In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.501 L flask at 1043 K. At equilibrium, the flask contains 0.169 mol of CO gas, 0.257 mol of H2 gas, and 0.255 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1043 K)? Enter to 4 decimal places. HINT: Look at sample problem 17.7 in the 8th ed Silberberg book. Write a balanced chemical equation. Write the K expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into K expression, solve for the unknown. (MTS 5/16/2018)

Answer:

Take 46.9 ml of the 12 M solution using a graduated cylinder and pour the liquid in a 250-ml volumetric flask. Add water until the mark.

Explanation:

To prepare this solution, you have to take a volume of the 12 M HCl solution and add water to 250 ml. What volume should you take?

The number of moles of HCl present in the volume you take from the concentrated solution will be the same as the number of moles in the final solution since you are only adding water. Then:

number of moles of HCl in the taken volume = number of moles in the final solution.

number of moles of HCl = concentration (in molarity) * volume

Then:

Ci * Vi = Cf * Vf

Where

Ci = the concentration of the solution from which you take the volume to prepare the more diluted solution.

Vi = the volume of the concentrated solution you have to take.

Cf = Concentration of the final solution

Vf = volume of the final solution

Replacing with the data:

12.0 M * Vi = 250.00 ml * 2.25M

Vi = 46.9 ml

Answer:

3,8×10⁻⁵ mol/L of potassium permanganate solution

Explanation:

To calculate concentration in mol/L you must convert the 3,8 umol to moles and 100 mL to liters, knowing 1 umol are 1×10⁻⁶mol and 1L are 1000 mL.

3,8 umol × (1×10⁻⁶mol / 1 umol ) = 3,8×10⁻⁶mol of potassium permanganate.

100 mL × ( 1L / 1000 mL) = 0,100 L

Thus, concentration in mol/L is:

3,8×10⁻⁶mol / 0,100 L = 3,8×10⁻⁵ mol/L of potassium permanganate solution

I hope it helps!

Final answer:

The concentration of the potassium permanganate solution is 3.8 x 10⁻⁵ M when rounded to two significant digits.

Explanation:

The concentration of a solution is calculated by dividing the number of moles of the solute by the volume of the solution in liters. To calculate the concentration of potassium permanganate (KMnO₄) in the chemist's solution, you need to use the equation:

C = n / V

where C is the molarity (concentration) in moles per liter (mol/L), n is the number of moles of KMnO₄, and V is the volume of the solution in liters.

In this case, the student already has 3.8 μmol (or 3.8 x 10⁻⁶ mol) of KMnO₄ and the total volume is 100 mL, which is equivalent to 0.1 L. Therefore, the molarity (C) of the solution is:

C = 3.8 x 10⁻⁶ mol / 0.1 L = 3.8 x 10⁻⁵ M

Thus, the concentration of the potassium permanganate solution is 3.8 x 10⁻⁵ M, which can be rounded to two significant digits as 3.8 x 10⁻⁵ M.

Answer: The mass of oxygen reacted is [tex]1.60\times 10^{3}g[/tex]

Explanation:

We are given:

Moles of propane = 9.98 mol  

For the given chemical equation:

[tex]C_3H_8(g)+5O2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]

By Stoichiometry of the reaction:

1 mole of propane reacts with 5 moles of oxygen.

So, 9.98 moles of propane will react with = [tex]\frac{5}{1}\times 9.98=49.9mol[/tex] of oxygen.

To calculate the mass of carbon dioxide, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of oxygen = 49.9 moles  

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation:

[tex]49.9mol=\frac{\text{Mass of oxygen}}{32g/mol}\\\\\text{Mass of oxygen}=(49.9mol\times 32g/mol)=1596.8g=1.60\times 10^{3}g[/tex]

Hence, the mass of oxygen reacted is [tex]1.60\times 10^{3}g[/tex]

Explanation:

Neutralization reaction -

The reaction of an acid and base to yield a salt and water , is a type of neutralization reaction .

