledge and observation. While this is true, the definition can be expanded. A hypothesis also includes an explanation of why the guess may be correct.

hope this helps :)

The true statement about hypothesis is: All these statements are true. The option (4) is correct.

A hypothesis is a testable statement that is formulated based on observations and background research.

It is typically written in an "if... then" form, which clearly states the relationship between two variables: the independent variable (the one at that the researcher manipulates) and the dependent variable (the one that the researcher measures).

A well-constructed hypothesis will also include a prediction regarding the dependent variable, which is an educated guess about the outcome of the experiment or study. Therefore, all the given statements about a hypothesis are true.

1. "It is written in an 'if...then' format." - This is true because a hypothesis often takes the form of ""If [independent variable], then [dependent variable]"" to clearly indicate the cause-and-effect relationship being tested.

2. "It is an educated guess based on background research." - This is true because a hypothesis is not just a random guess; it is formulated after conducting a thorough literature review and observing patterns or phenomena that suggest a certain relationship or outcome.

3. "It includes a prediction regarding the dependent variable." - This is true because a hypothesis predicts the effect that changing the independent variable will have on the dependent variable.

Since all three statements accurately describe the characteristics of a hypothesis, the correct answer is that all these statements are true.

The complete question is:

Which statement about a hypothesis is true?

1) It is written in an “if…then” format.

2) It is an educated guess based on background research.

3) It includes a prediction regarding the dependent variable.

4) All these statements are true.

Answer:

BeO

Explanation:

The chemical formula for beryllium oxide is BeO, following the combination of beryllium (Be2+) cations with oxide (O2-) anions in a 1:1 ratio, reflecting the stoichiometry of the compound and its ionic character.

The chemical formula for beryllium oxide is BeO. Beryllium oxide forms when beryllium, an alkaline earth metal with an atomic number of 4, combines with oxygen. Each beryllium atom releases two electrons, forming a Be2+ cation to Oxide2-. The resultant ionic compound, BeO, is a reflection of this 1:1 charge balance between beryllium and oxygen.

Beryllium compounds generally exhibit a significant degree of covalency. In particular, beryllium forms covalent bonds in different environments, as seen in compounds like beryllium hydride (BeH2) and basic beryllium acetate [tex](Be_{4} O(CH_{3}CO_{2} )_6)[/tex].

Although beryllium's tendency to form covalent bonds might suggest a preference for complex molecular structures, BeO is a simple, stable compound with each beryllium atom adopting an electron configuration similar to that of helium, the noble gas that precedes it in the periodic table.

The final temperature of the gas is 234K. As the volume of the increases to the given value, the temperature of the gas also increases.

Charle's law states that the volume of an ideal gas is directly proportional to the absolute temperature provided pressure is kept at constant.

It is expressed as;

V₁/T₁ = V₂/T₂

Given the data in the question;

Initial volume V₁ = 62.0mL = 0.062L

Initial temperature T₁ = 175K

Final volume V₂ = 82.9mL = 0.0829L

Final temperature T₂ = ?

To calculate the final temperature, we subtsitute our given values into the expression above.

V₁/T₁ = V₂/T₂

V₁T₂ = V₂T₁

T₂ = V₂T₁ / V₁

T₂ = ( 0.0829L × 175K ) / 0.062L

T₂ = 14.5075LK / 0.062L

T₂ = 234K

The final temperature of the gas is 234K. As the volume of the increases to the given value, the temperature of the gas also increases.

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Answer:

It's the middle one, B. H+ and OH-

Explanation:

I just did the question online

Answer:

81mmHg

Explanation:

Since the total pressure is 2700mmHg and we are told that nitrogen occupies 3% of that pressure in the Venus atmosphere. We have to find the pressure of nitrogen using its percentage.

It should be noted that the total pressure of the atmosphere is the sum total of the partial pressures of all the respective gases in the atmosphere.

