A 3.00 L flask contains 2.33 g of argon gas at 312 mm Hg What is the temperature of the gas

Final answer:

To determine the new concentrations of NH3 and NH4+ after the addition of HNO3, the moles of HNO3 added are calculated and the amounts of NH3 converted to NH4+ are accounted for. The final concentrations are found to be approximately 0.277 M for NH3 and 0.339 M for NH4+, after considering the change in total volume.

Explanation:

The student is asking to calculate the concentration of buffer components in a solution after the addition of a strong acid. To solve this, we have to determine how much the strong acid (HNO3) will neutralize the NH3 component of the buffer, and how it will change the concentrations of NH3 and NH4+ (the buffer components).

First, calculate the number of moles of HNO3 added:

Since NH3 and NH4+ are in a 1:1 molar ratio in the solution and they react with HNO3 in a 1:1 ratio, the added HNO3 will react with the NH3:

And NH4+ will increase by 0.009 moles (since each mole of NH3 reacts to form one mole of NH4+):

The new volume of the solution would be the initial volume of buffer solution plus the volume of HNO3 added, which totals 288.50 mL or 0.28850 L.

The new concentrations are therefore:

This change in concentrations due to the addition of HNO3 means the buffer will have adjusted to these new concentrations of NH3 and NH4+.

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c)  0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

[tex]H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O[/tex]

1 mole         3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

[tex]H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O[/tex]

10 ml            17.50 ml

(x) M              0.200 M

Molarity = [tex]\frac{0.2*17.5}{1000}[/tex]

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= [tex]0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}[/tex]

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =  [tex]\frac{mole \ \ of \ soulte }{ Volume \ of \ solution }[/tex]

Molar Concentration = [tex]\frac{0.00166 \ mole \ of \ H_3PO_4 }{10}*1000[/tex]

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

Answer:

C. adding more powder solute

Explanation:

The concentration of a mixture can be increased by removing solvent.

The concentration of a mixture can be increased in several ways. If we consider a distillery wanting to achieve a higher concentration of alcohol, they need to remove solvent, which in this case is water, from their product. This can be achieved through processes such as distillation, which separates alcohol from water due to their different boiling points. Another approach is the addition of a substance that reacts with the water, effectively removing water content from the mixture.

To increase the concentration of a solution in general, one can also add more solute (the substance being dissolved) into the solution or remove solvent (the substance dissolving the solute). It is important not to confuse the process of dilution, which involves adding solvent and decreases solute concentration, with the process of concentrating a solution, which involves removing solvent and increases solute concentration. Therefore, methods such as evaporation, reverse osmosis, or chemical reaction can be employed to increase the concentration of a solute in a solution.

The statement which are true for a neutral, aqueous solution at 25 °C are:

The statement which is/are true for a neutral, aqueous solution regardless of temperature is;

We must know that at standard temperature, 25°C, a neutral, aqueous solution has it's pH = pOH = 7. Additionally, the hydrogen ion concentration, [H+] is equal to its hydroxide ion concentration, [OH-]

Ultimately, the pH of a solution decreases with increase in temperature and vice versa.

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Answer:

1. Cu

2. Cu

3. 2 electrons.

Explanation:

Step 1:

The equation for the reaction is given below:

3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)

Step 2:

Determination of the change of oxidation number of each element present.

For Cu:

Cu = 0 (ground state)

Cu(NO3)2 = 0

Cu + 2( N + 3O) = 0

Cu + 2(5 + (3 x -2)) =0

Cu + 2 (5 - 6) = 0

Cu + 2(-1) = 0

Cu - 2 = 0

Cu = 2

The oxidation number of Cu changed from 0 to +2

For N:

HNO3 = 0

H + N + 3O = 0

1 + N + (3 x - 2) = 0

1 + N - 6 = 0

N = 6 - 1

N = 5

NO = 0

N - 2 = 0

N = 2

The oxidation number of N changed from +5 to +2

The oxidation number of oxygen and hydrogen remains the same.

Note:

1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1

2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1

Step 3:

Answers to the questions given above

From the above illustration,

1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.

