A geochemist in the field takes a sample of water from a rock pool lined with crystals of a certain mineral compound . He notes the temperature of the pool, , and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to . Then he filters it and evaporates all the water under vacuum. Crystals of are left behind. The researcher washes, dries and weighs the crystals. They weigh 3.00 g.

a. Using only the information above, can you calculate the solubility of X in water at 26 degrees Celsius.?
b. If yes than calculate the solubility of X. Round your answer to 3 significant digits

Answer:

ok

Explanation:

I have the same question

Answer:

a) Warmer

b) Exothermic

c) -10.71 kJ

Explanation:

The reaction:

KOH(s) → KOH(aq) + 43 kJ/mol

It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.

Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.

The enthalpy change for the dissolution of 14 g of KOH is:

[tex] n = \frac{m}{M} [/tex]

Where:

m: is the mass of KOH = 14 g

M: is the molar mass = 56.1056 g/mol

[tex] n = \frac{m}{M} = \frac{14 g}{56.1056 g/mol} = 0.249 mol [/tex]

The enthalpy change is:

[tex] \Delta H = -43 \frac{kJ}{mol}*0.249 mol = -10.71 kJ [/tex]

The minus sign of 43 is because the reaction is exothermic.

I hope it helps you!

The potential of this concentration cell can be calculated using the Nernst Equation, which provides an approximation of about 0.098 V for the potential of the cell. This corresponds to choice (e).

The potential for this concentration cell can be calculated using the Nernst equation, which incorporates the standard reduction potential and the concentrations of the reaction species in its equilibrium state. The Nernst Equation can be written as:

E = E0 - (0.05916/n) * logQ

Where E is the cell potential, E0 is the standard cell potential (+0.34 V for Cu2+/Cu), n is the number of moles of electrons exchanged in the reaction (which is 2 for this reaction), and Q is the reaction quotient, which is the ratio of concentration of products to reactants.

Since this cell involves Cu2+ being reduced to Cu, the concentration in the cell where [Cu2+]=2.0M would be considered the product, and the one where [Cu2+]=1.0×10–3M would be considered the reactant. Therefore, Q=[product]/[reactant]=2.0/ (1.0×10–3) = 2000.

Plugging these values into the Nernst Equation, we get:

E = 0.34V - (0.05916/2) * log2000

After calculation, gives us an E value of approximately V, which correlates to option (e).

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Answer:

hhgjghjfgjfghj

Explanation:

Answer:

Pollutants are typically the cause of major water quality degradation around the world. Globally, the most prevalent water quality problem is eutrophication, a result of high-nutrient loads (mainly phosphorus and nitrogen), which substantially impairs beneficial uses of water.

Explanation:

Answer:

Water pollution is caused by industrial waste, temperature rising, deforestation and many other factors. All these facts are affecting water and they happen all over the world in many large bodies of water. Thus, it is a global issue,

Explanation:

I explained it in the answer loll.

Answer:

check which reactant is totally consumed and which one remains in the mixture

Explanation:

Apart from doing calculations during an experiment, one can determine which reactant is limiting and which one is in excess by checking the resulting mixture for the presence of reactants.

A limiting reactant is one that determines the amount of product formed during a reaction. It is usually a reactant that is lower than stoichiometry amount.

On the other hand, an excess reactant is one that is present in more than the stoichiometrically required amount during a reaction.

Limiting reactants will be totally consumed in a reaction while excess reactant would still be seen present in mixture after the reaction has stopped.

Hence, apart from using stoichiometric calculation to determine which reactant is limiting or in excess during an experiment, one can just check the final mixture of the reaction for the presence of any of the reactants. The reactant that is detected is the excess reactant while the one without traces in the final mixture is the limiting reactant.