Hence partial pressure of nitrogen= 3.0%× 2700mmHg= 3.0/100× 2700

Partial pressure of nitrogen= 81mmHg

Answer:

The energy absorbed when 6.20 grams of aluminum is heated from 10.0°C to 65.0°C is 306.9 Joules.

Explanation:

[tex]Q=mc\Delta T=mc\times (T_2-T_1)[/tex]

Where:

Q = heat absorbed  or heat lost

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

[tex]T_1,T_2[/tex] : Initial and final temperature of the substance

We have mass of aluminium = m = 6.20 g

Specific heat of aluminium= c = 0.900 J/g°C

Initial and final temperature of the aluminium [tex]T_1=10.0^oC [/tex]

Final temperature of the aluminium = [tex]T_2=65.0^oC[/tex]

Heat absorbed by the aluminium:

[tex]Q=6.20 g\times 0.900J/g^oC\times (65.0^oC-10.0^oC)=306.9 J[/tex]

The energy absorbed when 6.20 grams of aluminum is heated from 10.0°C to 65.0°C is 306.9 Joules.

Answer:

Ecological Relationships are of two types - Oppositional Relationships and Symbiotic Relationships. Oppositional Relationships are of two types - Predation and Competititon. Symbiotic Relationships are of four types - Mutualism, Commensalism, Amensalism & Parasitism.

Explanation:

Answer:

0.925 mole

Explanation:

From the question given, the following were obtained:

Volume = 1.0L

Molarity = 0.925 M

Number of mole of Nickel (II) chloride =?

Molarity is simply defined as the mole of solute per unit litre of the solution.

It is represented mathematically as:

Molarity = mole /Volume

With the above equation, we can easily find the mole of Nickel (II) chloride present in the solution as follow:

Molarity = mole /Volume

0.925 = mole / 1

Mole = 0.925 x 1

Mole of Nickel (II) chloride = 0.925 mole

Most nonmetals do not lose electrons when bonding with metal atoms; instead, they tend to gain electrons, forming anions. They share electrons in covalent bonds with other nonmetals.

The characteristic that is not typical of most nonmetals is their atoms lose electrons when bonding with metal atoms. When bonding occurs, nonmetals usually do the opposite; they tend to gain electrons to achieve more stable electron configurations. This is often seen in ionic bonds, where metals lose electrons to become positively charged cations, and nonmetals gain electrons to become negatively charged anions. However, in the case of bonding with other nonmetals, they typically share electrons, resulting in covalent bonds. Furthermore, nonmetals do indeed readily form compounds with other elements, either through ionic or covalent bonding.

Final answer:

Most nonmetals gain electrons when bonding with metal atoms to achieve stability, and they can also share electrons with other nonmetal atoms.

Explanation:

The characteristic that is not a characteristic of most nonmetals is that their atoms lose electrons when bonding with metal atoms. Nonmetals tend to gain electrons when bonding with metal atoms in order to achieve a stable electron configuration. They can also share electrons with other nonmetal atoms to form covalent bonds. Additionally, nonmetals readily form compounds with other elements.

Answer:

Fatty acids have an even number of carbons because they are synthesized from a starting material (acetyl-CoACoA) that has an even number of carbons.

Explanation:

Fatty acids are carboxylic acids and are known to have even number of carbon atoms because they are synthesized from a 2 carbon atom acetyl Co-A molecules which are assembled together. It contains a carboxylate group covalently joined with an hydrophobic head of CH3-(CH2)n and may have an unsatireated CH=CH group within its CH2 chain. The synthesis of fatty acids from acetyly Co-A involves the activation reaction of an enzyme acetyl Co-A synthetase with an acid. It involves dehydrogebation, hydration, oxidation and thiolysis. Fatty acids vary in length of its chain, the number of carbon-carbon double bonds attached to its CH2 chain nd also the location of those double bonds in the carbon chain. Saturated fatty acids are those without carbon-carbon double bond in its chain while unsaturated have carbon- carbon double bonds. Monounsaturated fatty acids have one carbon-carbon double bonds and polyunsaturated fatty acids have two or more carbon-carbon double bonds.

