Ascorbic acid ( H2C6H6O2 ) is a diprotic acid with a1=8.00×10−5 and a2=1.60×10−12. Determine the pH of each solution.

A 0.190M ascorbic acid ( H2C6H6O2 ) solution.
pH=

A 0.190M sodium ascorbate ( Na2C6H6O2) solution.
pH=

Answers

Answer 1

Ascorbic acid, also known as Vitamin C, is a water-soluble vitamin that plays an important role in many biological processes in the human body

How do you find out the pH of the given ascorbic acid solution?

For the first part

Step 1: Write the dissociation reactions of H₂C₆H₆O₂ in water:

H₂C₆H₆O₂ ⇌ H⁺ + HC₆H₆O²⁻

HC₆H₆O²⁻ ⇌ H⁺ + C₆H₆O₂²⁻

Step 2: Write the equilibrium expressions for each dissociation reaction:

Kₐ₁= [H⁺][HC₆H₆O²⁻ ] / [H₂C₆H₆O₂]

Kₐ₂= [H⁺][C₆H₆O₂²⁻] / [ HC₆H₆O²⁻]

Step 3: Use the given values of Kₐ₁ and Kₐ₂ to set up an ICE table and solve for [H⁺] and pH.

Kₐ₁  = 8.00×10⁻⁵

Kₐ₂ = 1.60×10⁻¹²

[H₂C₆H₆O₂] = 0.190 M

Let x be the concentration of [H⁺] from the dissociation of H₂C₆H₆O₂

[H⁺] = x M

[HC₆H₆O²⁻] = x M

[C₆H₆O₂²⁻] = x(Kₐ₁/Kₐ₂) M

Now, we can substitute the values into the equilibrium expressions to get:

Kₐ₁ = (x)(x) / (0.190-x)

Kₐ₂ = (x)(x(Ka1/Ka2)) / x

Simplifying and solving for x, we get:

x = 7.62 × 10⁻⁴ M

pH = -log[H⁺] = 3.12

Therefore, the pH of a 0.190 M ascorbic acid solution is 3.12.

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Related Questions

True/False: Paraffin wax (d = 0.910 g/cm3) would float in ethanol (d = 789 kg/L)

Answers

The claim that paraffin wax would float in ethanol (d = 789 kg/L) is accurate.

In ethanol, does paraffin wax float?

A 40–50% aqueous solution would have a density that could be adjusted to be just below that of paraffin wax, while regular alcohol (ethanol) has a density of roughly 0.8. The wax would then begin to sink. Warming causes the wax's density to significantly decrease (more than ethanol does), causing it to float.

Does ethanol allow you to float?

Even with your lungs completely expanded, you cannot float in 80-proof (or 40%) alcohol since your body weighs more than the booze and will sink if you stop swimming.

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The isotope Tl-208 undergoes β decay with a half-life of 3.1 min.

What is the decay constant for this process?

a.)
4.47 min⁻¹
b.)
2.15 min⁻¹
c.)
0.224 min⁻¹
d.)
0.031 min⁻¹

Answers

The decay constant for this process is

c.) 0.224 min⁻¹

How to find the decay constant

The decay constant (λ) is related to the half-life (t1/2) by the following equation:

λ = ln(2) / t1/2

where

ln(2) is the natural logarithm of 2, which is approximately 0.693.

Substituting the given half-life of 3.1 min into the equation, we get:

λ = ln(2) / (3.1 min) ≈ 0.223 min^(-1)

Therefore, the decay constant for the β decay of Tl-208 is approximately 0.223 min^(-1).

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im struggling

What quantity of heat (in kJ) would be required to convert 13.4 g of ice to water at 0.00 °C? (∆Hfus = 6.01 kJ/mol for water)

Answers

Around 80.5 KJ

Multiply Heat of Fusion and Mass to get the q value.

Please help me
Define acid.
Mention four products of destructive distillation of coal.
In a tabular, highlight two differences between diamond and graphite.
List four types of salt.
Outline two physical properties of a base.

Answers

1. An acid is a substance that produces hydrogen ions, H +(aq), when dissolved in water


4. acidic salt, basic salt, neutral salt, and double salt

I only know the two questions not the rest sorry

How much heat is released when 60.0 g of ethanol cools from 70 °C to 43 °C?
a) 1,600 J
b) 1500 J
c) 810 J
d) 750 J

Answers

The heat released is 1600 joules, so the correct option is the first one.

How much heat will be released?

To calculate the heat released when 60.0 g of ethanol cools from 70 °C to 43 °C, we can use the formula for heat transfer:

q = m * C * ΔT

where:

q = heat transfer (in joules)m = mass of the substance (in grams)C = specific heat capacity of the substance (in J/(g°C))ΔT = change in temperature (in °C)

Given:

Mass of ethanol (m) = 60.0 g

Specific heat capacity of ethanol (C) = 1.0 J/(g°C) (at constant pressure)

Change in temperature (ΔT) = Final temperature - Initial temperature = 43 °C - 70 °C = -27 °C

Note that the negative sign in ΔT indicates that heat is being released (i.e., the substance is cooling).