Answer:

The grams of rust present in the bicycle  frame is  [tex]x = 8.50 g[/tex]

Explanation:

From the question we are told that

      The number of oxygen atom contained is  [tex]n = 9.62 *10^{22} \ atoms[/tex]

The molar mass of the compound is [tex]M_{Fe_3O_3} = 159.69 g[/tex]

At standard temperature and pressure the number of oxygen atom in one mole of  iron(III) oxide is  mathematically evaluated as

        [tex]N_o = 3 * Ne[/tex]

Where Ne is the avogadro's constant with a value [tex]N_e = 6.023 *10^{23} \ atoms[/tex]

      So

            [tex]N_o= 3 * 6.023*10^{23} \ atoms[/tex]

            [tex]N_o= 1.8069*10^{24} \ atoms[/tex]

So

   [tex]1.8069*10^{24} \ atoms[/tex] is  contained in [tex]M_{Fe_3O_3} = 159.69 g[/tex]

   [tex]9.62 *10^{22} \ atoms[/tex]   is contained in x

So

     [tex]x = \frac{159.69 * 9.62 *10^{22}}{1.8069 *10^{24}}[/tex]

    [tex]x = 8.50 g[/tex]

Final answer:

To calculate the grams of rust on the bicycle frame, we need to use stoichiometry. By converting the number of oxygen atoms to moles and then to grams using the molar mass of iron(III) oxide, we find that there are approximately 5.37x10^22 grams of rust on the bicycle frame.

Explanation:

To determine the grams of rust present on the bicycle frame, we need to calculate the molar mass of iron(III) oxide and then use stoichiometry to convert the number of oxygen atoms to grams of rust. The molar mass of Fe2O3 is 159.69 g/mol. Given that 1 mole of Fe2O3 contains 3 moles of oxygen atoms, we can calculate the moles of Fe2O3 using the given number of oxygen atoms and then convert it to grams using the molar mass.

9.62x10^22 oxygen atoms x (1 mol Fe2O3 / 3 mol O) x (159.69 g Fe2O3 / 1 mol Fe2O3) = 5.37x10^22 g Fe2O3

Therefore, there are approximately 5.37x10^22 grams of rust present on the bicycle frame.

Answer:

54 L

Explanation:

Given data

We have a concentrated NaCl solution and we will add water to it to obtain a diluted NaCl solution. We can find the volume of the final solution using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\V_2 = \frac{C_1 \times V_1}{C_2} = \frac{9.0M \times 450mL}{0.075M} = 5.4 \times 10^{4} mL = 54 L[/tex]

The volume of the final solution should be 54L.

Since

Initial concentration (C₁): 9.0 M

Initial volume (V₁): 450 mL

Final concentration (C₂): 0.075 M

We know that

C1 * V1 = C2 * V2

V2 = C1 * V1/ C2

= 9.0M * 450 mL/ 0.075 M

= 54 L

hence, The volume of the final solution should be 54L.

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Answer:an exoskeleton

Explanation:jus did it

Answer:

B

Explanation:

Answer:

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

Explanation:

Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF

0.3473 = m * 1.86

Solving, m = 0.187 m

Moles of HClO2 = mass / molar mass = 5.85 / 68.5 = 0.0854 mol

Molality = moles / mass of water in kg = 0.0854 / 1 = 0.0854 m

Initial molality

Assuming that a % x of the solute dissociates, we have the ICE table:

                 HClO2         H+    +   ClO2-

initial concentration:       0.0854                    0             0

final concentration:      0.0854(1-x/100)   0.0854x/100   0.0854x / 100

We see that sum of molality of equilibrium mixture = freezing point molality

0.0854( 1 - x/100 + x/100 + x/100) = 0.187

2.1897 = 1 + x / 100

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

Answer:

23388 years

Explanation:

Now we have to use the formula;

0.693/t1/2 = 2.303/t log No/N

Where

t1/2 = half life of the C-14 = 5700

t= age of the sample

No= radioactive material in a modern sample = No

N= radioactive material in a sample being studied. = 0.81No( since the amount N= No-0.19No)

Hence;

0.693/5700 = 2.303/t log No/0.81No

0.693/5700 = 2.303/t log 1/0.81

0.693/5700 = 2.303/t × 1.235

0.693/5700= 2.844/t

1.216×10^-4= 2.844/t

t= 2.844/1.216×10^-4

t= 23388 years

Answer:

7.5

Explanation:

pH + pOH = 14

pOH = 14 - 6.5

pOH = 7.5

The pOH of a solution, given a pH of 6.5 at temperature 25°C, is 7.5. This is calculated by subtracting the pH from 14.