Fatty acids are the building blocks of fat molecules. The fat when broken down into simpler forms, yields fatty acids and glycerol.

The correct answer is:

Option C. Fatty acids have an even number of carbons because they are synthesized from a starting material (acetyl-CoACoA) that has an even number of carbons.

The fatty acids are generally even numbers because of the even precursor.

The metabolism of fatty acids starts with the two molecules of the acetyl coenzyme A, which on assembling yields the even-numbered of fatty acids.

The acetyl coenzyme A is an even-numbered two-carbon molecule, which undergoes enzyme-catalyzed reactions to yield an even number of fatty acids.

The even number of fatty acids can be saturated (made up of single bonds) or can be unsaturated (presence of double or triple bonds).

Therefore, Option C is correct.

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Answer:

Explanation:

The dilution equation is:

Where C₁ and C₂ are concentrations and V₁ and V₂ are volumes of the concentrated and diluted solutions.

First, convert the 75. mL volume into liters:

Now, substitute the data into the dilution formula:

The molarity of the new 2.0 L solution made by diluting 75 mL of 17.6M acetic acid is 0.66M, calculated using the dilution equation M1 x V1 = M2 x V2.

To calculate the molarity of the diluted acetic acid solution, we can use the concept of dilution, which is described by the equation

M1 x V1 = M2 x V2, where:

Using the provided information, M1 = 17.6M, V1 = 75 mL, and V2 = 2,000 mL (as 2.0 L is equivalent to 2,000 mL). We need to solve for M2, the molarity of the new solution:

M1 x V1 = M2 x V2

17.6M * 75 mL = M2 * 2,000 mL

M2 = (17.6M * 75 mL) / 2,000 mL

M2 = 0.66M (rounded to two decimal places)

Therefore, the molarity of the diluted acetic acid solution is 0.66M.

Answer:

[tex]H_{rxn} = -93.52 \frac{KJ}{mol}[/tex]

The final temperature [tex]T_{2} = 50.21[/tex]  ° c

Explanation:

(a).

[tex]H_{rxn} = H_{products} - H_{reactents}[/tex]

[tex]H_{reactants}[/tex] = - 813.9 [tex]\frac{KJ}{mol}[/tex]

[tex]H_{Products}[/tex] = - 907.51  [tex]\frac{KJ}{mol}[/tex]

[tex]H_{rxn} = -907.51 + 813.9[/tex]

[tex]H_{rxn} = -93.52 \frac{KJ}{mol}[/tex]

(b).

[tex]H_{rxn} = -93.52 \frac{KJ}{mol}[/tex]

E = 93520 J

C = 3.5 [tex]\frac{KJ}{kg K}[/tex]

Initial temperature [tex]T_{1} = 25[/tex] ° c

Mass (m) = density × volume

m = 1060 × 1

m = 1060 gm

[tex]E = m C (T_{2} - T_{1} )[/tex]

93520 =1060 × 3.5 × ( [tex]T_{2}[/tex] - 298 )

[tex]T_{2} = 50.21[/tex]  ° c

This is the final temperature.

(c).

The density of sulfuric acid is more than the water. so when water is added to acid the mixing process does not takes properly. So we have to add acid in to the water.

Adding water is exothermic process so when we add water to acid than in that case more energy produces and wasted.

Answer:

94.8% is the percentage yield of this reaction.