Plugging in the given values into the formula:

q = 60.0 g *1.0 J/(g°C) * (-27 °C)

q ≈ -1600 J

The negative sign is for notation, here we can see that the amount of heat is 1600 joules, so the correct option is the first one.

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A balloon holds 60.0 kg of helium. What is the volume of the balloon if the final pressure is 1.20 atm and the temperature is 22°C?

Answers

Answer:

Explanation:

Use the formula PV=nRT

P is pressure in atm

V is volume in whatever unit you're working in as long as everything is in that unit (anything volume related)

n is the number of moles

R is the constant so 0.08206

and T is temperature and this MUST be in Kelvin which is 173.15 + C

the equation can be shifted depending on what you need to solve

20. Calculate the mole fractions (X) of each compound in each of the following solutions:

a. 19.4 g of H2SO4 in 0.251 L of H20 (density of water is 1.00 g/mL)

b.35.7 g of KBr in 16.2 g of water

C.233 g of CO2 in 0.409 L of water

Answers

[tex]CO_{2}[/tex]The following compounds' mole fractions (X) are (a)0.986 (b)0.750 (c)0.811 for the given solutions.

How can the mole fraction of 19.4 g of H2SO4 in 0.251 L of water be determined?

[tex]H_{2}SO_{4}[/tex] mass is 19.4 g.

[tex]H_{2}SO_{4}[/tex]'s molecular weight is 98.08 g/mol.

It's molecular weight is 19.4 g/98.08 g/mol, or 0.1979 mol.

Density times volume is 1.00 g/mL times 0.251 L and 251 g for water mass.

[tex]H_{2} O[/tex] has a molecular weight of 18.02 g/mol.

Water moles are equal to 251 g / 18.02 g/mol, or 13.93 mol.

The solution's total moles are equal to 0.1979 mol plus 13.93 mol, or 14.13 mol.

Sulphuric Acid's mole fraction is equal to 0.1979 mol/14.13 mol, or 0.014.

Water mole fraction is equal to 13.93 mol / 14.13 mol, or 0.986 mol.

How can the mole fraction of 35.7 g of KBr in 16.2 g of water be determined?

KBr's mass is 35.7 g.

KBr has a molecular weight of 119 g/mol.

The formula for KBr is 35.7 g/119 g/mol, which equals 0.300 mol.

16.2 g of water in mass

Water has a molecular weight of 18.02 g/mol.

Water moles are equal to 16.2 g / 18.02 g/mol, or 0.899 mol.

The solution has a total of 1.199 moles (0.300 mol + 0.899 mol).

The mole fraction of KBr is equal to 0.300 mol/1.199 mol, or 0.250

Water mole fraction is equal to 0.899 mol / 1.199 mol, or 0.750 moles.

How can the mole fraction of 233 g of CO2 in 0.409 L of water be determined?

[tex]CO_{2}[/tex] mass = 233 g

It has a molecular weight of 44.01 g/mol.

Its moles are equal to 233 g / 44.01 g/mol, or 5.291 mol.

Water volume equals 0.409 L.

Water has a molecular weight of 18.02 g/mol.

(density × volume) / molecular weight (1.00 g/mL 409 mL) / 18.02 g/mol = 22.71 mol = number of moles of water

The solution's total moles are equal to 5.291 mol plus 22.71 mol, or 28.00 mol.

[tex]CO_{2}[/tex] mole fraction = 5.291 moles / 28.00 moles = 0.189

[tex]H_{2} O[/tex] mole fraction is 22.71 mol/28.00 mol, or 0.811 moles.

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If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?


2.94 x 10⁴ °C

0.00715 °C

7.15 °C

7.15 x 10³ °C

Answers

[tex] \ddots[/tex] The heat energy can be deduced as -

[tex] \odot\sf \footnotesize{Heat \:energy = Mass\: of\: substance\times Specific \:heat\times Change\: in \:temperature}\\[/tex]

[tex] \qquad :\implies\sf \boxed{\sf Q = mS\Delta T}\\[/tex]

Where-

Q = Heat energy (Joules, J)

m = Mass of a substance (g or,kg)

S = Specific heat ( J/g∙°C or, J/kg.°C)For liquid water, this value is 4.18J/g∙°C∆ is a symbol meaning "the change in"

∆T = Change in temperature (Kelvins, K)

In this instant, we are given -

Heat energy,Q = 14.5 KJ = 14500JMass of substant, m = 485 g

[tex] \ddots[/tex] Now that we have all the required values except ∆T,so we can plug the rest of the known values into the formula and solve for ∆T -

[tex] \qquad :\implies\sf \underline{Q = mS\Delta T}\\[/tex]

[tex] \qquad :\implies\sf 14500 = 485 \times 4.18 \times \Delta T\\[/tex]

[tex] \qquad :\implies\sf 14500 = 2027.3\times \Delta T\\[/tex]

[tex] \qquad :\implies\sf \Delta T = \dfrac{14500}{2027.3}\\[/tex]

[tex] \qquad :\implies\sf \Delta T = 7.152370........\:°C\\[/tex]

[tex] \qquad :\implies\sf \underline{\boxed{\sf \Delta T=7.15\:°C}}\\[/tex]

[tex] \ddots[/tex]Correct answer - [tex]\boxed{\sf \Delta T=7.15\:°C}.[/tex]

The water would increase its temperature by approximately 7.15°C if 14.5 kJ of heat were added. The third option is correct.

This is an exercise in specific heat and thermal conductivity which are two important physical properties that describe how materials interact with heat. Specific heat refers to the amount of energy required to raise the temperature of a material by a given amount, while thermal conductivity refers to a material's ability to transfer heat through itself.

The formula for specific heat is Q = mcΔT, where Q is the amount of heat transferred, m is the mass of the material, c is the specific heat, and ΔT is the change in temperature. The unit of measure for specific heat is J/(g*°C).

On the other hand, thermal conductivity is measured in terms of the amount of heat that is transferred through a material per unit time and area, given a temperature difference. It is expressed as the amount of heat transferred per second, per square meter, per meter of material thickness, when the temperature difference between the extremes is one Kelvin. Its formula is Q/t = -kA(∆T/∆x), where Q/t is the heat transfer rate, k is the thermal conductivity, A is the cross-sectional area, ∆T is the temperature difference, and ∆ x is the thickness of the material.

These properties are useful for understanding how materials interact with heat in a variety of situations, from building design to heating and cooling equipment manufacturing.

We solve the exercise, for the temperature change:

Now to calculate the temperature rise of 485 g of liquid water when 14.5 kJ of heat is added to it, we can use the formula:

Q = mcΔT

We must know that it has a quantity of heat of 14.5 Kj, with a mass of 485 g. The specific heat capacity of water is 4.18 J/(g °C).

First, we need to convert the heat added to joules:

Q = 14.5 KJ × (1000 J/1 KJ)

Q = 14500 J

We can then solve for ΔT. We clear the formula.

ΔT = Q / (m × c)

We substitute our data in the formula and solve the temperature change:

ΔT = Q / (m × c)

ΔT = (14500 J)/(485 g × 4.18 J/(g·°C))

ΔT ≈ 7.15 °C

The water would increase its temperature by approximately 7.15°C if 14.5 kJ of heat were added.

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6. What is the pH of a 0.25 M solution of NH4Cl? [Kb(NH3) = 1.8  10–5

Answers

The Ammonium Chloride solution at 0.25 M has a pH of 2.67.

Why is the pH of Ammonium Chloride below 7?

As a result, the weak basic (Chlorine) in the solution is overpowered by the conjugate acid (Ammonium cation), making the solution mildly acidic. According to the equation pH =log[Hydrogen ion], an acidic solution has a pH lower than 7. Aqueous ammonium chloride solution has a pH that is less than 7.

Ammonium cation + Water ⇌ Nitrogen trihydride + Hydronium ion

Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]

[Nitrogen trihydride] = [Hydronium ion] = x

[Ammonium cation] = 0.25 - x

Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]

1.8 × 10–5 = x² / (0.25 - x)

1.8 × 10–5 = x² / 0.25

x² = 4.5 × 10–6

x = 2.12 × 10–3

pH = -log[Hydronium ion] = -log(2.12 × 10–3) = 2.67

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1-A tennis ball travelling at a speed of 46 m/s with a mass of 58 g. Calculate its kinetic energy Ek=0.5mv2

2-A plane at a speed of 255 m/s with a mass of 2.15 × 105 kg. calculate its kinetic energy

3-A hot air balloon with a kinetic energy of 76 550 J and a mass of 1890 kg. Calculate its velocity

Answers

The tennis ball has a kinetic energy of around 56.8 J. The aircraft has a kinetic energy of around 4.43 x 10⁹ J. The hot air balloon travels at a speed of around 9.0 m/s.

A 750 kilogramme automobile travelling at a speed of 50.0 km/h has how much kinetic energy?

How much effort must be put into slowing down a 750 kg automobile from 100 km/h to 50 km/h. We know that the of this automobile at 50.0 km/h is 72,300 Joules from the last example problem.

Ek = 0.5 x 0.058 kg x (46 m/s)²

Ek = 0.5 x 0.058 kg x 2116 m²/s²

Ek = 56.8468 J

Ek = 0.5mv²

Ek = 0.5 x 2.15 x 10⁵ kg x (255 m/s)²

Ek = 0.5 x 2.15 x 10⁵ kg x 65025 m²/s²

Ek = 4.433 x 10⁹ J

Ek = 0.5mv²

v = √(2Ek/m)

v = √(2 x 76550 J / 1890 kg)

v = √(81.011 J/kg)

v = 9.0 m/s (approx.)

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C6H12O6 + 6 O₂ ---> 6CO₂ + 6 H₂O
How many moles of oxygen are needed to burn 5.00 moles of glucose (C6H12O6)?

Answers

Answer is 30 mol

Explanation
Ratio of C6H12O6:O2 IS 1:6
Mole of O2 = 5x6=30 mol

CHALLENGE The circles below represent of the large circle, and multiply it by 30. That Earth and the moon. Measure the diameter would be the correct distance from Earth to the moon at this scale. Draw the two circles in the space provided. Use the correct distance you found.● = Earth ●=moon ​

Answers

To draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.

What is the explanation for the above response?

If we assume that the larger circle represents the Earth, then the diameter of the Earth would be 30 times the diameter of the smaller circle representing the moon. Let's say that the diameter of the smaller circle is x. Then the diameter of the larger circle (Earth) would be 30 times x or 30x.

To find the correct distance from Earth to the moon at this scale, we need to know the actual distance from Earth to the moon, which is approximately 238,855 miles or 384,400 kilometers. If we divide this distance by the scale factor of 30, we get:

238,855 miles / 30 = 7,961.8 miles

Therefore, the diameter of the smaller circle (moon) would be approximately 7,961.8 miles / π = 2,532.5 miles (rounded to one decimal place). And the diameter of the larger circle (Earth) would be 30 times that or 75,974.4 miles

So, to draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.

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C) A solution containing NaOH and Na2CO3 was titrated with 0.1202 M HCl. Two titration measurements were carried out using different indicators to determine the concentration of sodium hydroxide and sodium carbonate in the solution. In the first titration 25.00 mL of this solution required 36.42 mL of HCl with bromocresol green as indicator. In the second titration, 25.00 mL of this solution required and 29.64 mL of HCl with phenolphthalein as the indicator. Calculate the concentration of each solute in mg/mL of solution

Answers

When sodium carbonate is titrated against HCl in the presence of the indicator phenolphthalein, it is transformed to NaCl.

When phenolphthalein was used to titrate a combination of NaOH and Na2CO3 with HCl?

To decolorize phenolphthalein, 50 mL of a combination of NaOH and Na2CO3titrated with N10 HCl using phenolphthalein indicator required 50 mL of HCl. At this point, methyl orange was added, and the acid addition was continued. The second endpoint was obtained when another 10 ml of N10 HCl was added.

You can use more than one indicator since the interaction between sodium carbonate and hydrochloric acid occurs in two phases. The first stage is better served by phenolphthalein, whereas the second is best served by methyl orange.

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For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species.

For the reaction

N2(g)+3H2(g)↽−−⇀2NH3(g)

the standard change in Gibbs free energy is Δ°=−32.8 kJ/mol
. What is ΔG for this reaction at 298 K when the partial pressures are N2=0.350 atm
, H2=0.300 atm
, and NH3=0.750 atm
?

Answers

We can use the following equation to calculate the Gibbs free energy change at non-standard conditions:

ΔG = Δ°G + RT ln(Q)

where Δ°G is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol-K), T is the temperature in Kelvin, and Q is the reaction quotient.

First, we need to calculate Q for the given partial pressures:

Q = (P(NH3))^2 / (P(N2) * P(H2)^3)
= (0.750 atm)^2 / (0.350 atm * 0.300 atm^3)
= 4.08

Next, we can substitute the values into the equation:

ΔG = -32.8 kJ/mol + (8.314 J/mol-K * 298 K) * ln(4.08)
= -32.8 kJ/mol + (2471 J/mol) * 1.407
= -32.8 kJ/mol + 3476 J/mol
= -29.3 kJ/mol

Therefore, the Gibbs free energy change for the reaction at 298 K and the given partial pressures is -29.3 kJ/mol.

If the volume of a gas at -40°C is double to 80 L what is the final temperature in degrees Celsius?

Answers

The final temperature is -160°C

To solve this problem

We can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

(P₁V₁)/T₁ = (P₂V₂)/T₂

Where

P₁, V₁, and T₁ are the initial pressure, volume, and temperature of the gas, and P₂, V₂, and T₂ are the final pressure, volume, and temperature of the gas

In this case, we can assume that the pressure of the gas is constant, since it is not given in the problem statement. So we can simplify the equation to:

(V₁/T₁) = (V₂/T₂)

Where

V₁ and T₁ are the initial volume and temperature V₂ and T₂ are the final volume and temperature

We are given that the initial volume (V₁) is 80 L and the final volume (V₂) is twice that, or 160 L. We are also given that the initial temperature (T₁) is -40°C. To find the final temperature (T₂), we can plug these values into the equation:

(V₁/T₁) = (V₂/T₂)

(80 L)/(-40°C) = (160 L)/T₂

Simplifying:

-2 L/°C = (160 L)/T₂

Multiplying both sides by -1°C/2 L (the reciprocal of -2 L/°C):

1/2 = (T₂)/(160 L) x (-1°C/2 L)

1/2 = -T₂/320

Multiplying both sides by -1 to isolate T₂:

-1/2 = T₂/320

T₂ = -160°C

Therefore, the final temperature is -160°C.

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In the Periodic Table below, shade all the elements for which the neutral atom has an outer electron configuration of ms2nd2, where n and m are integers, and =m+n1.

Answers

The elements that have an outer electron configuration of ms2nd2 are located in the d-block of the periodic table and include some of the transition metals and lanthanides.

What is the periodic table?

To determine which elements in the periodic table have this outer electron configuration, you can look at the position of the d-block elements in the table. The d-block elements are located in the middle of the table and include the transition metals. These elements have partially filled d orbitals, which can accommodate up to 10 electrons.

Elements in the d-block with an atomic number of 21 through 30 (scandium through zinc) have an outer electron configuration of d10s2 and do not fit the ms2nd2 configuration. However, elements in the d-block with an atomic number of 39 through 48 (yttrium through cadmium) have an outer electron configuration of d10s2p1 and can have the ms2nd2 configuration by removing the single electron in the p orbital. Elements in the d-block with an atomic number of 57 through 80 (lanthanum through mercury) also have the possibility of having an outer electron configuration of ms2nd2.

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Strong Acid/Strong Base Titration.
100 mL of 0.10M HNO3 titrated with 0.10M NaOH
a. Initial pH, 0.0 mL of NaOH
Circle one: Type of solution (strong acid, weak acid, strong base, weak base,
buffer, neither)
Reaction:

Answers

The compound NaOH as shown is a strong base.

Is NaOH a strong base?

NaOH (sodium hydroxide) is considered a strong base. A strong base is a base that dissociates completely in water to form hydroxide ions (OH-) and cations. NaOH is highly soluble in water and, when added to water, it completely dissociates into Na+ and OH- ions, which makes it a strong base.

The strength of a base depends on the extent of its dissociation in water. Strong bases dissociate completely in water, while weak bases dissociate only partially. The dissociation of a base is usually represented by its base dissociation constant (Kb), which is the equilibrium constant for the reaction of the base with water to form hydroxide ions.

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Please THANKS FOR WHO EVER WILL ANSWER THESE 14

Answers

The above is about the movement of lithospheric plates. See explanation and attached image for details.

What are the process of lithospheric plates movement?

The movement of lithospheric plates is driven by convection currents in the Earth's mantle, which are caused by heat generated from the Earth's core. These currents cause the lithospheric plates to move, and the motion can result in a variety of geological phenomena, including earthquakes, volcanic activity, and the formation of mountain ranges and oceanic trenches.

There are three main types of plate boundaries, and each one results in a different type of movement of lithospheric plates:

Divergent Boundaries: At divergent boundaries, lithospheric plates move away from each other. This movement is caused by the upwelling of hot material from the mantle, which pushes the plates apart. As the plates move away from each other, magma rises up to fill the gap between them, creating new crust. Divergent boundaries are where new oceanic crust is formed.

Convergent Boundaries: At convergent boundaries, lithospheric plates move towards each other. There are three types of convergent boundaries: oceanic-oceanic, oceanic-continental, and continental-continental. At oceanic-oceanic and oceanic-continental convergent boundaries, one plate is forced beneath the other, creating a subduction zone. This movement is caused by the sinking of a denser plate beneath a less dense plate. As the denser plate sinks, it melts and can trigger volcanic activity. At continental-continental convergent boundaries, the plates are too buoyant to subduct, so they instead buckle and push up, forming mountain ranges.

Transform Boundaries: At transform boundaries, lithospheric plates move past each other. This movement is caused by the lateral movement of the convection currents in the mantle. Transform boundaries can create large faults, which can lead to earthquakes.

Overall, the movement of lithospheric plates is a complex and dynamic process, driven by the movement of material within the Earth's mantle.

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At 25 ∘C
, the equilibrium partial pressures for the reaction

A(g)+2B(g)↽−−⇀C(g)+D(g)

were found to be A=5.63
atm, B=5.00
atm, C=5.47
atm, and D=5.63
atm.

What is the standard change in Gibbs free energy of this reaction at 25 ∘C
?

Answers

The standard change in Gibbs free energy of the reaction at 25 ∘C is -1.69 kJ/mol.

What is standard change?

To find the standard change in Gibbs free energy of the reaction, we need to use the following equation:

ΔG° = -RT ln(K)

where ΔG° is the standard change in Gibbs free energy, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (25 °C = 298 K), and K is the equilibrium constant.

To find K, we need to use the equilibrium partial pressures:

K = (PC × PD) / (PA × PB²)

where PA, PB, PC, and PD are the equilibrium partial pressures of A, B, C, and D, respectively.

Substituting the values, we get:

K = (5.47 atm × 5.63 atm) / (5.63 atm × (5.00 atm)²)

K = 0.6176

Now we can calculate the standard change in Gibbs free energy:

ΔG° = -RT ln(K)

ΔG° = -(8.314 J/mol·K) × (298 K) × ln(0.6176)

ΔG° = -1,690 J/mol or -1.69 kJ/mol

Therefore, the standard change in Gibbs free energy of the reaction at 25 ∘C is -1.69 kJ/mol.

What is free energy?

Free energy, also known as Gibbs free energy, is a thermodynamic quantity that represents the amount of energy in a system that is available to do work at a constant temperature and pressure. It is denoted by the symbol G and is expressed in units of joules (J) or calories (cal).

In simple terms, free energy is the energy that can be used to do work. It is defined by the equation:

ΔG = ΔH - TΔS

where ΔH is the change in enthalpy (heat content) of the system, ΔS is the change in entropy (disorder) of the system, and T is the absolute temperature in Kelvin.

If ΔG is negative, the reaction is spontaneous and can proceed without the input of external energy. If ΔG is positive, the reaction is non-spontaneous and requires energy input to proceed. If ΔG is zero, the system is at equilibrium.

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Phosphorus (P4) and oxygen react to produce diphosphorus pentoxide. How many grams of diphosphorus pentoxide are produced from the reaction of 38.76 grams of oxygen with excess phosphorus? Round your answer to the hundredths of a gram. Do not include units. Use the atomic masses in this PERIODIC TABLE. DO NOT ROUND THESE MASSES IN YOUR CALCULATION.

Answers

The mass of diphosphorus pentoxide produced is: 549.47 g

What is mass?

Mass is a measure of the amount of matter in an object. It is typically measured in units of grams (g) or kilograms (kg). Mass is a scalar quantity and is different from weight, which is the force exerted on an object due to gravity and depends on both the mass of the object and the acceleration due to gravity.

The balanced chemical equation for the reaction between phosphorus and oxygen to produce diphosphorus pentoxide is:

[tex]4P[/tex] + [tex]5O_{2}[/tex] → [tex]2P_{4}O_{10}[/tex]

To solve the problem, we need to first calculate the amount of phosphorus required to react with 38.76 grams of oxygen. Since phosphorus is in excess, the amount of diphosphorus pentoxide produced will be limited by the amount of oxygen.

The molar mass of oxygen is 15.999 g/mol. Therefore, the number of moles of oxygen in 38.76 grams is:

38.76 g / 15.999 g/mol = 2.422 mol

According to the balanced equation, 5 moles of oxygen react with 4 moles of phosphorus to produce 2 moles of diphosphorus pentoxide. Therefore, the number of moles of phosphorus required is:

4/5 × 2.422 mol = 1.9376 mol

The molar mass of diphosphorus pentoxide is 283.886 g/mol. Therefore, the mass of diphosphorus pentoxide produced is:

1.9376 mol × 283.886 g/mol = 549.47 g

Rounding to the hundredths of a gram, the answer is:

549.47 g → 549.47

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write the correct IUPAC for this molecule.
I need help please

Answers

I think it is 2,4-dimethylpentane

Consider the reaction described by the chemical equation shown.

C2H4(g)+H2O(l)⟶C2H5OH(l)Δ∘rxn=−44.2 kJ

Use the data from the table of thermodynamic properties to calculate the value of Δ∘rxn
at 25.0 ∘C.


Δ∘rxn= ? J⋅K−1

Calculate Δ∘rxn.

Δ∘rxn= ? kJ


In which direction is the reaction, as written, spontaneous at 25 ∘C
and standard pressure?
reverse
both
neither
forward

Answers

The direction of the reaction, as written, spontaneous at 25 ∘C and standard pressure is reverse.

What is the direction of the reaction?

To calculate the value of Δ∘rxn at 25.0 ∘C, we can use the equation:

Δ∘rxn(T2) = Δ∘rxn(T1) + ΔH∘(products) - ΔH∘(reactants)

where;

T2 is the desired temperature (25.0 ∘C), T1 is the standard temperature (usually 25 ∘C), ΔH∘(products) is the enthalpy change of formation of the products, and ΔH∘(reactants) is the enthalpy change of formation of the reactants.

Using the data from the table of thermodynamic properties, we can look up the enthalpy change of formation values for C2H4(g), H2O(l), and C2H5OH(l):

ΔH∘f(C2H4(g)) = 52.26 kJ/mol

ΔH∘f(H2O(l)) = -285.83 kJ/mol

ΔH∘f(C2H5OH(l)) = -277.69 kJ/mol

Substituting these values into the equation, we get:

Δ∘rxn(25.0 ∘C) = -44.2 kJ + (-277.69 kJ/mol) - (-52.26 kJ/mol)

Δ∘rxn(25.0 ∘C) = -44.2 kJ - (-277.69 kJ/mol) + 52.26 kJ/mol

Δ∘rxn(25.0 ∘C) = -44.2 kJ + 277.69 kJ/mol + 52.26 kJ/mol

Δ∘rxn(25.0 ∘C) = 233.23 kJ/mol

So the value of Δ∘rxn at 25.0 ∘C is 233.23 kJ/mol.

In which direction is the reaction, as written, spontaneous at 25 ∘C and standard pressure?

Since the value of Δ∘rxn at 25.0 ∘C is positive (233.23 kJ/mol), the reaction as written is not spontaneous at this temperature and standard pressure. The correct answer is "reverse."

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Calculate the density of Sulfur dioxide gas at a temperature of 15oC and pressure of 300 torr. Convert to atm

Answers

The density of sulfur dioxide gas at a temperature of 15°C and pressure of 300 torr is 0.001022 g/cm³, or 0.001022 g/mL, or 1.022 kg/m³, or 0.01022 g/L when converted to atm.

What is density?

To calculate the density of sulfur dioxide gas at a temperature of 15°C and a pressure of 300 torr, we can use the ideal gas law:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.08206 L·atm/(mol·K)), and T is the temperature in Kelvin.

First, we need to convert the given temperature of 15°C to Kelvin:

T = 15°C + 273.15 = 288.15 K

Next, we can rearrange the ideal gas law to solve for the number of moles:

n = PV/RT

where we can use the given pressure of 300 torr and convert it to atm by dividing by 760 torr/atm:

P = 300 torr / 760 torr/atm = 0.3947 atm

Substituting the values into the equation, we get:

n = (0.3947 atm) V / (0.08206 L·atm/(mol·K) × 288.15 K)

Now, we can use the molar mass of sulfur dioxide, which is 64.06 g/mol, to convert the number of moles to mass:

mass = n × molar mass

Finally, we can calculate the density of sulfur dioxide gas using the mass and volume:

density = mass / V

To convert the density from g/L to g/cm³, we divide by 1000.

Putting it all together, we get:

n = (0.3947 atm) V / (0.08206 L·atm/(mol·K) × 288.15 K)

n = 0.01595 V

mass = n × molar mass = 0.01595 V * 64.06 g/mol = 1.022 gV

density = mass / V = 1.022 gV / V = 1.022 g/L = 0.001022 g/cm³

Therefore, the density of sulfur dioxide gas at a temperature of 15°C and pressure of 300 torr is 0.001022 g/cm³, or 0.001022 g/mL, or 1.022 kg/m³, or 0.01022 g/L when converted to atm.

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Complete question is: The density of Sulfur dioxide gas at a temperature of 15oC and pressure of 300 torr is 0.01022 atm.

When you balance the equation Ca(OH)₂ + H₃PO₄ ---> Ca₃(PO₄)₂ + H₂O, what is the coefficient of calcium phosphate?


a.)
1
b.)
3
c.)
2
d.)
6

Answers

The answer is A --------

Question 5(Multiple Choice Worth 3 points)
(07.02 LC)

The substances below are listed by increasing specific heat capacity value. Starting at 30.0 °C, they each absorb 100 kJ of thermal energy. Which one do you expect to increase in temperature the least?

a) Cadmium, 0.230 J/(g °C)
b) Sodium, 1.21 J/(g °C)
c) Water, 4.184 J/(g °C)
d) Hydrogen, 14.267 J/(g °C)

Answers

Component form of the vector v is as follows: 4 3 1.5 1 Using the standard basis vectors I and j), express the vector w as follows: 3 two 1 4 pp . 1 3 w 3.5 C. V plus w= d. Determine the vector v's magnitude

What does "vector" mean?

Latin word for "carrier" is "vector." Point A is transported to point B by vectors. The orientation of the vectors AB is the direction in which point A is moved in relation to point B, and the amplitude of the vector is the width of the line connecting the two locations A and B. The terms Euclidean vectors and spatial vectors are also used to refer to vectors.

A vector space is what?

A vector space, also known as a linear space, is a collection of things called vectors that can be added to and multiplied ("scaled") by figures called scalars in the fields of mathematics, physics, and engineering.

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What is the Molality of a solution in which
25 g of sodium chloride is dissolved in 2.0
kg of water?

Answers

The molality of a solution is determined by the amount of solute (in moles) and the mass of the solvent (in kilograms). To convert the mass of NaCl to moles, the molar mass of NaCl is 58.44 g/mol. The number of moles of NaCl is 25 g / 58.44 g/mol = 0.427 mol. The molality of the solution is 0.213 mol/kg.

What is molality?

The amount of a solute dissolved in a solvent is indicated by the chemical term "molality," which is commonly defined in terms of moles of solute per kilogramme of solvent. Because it takes into account variations in the volume of the solution owing to temperature and pressure, it differs from molarity, which quantifies the quantity of a solute in moles per litre of solution.

To calculate the molality of a solution, we need to know the amount of solute (in moles) and the mass of the solvent (in kilograms).

In this case, we are given:

Mass of solute (NaCl) = 25 g

Mass of solvent (water) = 2.0 kg

To calculate the amount of solute in moles, we need to convert the mass of NaCl to moles using its molar mass:

Molar mass of NaCl = 58.44 g/mol

Number of moles of NaCl = (25 g) / (58.44 g/mol) = 0.427 mol

Now we can calculate the molality of the solution:

Molality = (number of moles of solute) / (mass of solvent in kg)

Molality = (0.427 mol) / (2.0 kg) = 0.213 mol/kg

Therefore, the molality of the solution is 0.213 mol/kg.

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Consider the following reaction at 298 K.

C(graphite)+2Cl2(g)⟶CCl4(l)Δ∘=−139 kJ

Calculate the following quantities. Refer to the standard entropy values as needed.

Δsys= ? J/K

Δsurr= ? J/K

Δuniv= ? J/K

Answers

System entropy = -80.8 J/K Surrounding entropy = 253.7 J/K The universe's entropy is 172.9 J/K. The reaction is unplanned. The procedure is a natural one.

In layman's terms, what is entropy?

Entropy is a measure of energy quality in the way that as lower the entropy, the more desirable the energy. Energy stored in a well-organized manner (the efficient library has a lower entropy. The entropy of energy contained in a chaotic manner (the random-pile library) is high.

What is another name for entropy?

Entropy is a gauge of a system's randomness or disorder. Entropy is greater in gases than in liquids, and greater in liquids than in solids. Order and disorder are important concepts in physical systems also known as randomness.

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2AI + 6HCI=2AlCl3 + 3H₂
3. Aluminum reacts with HCI to produce aluminum chloride (AICI3) and hydrogen gas (H₂).
Calculate the number of moles of HCI required to react with 0.62 moles of Al.

Answers

3.0 moles of [tex]Al[/tex] can fully react with hydrogen chloride to produce 4.5 moles of [tex]H_{2}[/tex]. Thus, 0.93 moles will be produced by 0.62 moles of [tex]Al[/tex].

STOICHIOMETRYBased on this inquiry, how does aluminum react with hydrogen chloride to produce aluminum chloride and hydrogen gas[tex]Al +6HCl= AlCl_{3} +3H_{2}[/tex]According to this equation, 3 moles of hydrogen gas are produced during the reaction of 2 moles of aluminum ([tex]Al[/tex]).As a result, 3 moles of aluminum will result in 3 3 2 = 4.5 moles of hydrogen gas.As a result, the entire reaction of 3.0 moles of [tex]Al[/tex]with hydrogen chloride can produce 4.5 moles of [tex]H_{2}[/tex].The proportion of reactants to products before, during, and after chemical processes is known as stoichiometry.

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Which state of matter - solid, liquid, or gas- tends to have unique factors (different from the other two) to consider when discussing solubility

Answers

The state of matter that tends to have unique factors to consider when discussing solubility compared to the other two states (solid and gas) is the liquid state.

Which state has unique factors?

Solubility refers to the ability of a substance (solute) to dissolve in a particular solvent to form a homogeneous mixture (solution). Various factors can affect the solubility of a substance, including temperature, pressure, and the nature of the solute and solvent.

In the case of liquids, the unique factor to consider when discussing solubility is often temperature. The solubility of many solid solutes in liquids generally increases with increasing temperature. This is because higher temperatures provide more energy to break the intermolecular forces between solute particles, allowing them to disperse more evenly throughout the solvent.

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The table shows the number of charged particles in an ion.


Charged Particles
Charge on Particle Number of Particles
Positive 3
Negative 2

A negatively charged substance is brought near the ion. What will most likely happen?
The negatively charged ion will repel the substance.
The negatively charged ion will attract the substance.
The positively charged ion will repel the substance.
The positively charged ion will attract the substance.

Answers

The answer is D. It’s positively charged because there are more positive than negative ions and remember opposites always attract because it wants to be balanced, attracting the negative ion.

Answer: three

Explanation:

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