2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.

3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.

In the given chemical reaction, the element oxidized and the reducing agent is Copper (Cu). The number of electrons transferred from the reducing agent to the oxidizing agent is six.

In the reaction equation, 3Cu(s) + 8HNO3(aq) --> 2NO(g) + 3Cu(NO3)2(aq) + 4H2O(l), the element oxidized is Cu (Copper), as Copper goes from an oxidation state of 0 in Cu(s) to +2 in Cu(NO3)2. Hence, the reducing agent is also Copper (Cu). Oxidation is the process of losing electrons, and in this chemical reaction, each copper atom loses 2 electrons (going from 0 to +2). Since the equation shows the reaction of 3 copper atoms, therefore, the total number of electrons transferred from reducing to oxidizing agent in the equation, as written, is 6.

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Final answer:

The boiling point of the student's solution with 12.5g of NaCl in 0.750L of water would be approximately 100.04 °C. Given the slight elevation, salt's effect on boiling temperature is negligible, and the water will practically boil at standard boiling temperature of pure water (100 °C).

Explanation:

The student has prepared a solution by adding 12.5g of NaCl to 0.750L of water and wants to know at what temperature the solution will boil. Generally, pure water boils at 100 °C at standard atmospheric pressure. However, when salt (NaCl) is added to water, the boiling point elevation can occur due to the colligative properties of the solution. The boiling point elevation (ΔTb) depends on the molality (m) of the solution and the ebullioscopic constant (kb) of the solvent, which for water is 0.512 °C/m.

Given this information, we can approximate that the boiling point of water will be slightly elevated, by about 0.04 °C, due to the addition of NaCl. However, the exact boiling point will depend on the concentration of the NaCl solution. For the student's solution, the precise boil elevation calculation would require the formula ΔTb = i * kb * m, where i is the Van't Hoff factor for NaCl (approximately equal to 2, because NaCl dissociates into two ions: Na+ and Cl-).

Answer for the Student's Question:

In this educational context, the increase in boiling temperature is minimal, and the solution in the pot will boil at approximately 100.04 °C, or simply 100 °C when rounded to three significant figures. Adding salt to water used for cooking pasta causes a negligible effect on the boiling temperature and consequently on the cooking time.

Answer:

90 g CH₃NH₃Cl

Explanation:

It appears your question lacks the values required to solve this problem. However, an online search tells me these are the values. Be aware if your values are different your answer will also be different, but the methodology remains the same.

" A chemistry graduate student is given 450 mL of a 1.70 M methylamine solution. Methylamine is a weak base with Kb=4.4x10⁻⁴. What mass of CH₃NH₃Cl should the student dissolve in the methylamine solution to turn it into a buffer with pH = 10.40 ? You may assume that the volume of the solution doesn't change when the is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits. "

We'll use the Henderson-Hasselbach equation:

So first we use Kb to calculate Ka and then pKa:

Now we can calculate the concentration of the salt, CH₃NH₃Cl:

Now we use the final volume to calculate the moles of CH₃NH₃Cl and finally its mass, using its molecular weight:

Which when rounded to 2 significant digits becomes 90 g CH₃NH₃Cl

Answer:

Amylase.

Explanation:

The process of digestion begin to start in mouth when food mix with saliva. An enzyme is released which is called Amylase help in digestion of carbohydrates.

Answer:

[tex]224.5 m^3[/tex]

Explanation:

By using the first law of thermodynamics, we can find the work done by the gas:

[tex]\Delta U=Q-W[/tex]

where in this problem:

[tex]\Delta U=-69 kJ[/tex] is the change in internal energy of the gas

[tex]Q=+260 kJ[/tex] is the heat absorbed by the gas

W is the work done by the gas (positive if done by the gas, negative otherwise)

Therefore, solving for W,

[tex]W=Q-\Delta U=+260-(-69)=+329 kJ = +3.29\cdot 10^5 J[/tex]

So, the gas has done positive work: it means it is expanding.