Explanation:

Mass of aluminium = 3.8 g

Moles of aluminium = [tex]\frac{3.8 g}{27 g/mol}=0.1407 mol[/tex]

[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]

According to reaction, 2 moles of aluminum gives 2 moles of aluminium chloride, then 0.1407 moles of aluminium will give:

[tex]\frac{2}{2}\times 0.1407 mol=0.01407 mol[/tex] aluminium chloride

Mass of 0.1407 moles of aluminum chloride:

= 0.1407 mol × 133.5 g/mol = 18.78 g

Theoretical yield of aluminum chloride = 18.78 g

Experimental yield of aluminum chloride = 17.8 g

The percentage yield of reaction:

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

[tex]=\frac{17.8 g}{18.78 g}\times 100=94.8\%[/tex]

94.8% is the percentage yield of this reaction.

Final answer:

To produce 22 g of CO₂, we calculate the moles of CO₂ and use the 1:1 molar ratio with O₂ to find that 16 g of oxygen gas is required, according to the balanced chemical equation for the reaction.

Explanation:

Calculating Mass of Oxygen Required to Produce Carbon Dioxide

To determine the mass of oxygen gas required to produce 22 g of carbon dioxide in the reaction C(s) + O₂(g)
ightarrow CO₂(g), we will first consider the molar masses of the substances involved. The molar mass of carbon dioxide (CO₂) is 44.009 g/mol, which means 1 mole of CO₂ weighs 44.009 grams. Since carbon is a pure element in its solid form, its molar mass is equivalent to its atomic mass, which is 12.0 g/mol. The molar mass of oxygen (O₂) is 32.0 g/mol because one mole of O₂ consists of 2 atoms of oxygen and the atomic mass of oxygen is 16.0 g/mol per atom.

Using the stoichiometry from the balanced chemical equation, we know that 1 mole of C will react with 1 mole of O₂ to produce 1 mole of CO₂. Therefore, if we have 22 g of CO₂, we can calculate the moles of CO₂ and then the moles of O₂ required:

Calculate moles of CO₂ produced: (22 g CO₂) / (44.009 g/mol) = 0.5 mol CO₂.

We need 0.5 mol of O2 since the mole ratio is 1:1.

Calculate mass of O₂ required: (0.5 mol O₂) times (32.0 g/mol) = 16 g O₂.

So, the mass of oxygen gas required to produce 22 g of carbon dioxide is 16 g.

Answer:

1.5 litre of C5H12 produce 9 litre of water vapour;

Explanation:

First balance the combustion reaction;

[tex]C_5H_12 +8O_2[/tex] → [tex]5CO_2 +6H_2O[/tex]

from the balance reacion it is clearly that,

one mole of C5H12 reacts completely with 8 mole of O2 gas;

Hence

one litre of C5H12 reacts completely with 8 litre of O2 gas;

there fore 1.5 litre C5H12 needs 12 litre oxygen gas but we have 15 litre.

so,

C5H12 is a limiing reactant and O2 is a excess reactant.

so quantity of H2O depends on limiting reactant;

one litre of C5H12 produce 6 litre of water vapour;

therefore,

1.5 litre of C5H12 produce 9 litre of water vapour;

Answer:

The answer is 1.5 moles.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law:

P * V = n * R * T

where n is the number of moles and R is the molar constant of the gases.

In this case:

Replacing:

1 atm*33.6 L= n* 0.082 [tex]\frac{atm*L}{mol K}[/tex] *273 °K

Solving:

[tex]n=\frac{1 atm* 33.6 L}{0.082\frac{atm*L}{mol K}*273K }[/tex]

n= 1.5 moles

So, the answer is 1.5 moles.

Answer:

1.5 moles

Explanation:

Answer:

Your answer is ANION

Answer:

500.65mL

Explanation:

The following information were obtained from the question:

V1 (initial volume) = 461 mL

P1 (initial pressure) = stp = 101325Pa

P2 (final pressure) = 93.3 kPa

Recall: 1KPa = 1000Pa

Therefore, 93.3 kPa = 93.3x1000 = 93300Pa

V2 (final volume) =?