The reaction of  potassium hydroxide and phosphoric acid  is a type of neutralization reaction ,

Hence , the reaction is as follows -

KOH (aq) + H₃PO₄ (aq) ----> K₃PO₄ (aq) + 3H₂O (l)

The reaction after balancing the atoms on the reactant side and on the product side is -

3 KOH (aq) + H₃PO₄ (aq) ----> K₃PO₄ (aq) + 3H₂O (l)

Answer:

The runner should run 51 minutes.

Explanation:

Distance wished by runner to cover = d = 10.4 km

Time taken by the runner to cover 10.4 km = T

Speed of the runner  = 7.6 mile/hour

1 mile = 1.60934 km

1 hour = 60 min

[tex]7.6 mile/Hour=\frac{7.6\times 1.60934 km}{1\times 60 min}=0.2038 km/min[/tex]

[tex]Speed=\frac{Distance}{Time}[/tex]

[tex]T=\frac{10.4 km}{0.2038 km/min}=51.0179 min\approx 51 minutes[/tex]

The runner should run 51 minutes.

Answer: The mass of alcohol is 790 grams.

Explanation:

To calculate the mass of alcohol, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Volume of alcohol = 1.0 L = 1000 mL    (Conversion factor:  1 L = 1000 mL)

Density of alcohol = 0.79 g/mL

Putting values in above equation, we get:

[tex]0.79g/mL=\frac{\text{Mass of alcohol}}{1000mL}\\\\\text{Mass of alcohol}=790g[/tex]

Hence, the mass of alcohol is 790 grams.

Final answer:

The mass of 1.0 L of alcohol is 790 grams.

Explanation:

To calculate the mass in grams of 1.0 L of alcohol, you can use the density of alcohol.

The given density is 0.79 g/mL.

This means that for every 1 mL of alcohol, there is a mass of 0.79 grams.

Since 1 L is equal to 1000 mL, you can multiply the density by the volume to find the mass: 0.79 g/mL × 1000 mL = 790 grams of alcohol.

Answer: The molar solubility of carbon dioxide gas is 0.003 M

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{liquid}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]3.7\times 10^{-2}mol/L.atm[/tex]

[tex]p_{CO_2}[/tex] = partial pressure of carbonated drink = 0.71 atm

Putting values in above equation, we get:

[tex]C_{CO_2}=3.7\times 10^{-2}mol/L.atm\times 0.71atm\\\\C_{CO_2}=2.637\times 10^{-2}mol/L=0.003M[/tex]

Hence, the molar solubility of carbon dioxide gas is 0.003 M

Answer: The molar solubility of carbon dioxide is [tex]2.63\times 10^{-2}M[/tex]

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]3.7\times 10^{-2}mol/L.atm[/tex]

[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas = ?

[tex]p_{CO_2}[/tex]  = partial pressure of carbon dioxide gas = 0.71 atm

Putting values in above equation, we get:

[tex]C_{CO_2}=3.7\times 10^{-2}mol/L.atm\times 0.71atm\\\\C_{CO_2}=2.63\times 10^{-2}M[/tex]

Hence, the molar solubility of carbon dioxide is [tex]2.63\times 10^{-2}M[/tex]

Answer and Explanation:

Interstitial diffusion allows the purity atom like (C, O, H) to occupy the interstitial sites in the atom which are abundantly present as a result of weaker bond formation with the surrounding atom.

The interstitial diffusion of iron in Body Centered Cubic (BCC) crystal lattice is faster than Face Centered Cubic (FCC) and the reason being that BCC structure is more open with more interstitial spaces than FCC structure.

Also the packing fraction of BCC lattice is less than that of FCC lattice.

For BCC is 0.68 whereas for FCC, it is 0.74

Answer:

Percentage of the isotope left is 75.87 %.

Explanation:

Initial mass of the isotope = 1 mg

Time taken by the sample, t = [tex]t_{\frac{1}{2}}=123 years[/tex]

Formula used :

[tex]N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]

where,

[tex]N_o[/tex] = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

[tex]t_{\frac{1}{2}}[/tex] = half life of the isotope

[tex]\lambda[/tex] = rate constant

[tex]\lambda =\frac{0.693}{123 year}=0.005635 year^{-1}[/tex]

[tex]N=N_o\times e^{-\lambda \times t}[/tex]

Now put all the given values in this formula, we get

[tex]N=1 mg\times e^{-0.005634 year^{-1}\times 49 years}[/tex]

[tex]N=0.7587 mg[/tex]

Percentage of the isotope left:

[tex]\frac{N}{N_o}\times 100[/tex]

=[tex]\frac{0.7587 mg}{1 mg}\times 100[/tex]

Percentage of the isotope left is 75.87 %.