Then we can rewrite the work done by the gas as

[tex]W=p(V_f-V_i)[/tex]

where:

[tex]p=7400 Pa[/tex] is the pressure of the gas

[tex]V_i=180 m^3[/tex] is the initial volume of the gas

[tex]V_f[/tex] is the final volume

And solving for Vf, we find

[tex]V_f=V_i+\frac{W}{p}=180+\frac{3.29\cdot 10^5}{7400}=224.5 m^3[/tex]

Final answer:

To determine the final volume of the gas, the first law of thermodynamics was used, considering the process as isobaric due to constant pressure. The final volume was calculated to be approximately 45045.9 m³ after applying the formula and accounting for the work done and heat supplied.

Explanation:

To find the final volume of the gas, we can use the first law of thermodynamics, which is given by ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. Since the question states the pressure remains constant (isobaric process), the work done by the gas can be found by W = pΔV, where p is the pressure and ΔV is the change in volume.

From the given data, we have:

Using ΔU = Q - W and substituting W with pΔV, we get:

-69 kJ = 260 kJ - (7400 Pa × ΔV)

Convert kJ to J by multiplying by 1000 (since 1 kJ = 1000 J):

-69000 J = 260000 J - (7400 Pa × ΔV)

We can then solve for ΔV:

ΔV = (260000 J + 69000 J) / 7400 Pa

ΔV = 44865.9 m³ (rounded to one decimal place)

To find the final volume (V₂), we add the change in volume to the initial volume:

V₂ = V₁ + ΔV

V₂ = 180 m³ + 44865.9 m³

V₂ = 45045.9 m³

So, the final volume of the gas after the process is approximately 45045.9 m³.

Answer: 112.5moles

Explanation:

P= 0.6atm, V= 4000L, R= 0.082, T= 260K

Applying PV = nRT

0.6×4000= n×0.082×260

Simplify n= (0.6× 4000)/(0.082×260)

n= 112.5moles

Using the Ideal Gas Law, we calculated that the weather balloon near the top of Mt. Rainier, with given conditions, contains approximately 113.1 moles of helium.

The Ideal Gas Law formula is PV = nRT, where:

We can use this formula to calculate the number of moles of helium in the weather balloon.

Given:

Using the Ideal Gas Law:

Plugging in the given values:

Thus, the number of moles of helium in the weather balloon is approximately 113.1 moles.

Answer:

The correct answer is option C. The unknown solution definitely has Hg22+ present.

Explanation:

In the analysis of group 1 metal cation, the unknown solution is treated with sufficient quantity of 6 M HCl solution and if group 1 metal cations are present then white precipitate of Agcl, PbCl2 or Hg2Cl2 is formed. The precipitate of PbCl2 is soluble in hot water but the other two remains insoluble after treating with hot water. Precipitate of AgCl disappears upon treatment of NH3 solution but Hg2Cl2 becomes black in the reaction with NH3. The black Colour appears due to the formation of metallic Hg.

Balanced chemical equation of the reation is -

Hg2Cl2 + 2NH3  ---------> HgNH2Cl (white ppt.) + Hg (black ppt.) + NH4Cl

Therefore, from the given information the conclusion which can be drawn is that the unknown solution definitely has Hg22+ present.

Answer:

C-  the unknown solution definitely has Hg22+ present.

Explanation:,

Answer:

Explanation:

*Since the titration is between the strong acid HCl and the strong base Ca(OH)2, the pH at the equivalent point should be 7. On interpolation, we will obtain that 9.50mL and 9.82 mL of HCl is required to completely neutralized the given Ca(OH)2 solution.

*pH at the equivalence point =7

we know that pH + pOH = 14

Hence pOH= 14-7=7

pOH= -log(OH-)

The concentration of OH-= 10-pH= 1X10-7 M

One reason for the low solubility may be the higher reaction temperature, Another reason is the common ion effect.

The titration is between the strong acid HCl and the strong base Ca(OH)2, the pH at the equivalent point should be 7.

The process of determining the quantity of a substance A by adding measured increments of substance B, the titrant, with which it reacts until exact chemical equivalence is achieved (the equivalence point).