The pH and pOH of a solution are related by the equation pH + pOH = 14 at 25°C. Given that the pH of the solution is 6.5, to find the pOH you simply subtract the pH from 14. So, pOH = 14 - pH. In this case, pOH = 14 - 6.5 = 7.5. Therefore, the pOH of a solution with a pH of 6.5 is 7.5.

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Answer:

Check the explanation

Explanation:

VOLUME OF newly freed EDTA

=(VOLUME OF Ca2* STRENGTH OF Ca2)/ STRENGTH OF EDTA

=(24.9*0.0220)/ 0.0700

=7.825mL

STRENGTH OF Cd2

=( VOLUME OF newly freed EDTA * STRENGTH OF EDTA) / VOLUME OF SAMPLE

=(7.825*0.0700)/50

=0.0109 M

VOLUME OF excess unreacted EDTA

=(VOLUME OF Ca2* STRENGTH OF Ca2)/ STRENGTH OF EDTA

=(19.5*0.0270)/ 0.0400

=13.16mL

VOLUME OF EDTA REQUIRED FOR SAMPLE CONTAINING Cd2 AND Mn2= (17.1-13.16) mL

      =3.94 mL

VOLUME OF EDTA REQUIRED FOR Mn2

= (VOLUME OF EDTA REQUIRED FOR SAMPLE CONTAINING Cd2 AND Mn2 - VOLUME OF newly freed EDTA )

=40.02-7.825 mL

=32.20 mL

STRENGTH OF Mn2

=( VOLUME OF EDTA REQUIRED FOR Mn2* STRENGTH OF EDTA) / VOLUME OF SAMPLE

=(32.20*0.0700)/50

=0.045 M

The anion that would bond with K+ in a 1:1 ratio to form a neutral ionic compound is F- (fluoride ion)

The anion that would bond with K+ in a 1:1 ratio to form a neutral ionic compound is B. F- (fluoride ion).

Potassium (K+) is a cation with a +1 charge, and to form a neutral ionic compound, it needs to bond with an anion with a -1 charge. The fluoride ion (F-) has a -1 charge, and therefore, it can form a 1:1 ratio with K+ to create a neutral compound.

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Answer:

See explanation below

Explanation:

A claisen reaction, is often used between two esters (One of them, usually having alpha hydrogens atoms) to form Beta Keto esters. Depending of the reagents, this reaction is taking place with base where one ester acts as nucleophile and the other as electrophile.

Now, we have here ethyl propanoate and ethyl formate. The ethyl propanoate has two alpha hydrogens, while ethyl formate do not have that. So, the base will react with the ethyl propanoate first, and then, it will react as nucleophile with the formate.

Now, the base to be used in this case will have to be a relatively strong base such sodium ethanoate, because if we use a strong base such NaOH, this will cause the saponification of the ester and not the condensation. So, in order to promove the claisen condensation, we need to use a base not too strong.

In the picture attached you have the mechanism and structure of the final product.

Answer:

4585 KJ

Explanation:

Data obtained from the question. This includes following:

Number of mole (n) of glycerin = 50 moles

Heat of fusion glycerin (Hf) = 91.7 KJ/mol

Heat (Q) released =?

The heat released can be obtained as follow:

Q = n•Hf

Q = 50 x 91.7

Q = 4585 KJ

Therefore, 4585 KJ of heat is released.

Answer:

The new volume is 87.5 ml.

Explanation:

We have,

Volume of sample of neon is 175 ml

The pressure changed from 75 kPa to 150 kPa.

We need to find new volume.

It is based on the concept of Boyle's law. Let V₂ is new volume. So,

[tex]P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{175\times 75}{150}\\\\V_2=87.5\ ml[/tex]

So, new volume is 87.5 ml.