Answer:

In 3.5 x 10^(15) J of energy there are 9*10^(33) photons.

Explanation:

To solve this problem, we need two equations.

The equation of light velocity, wich is a relation between wavelenght and frecuency.

                                                      c=λν            (1)

where:

The Photoelectric Effect equation, that refers to the energy absorbed or emanate by ONE photon.

                                                             E = hν            (2)

where:

The first we do is to calculate the frecuency of the flash using equation (1). The wavelenght of the flash is 510 nm = 510 * 10^(-9) m

c=λν........................ ν= c/λ = 3 × 10^8 [m/s]/  510 * 10^(-9) m  = 5,88 * 10^(14) 1/s

Note: small wavelenghts always have big frequencies

Now, we use the photoelectric effect equation to calculate the amount of energy that ONE  photon can abosrb.

 E = hν  ..................... E = 6,626*10^{-34} [J.s] * 5,88 * 10^(14) 1/s =3,9 * 10^(-19) J

To know the number of photons, we just have to divide the TOTAL amount of energy between the energy of ONE photon. So:

# photons = 3.5 x 10^(15) J / 3,9 * 10^(-19) J = 9*10^(33) photons.

Answer:

Carbohydrates are the biological molecules having a general emperical formula [tex]C_{m} (H_{2} O)_{n}[/tex].  

There are various important biological functions of carbohydrates. Carbohydrates serve as structural components, component of coenzymes, and backbone of RNA. Carbohydrates provide and store energy, and also plays important role in blood clotting, development, immune system, and preventing pathogenesis.

Burning 1 kg of coal with a 50% carbon content produces 1.833 kg of CO2 after performing a stoichiometric calculation based on the molar masses of carbon and CO2.

To calculate the amount of CO2 released when 1 kg of coal is burned, we first consider that the coal is 50% carbon by mass. This means that 0.5 kg (or 500 g) of that coal is carbon. Using the stoichiometry of the combustion reaction (C + O2 -> CO2), we find that each 12 grams of carbon (C) will produce 44 grams of CO2 (since the molar mass of carbon is 12 g/mol, and the molar mass of CO2 is 44 g/mol).

Therefore, to find out how much CO2 is produced from the carbon in coal, you can use the ratio (44 g CO2 / 12 g C). Calculating the amount of CO2 produced from 500 g of carbon:

500 g C x (44 g CO2 / 12 g C) = 1833.33 g CO2

Converting this to kilograms:

1833.33 g CO2 x (1 kg / 1000 g) = 1.833 kg CO2

Answer:

The rate constant [tex]k_{2}[/tex] at 84.8°C is [tex]k_{2}=6.423sec^{-1}[/tex]

Explanation:

Taking the Arrhenius equation we have:

[tex]ln\frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})[/tex]

Where [tex]k_{2}[/tex] is the rate constant at a temperature 2, [tex]k_{1}[/tex] is the rate constant at a temperature 1; [tex]T_{1}[/tex] is the temperature 1, [tex]T_{2}[/tex] is the temperature 2, R is the gas constant and [tex]E_{a}[/tex] is the activation energy.

Now, we need to solve the equation for [tex]k_{2}[/tex], so we have:

[tex]ln\frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})[/tex]

[tex]ln({k_{2})-ln(k_{1})=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})[/tex]

[tex]ln(k_{2})=E_{a}(\frac{1}{T_{1}}-\frac{1}{T_{2}})+ln(k_{1})[/tex]

Then we need to make sure that we are working with the same units, so:

[tex]R=8.314\frac{J}{mol.K}[/tex]

[tex]T_{1}=16.6^{o}C+273.15=289.75K[/tex]

[tex]T_{2}=84.4^{o}C+273.15=357.95K[/tex]

And now we can replace the values into the equation:

[tex]ln(k_{2})=\frac{22000\frac{J}{mol}}{8.314\frac{J}{mol.K}}(\frac{1}{289.75K}-\frac{1}{357.95K})+ln(1.868sec^{-1})[/tex]

[tex]ln(k_{2})=2646.139K(0.003451K^{-1}-0.002794K^{-1})+0.6249[/tex]

[tex]ln(k_{2})=2.363sec^{-1}[/tex]

To solve the ln we have to apply e in both sides of the equation, so we have:

[tex]e^{ln(k_{2})}=e^{2.363}sec^{-1}[/tex]

[tex]k_{2}=6.423sec^{-1}[/tex]

Answer:

10.37 s-1

Explanation:

From

k= A e-^Ea/RT

Given

Ea=22KJmol-1

T=16.6+273= 289.6K

k= 1.868 sec-1

R= 8.314JK-1mol-1

A???

Hence

A= k/e^-Ea/RT

A= 1.868/e-(22000/8.314×289.6)

A= 1.7 ×10^4

Substitute into to find k at 84.8°C

k= 1.7×10^4× e-(22000/8.314×357.8)

k=10.37 s-1

To determine the amount of sodium ions [tex]\rm (Na^+)[/tex]in the final solution, we need to take into account the contribution of sodium ions from both sodium chloride (NaCl) and sodium sulfate ([tex]\rm Na_2SO_4[/tex]). The volumes can be considered as individual contributions and then added together because they are additive.

Sodium chloride (NaCl):

Given volume of NaCl solution = 3.58 ml

Molarity of NaCl = 0.288 M

Number of moles of NaCl = Molarity × Volume (in liters)

So, the number of moles of NaCl = 0.288 M × (3.58 / 1000) L = 0.00082704 moles

The amount of sodium ions from NaCl is also 0.00082704 moles because each molecule of sodium chloride dissociates into one sodium ion ([tex]\rm Na^+[/tex]).

Sodium sulfate (Na₂SO₄):

Given volume of Na₂SO₄ solution = 500 ml

Molarity of Na₂SO₄ = 6.51 * 1/1000 M = 0.00651 M

Number of moles of Na₂SO₄ = Molarity × Volume (in liters)

So, the number of moles of Na₂SO₄ = 0.00651 M × (500 / 1000) L = 0.003255 moles

Since [tex]\rm Na_2SO_4[/tex] gives 2 sodium ions ([tex]\rm Na^+[/tex]) per formula unit, the amount of sodium ions from [tex]\rm Na_2SO_4[/tex] is 2 * 0.003255 = 0.00651 moles.

Adding the moles of sodium ions from both sources:

Total moles of sodium ions = Moles from NaCl + Moles from Na₂SO₄

Total moles of sodium ions = 0.00082704 + 0.00651 = 0.00733704 moles

So, the molarity of sodium ions in the final solution:

Molarity of sodium ions = Total moles of sodium ions / Total volume (in liters)

Molarity of sodium ions = 0.00733704 moles / (500 / 1000 + 3.58 / 1000) L ≈ 0.0162 M

Therefore, the molarity of sodium ions in the final solution is approximately 0.0162 M.

Learn more about molarity, here:

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The molarity of sodium ions in the solution made by mixing 3.58 mL of 0.288 M sodium chloride and 500 mL of 6.51/1000 M sodium sulfate is 0.00851 M.

To calculate the molarity of sodium ions in the mixture, we need first to calculate the moles of sodium in each solution and then sum those up. This can then be divided by the total volume of the mixture in liters.

For sodium chloride:

moles = Molarity * volume = 0.288 mol/L * 3.58 mL * (1 L/1000 mL) = 0.00103 mol

For sodium sulfate:

moles = Molarity * volume = 6.51/1000 mol/L * 500 mL * (1 L/1000 mL) = 0.00326 mol

Total moles of sodium ions = moles from sodium chloride + moles from sodium sulfate = 0.00103 mol + 0.00326 mol = 0.00429 mol

Total volume = volume of sodium chloride + volume of sodium sulfate = 3.58 mL + 500 mL = 503.58 mL = 0.50358 L

Therefore, molarity = moles / volume = 0.00429 mol / 0.50358 L = 0.00851 M

So, the molarity of sodium ions in the solution is 0.00851 M.