It is possible titrations are affected by temperature. The pH range for indicators can change with temperature. strong base

Since the titration is between the strong acid HCl and the strong base Ca(OH)2, the pH at the equivalent point should be 7.

On interpolation, we will obtain that 9.50mL and 9.82 mL of HCl is required to completely neutralized the given Ca(OH)2 solution.

*pH at the equivalence point =7

we know that pH + pOH = 14

Hence pOH= 14-7=7

pOH= -log(OH-)

The concentration of [tex]OH^-= 10^{-pH}= 1*10^{-7} M[/tex].

One reason for the low solubility may be the higher reaction temperature, Another reason is the common ion effect.

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Answer:  Option A and then Option c

Explanation: Because Concentrated Sulphuric acid is a very strong acid and  corrosive, when spilled on the skin, can affect the surface and internal skin. It is necessary to First rinse the affected area with copious amount of water to reduce effect of the acid  and Secondly to treat the area with aqueous sodium bicarbonate solution since it will further neutralize and reduce stinging pain from the acid spill by acting as an alkali or base,

Rinsing with a copious amount of water is the first step to dilute concentrated sulfuric acid before neutralizing it with sodium bicarbonate. For a spill of 27.6 mL of 6.25 M H2SO4 solution, one would need 28.96 grams of sodium bicarbonate.

In the event of a concentrated sulfuric acid spill, the first step is to rinse the affected area with a copious amount of water to dilute the acid. This is important because adding water to concentrated sulfuric acid can lead to a highly exothermic reaction which can cause the sulfuric acid to splatter and cause further damage or injury. Therefore, careful dilution with water is essential. Once the area is rinsed and the acid is diluted, the next step is to neutralize the acid. This is where sodium bicarbonate (NaHCO3) comes into play.

For the reaction between sodium bicarbonate and sulfuric acid:
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) + 2CO2(g),
we can calculate the moles of sodium bicarbonate needed. If 27.6 mL of a 6.25 M H2SO4 solution were spilled, this is equivalent to 27.6 mL × 6.25 mmol/mL = 172.5 mmol of H2SO4. Since the stoichiometry of the reaction requires 2 moles of NaHCO3 for every 1 mole of H2SO4, we need 2 × 172.5 mmol = 345 mmol of NaHCO3. To convert moles to grams, we use the molar mass of NaHCO3, which is approximately 84.01 g/mol. Hence, 345 mmol × 84.01 g/mol = 28.96 g of NaHCO3 would be required to neutralize the spill.

Answer:

Acetate ion

Explanation:

CH₃COOH  ↔ CH₃COO⁻ + H⁺

HCl -> H⁺ + Cl⁻

The H+ ions from added HCl reacts with acetate ions and more amount of acetic acid will be formed due to common ion effect.

The pH of the buffer solution does not decrease drastically because the added HCl reacts with the sodium acetate present in the buffer, maintaining the buffer's capacity to neutralize added protons.

Sodium acetate acts as a conjugate base in the buffer system, reacting with added HCl to form acetic acid and water. This reaction prevents the pH from decreasing drastically because the added protons from HCl are consumed in the formation of acetic acid, thereby maintaining the buffer capacity.

A buffer system works to maintain a relatively constant pH upon the addition of acids or bases, demonstrating its capacity to minimize changes in the hydrogen ion concentration of a solution.

Answer:

The pH is 3.08

Explanation:

Step 1:

The equation for the reaction.

AH ⇌ A- + H+

Step 2:

Data obtained from the question.