Using Boyle's Law, the new volume of the neon gas when the pressure is changed from 75 kPa to 150 kPa is calculated to be 87.5 mL.

To find the new volume of a gas when its pressure is changed, we use Boyle's Law.

Boyle's Law states that the pressure and volume of a gas are inversely proportional when temperature and the number of gas molecules remain constant.

The formula is expressed as:

P₁V₁ = P₂V₂

In this problem:

We need to find the final volume, V₂. Rearranging Boyle's Law, we get:

V₂ = (P₁V₁) / P₂

Plugging in the values:

V₂ = (75 kPa * 175 mL) / 150 kPa

V₂ = 13125 mL / 150 kPa

V₂ = 87.5 mL

Therefore, the new volume of the neon gas sample is 87.5 mL.

Answer:

When you begin to add the molecules, the temperature inside the container is  300  K. After all 250 molecules have been added, the temperature inside the container is  300  K.

Answer:

see explanation below

Explanation:

In this case, we need to write the overall reaction:

HC₃H₅O₃ + H₂O <-------> C₃H₅O₃⁻ + H₃O⁺

The lactic acid is a weak acid, so, when it dissociates in it's ions, part of the acid is dissociated. This depends of it's Ka to know which quantity was dissociated.

To calculate Ka, let's write an ICE chart first:

     HC₃H₅O₃ + H₂O <-------> C₃H₅O₃⁻ + H₃O⁺      Ka = ?

i)       0.045                                  0            0

c)          -y                                     +y           +y

e)   0.045 - y                                 y              y

Writting the Ka expression we have:

Ka = [C₃H₅O₃⁻] [H₃O⁺] / [HC₃H₅O₃]

Now, to calculate Ka we need the values of [C₃H₅O₃⁻] and [H₃O⁺] in equilibrium. Fortunately, we have the value of the pH, which is 2.63 and with this we can get the value of [H₃O⁺] and then, the value of y. With that value, we replace it in the Ka expression to calculate Ka:

[H₃O⁺] = 10^(-pH)

[H₃O⁺] = 10^(-2.63)

[H₃O⁺] = [C₃H₅O₃⁻] = x = 2.34x10⁻³ M

Now, let's replace this value in the Ka expression:

Ka = (2.34x10⁻³)² / (0.045 - 2.34x10⁻³)

Ka = 1.28x10⁻⁴

b) Now, let's calculate the pH with the obtained value of Ka. We will use the same expression of Ka so:

1.28x10⁻⁴ = y² / (0.045-y)    

1.28x10⁻⁴ (0.045 - y) = y²

5.76x10⁻⁶ - 1.28*10⁻⁴y = y²

y² + 1.28x10⁻⁴y - 5.76x10⁻⁶ = 0

From here, we'll use the quadratic equation general formula, for solving y:

y = -1.28x10⁻⁴ ±√(1.28x10⁻⁴)² + 4 * 1 * 5.76x10⁻⁶ / 2

y =  -1.28x10⁻⁴ ±√2.31x10⁻⁵ / 2

y = -1.28x10⁻⁴ ± 4.8x10⁻³ / 2

y₁ = 2.34x10⁻³ M

y₂ = -2.464x10⁻³ M

Now, as y₁ is positive this is the value that we will take.

This value would be the [H₃O⁺] in equilibrium.

The value of pH would be:

pH = -log[H₃O⁺]

pH = -log(2.34x10⁻³)

pH = 2.631

According to this value of pH we can actually expect an inprovement in the accuracy, basically because we obtain a value with more significant figures, and this are relationed with accuracy.

Answer:

the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]

it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C

Explanation:

Given that :

[tex]k_1 = 7.21*10^{-3} s^{-1} \\ \\ E_a = 247 kJ/mol \ \ \ = 247*10^3 \ J/mol \\ \\ T_1 = 634 ^ {^0} C= (273 + 634) K = 907 \ K \\ \\ T_2 = 741^{^0 } C = (273+ 741) K = 1014 \ K \\ \\ R =8.314 \ \ J/mol/K[/tex]

a)

According to Arrhenius Equation ;

[tex]In\frac{k_2}{k_1} = -\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = 3.456412[/tex]

[tex]\frac{k_2}{7.21*10^{-3}} =e^{ 3.456412[/tex]

[tex]k_2 = 31.70302 * 7.21*10^{-3}[/tex]

[tex]k_2 = 0.22858 \ s^{-1}[/tex]

Therefore ,  the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]

b) Given that  :

[tex]t_{(1/2)} = 96.1 \ s[/tex]

[tex]k_1 = \frac{0.693}{ t_{(1/2)}}[/tex]

[tex]k_1 = \frac{0.693}{ 96.1 \ s}[/tex]

[tex]k_1 = 7.211*10^{-3} \ s^{-1}[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = 3.456412[/tex]

[tex]\frac{k_2}{7.211*10^{-3}} =e^{ 3.456412[/tex]

[tex]k_2 =e^{ 3.456412 * 7.21*10^{-3}[/tex]

[tex]k_2 = 0.22862 \ s^{-1}[/tex]

[tex]t_{(1/2)_2}} = \frac{0.693}{k_2}[/tex]

[tex]t_{(1/2)_2}} = \frac{0.693}{0.22862}[/tex]

[tex]t_{(1/2)_2}} =3.0313 \ s[/tex]

it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C

Explanation:

(a) The given data is as follows.

            B = [tex]2.180 \times 10^{-18} J[/tex]

            Z = 4 for Be

Now, for the first excited state [tex]n_{f}[/tex] = 2; and [tex]n_{i} = \infinity[/tex] if it is ionized.

Therefore, ionization energy will be calculated as follows.

         I.E = [tex]\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})[/tex]

              = [tex]8.72 \times 10^{-18} J[/tex]

Converting this energy into kJ/mol as follows.

           [tex]8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol[/tex]  

           = 5249 kJ/mol

Therefore, the ionization energy of the [tex]Be^{3+}[/tex] ion in its first excited state in kilojoules per mole is 5249 kJ/mol.

(b) Change in ionization energy is as follows.

         [tex]\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}[/tex]

   [tex]\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}[/tex]                

        [tex]\lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}[/tex]

                     = [tex]303.7 \times 10^{-10} m[/tex]

or,                 = [tex]303.7^{o}A[/tex]

Therefore, wavelength of light given off from the [tex]Be^{3+}[/tex] ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels [tex]303.7^{o}A[/tex].

The ionization energy of the Be³+ ion in its first excited state is approximately 5250 kJ/mol. The wavelength of light emitted when an electron transitions from n=4 to n=2 in a Be³+ ion is approximately 75.9 nm.

Ionization Energy and Wavelength Calculation for Be³+

The problem involves calculating the ionization energy of a Be³+ ion in its first excited state and the wavelength of light emitted during an electron transition.

a) Ionization Energy

For a one-electron species like Be³+, the energy in the nth state is given by:

E = –BZ²/n², where Z is the charge on the nucleus and B = 2.180 × 10⁻¹⁸ J.

In the first excited state (n = 2) for Be³+ (Z = 4), we get:

En = –B(4)²/2² = –2.180 × 10⁻¹⁸ J × 16/4 = –8.72 × 10⁻¹⁸ J.

To find the ionization energy, we need the energy to liberate 1 mole of Be³+ ions:

Ionization Energy = |En| × Avogadro's Number = 8.72 × 10⁻¹⁸ J × 6.022 × 10²³ mol⁻¹ = 5.25 × 10⁶ J/mol = 5250 kJ/mol.

b) Wavelength of Light

The energy difference between the fourth (n=4) and second (n=2) energy levels is calculated using:

ΔE = E₄ - E₂ = -BZ² / 4² - (-BZ² / 2²) = BZ² (1/4² - 1/2²)

ΔE = 2.180 × 10⁻¹⁸ J × 16 × (1/16 - 1/4) = 2.180 × 10⁻¹⁸ J × 16 × (-¾) = -2.616 × 10⁻¹⁷ J

The emitted photon's wavelength is given by:

λ = hc /ΔE = (6.626 × 10⁻³⁴ Js × 3.00 × 10⁸ m/s) / 2.616 × 10⁻¹⁷ J ≈ 7.59 × 10⁻⁸ m = 75.9 nm

Answer:

Explanation:

From the data it is clear that after each .80881 hours , the concentration becomes half . so its decomposition appears to be first order .

k = 1 / t ln A₀ / A

= (1/.80881) ln 2

= .857

for the given case

A = .0161 , A₀ = .479

t = ?

t = 1/k ln( .479 / .0161)

= (1 / .857 )ln (.479 / .0161 )

= (1 / .857 ) x ln 29.75

= 3.959 hours

Complete Question

Despite the destruction from Hurricane Katrina in Sept 2005,the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered at atmospheric pressure of 88.2 kPa on Oct 19, 2005,about 2 kPa lower than Hurricane Katrinia. What was the difference between the two hurricanes in:

millimeters of Hg? ___ mm Hg

atmospheres? ___ atm

millibars? ___ mb

Answer:

The difference in pressure in mm of  Hg is  [tex]Pg = 15.05 \ mm\ Hg[/tex]

The difference in pressure in atm is   [tex]P_{atm} = 0.0198 \ atm[/tex]

The difference in pressure in millibar is   [tex]P_{bar} = 20 \ mbar[/tex]

Explanation:

From the question we are told that

      The pressure of Hurricane Wilma is  [tex]P = 88 kPa[/tex]

      The difference is pressure is  [tex]\Delta P =2\ k Pa[/tex]

Generally

            [tex]1\ atm = 1.01 *10^5 \ Pa[/tex]

            [tex]1 \ atm = 760\ mmHg[/tex]

The pressure difference in mm Hg is mathematically  evaluated as

             [tex]Pg = \Delta P * \frac{1000 Pa}{1kPa} * \frac{1\ atm}{1.01 *10^5} * \frac{760 mm Hg}{1 \ atm}[/tex]

Substituting the value

            [tex]Pg = 2 kPa * \frac{1000 Pa}{1kPa} * \frac{1\ atm}{1.01 *10^5} * \frac{760 mm Hg}{1 \ atm}[/tex]

            [tex]Pg = 15.05 \ mm\ Hg[/tex]

  The pressure difference  in atm is  

                                      [tex]P_{atm } = 2kPa * \frac{1000}{1\ kPa} *\frac{1 \ atm}{1.01 *10^5 Pa}[/tex]

                                               [tex]= 0.0198 \ atm[/tex]

  The pressure in millibars is  

                                            [tex]= 2kPa * \frac{1000Pa}{1 kPa} * \frac{1\ bar }{1*10^5 Pa} * \frac{1000 \ milli bar}{1 bar}[/tex]

                                            [tex]= 20 \ m bar[/tex]

Final answer:

The pressure difference between Hurricane Wilma and Hurricane Katrina was 2.0 kPa, with Wilma's pressure being 88.2 kPa and Katrina's being 90.2 kPa. The pressure drop during a hurricane on the northeast U.S. coast in torr is 50.8 torr. A nonvolatile liquid is used in barometers and manometers because volatility would affect the accuracy of pressure measurements.

Explanation:

The difference in atmospheric pressure between Hurricane Wilma and Hurricane Katrina can be calculated using the information provided. If Wilma's pressure was 88.2 kPa and that was 2.0 kPa lower than Katrina's, then Katrina's pressure was 88.2 kPa + 2.0 kPa = 90.2 kPa. Therefore, the difference in pressure between the two hurricanes was 2.0 kPa.

To calculate the drop in pressure in torr during a hurricane on the northeastern United States coast: the pressure usually at 30.0 in. Hg can drop to 28.0 in. Hg. Since 1 in. Hg is equivalent to 25.4 mm Hg, and 1 mm Hg is the same as 1 torr, the drop in pressure is (30.0 in. Hg - 28.0 in. Hg) x 25.4 mm Hg/in. Hg = 50.8 torr.

It is necessary to use a nonvolatile liquid in a barometer or manometer because a volatile liquid would evaporate and form a vapor that would affect the pressure measurement, leading to inaccurate readings.

Answer:

[tex]m_{w} = 1755.323\,g[/tex]

Explanation:

The energy stored in the hamburger is:

[tex]Q = (0.75\,pd)\cdot \left(\frac{453\,g}{1\,pd} \right)\cdot \left(3.3\,\frac{kcal}{g} \right)\cdot \left(4.186\,\frac{kJ}{kcal} \right)[/tex]

[tex]Q = 4693.239\,kJ[/tex]

The amount of water perspired is derived from the following formula:

[tex]Q = m_{w}\cdot (c_{p}\cdot \Delta T + L_{v})[/tex]

[tex]m_{w} = \frac{Q}{c_{p}\cdot \Delta T + L_{v}}[/tex]

[tex]m_{w} = \frac{4693.239\,kJ}{\left(4.186\times 10^{-3}\,\frac{kJ}{g\cdot ^{\circ}C} \right)\cdot (100^{\circ}C-37^{\circ}C)+2.41\,\frac{kJ}{g} }[/tex]

[tex]m_{w} = 1755.323\,g[/tex]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = 4.82 grams benzene

Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

Mass CO2 = 9.15 grams

Answer:

a

The  standard Gibbs free energy change for the reaction at 298 K is

               [tex]\Delta G^o_{re} = -800.99 kJ/moles[/tex]

b

  The  energetic (ΔH)  is  [tex]\Delta H^o _{re} = -802.112 \ kJ/mole[/tex]

    The  entropic contributions  is  [tex]T \Delta S = -1.126 \ kJ/mole[/tex]

Energetic is the dominant contribution

c

  The equilibrium constant at 298 K  is  [tex]K = 2.53[/tex]

Explanation:

From the question we are told that

    The chemical reaction is  

              [tex]CH_4 + 2 O_2 ----> 2 H_2 O + CO_2[/tex]

Generally ,

The free energy of formation of  [tex]CH_4[/tex]  is a constant with a value  

          [tex]\Delta G^o_f __{CH_4}} = -50.794 \ kJ / moles[/tex]

The free energy of formation of  [tex]O_2[/tex]  is a constant with a value  

        [tex]\Delta G^o_f __{O_2}} = 0 \ kJ / moles[/tex]

The free energy of formation of  [tex]H_2O[/tex]  is a constant with a value  

            [tex]\Delta G^o_f __{H_2O}} = -228.59 \ kJ / moles[/tex]

The free energy of formation of  [tex]CO_2[/tex]  is a constant with a value  

            [tex]\Delta G^o_f __{H_2O}} = -394.6 \ kJ / moles[/tex]

The Enthalpy  of  formation of  [tex]CH_4[/tex] at standard condition i  is a constant with a value  

             [tex]\Delta H^o_f __{CH_4}} = -74.848 \ kJ / moles[/tex]

The Enthalpy  of   formation of  [tex]CO_2[/tex] at standard condition is a constant with a value  

              [tex]\Delta H^o_f __{CO_2}} = -393.3 \ kJ / moles[/tex]

The Enthalpy  of  formation of [tex]O_2[/tex] at standard condition is a constant with a value  

              [tex]\Delta H^o_f __{O_2}} = 0 \ kJ / moles[/tex]

The Enthalpy  of   formation  of  [tex]H_2O[/tex] at standard condition is a constant with a value  

              [tex]\Delta H^o_f __{H_2O}} = -241.83 \ kJ / moles[/tex]

The standard Gibbs free energy change for the reaction at 298 K is mathematically represented as

      [tex]\Delta G^o_{re} = (\Delta G^o_f __{H_2O}} + (2 * \Delta G^o_f __{H_2O}} )) - ((\Delta G^o_f __{CH_4}} + (2 * \Delta G^o_f __{O_2}}))[/tex]

Substituting values

 [tex]\Delta G^o_{re} =\Delta G= ( (-394.6 ) + (2 * (-228.59)) ) - ((-50.794) +(2* 0))[/tex]

 [tex]\Delta G^o_{re} = -800.99 kJ/moles[/tex]

The Enthalpy  of   formation  of the reaction is

[tex]\Delta H^o _{re} =( \Delta H^o_f __{CH_4}} + (2 * (\Delta H^o_f __{H_2O}} ))) - ( \Delta H^o_f __{CH_4}} + (2 * \Delta H^o_f __{O_2}}))[/tex]

Substituting values

  [tex]\Delta H^o _{re} = \Delta H = ((-393.3) + 2 * ( -241.83)) - ( -74.848 + (2 * 0))[/tex]

 [tex]\Delta H^o _{re} = -802.112 \ kJ/mole[/tex]

 The entropic contributions is mathematically represented as

    [tex]T \Delta S = \Delta H -\Delta G[/tex]

 Substituting values

     [tex]T \Delta S =-802 .112-(-800.986)[/tex]

    [tex]T \Delta S = -1.126 \ kJ/mole[/tex]

Comparing the values of  [tex]T \Delta S \ and \ \Delta G[/tex] we see that  energetic is the dominant contribution

The standard Gibbs free energy change for the reaction at 298 K can also be represented mathematically  as

         [tex]\Delta G = -RT lnK[/tex]

Where  R  is the gas constant with as value of  [tex]R = 8.314 *10^{-3} kJ/mole[/tex]

   K is the equilibrium constant

   T is the temperature with a given value  of [tex]T = 298K[/tex]

Making K the subject we have

      [tex]K = e ^{- \frac{\Delta G }{RT} }[/tex]

Substituting values  

      [tex]K = e ^{- \frac{-800.99 }{(8.314 *10^{-3} ) * (298)} }[/tex]

       [tex]K = 2.53[/tex]

Final answer:

To calculate the standard Gibbs free energy change for the combustion of methane, use the formula ΔG = ΔH - TΔS, where ΔG is the standard Gibbs free energy change, ΔH is the standard enthalpy change, T is the temperature in Kelvin, and ΔS is the standard entropy change.

Explanation:

The standard Gibbs free energy change for a reaction can be calculated using the formula:

ΔG = ΔH - TΔS

where ΔG is the standard Gibbs free energy change, ΔH is the standard enthalpy change, T is the temperature in Kelvin, and ΔS is the standard entropy change.

In this case, we are given the thermochemical equation for the combustion of methane as: CH4 + 2O2 → 2H2O + CO2

To calculate the standard Gibbs free energy change, we need the values of ΔH and ΔS for this reaction at 298 K. Once we have these values, we can plug them into the formula to get the answer.

Answer:

A) A solid salt dissolves in water.

Explanation:

A solid, like a salt, dissociates into ions as it dissolves in liquid. The particles (ions) become more spaced apart and with greater randomness. This is increasing entropy.

Answer: B-- the length of the chain and degree of cross-linking.

Explanation: Silicones can be oils or rubber-like materials depending on ___the length of the chain and degree of cross-linking.____

Silicone rubber is an elastomer which is composed of silicone containing silicon together with carbon, hydrogen, and oxygen. silicone rubber can be found in a wide variety of products because of its ability to withstand very high temperatures and pressures. eg cooking and food storage products; clothing apparels , automotive devices etc   while

Silicone oil is a linear chain siloxane repeating units Si--0 with radical side group as such as methyl, phenyl etc  bonded to it. They are oils because of they are  viscose  and have ability to  repel. Silicone oils are mostly employed in medicine for surgical tools.

Answer: it’s 15.0 L

Explanation: Trust me

Answer:

Check the explanation

Explanation:

Kindly check the attached image below for the step by step explanation to the question above.