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Answer: The mass of zinc oxalate, the chemist has added is [tex]7.6\times 10^5mg[/tex]

Explanation:

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]

We are given:

Molarity of solution = [tex]1.08\times 10M=10.8M[/tex]

Molar mass of zinc oxalate = 155.4 g

/mol

Volume of solution = 0.45 L

Putting values in above equation, we get:

[tex]10.8M=\frac{\text{Mass of zinc oxalate}}{155.4g/mol\times 0.45L}\\\\\text{Mass of zinc oxalate}=(10.8mol/L\times 155.4g/mol\times 0.45L)=7.6\times 10^2g[/tex]

To convert the calculated mass into milligrams, we use the conversion factor:

1 g = 1000 mg

So, [tex]7.6\times 10^2g\times \frac{1000mg}{1g}=7.6\times 10^5mg[/tex]

Hence, the mass of zinc oxalate, the chemist has added is [tex]7.6\times 10^5mg[/tex]

Answer:

C : CH₃OH

Explanation:

According to the concept of Bronsted - Lowry -

An acid is a substance , that can give a proton .

A base is a substance , that can take a proton .

According to the reaction given in the question ,

CH₃OH   +   HI   -->    CH₃OH₂⁺   +  I⁻

From , the above reaction ,

It is visible that , the reactant CH₃OH accepts a proton and forms CH₃OH₂⁺  , thereby acting as a base ,

And ,  HI act as an acid , as is losses a proton and becomes  I⁻ .

Hence ,

In the above reaction ,  CH₃OH act as a base .

Answer: The chemical formula for magnesium sulfite is [tex]MgSO_3[/tex]

Explanation:

The given compound is formed by the combination of magnesium and sulfite ions. This is an ionic compound.

Magnesium is the 12th element of periodic table having electronic configuration of [tex][Ne]3s^2[/tex].

To form [tex]Mg^{2+}[/tex] ion, this element will loose 2 electrons.

Sulfite ion is a polyatomic ion having chemical formula of [tex]SO_3^{2-}[/tex]

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for magnesium sulfite is [tex]MgSO_3[/tex]

Answer:

MgSO3

Explanation:

Answer:

(4.29×10¹⁵)⋅(1.96×10⁻⁴) = 8.40 × 10¹¹, has three significant digits.

Explanation:

To solve: (4.29×10¹⁵)⋅(1.96×10⁻⁴)

According to the product rule of exponents, when exponents having the same base are multiplied, the base is kept the same and the powers are added.

Therefore,

(4.29×10¹⁵)⋅(1.96×10⁻⁴) = (4.29 × 1.96) · 10⁽¹⁵⁻⁴⁾ = 8.40 × 10¹¹

The number, 8.40 × 10¹¹ has three significant digits.

To calculate the product of (4.29×1015) and (1.96×10−4), multiply the significant figures to get 8.4084, then add the exponents to get 1011, and combine them to express the product in scientific notation as 8.41×1011, rounded to three digits.

Explanation:

To compute the product of (4.29×1015) and (1.96×10−4), you multiply the significant figures and then add the exponents of 10. First, multiply the significant figures:

4.29 × 1.96 = 8.4084.

Next, add the exponents:

1015 × 10−4 = 1015−4 = 1011.

Combine the significant figure product with the exponent sum to express the answer in scientific notation:

8.4084 × 1011 → 8.41×1011 (rounded to three digits).

Answer:

9.23 μm = 0.000363 in

Explanation:

In order to convert 9.23 μm into inches, we need to keep in mind two conversion factors:

Now we proceed to calculate, keeping in the denominator the unit we want to convert, and in the numerator the unit that we wish to convert to:

[tex]9.23um*\frac{1cm}{10000um}*\frac{1in}{2.54cm}  =0.0363 in[/tex]

Thus 9.23 μm are equal to 0.000363 inches.