Initial concentration of AH, [AH] = 0.97 M

Dissociation constant, Ka = 7.23x10^-7

Step 3:

Determination of the concentration of A-, [A-], the concentration of H+, [H+] and the concentration of AH, [AH] after the reaction. This is illustrated below:

Before the reaction:

[AH] = 0.97 M

[A-] = 0

[H+] = 0

During the reaction:

[AH] = -y

[A-] = +y

[H+] = +y

After the reaction

[AH] = 0.97 - y

[A-] = y

[H+] = y

Step 4:

Obtaining the value of y. This is illustrated below:

Ka = [A-] [H+] / [AH]

Ka = 7.23x10^-7

[AH] = 0.97

[A-] = y

[H+] = y

Ka = [A-] [H+] / [AH]

7.23x10^-7 = y x y / 0.97

Cross multiply to express in linear form

7.23x10^-7 x 0.97 = y^2

7.0131x10^-7 = y^2

Take the square root of both side

y = √(7.0131x10^-7)

y = 8.37x10^-4

Therefore [H+] = y = 8.37x10^-4 M

Step 5:

Determination of the pH.

pH = - log [H+]

[H+] = 8.37x10^-4 M

pH = - log [H+]

pH = - log 8.37x10^-4

pH = 3.08

Nutrients are used by the body for energy, building materials, and maintaining physiological functions. Carbohydrates and lipids are primarily energy-yielding, while proteins provide amino acids for the growth and repair of tissues.

Nutrients are the substances obtained from food that are vital for growth and maintenance of a healthy body throughout life. Our body uses nutrients for energy, building materials, and sustaining bodily functions. Among these nutrients, carbohydrates and lipids are primarily known as the energy-yielding nutrients. Proteins, while not primarily used for energy, supply essential amino acids that serve as building blocks for the repair and growth of tissues.

Carbohydrates are broken down into glucose, which is used in the metabolic processes to create adenosine triphosphate (ATP), the energy currency of the cell. Lipids are utilized for energy storage, insulation, and cellular structure. Proteins contribute to various functions, including enzyme reactions, transport mechanisms, and cell signaling.

Micronutrients, such as vitamins and minerals, do not provide energy but are crucial in other physiological processes, including metabolism regulation, bone health, and immune system function. Water, despite not providing energy, is essential for nutrient transportation, temperature regulation, and waste excretion.

Overall, a balanced diet with the appropriate intake of macronutrients and micronutrients is vital for maintaining health and supporting the body's various functions.

Explanation:

It is the boiling point because is the point when liquid starts to boil and changes to gases.

Answer:

[tex]\large \boxed{\text{12.0 atm}}[/tex]

Explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ = 9.00 atm; T₁ =   28.0 °C

p₂ = ?;              T₂ = 129.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ =   (28.0 + 273.15) K = 301.15

T₂ = (129.0 + 273.15) K = 402.15

2. Calculate the new pressure

[tex]\begin{array}{rcl}\dfrac{9.00}{301.15} & = & \dfrac{p_{2}}{402.15}\\\\0.02989 & = & \dfrac{p_{2}}{402.15}\\\\0.02989 \times 402.15 &=&p_{2}\\p_{2} & = & \textbf{12.0 atm}\end{array}\\\text{The new pressure is $\large \boxed{\textbf{12.0 atm}}$}[/tex]

Answer:

a)

The overall  balanced combustion  reaction is written as :

[tex]0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2[/tex]

[tex](F/A)_{stoichiometric} = 0.0424[/tex]

[tex](A/F)_{stoichiometric} = 23.562[/tex]

b)

the higher heating values [tex](HHV)_f[/tex] per unit mass of LPG = 49.9876 MJ/kg

the lower heating values [tex](LHV)_f[/tex] per unit mass of LPG = 46.4933 MJ/kg

Explanation:

a)

The stoichiometric equation can be expressed as :

[tex]0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2 \ + \ bH_2O \ + \ cN_2[/tex]

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

[tex]x = \frac{2a+b}{2} \\ \\ x = \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95[/tex]

Equating the coefficient of Nitrogen

[tex]c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612[/tex]

Therefore, The overall  balanced combustion  reaction can now be written as :

[tex]0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2[/tex]

Now;  To determine the stoichiometric F/A and A/F ratios; we have:

[tex](F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\ (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424[/tex]

[tex](A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\ (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562[/tex]

b)

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel [tex]M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})[/tex]

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

[tex]m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8} = 0.7 \\ \\ \\ m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06 \\ \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6} = 0.24[/tex]