Using the Boyle's law equation P1V1 = P2V2, the final volume of the gas can be obtained as follow:

P1V1 = P2V2

461 x 101325 = 93300 x V2

Divide both side by 93300

V2 = (461 x 101325)/93300

V2 = 500.65mL

Therefore, the volume of Neon at 93.3 kPa is 500.65mL

Answer:

Explanation:

M

The term associated with the solution is saturated.

The term associated with a solution that is saturated at 25°C and then slowly cooled to 20°C with no change to its appearance is saturated.

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Answer:200 c

Explanation:

Final answer:

To raise the temperature of 20 grams of water from 30°C to 40°C, 200 calories of heat are required.

Explanation:

To calculate the amount of heat required to raise the temperature of water, we can use the equation:

heat = mass * specific heat * temperature change

In this case, the mass of water is 20 grams, the specific heat of water is 1 calorie/gram °C, and the temperature change is 10 °C. Substituting these values into the equation:

heat = 20 g * 1 calorie/g °C * 10 °C = 200 calories

Therefore, 200 calories of heat are required to raise the temperature of 20 grams of water from 30 °C to 40 °C.

Answer:

The specific heat of alloy  [tex]C_{alloy} = 1.007 \frac{KJ}{Kg K}[/tex]

Explanation:

Mass of the alloy = 30.5 gm = 0.0305 kg

Initial temperature = 93 °c = 366 K

Mass of water = 50 gm = 0.05 kg

Initial temperature = 22 °c = 295 K

Final temperature of the mixture = 31.1 °c = 304.1 K

From the energy conservation principal the heat lost by the alloy is equal to heat gain by the water.

Heat lost by alloy

[tex]Q_{alloy} = m C (T_{f}- T_{i} )[/tex]

[tex]Q_{alloy} = (0.0305) C_{alloy} (366-304.1)[/tex]

[tex]Q_{alloy} = (1.88795) C_{alloy}[/tex]  ------- (1)

Heat gain by water

[tex]Q_{w} = (0.05) (4.18) (304.1 - 295)[/tex]

[tex]Q_{w} = 1.9019 \frac{KJ}{kg K}[/tex]  ------- (2)

Equation (1) = Equation (2)

[tex](1.88795) C_{alloy} = 1.9019[/tex]

[tex]C_{alloy} = 1.007 \frac{KJ}{Kg K}[/tex]

This is the specific heat of alloy.

Taking into account the definition of calorimetry,  the specific heat capacity of the alloy is 1.063 [tex]\frac{kJ }{kgK}[/tex]

In first place, calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

In this way, between heat and temperature there is a direct proportional relationship.

The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat by the mass of the body.

So, the equation that allows to calculate heat exchanges is:

Q = C× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance C and where ΔT is the temperature variation.

In this case, you know:

for alloy:

for water:

Replacing in the expression to calculate heat exchanges:

for alloy:

for water:

On the other hand, the heat of the calorimeter can be expressed as:

Qcalorimeter= Ccalorimeter×ΔT

Being:

and replacing you get:

Qcalorimeter= 0.0092 [tex]\frac{kJ}{K}[/tex]× 9.2 K

It should be taken into account that a system at different temperatures evolves spontaneously towards a state of equilibrium in which all bodies have the same temperature. Then, mixing two quantities of liquids at different temperatures generates an energy transfer in the form of heat from the hottest to the coldest. Said energy transit is  held until temperatures equalize, when it is said to have reached thermal equilibrium.

So the heat released by the sample is absorbed by the calorimeter and the water.

Qalloy= Qcalorimeter + Qwater

Replacing the corresponding expressions and solving:

Calloy× 0.0305 kg× 61.9 K= 0.0092 [tex]\frac{kJ}{K}[/tex]× 9.2 K + 4.18 [tex]\frac{kJ}{kgK}[/tex]× 0.050 kg× 9.2 K

Calloy× 1.88795 kg× K= 0.08464 kJ + 1.9228 kJ

Calloy× 1.88795 kg× K= 2.00744 kJ

[tex]Calloy=\frac{2.00744 kJ }{1.88795 kgK}[/tex]

Calloy= 1.063 [tex]\frac{kJ }{kgK}[/tex]

Finally, the specific heat capacity of the alloy is 1.063

Learn more:

The standard potential for a galvanic cell consisting of nickel and copper electrodes at 25 °C is calculated using the standard reduction potentials for each electrode, resulting in a cell potential of +0.77 V.

To calculate the standard potential for the galvanic cell at 25 °C, you need to know the standard reduction potentials for the nickel and copper electrodes. These potentials can be found in a table of standard reduction potentials. The cell potential (Ecell) is the difference between the reduction potential of the cathode and the anode.

In our case, for a cell composed of nickel and copper electrodes, the standard reduction potential for the nickel(II) ion to nickel reaction is approximately -0.25 V, and for the copper(I) ion to copper reaction, it is +0.52 V.

Considering that the copper electrode undergoes reduction (acts as cathode) and the nickel electrode undergoes oxidation (acts as anode), we use the following formula:

[tex]E_{cell} = E_{cathode} - E_{anode[/tex]

Therefore: Ecell = (+0.52 V) - (-0.25 V) = +0.52 V + 0.25 V = +0.77 V

The standard potential for this galvanic cell at 25 °C is +0.77 V.

Answer:

P₃ = 594.1 mmHg

Explanation:

Given data:

Total pressure of container = 1439 mmHg

Partial pressure of 1st gas = 523.3 mmHg

Partial pressure of 2nd gas = 509.8 mmHg

Partial pressure of 3rd gas = ?

Solution:

According to Dalton law of partial pressure,

The total pressure inside container is equal to the sum of partial pressures of individual gases present in container.

Mathematical expression:

P(total) = P₁ + P₂ + P₃+ ............+Pₙ

Now we will solve this  problem by using this law.

P(total) = P₁ + P₂ + P₃

439 mmHg =  523.3 mmHg +  509.8 mmHg + P₃

439 mmHg =  1033.1 mmHg + P₃

P₃ = 1033.1 mmHg -439 mmHg

P₃ = 594.1 mmHg

In the precipitation reaction, the ion that would NOT be present in the complete ionic equation is the iodide ion (I-).

In the precipitation reaction Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq), the ions Pb2+ and NO3- are present in the complete ionic equation because they are soluble in water and dissociate into ions. However, the solid PbI2 that forms is insoluble in water and does not dissociate into ions. Therefore, the iodide ion (I-) is the ion that would NOT be present in the complete ionic equation.

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In the given precipitation reaction, all ions mentioned would be present in the complete ionic equation because soluble compounds turn into ions in such reaction. However, the potassium ion (K+) and the nitrate ion (NO3-) do not partake in the formation of the solid lead iodide, so they are omitted from the net ionic equation.

The question is asking about a precipitation reaction, which is when an insoluble compound forms in an aqueous solution. For the reaction given (2KI(aq) + Pb(NO3)2(aq) → PbI₂ (s) + 2KNO3(aq)), the insoluble compound that forms is lead iodide (PbI₂), which is why it is denoted with an (s) for solid. Soluble compounds break down into ions in aqueous solutions, meaning Pb(NO3)2(aq) becomes Pb²⁺ + 2NO3⁻, and 2KI(aq)  becomes 2K⁺ + 2I⁻. The products would then be K⁺ + NO3⁻ from KNO3(aq) and PbI₂(s), which stays unbroken because it is insoluble. Therefore, the complete ionic equation would be Pb²⁺(aq) + 2NO3⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO3⁻(aq).

In such an equation, the ions that are on both sides of the equation (i.e., do not participate in generating the solid) are spectator ions and can be omitted to achieve the net ionic equation. For our reaction, the spectator ions are K⁺ and NO3⁻. Thus, they would NOT be present in the net ionic equation, but they would still be present in the complete ionic equation.

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