Answer:

Degree of ionization = 0.0377

pH of the solution = 1.769

Explanation:

Initial concentration of HF = 0.45 M

[tex]K_a = 6.8 \times 10^{-4}[/tex]

                     [tex]HF \leftrightharpoons  H^+ + F^-[/tex]

Initial        0.45                             0          0

At equi      0.45 - x                      x           x

Equilibrium constant = [tex]\frac{[H^+][F^-]}{HF}[/tex]

                   [tex]6.8 \times 10^{-4}= \frac{[x][x]}{0.45 - x}[/tex]

           [tex]x^2 + 6.8 \times 10^{-4} x -  6.8 \times 10^{-4} \times 4.5 = 0[/tex]

x = 0.017 M

x = Cα

α = Degree of ionization

C = Concentration

Degree of ionization = [tex]\frac{0.017}{0.45} = 0.0377[/tex]

[tex]pH = -log[H^+][/tex]

[H^+]=0.017 M

[tex]pH = -log[0.017][/tex]

             = 1.769

Final answer:

The acid dissociation reaction for hydrofluoric acid in water is [tex]HF (aq) + H_2O (l)[/tex] ⇌ [tex]H_3O^+ (aq) + F- (aq)[/tex]. The concentration of a 0.1 L solution containing 0.05 g of HF is 0.025 M. To find the pH for such a solution using the given Ka, the ICE table method can be utilized.

Explanation:

To answer your questions regarding hydrofluoric acid (HF) and its properties, we can proceed as follows:

a) Write out the acid dissociation reaction for hydrofluoric acid. Label the conjugate acid/base pairs.

Hydrofluoric acid dissociates in water as follows:

[tex]HF (aq) + H_2O (l)[/tex] ⇌ [tex]H_3O^+ (aq) + F- (aq)[/tex]

In this reaction, HF is the conjugate acid and F- is the conjugate base.

b) What is the concentration (M) of a solution containing 0.05 g of HF in 0.1 L H2O?

The molecular weight of HF is approximately 20.01 g/mol. To find the molarity, first convert grams to moles:

0.05 g HF × (1 mol HF/20.01 g HF) = 0.0025 mol HF

Then, divide the moles of HF by the volume of the solution in liters:

0.0025 mol HF / 0.1 L = 0.025 M

c) Using the given Ka value, calculate the pH of the solution from part b

Since HF is a weak acid, and given that Ka = 7.2 × [tex]10^-^4[/tex], you can use the ICE table method to find the concentration of H3O+ and then calculate the pH.

Answer:

Ba(OH)2 (aq) + H2SO4(aq) → BaSO4(s) + 2 H2O(l)

That's the right one.

Explanation:

You should see that this equation is balanced, not as

Ba(OH)2 (aq) + H2SO4(aq) → BaSO4(s) + H2O(l)

(on reactive we have 4 H, on products, we have only 2)

Ba(OH)2 (aq) + H2SO4(aq) → H2Ba(s) + SO4(OH)2(l)

(this is impossible, it's a nonsense)

BaSO4(s) + 2 H2O(l) → Ba(OH)2 (aq) + H2SO4(aq)

(it is the same with the right one but is the other way around. The statement says, reaction of Ba(OH)2 with H2SO4, not BaSO4 with water. Also, it is not a chemical balance.

Answer : The energy required is, 574.2055 KJ

Solution :

The conversions involved in this process are :

[tex](1):H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(95^oC)[/tex]

Now we have to calculate the enthalpy change or energy.

[tex]\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]

where,

[tex]\Delta H[/tex] = energy required = ?

m = mass of ice = 1 kg  = 1000 g

[tex]c_{p,s}[/tex] = specific heat of solid water = [tex]2.09J/g^oC[/tex]

[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]

n = number of moles of ice = [tex]\frac{\text{Mass of ice}}{\text{Molar mass of ice}}=\frac{1000g}{18g/mole}=55.55mole[/tex]

[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

Now put all the given values in the above expression, we get

[tex]\Delta H=[1000g\times 4.18J/gK\times (0-(-10))^oC]+55.55mole\times 6010J/mole+[1000g\times 2.09J/gK\times (95-0)^oC][/tex]

[tex]\Delta H=574205.5J=574.2055kJ[/tex]     (1 KJ = 1000 J)

Therefore, the energy required is, 574.2055 KJ

Answer:

Mass of 246 lb object is 7.646 slugs.

Explanation:

Mass of the object = m

Weight of the object = W = 246 lb

Acceleration due to gravity= g = [tex]32.174 ft/s^2[/tex]

[tex] Weight=mass \times g[/tex]

[tex]W=mg[/tex]

[tex]1 lb=slugs\times 1 ft/s^2[/tex]

[tex]1 slugs=\frac{lb\times s^2}{ft}[/tex]

[tex]m=\frac{W}{g}=\frac{246 lb}{32.174 ft/s^2}=7.646 lb s^2/ft=7.646 slugs[/tex]

Mass of 246 lb object is 7.646 slugs.

The enthalpy of vaporization for pyridine can be calculated using the Clausius-Clapeyron equation. Given that the vapor pressure of pyridine is 50.0 kPa at 365.7 K, and its boiling point is 388.4K, we can substitute these values into the equation to find the enthalpy of vaporization.

The question is asking for the enthalpy of vaporization of pyridine. We first need to apply the Clausius-Clapeyron equation which is

ln(P2/P1) = -ΔHvap/R *(1/T1 - 1/T2), where P2 is the vapor pressure at the boiling point (1.00 atm or 101.3 kPa), P1 is the given vapor pressure (50.0 kPa), ΔHvap is the enthalpy of vaporization which we want to find, R is the gas constant (8.314 J/K.mol), T1 is 365.7 K, and T2 is the boiling point (388.4 K). By rearranging and substituting the values into the equation, one can solve for ΔHvap to find its value. Remember always to convert the pressure units into the same, in this case we used kilopascal.

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Answer:

2.4E-4

Explanation:

Hello,

By applying the following mass-mole relationship, the concentration could be computed as follows (assuming molarity as long as it isn't specified), since in the silver carbonate two silver molecules are present:

[tex][Ag]=1.2x10^{-4}\frac{molAg_2CO_3}{L} *\frac{2mol Ag}{1 mol Ag_2CO_3}=2.4x10^{-4}\frac{mol Ag}{L}[/tex]

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Answer: Wavelength of ultraviolet light is 54 nm.

Explanation:

Electromagnetic wave is defined as the wave which is associated with both electrical and magnetic component associated with them. They can travel in vacuum as well and travel with the speed of light.

The relationship between wavelength and frequency of the wave follows the equation:

[tex]\nu=\frac{c}{\lambda}[/tex]

where,

[tex]\nu[/tex] = frequency of the wave  =[tex]5.5\times 10^{15}Hz[/tex]

c = speed of light  =[tex]3\times 10^8ms^{-1}[/tex]

[tex]\lambda [/tex] = wavelength of the wave  

Putting in the values we get:

[tex]5.5\times 10^{15}s^{-1}=\frac{3\times 10^8ms^{-1}}{\lambda}[/tex]

[tex]\lambda=0.54\times 10^{-7}m[/tex]

[tex]1nm=10^{-9}m[/tex]

[tex]\lambda=54nm[/tex]

Thus wavelength of ultraviolet light is 54 nm.

Answer:

The sum of the coefficient is: 1 + 2 +2 + 1 = 6 ( option c)

Explanation:

First we will balance on both sides Na

On the right side we have 2x Na but on the left side we have only 1x Na. So we have to multiply NaOH on the left side by 2.

This will give us:

H2SO4 + 2 NaOH → H2O + Na2SO4

Now we have on both sides 2x Na

We see that on the left side we have 4x H ( 2x H of H2SO4 and 2x H of NaOH), but on the right side we only have 2x H. So, we have to multiply H2O on the right side by 2.

This will give us:

H2SO4 + 2 NaOH → 2 H2O + Na2SO4

Now we have on both sides 2x Na and 4x H.

Also the number of O is on both sides equal, due to this. ( Both sides have 6x O).

Finally, we have this reaction:   H2SO4 + 2 NaOH → 2 H2O + Na2SO4

The sum of the coefficient is: 1 + 2 +2 + 1 = 6 ( option c)

Final answer:

The sum of the coefficients when the equation H2SO4 + NaOH → H2O + Na2SO4 is balanced is 6. The balanced equation is 1 H2SO4 + 2 NaOH → 2 H2O + 1 Na2SO4.

Explanation:

The sum of the coefficients when the chemical equation H2SO4 + NaOH → H2O + Na2SO4 is balanced is the total of the numbers that are used to balance the equation.

To balance the equation, we need to ensure that there is the same number of each type of atom on both the reactant and product sides of the equation. In this case, we balance the equation as follows: 1 H2SO4 (aq) + 2 NaOH (aq) → 2 H2O (l) + 1 Na2SO4 (aq). So, the coefficients in the balanced equation are 1, 2, 2, and 1 respectively.

Adding these coefficients, we get:

1 (for H2SO4) + 2 (for NaOH) + 2 (for H2O) + 1 (for Na2SO4) = 6.

The answer is option c: 6.

The substances Ar, Cl2, CH4, and CH3COOH can be ranked in increasing order of strength of intermolecular attractions as CH4 < Ar < Cl2 < CH3COOH.

The intermolecular forces in the given substances can be ranked from weakest to strongest as follows:

Answer:

Pressure in duct = 799.75 mmHg

Atmospheric pressure = 774.75 mmHg

Gauge pressure = 24.99 mmHg

Explanation:

First of all, it is needed to set a pressure balance (taking in account that diameter of manometer is constant) in the interface between the air of the duct and the fluid mercury.

From the balance in the sealed-end manometer, we have the pressure of air duct as:

[tex]P = \rho g h_1[/tex]

We have that ρ is density of mercury and g is the gravity

[tex]\rho = 13600 kg/m^{3}[/tex]

[tex]g = 9.8 m/s^{2}[/tex]

So, replace in the equation:

[tex]P = (13600 kg/m^{3} )(9.8 m/s^{2})(800 mmHg)(\frac{1 mHg}{1000 mmHg})[/tex]

[tex]P = 106624.0 \frac{kg}{s^{2}} = 106624.0 Pa[/tex]

Transforming from Pa to mmHg

[tex]P =  106624.0 Pa (\frac{760 mmHg}{101325 Pa}) = 799.7 mmHg[/tex]

From the balance in the open-end manometer, we have the pressure of air duct as:

[tex]P = \rho g h_2 + P_atm[/tex]

Isolate [tex]P_atm[/tex]:

[tex]P_atm = P - \rho g h_2[/tex]

Calculating:

[tex]P_atm = 799.75 mmHg - (13600 kg/m^{3} )(9.8 m/s^{2})(25 mmHg)(\frac{1 mHg}{1000 mmHg})(\frac{760 mmHg}{101325 Pa} )[/tex]

[tex]P_atm = 774.75 mmHg[/tex]

Finally, gauge pressure is the difference between duct pressure and atmospheric pressure, so:

[tex]P_gau = P - Patm[/tex]

[tex]P_gau = 799.75 mmHg - 774.75 mmHg[/tex]

[tex]P_gau = 24.99 mmHg[/tex]

End.

Answer: The formula mass of caffeine is 97 amu and molar mass of caffeine is 194 g/mol

Explanation:

Formula mass is defined as the sum of the mass of all the atoms each multiplied its atomic masses that are present in the empirical formula of a compound. It is expressed in amu.

Molar mass is defined as the sum of the mass of all the atoms each multiplied its atomic masses that are present in the molecular formula of a compound. It is expressed in g/mol.

Empirical formula is defined as the formula in which atoms in a compound are present in simplest whole number ratios.

The molecular formula of caffeine is [tex]C_8H_{10}N_4O_2[/tex]

Dividing each number of atoms by '2', we will get the empirical formula of caffeine. The empirical formula of caffeine is [tex]C_4H_5N_2O[/tex]

We know that:

Atomic mass of carbon = 12 amu

Atomic mass of hydrogen = 1 amu

Atomic mass of nitrogen = 14 amu

Atomic mass of oxygen = 16 amu

Formula mass of caffeine = [tex](4\times 12)+(5\times 1)+(2\times 14)+(1\times 16)]=97amu[/tex]

Molar mass of caffeine = [tex](8\times 12)+(10\times 1)+(4\times 14)+(2\times 16)]=194g/mol[/tex]

Hence, the formula mass of caffeine is 97 amu and molar mass of caffeine is 194 g/mol