Finally; calculating the higher heating values [tex](HHV)_f[/tex] per unit mass of LPG; we have:

[tex](HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg[/tex]

calculating the lower heating values [tex](LHV)_f[/tex] per unit mass of LPG; we have:

[tex](LHV)_f = (HHV)_f - \delta H_w \\ \\ (LHV)_f = (HHV)_f - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f = 49.9876 \ MJ/kg - [\frac{3.8*18}{44.2}*2.258 \ MJ/kg] \\ \\ (LHV)_f = 46.4933 \ M/kg[/tex]

The higher heating value (HHV) per unit mass of LPG is approximately 49.91 MJ/kg, and the lower heating value (LHV) per unit mass of LPG is approximately 44.92 MJ/kg.

To solve this problem, we will first determine the overall combustion reaction for stoichiometric combustion of 1 mole of LPG with air. Then, we will calculate the stoichiometric fuel-to-air (F/A) and air-to-fuel (A/F) ratios. Finally, we will find the higher and lower heating values per unit mass of LPG.

Step 1: Determine the overall combustion reaction

First, we need to calculate the moles of each component in the LPG mixture based on the given volume percentages:

- Propane (C3H8): 70% of the mixture

- Butane (C4H10): 5% of the mixture

- Propene (C3H6): 25% of the mixture

Since the mixture is on a volume basis, we can directly use the molar percentages for the calculations. We will assume 1 mole of LPG for simplicity, which means:

- Moles of propane: 0.70 moles

- Moles of butane: 0.05 moles

- Moles of propene: 0.25 moles

Next, we write the balanced combustion reactions for each component:

Propane: C3H8 + 5O2 → 3CO2 + 4H2O

Butane: C4H10 + 6.5O2 → 4CO2 + 5H2O

Propene: C3H6 + 4.5O2 → 3CO2 + 3H2O

To find the overall reaction, we sum up the reactions for each component, taking into account their respective moles in the mixture:

Overall reaction = (0.70 * Propane reaction) + (0.05 * Butane reaction) + (0.25 * Propene reaction)

This gives us:

Overall reaction = (0.70 * (C3H8 + 5O2 → 3CO2 + 4H2O)) + (0.05 * (C4H10 + 6.5O2 → 4CO2 + 5H2O)) + (0.25 * (C3H6 + 4.5O2 → 3CO2 + 3H2O))

Combining the terms, we get:

Overall reaction = (0.70C3H8 + 0.05C4H10 + 0.25C3H6) + (3.5O2 + 0.325O2 + 1.125O2) → (2.1CO2 + 0.2CO2 + 0.75CO2) + (2.8H2O + 0.25H2O + 0.75H2O)

Simplifying the coefficients by multiplying by the least common multiple to get whole numbers, we have:

Overall reaction = (7C3H8 + C4H10 + 2.5C3H6) + (35O2 + 3.25O2 + 11.25O2) → (21CO2 + 2CO2 + 7.5CO2) + (28H2O + 2.5H2O + 7.5H2O)

Further simplifying, we get:

Overall reaction = (7C3H8 + C4H10 + 2.5C3H6) + 50O2 → (29CO2 + 2.5CO2) + (35H2O + 2.5H2O)

Finally, the overall balanced reaction is:

Overall reaction = (7C3H8 + C4H10 + 2.5C3H6) + 50O2 → 31.5CO2 + 37.5H2O

Step 2: Calculate the stoichiometric F/A and A/F ratios

The stoichiometric F/A ratio is the ratio of the mass of fuel to the mass of air required for complete combustion. The stoichiometric A/F ratio is the inverse of the F/A ratio.

First, we need to calculate the molar mass of the LPG mixture:

Molar mass of LPG = (0.70 * Molar mass of C3H8) + (0.05 * Molar mass of C4H10) + (0.25 * Molar mass of C3H6)

Next, we calculate the mass of air required for the combustion of 1 mole of LPG. The stoichiometric combustion of hydrocarbons with air can be represented as

For 1 mole of LPG, the mass of oxygen required is:

The mass of air required is the mass of O2 multiplied by the ratio of the molar mass of air to the molar mass of O2 (approximately 28.97/32):

Now, we can calculate the F/A and A/F ratios:

Step 3: Calculate the higher and lower heating values per unit mass of LPG

The higher heating value (HHV) of the LPG mixture is calculated by taking the weighted average of the HHVs of its components:

The lower heating value (LHV) is typically about 10% less than the HHV due to the energy used to vaporize the water produced during combustion. We can estimate the LHV as follows:

Final

The overall combustion reaction for stoichiometric combustion of 1 mole of LPG with air is:

The stoichiometric F/A and A/F ratios are approximately 0.1533 and 6.52, respectively.

The higher heating value (HHV) per unit mass of LPG is approximately 49.91 MJ/kg, and the lower heating value (LHV) per unit mass of LPG is approximately 44.92 MJ/kg.

Answer:

q(problem 1) = 25,050 joules;  q(problem 2) = 4.52 x 10⁶ joules

Explanation:

To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...

=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:

Heat of fusion (ΔHₓ) = 80 cal/gram

Heat of vaporization (ΔHv) = 540 cal/gram

specific heat of ice [c(i)] = 0.50 cal/gram·°C

specific heat of water [c(w)] = 1.00 cal/gram·°C

specific heat of steam [c(s)] = 0.48 cal/gram·°C

Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ...   Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.

q(warming ice) =  m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal

q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal

q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal

q(evaporation of water) =  m·ΔHv = (10g)(540cal/g) = 5400 cal

q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal

Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.

Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).    

Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.

Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.

Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).

Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.

Problems containing a temperature change:

NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve.  Good luck on your efforts. Doc :-)

Answer:

See the attached file for the structure

Explanation:

See the attached file for the explanation

Explanation:

Given that,

The wavelength of high‑frequency radio wave, [tex]\lambda=0.25\ km=250\ m[/tex]

We need to find the energy of exactly one photon of this radio wave radiation. It is given by :

[tex]E=\dfrac{nhc}{\lambda}[/tex]

Here, n = 1

[tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{250}\\\\E=7.95\times 10^{-28}\ J[/tex]

So, the energy of exactly one photon of this radio wave radiation is [tex]7.95\times 10^{-28}\ J[/tex].

The energy of exactly one photon of this radio wave radiation is [tex]7.95 \times 10^{-28}J[/tex].

Given that,

Based on the above information, the calculation is as follows:

We know that

[tex]E = nhc \div \lambda\\\\= (6.63 \times 10^{-34} \times 3 \times 10^8) \div 250[/tex]

= [tex]7.95 \times 10^{-28}J[/tex]

Therefore we can conclude that the energy of exactly one photon of this radio wave radiation is [tex]7.95 \times 10^{-28}J[/tex].

Learn more: brainly.com/question/23334479

Answer:

CHECK THE ATTACHMENT TO SEE THE STRUCTURAL DIAGRAM

Explanation:

The spectral data suggests the product is ethylbenzene, which can be synthesized from benzene using ethylchloride and a catalyst like AlCl3 via a Friedel-Crafts alkylation reaction.

The spectral data provided corresponds to the molecule ethylbenzene. The presence of a 13C NMR signal at 142 ppm indicates a carbon directly invested in a aromatic system. The signals at 38 ppm, 24 ppm and 13 ppm are indicative of the ethyl side chain. The IR peaks at 3100 cm-1 and 2900 cm-1 indicate C-H stretching of aromatic and aliphatic hydrogens respectively. The absence of an absorbance near 1700 cm-1 indicates absence of carbonyl group. The lack of a M+2 peak in MS data suggests no halogens are part of the structure. To obtain ethylbenzene from benzene, the reagents used would include ethylchloride and aluminum chloride (AlCl3) in a Friedel-Crafts alkylation reaction.Benzene, ethylchloride and AlCl3 are the key components to this reaction.

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Answer:

Explanation:

Since it first order, we use order rate equation

In ( [tex]\frac{A1}{A0}[/tex]) = -kt where A1 is the final quality = 0.8 (80%), A0 is the initial quality = 1 ( 100%)

also, t half life = [tex]\frac{In2}{k}[/tex] where k is rate constant

k = [tex]\frac{In 2}{45 days}[/tex] = 0.0154

In ( [tex]\frac{0.8}{1}[/tex]) = - 0.0154 t

-0.223 / -0.0154 = t

t = 14.49 approx 14.5 days from the date the yogurt was packaged

Answer:

B) I2(s) is a stronger oxidizing agent than Al3 (aq), and Al(s) is a stronger reducing agent than I-(aq).

Explanation:

An oxidizing agent accepts electrons in a redox reaction and become reduced while a reducing agent looses electrons and become oxidized.

Hence in a redox reaction, oxidizing agents are reduced while reducing agents are oxidized.

Looking at I2 and Al3+, I2 is a better oxidizing agent since it has a reduction potential of +0.54V compared to -1.66V for Al3+.

Given the statements above, the converse must be true, that is; All is a better reducing agent compared to I-

Answer:

Correct Answer: B. 198 kJ/mol

Explanation:

In this reaction, four C–H bonds and two O–H bonds are broken, so the total energy absorbed is 4 · 413 + 2 · 463 = 1,652 + 926 = 2,578 kJ/mol. For the products, one C=O bond and three H–H bonds are formed, so the energy released is 1,072 + 3 · 436 = 1,072 + 1,308 = 2,380 kJ/mol. The total enthalpy change is equal to the energy absorbed minus the energy released, so we have 2,578 – 2,380 = 198 kJ/mol. The enthalpy change is positive, so the reaction is endothermic.

Final answer:

Bottle A contains barium hydroxide, Bottle B contains hydrochloric acid, and Bottle C contains sodium carbonate. When combined, they form a white precipitate in one case, release a gas in another case, and give off heat in the last case.

Explanation:

The bottles contain:

The chemical changes observed are:

Final answer:

Water molecules consist of polar covalent bonds within the molecule and hydrogen bonds between molecules. Polar covalent bonds are formed due to the oxygen atom sharing electrons with two hydrogen atoms and attracting them more strongly. Hydrogen bonds occur between the positive portions of hydrogen and negative oxygen atoms of adjacent water molecules.

Explanation:

Water molecules contain two types of bonds: polar covalent bonds and hydrogen bonds. Each water molecule is composed of one oxygen atom and two hydrogen atoms. The oxygen atom shares a pair of valence electrons with each hydrogen atom, forming two polar covalent bonds within the molecule. This happens because oxygen has a higher electronegativity than hydrogen, pulling the shared electrons closer and creating partial charges within the molecule. The result is that the oxygen atom acquires a partial negative charge while the hydrogen atoms have partial positive charges.

The hydrogen bonds occur between different water molecules. A hydrogen bond is a dipole interaction where the partially positive hydrogen atom of one water molecule is attracted to the electronegative oxygen atom of a nearby water molecule. While hydrogen bonds are much weaker than covalent bonds, they are strong enough in water to hold adjacent molecules together. The bridging of molecules by hydrogen bonds gives water many of its unique characteristics. Diagrams illustrating hydrogen bonds often represent them with dotted lines to indicate their relative weakness compared to solid lines representing covalent bonds.

Complete question:

The student performs a second titration using the 0.10MNaOH(aq) solution again as the titrant, but this time with a 20.mL sample of 0.20MHCl(aq) instead of 0.10MHCl(aq).

1. The box below to the left represents ions in a certain volume of 0.10MHCl(aq). In the box below to the right, draw a representation of ions in the same volume of 0.20MHCl(aq). (Do not include any water molecules in your drawing.)

Answer:

The concentration of H+ and Cl- will be doubled.

Explanation:

See the attached photo for the representation of ions in the same volume.

Final answer:

Balancing chemical reactions involves trial and error, with the aim of achieving the same number of atoms for each element on both sides of the equation. Atoms that are in only one of the reactants and only one of the products should be done last, and the final coefficients should be the biggest numbers possible.

Explanation: