A solution has a [OH−] of 1 x 10−9. What is the pOH of this solution?

a) 5

b) 7

c) 9

d) 12

Explanation:

It is known that specific heat of water is 4.184 [tex]J/g^{o}C[/tex] and atomic mass of tin is 118.7 g/mol. For the given situation,

                 [tex]Q_{lost} = Q_{gained}[/tex]

Let us assume that,

               [tex]m_{1}[/tex] = mass of Sn

               [tex]m_{2}[/tex] = mass of [tex]H_{2}O[/tex]  

Therefore, heat energy expression for heat lost and gained is as follows.

           [tex]Q_{lost} = Q_{gained}[/tex]

      [tex]m_{1}C_{1}(T_{2} - T_{1}) = m_{2}C_{2}(T_{1} - T_{2})[/tex]

   [tex]100 g \times C_{1} \times (100^{o}C - 27.4^{o}C) = 150 g \times 4.184 /g^{o}C \times (27.4^{o}C - 25^{o}C)[/tex]

           [tex]7260C_{1} = 150 \times 4.184 \times 2.4[/tex]

                 [tex]C_{1} = \frac{1506.24}{7260}[/tex]

                              = 0.207 [tex]J/g^{o}C[/tex]

For, 118.7 g the specific heat of tin will be calculated as follows.

               [tex]C_{1} = 0.207 J/g^{o}C \times 118.7 g[/tex]

                          = 24.5 [tex]J/mol^{o}C[/tex]

Thus, we can conclude that specific heat of tin is 24.5 [tex]J/mol^{o}C[/tex].

To calculate the specific heat capacity of the metal, we can use the equation q = m * c * ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. By using the known temperatures and masses of the metal and water, we can solve for the specific heat capacity of the metal.

To calculate the specific heat capacity of the metal, we can use the equation q = m * c * ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we know the initial and final temperatures of the metal and water, as well as the masses of the metal and water. We can rearrange the equation to solve for the specific heat capacity of the metal, c.

We can start by calculating the heat transferred from the metal to the water. The heat transferred to the water can be calculated using the equation q_water = m_water * c_water * ΔT_water, where m_water is the mass of the water, c_water is the specific heat capacity of water, and ΔT_water is the change in temperature of the water.

Next, we can calculate the heat transferred from the metal to the water using the equation q_metal = m_metal * c_metal * ΔT_metal, where m_metal is the mass of the metal, c_metal is the specific heat capacity of the metal, and ΔT_metal is the change in temperature of the metal.

Since the metal and water reach the same final temperature, we can set q_water equal to q_metal and solve for the specific heat capacity of the metal, c_metal.

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Answer:

Consider the reaction of hydrogen peroxide with iodine: H2O2(aq)+I2(aq)⇌OH−(aq)+HIO(aq) The reaction is first order in I2 and second order overall.

What is the rate law?

If the concentration of H2O2 is increased by half and the concentration of I2 is quadrupled, by what factor does the reaction rate increase?

Rate law = k[H₂O₂] [I₂]

The new rate is six times that of the old rate

Explanation:

Rate law = k[H₂O₂] [I₂]

lets input

[H₂O₂] = x

[I₂] = y

Rate = kxy

[H₂O₂] = x + x/2

= 3x/2

[I₂] = 4y

New rate = [tex]k * \frac{3x}{2} *4y[/tex]

New rate = 6

The new rate is six times that of the old rate

Answer:

temperature = stdin.nextDouble();

Explanation: since the temperature has already been declared as a double variable, the above statement will read so.

As temperature equal standard input dot next double variable.

To read a floating point value from standard input into a double variable, use the scanf function with %lf format specifier.

To read a floating point (real) value from standard input into a double variable named temperature, you can use the scanf function in the C programming language. The scanf function allows you to read formatted input, and the %lf format specifier is used to read a double value. Here's an example:

This line of code will read a floating point value from the standard input and store it in the temperature variable declared as a double.

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Answer:

macro-nutrients , micro-nutrients

Explanation:

For plant growth soil needs some basic elements which is basic need for plant growth and metabolic synthesis.

plant need approx 17 essential nutrients for growth and for completing its life cycle among which some are found in water and air. Nitrogen, carbon, hydrogen and oxygen needed the most and in large quantity as nitrogen helps in rapid growth and green color and is mobile and its deficiency lead to reduced growth and yellow foliage.

Answer:

16. Option c.

Explanation:

The reaction is:

C₈H₁₈ + O₂  →  CO₂ + H₂O

Let's count, we have 8 C, 18 H and 2 O in reactant side

We have 1 C, 2 H and 3 O  in product side

In order to balance the C, we add 8 to CO₂ and to balance the H, we add a 9 to H₂O

Now we have 8 C on both sides and 18 H On both sides. Finally we have 25 O on product side.

C₈H₁₈ + O₂  →  8CO₂ + 9H₂O

To balance the O in product side we must add 25/2, but as it is a rational number, we must multiply x2 to get an integer number (x2 in all the stoichiometry). The balanced reaction is :

2C₈H₁₈ + 25O₂  →  16CO₂ + 18H₂O

with 16 C, 36 H and 50 O, on both sides

Answer:

Kc → 41.9

Explanation:

This is the equilibrium:

2NO₂(g) ⇌ N₂O₄(g)

So the expression for Kc will be:

Kc = [N₂O₄] / [NO₂]²

We prospose the situations:

Initially we have 1.2 moles of NO₂ and N₂O₄

X amount has reacted. As stoichiometry is 2:1, we have produced x/2 of the product during the reaction

Finally In equilibrium we have, 0.38 NO₂

                2NO₂(g)     ⇌       N₂O₄(g)

Initially        1.2                        1.2

React            x                         x/2

Eq         (1.2 - x) = 0.38          1.2 + x/2

As we have [NO₂] in the equilibrium, we can determine x (the amount that has reacted) to solve and determine, the [N₂O₄] in the equilibrium

1.2-0.38 = x → 0.82

1.2 + 0.82/2 = 2.02 → [N₂O₄]

For Kc, we need Molar concentration, so we have to divide [N₂O₄] and [NO₂] by the volume

[N₂O₄] → 2.02 mol/3L = 0.673 M

[NO₂] → 0.38 mol/3L = 0.127 M

Now we can replace the Kc expression:

Kc →  [N₂O₄] / [NO₂]² → 0.673 / 0.127² = 41.9

Remember that Kc has no UNITS

To calculate the moles of O2(g) present at equilibrium, use the equilibrium constant expression and the given concentrations of H2O(g) and H2(g). The moles of O2(g) present at equilibrium is approximately 7.23 × 10-3 mol.

To calculate the moles of O2(g) present at equilibrium, we can use the equilibrium constant expression and the given concentrations of H2O(g) and H2(g). The equilibrium constant expression for the given reaction is K = [H2]^2/[H2O]. Since we have the values for [H2] and [H2O], we can rearrange the expression to solve for [O2], which is the unknown.

By substituting the known values into the expression and solving for [O2], we get [O2] ≈ 2.67 × 10-3 M. This represents the concentration of O2(g), which we can then convert to moles of O2 using the volume of the container.

Given that the container has a volume of 2.7 L, we can multiply the concentration of O2 by the volume to get the moles of O2(g). In this case, the moles of O2(g) present at equilibrium would be approximately 7.23 × 10-3 mol.

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To calculate the moles of O2(g) present under the given conditions, divide the concentration of H2(g) by 2 and use the ideal gas law to find the number of moles in the 2.7 L container. The number of moles of O2(g) in the 2.7 L container is equal to the number of moles in any other volume multiplied by the ratio of the volumes.

To calculate the moles of O2(g) present under the given conditions, we can first start by calculating the concentration of O2(g) using the given equilibrium concentrations of H2O(g) and H2(g). Since the reaction is 2 H2O(g) <=> 2 H2(g) + O2(g), the concentration of O2(g) is equal to half the concentration of H2(g). Therefore, the concentration of O2(g) is 2.5 x 10^-2 M / 2 = 1.25 x 10^-2 M.

Next, we can use the ideal gas law to calculate the number of moles of O2(g) in the 2.7 L container. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin. In this case, we don't have the pressure or the temperature, so we can't directly calculate the number of moles. However, since the pressure and temperature are not changing, we can assume that the pressure and temperature are constant, and therefore the number of moles is directly proportional to the volume. So, if we know the number of moles of O2(g) in the 2.7 L container, we can calculate the number of moles in any other volume using the equation: n1V1 = n2V2.

Therefore, if we let n1 be the number of moles of O2(g) in the 2.7 L container and n2 be the number of moles of O2(g) in another volume, we can set up the equation: n1 x 2.7 = n2 x V2. Substituting the values, we get: n1 x 2.7 = 1.25 x 10^-2 x V2. Solving for n2, we get: n2 = n1 x 2.7 / (1.25 x 10^-2). Plugging in the value of n1 as 1.25 x 10^-2 M x 2.7 L, we can calculate n2.

Therefore, the moles of O2(g) present under the given conditions can be calculated as follows: n2 = (1.25 x 10^-2 M) x (2.7 L) / (1.25 x 10^-2) = 2.7 moles.

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Answer:

Explanation:

Given parameters:

Volume of AgNO₃ = 1.3L

Molarity or concentration AgNO₃   = 0.245M

Unknown:

Mass of AgCl produced = ?

Solution:

To solve this problem, we have to work from the known compound to the unknown one using the mole concept.

The known compound is the one that we can obtain the value of the number of moles from. Here it is the given AgNO₃ that can furnish us with this piece of information;

     Now let us establish the balanced reaction equation;

           2AgNO₃     +    CaCl₂    →     2AgCl    +   Ca(NO₃)₂

Now;

Find the number of moles of AgNO₃;

 Number of moles of AgNO₃  = concentration x volume  

                                                  = 1.3 x 0.245

                                                   = 0.32moles

From the balanced reaction equation;

        2 mole of AgNO₃  produced 2 moles of AgCl

        0.32 moles of AgNO₃ will produce 0.32moles of AgCl

Now that we know the number of moles of the AgCl, we can find the mass;

         Mass of AgCl  = number of moles x molar mass

Molar mass of AgCl  = 107.9 + 35.5  = 143.4g/mol

  Mass of AgCl  = 0.32 x 143.4  = 45.89g

Answer:

The calculated concentration of HCl will be less than actual.

Explanation:

Suppose during titration, the HCl was taken in burette and the NaOH in the volumetric flask.

Now we will use equivalence formula for the calculation of concentration of HCl.

             [tex]N_{1} V_{1} = N_{2} V_{2}[/tex]

Where L.H.S is for hydrochloric acid and R.H.S is for sodium hydroxide. The terms N and V represent normality and volume respectively.

If we calculate for

                              [tex]N_{1} = \frac{N_{2}V_{2} }{V_{1} }[/tex]

We see that if the volume of the HCl is greater then the concentration of the HCl will be reduced.

The error will cause the concentration of the hydrochloric acid to be underestimated.

The concentration of a solution is calculated from the ratio of the number of moles of the solutes and that of the volume of the solution.

Mathematically; concentration = mole/volume

Thus, with the number of moles being constant, the higher the volume of the solution, the lower the concentration that would be derived and vice versa.

This means that any volume that exceeds that of the accurate endpoint will cause the concentration to be underestimated and below the endpoint, the concentration would be overestimated.

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Answer:

67.3 s

Explanation:

The equation for the reaction can be represented as:

N₂O₅               ⇄                     NO₂        +          O₂

Rate (k) = [tex]5.89 * 10^{-3[/tex]

Rate law for first order is expressed as:

In [A] = -kt + In [A]₀

Given that:

[A] = Final Concentration = 1.85 M

[A]₀ = Initial Concentration = 2.75 M

time-taken  = ???

substituting our given data; we have:

In[1.85] = -[5.89 × 10⁻³](t) + In [2.75]

t = [tex]\frac{In(2.75)-In(1.85)}{(5.89*10^{-3}}[/tex]

t = [tex]\frac{0.3964}{5.89*10^{-3}}[/tex]

t = 67.3 s

Explanation:

it's is d cause bromine can even react with water so

Bromine atoms have more electrons in their outer shell than calcium atoms. So the correct option is D.

In chemistry, valence, usually spelled valency, is the characteristic of an element that establishes the maximum number of other atoms that one atom of the element may combine with. The phrase, which was first used in 1868, is used to indicate both the broad power of a combination of elements as well as its numerical value.

For 19th-century chemists, the explanation and systematization of valence posed significant difficulties. The majority of the work was focused on developing empirical guidelines for finding the valences of the elements because there was no good hypothesis for its origin.

The amount of hydrogen atoms that an element's atom may mix with or replace in a compound is used to quantify the characteristic valences of the various elements. But it soon became clear that various compounds had varied valences for numerous elements.

The identification of the chemical bond of organic compounds with a pair of electrons held jointly by two atoms and acting to hold them together in 1916 by the American chemist G.N. Lewis represented the first significant step in the development of a satisfactory explanation of valence and chemical combination. The German scientist W. Kossel studied the nature of the chemical connection between electrically charged atoms (ions) in the same year.

The theory of valence was reconstructed in terms of electronic structures and interatomic forces following the advent of the precise electronic theory of the periodic system of the elements. In response to the many ways that atoms interact, new notions such as ionic valence, covalence, oxidation number, coordination number, and metallic valence were developed.

Therefore the correct option is D.

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Answer:

decreases and increases

Explanation:

the addition of solutes can alter the freezing point and the boiling point .the changes in solute content where there is an addition of solutes results in the dropping of freezing point and increase in boiling point.this is known as freezing point depression and boiling point depression.

The molar mass of the unknown gas can be determined by comparing its effusion rate to the effusion rate of a known gas, such as nitrogen gas. According to Graham's law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass of the gas. Let's set up a proportion to find the molar mass of the unknown gas.

The molar mass of the unknown gas can be determined by comparing its effusion rate to the effusion rate of a known gas, such as nitrogen gas. In this case, it took 60 seconds for the unknown gas to effuse, whereas nitrogen gas required 48 seconds. According to Graham's law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass of the gas. Using this information, we can set up a proportion to find the molar mass of the unknown gas.

Let's assume the molar mass of the unknown gas is represented by 'X'. The proportion can be set up as follows:

(rate of unknown gas)/(rate of nitrogen gas) = sqrt(molar mass of nitrogen gas)/sqrt(molar mass of unknown gas)

Substituting the known values, we have:

(60 s)/(48 s) = sqrt(28.01 g/mol)/sqrt(X)

Solving for X, the molar mass of the unknown gas is approximately 37.07 g/mol.

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Answer:

Explanation:

(A) Phenol gives violet color by complexation with Fe3+ solution . It is best identification...

(B) For b , same phenolic test can be done ...But other esterification is also possibility...

(C) Picric acid has a particular identification test ,  

(D) Here , like the first one , phenolic test with FeCl3 gives violet color for 4-ethylphenol and no color for ethyl phenyl ether...

Answer:

Explanation:

The atomic unit of mass (a.m.u) is the unit used to compare the relative masses of atoms. An atomic mass unit is a twelfth of the mass of a carbon-12 atom.

The calculation of the molecular mass of a compound is done by adding the relative atomic masses of the atoms of the formula of said substance, taking into account its abundance in the compound. The atomic mass of the atoms that make up the molecule are obtained in the Periodic Table.

So,  in these cases you have:

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

Br: 79.9 amu

N: 14 amu

I take into account the abundance of each element in the compound you get:

BrN₃=79.9 amu + 3* 14 amu= 121.9 amu

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

C: 12 amu

H: 1 amu

I take into account the abundance of each element in the compound you get:

C₂H₆= 2*12 amu + 6*1 amu= 30 amu

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

N: 14 amu

F: 19 amu

I take into account the abundance of each element in the compound you get:

NF₂= 14 amu + 2*19 amu= 52 amu

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

Al: 27 amu

S: 32 amu

I take into account the abundance of each element in the compound you get:

Al₂S₃= 2*27 amu + 3*32 amu= 150 amu

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

Fe: 55.85 amu

N: 14 amu

O: 16 amu

I take into account the abundance of each element in the compound you get:

Fe(NO₃)₃= 55.85 amu + 3*(14 amu + 3*16 amu)= 241.85 amu

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

N: 14 amu

Mg: 24 amu

I take into account the abundance of each element in the compound you get:

Mg₃N₂= 3*24 amu + 2*14 amu= 100 amu

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

N: 14 amu

H: 1 amu

C: 12 amu

O: 16 amu

I take into account the abundance of each element in the compound you get:

(NH₄)₂CO₃: 2*(14 amu + 4*1 amu) + 12 amu+3*16 amu= 96 amu

The molecular mass or formula mass of the listed substances have been calculated using atomic masses from the periodic table.

The molecular mass or formula mass of a substance can be calculated by summing up the atomic masses of all atoms in the substance. Using the atomic masses from the periodic table, we have:

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Answer: All of these statements are true

Explanation:

Melting point help us to determine if a mixture is pure or has impurities by the virtues of it melting range..

Answer : The enthalpy of this reaction is, 1.06 kJ/mol

Explanation :

First we have to calculate the heat produced.

[tex]q=m\times c\times (T_2-T_1)[/tex]

where,

q = heat produced = ?

m = mass of solution = 137 g

c = specific heat capacity of water = [tex]4.18J/g^oC[/tex]

[tex]T_1[/tex] = initial temperature = [tex]21.00^oC[/tex]

[tex]T_2[/tex] = final temperature = [tex]24.70^oC[/tex]

Now put all the given values in the above formula, we get:

[tex]q=137g\times 4.18J/g^oC\times (24.70-21.00)^oC[/tex]

[tex]q=2118.842J=2.12kJ[/tex]

Now we have to calculate the enthalpy of this reaction.

[tex]\Delta H=\frac{q}{n}[/tex]

where,

[tex]\Delta H[/tex] = enthalpy change = ?

q = heat released = 2.12 kJ

n = moles of compound = 2.00 mol

Now put all the given values in the above formula, we get:

[tex]\Delta H=\frac{2.12kJ}{2.00mole}[/tex]

[tex]\Delta H=1.06kJ/mol[/tex]

Thus, the enthalpy of this reaction is, 1.06 kJ/mol

Answer:

Effects of Calcium channel blockers on the myocardial cells include lower excitability of the cells due lower influx of calcium ions during an action potential. This effect helps to regulate abnormal rapid heart rhythms (arrhythmias).

Calcium channel blockers also reduce blood pressure by dilating the arteries thereby reducing pressure in the arteries. This in turn also reduces the workload on the heart and therefore allows for adequate oxygen supply to heart in the treatment of Angina.

Examples of calcium ion channel blockers are amlodipine, nimodipine, verapamil, diltiazem etc.

Here is the full question.

Sodium carbonate is often added to laundry detergents to soften hard water and make the detergent more effective. Suppose that a particular detergent mixture is designed to soften hard water that is 3.8×10−3M in Ca2+ and 1.1×10−3M in Mg2+ and that the average capacity of a washing machine is 24.5 gallons of water. 1gallon=3.785L

If the detergent requires using 0.61 kg detergent per load of laundry, determine what percentage (by mass) of the detergent should be sodium carbonate in order to completely precipitate all of the calcium and magnesium ions in an average load of laundry water.

Answer:

7.90%

Explanation:

The equation for the reaction can be written as:

[tex]Na_2CO_3+Ca^{2+} ---------> CaCO_3(s) +2Na^+[/tex]

[tex]Na_2CO_3+Mg^{2+} ---------> MgCO_3(s) +2Na^+[/tex]

Molar mass of [tex]Na_2CO_3[/tex] = 106 g/mol

1.00 gallon of water = 3.785 L

∴ 24.5 gallons of water = 24.5 × 3.785

= 92.7325 L

mass precipitate = [tex][(3.8*10^{-3})+(1.1*10^{-3})]\frac{mol}{L} *92.7325*\frac{106g}{mol}[/tex]

mass precipitate = 48.162605

mass precipitate = 48.2 g

mass % = [tex]\frac{mass ofprecipitate}{mass of detergent} *100%[/tex]%

mass % = [tex]\frac{48.2g}{0.61kg}*\frac{1kg}{1000g}*100[/tex]%

mass % = 7.901639344

mass % = 7.90 %

The percentage (by mass) of the detergent should be 7.90%.

Since 3.8×10−3M in Ca2+ and 1.1×10−3M in Mg2+ and that the average capacity of a washing machine is 24.5 gallons of water. 1gallon=3.785L

The molar mass of Na_2CO_3 is 106g/mol

Since 1.00 gallon of water = 3.785 L

so, for 24.5 gallons of water, it is

= 24.5 × 3.785

= 92.7325 L

Now mass precipitate is

= ((3.8*10^-3) * (1.1*10^-3) * 92.7325 * 106g/mol

= 48.162605

= 48.2g

Now the mass percentage is

= mass of precipitate / mass of detergent

= 48.3g / 0.61 * 1kg / 1000g

= 7.90%

This is an incomplete question. Please find the full question below.

Sodium carbonate is often added to laundry detergents to soften hard water and make the detergent more effective. Suppose that a particular detergent mixture is designed to soften hard water that is 3.8×10−3M in Ca2+ and 1.1×10−3M in Mg2+ and that the average capacity of a washing machine is 24.5 gallons of water. 1gallon=3.785L

If the detergent requires using 0.61 kg detergent per load of laundry, determine what percentage (by mass) of the detergent should be sodium carbonate in order to completely precipitate all of the calcium and magnesium ions in an average load of laundry water.

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Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

The rate constant at 48.5°C if the activation energy is 116 kJ/mol is [tex]1.017 E-8 s^{-1}[/tex]

CH₃Cl + H₂O → CH₃OH + HCl

At T₁ = 25°C (298 K) ⇒ K₁ = 3.32 E-10 s-1

At T₂ = 48.5°C (321.5 K) ⇒ K₂ = ?

According to Arrhenius equation:

[tex]K(T) = A*e^{(-Ea/RT)}\\\\ln K = ln(A) - [(Ea/R)(1/T)][/tex]

where,

A = frequency factor

R = 8.314 E-3 KJ/K.mol

[tex]ln K_1 = ln(A) - [Ea/R)*(1/T_1)][/tex]..........(1)

[tex]ln K_2 = ln(A) - [(Ea/R)*(1/T_2)][/tex]...........(2)

On dividing equation 1 by 2:

ln (K₁/K₂) = (Ea/R)* (1/T₂ - 1/T₁)

ln (K₁/K₂)  = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

ln (K₁/K₂)  = (13952.37 K)*(- 2.453 E-4 K-1)

ln (K₁/K₂)  = - 3.422

(K₁/K₂)= [tex]e^{(-3.422)}[/tex]

[tex](3.32 E-10 s^{-1})[/tex] / K₂ = 0.0326

K₂ = [tex](3.32 E-10 s^{-1})[/tex] /0.0326

K₂ = [tex]1.017 E-8 s^{-1}[/tex]

Thus, the rate constant at 48.5°C if the activation energy is 116 kJ/mol is 1[tex]0.017 E-8 s^{-1}[/tex]

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Explanation:

Answer:

0.89 g of ethane

Explanation:

The balanced reaction equation is

C2H6(g) + 7/2 O2(g) -----------> 2CO2(g) + 3H2O(g)

From this balanced reaction

30g of ethane yields 54g of water

Therefore mass of ethane necessary to obtain 1.61g of water we have:

30 × 1.61/54 = 0.89 g of ethane

Here is the full question

How many moles of sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00? Ignore the volume change due to the addition of sodium acetate.  Ka of acetic acid is 1.7×10-5.

Answer:

0.3396 moles

Explanation:

pH = pKA + log[tex]\frac{[acetate]}{[aceticacid]}[/tex]

[acetic acid] = 0.1 M

pH = 5.0

[tex]K_a[/tex] = [tex]1.7*10^{-5[/tex]

5.0 = -log ([tex]1.7*10^{-5[/tex]) + log [tex]\frac{[acetate]}{0.1}[/tex]

5.0 = 4.77 + log [acetate] + 1

5.0 = 5.77 + log [acetate]

- log [acetate] = 5.77 - 5.0

- log [acetate]  = 0.77

log [acetate]  = -0.77

[acetate]  = log⁻¹ (-0.77)

[acetate]  =  0.1698

∴ for 2L = 2 × 0.1698

= 0.3396 moles

The sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00 - 0.3396 moles

Given:

The concentration of acetic acid = 0.10 M

pH = 5.0  

[tex]K_a[/tex] = [tex]1.7\times10^{-5[/tex]

Solution:

According to Hasselbalch equation:

[tex]pH=pKa + log\frac{[A^-]}{[HA]}\\\\or\\\\pH = pKa + log\frac{(acetate)}{(acetic acid)}[/tex]

5.0 = -log [tex](1.7\times10^{-5})[/tex] + log [tex]\frac{(acetate)}{(0.1)}[/tex]

5.0 = 4.77 + log [acetate] + 1

5.0 = 5.77 + log [acetate]

- log [acetate] = 5.77 - 5.0

- log [acetate]  = 0.77

log [acetate]  = -0.77

[acetate]  = log⁻¹ (-0.77)

[acetate]  =  0.1698

∴ for 2L = 2 × 0.1698

= 0.3396 moles

Thus, the sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00 - 0.3396 moles

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The enthalpy change [tex](\( \Delta H \))[/tex] for this reaction per mole of [tex]\( X \)[/tex] is approximately 88.27 kj/mol.

To calculate the enthalpy change[tex](\( \Delta H \))[/tex] for the reaction per mole of  X , we can use the formula:

[tex]\[ \Delta H = \frac{q}{n} \][/tex]

where:

- [tex]\( q \)[/tex] is the heat absorbed or released by the reaction,

- [tex]\( n \)[/tex]  is the moles of [tex]\( X \).[/tex]

First, let's find the moles of \( X \):

[tex]\[ \text{moles of } X = \frac{\text{mass}}{\text{molar mass}} \][/tex]

[tex]\[ \text{moles of } X = \frac{1.20 \, \text{g}}{54.0 \, \text{g/mol}} \][/tex]

[tex]\[ \text{moles of } X \approx 0.0222 \, \text{mol} \][/tex]

Now, we need to find the heat [tex](\( q \))[/tex] absorbed or released by the reaction. The heat can be calculated using the formula:

[tex]\[ q = mc\Delta T \][/tex]

where:

- m  is the mass of the solution [tex](water + \( X \)),[/tex]

- c  is the specific heat of the solution,

- [tex]\( \Delta T \)[/tex]is the change in temperature.

First, calculate the mass of the solution:

[tex]\[ \text{mass of solution} = \text{mass of water} + \text{mass of } X \][/tex]

[tex]\[ \text{mass of solution} = 32.0 \, \text{g} + 1.20 \, \text{g} \][/tex]

[tex]\[ \text{mass of solution} = 33.20 \, \text{g} \][/tex]

Now, calculate [tex]\( \Delta T \)[/tex]:

[tex]\[ \Delta T = T_f - T_i \][/tex]

[tex]\[ \Delta T = 29.0^\circ \text{C} - 15.0^\circ \text{C} \][/tex]

[tex]\[ \Delta T = 14.0^\circ \text{C} \][/tex]

Now, plug in the values into the heat formula:

[tex]\[ q = (33.20 \, \text{g}) \times (4.18 \, \text{J/(g} \cdot ^\circ \text{C})) \times (14.0 \, ^\circ \text{C}) \][/tex]

[tex]\[ q \approx 1961.12 \, \text{J} \][/tex]

Now, use the first formula to find [tex]\( \Delta H \):[/tex]

[tex]\[ \Delta H = \frac{1961.12 \, \text{J}}{0.0222 \, \text{mol}} \][/tex]

[tex]\[ \Delta H \approx 88272 \, \text{J/mol} \][/tex]

Therefore, the enthalpy change[tex](\( \Delta H \))[/tex] for this reaction per mole of X is approximately[tex]\( 88.27 \, \text{kJ/mol} \).[/tex]

Calculate the heat absorbed using q = mcΔT and convert to kJ. Determine the moles of substance X, then compute the enthalpy change per mole. The enthalpy change (ΔH) for the reaction per mole of X is 84.34 kJ/mol.

To calculate the enthalpy change (ΔH) for the reaction per mole of substance X, follow these steps:

Answer:

2.6×10^-3 mols-1

Explanation:

From

k= 2.303/t × log[A]o/[A]t

Time taken t= 30 seconds

Initial concentration [A]o= 0.427M

Concentration at time (t) [A]t= 0.395M

k= 2.303/30 ×log[0.427]/[0.395]=2.6×10^-3 mols-1

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

[tex]R_{in}[/tex] [tex]= \frac{0.04}{100}*2000[/tex]

[tex]R_{in}[/tex] = 0.8

The rate-out

[tex]R_{out}[/tex] = [tex]\frac{A}{6000}*2000[/tex]

[tex]R_{out}[/tex] = [tex]\frac{A}{3}[/tex]

We can say that:

[tex]\frac{dA}{dt}=[/tex] [tex]0.8-\frac{A}{3}[/tex]

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

[tex]\frac{dA}{dt} +\frac{A}{3} =0.8[/tex]

Integration of the above linear equation =

[tex]e^{\int\limits \frac {1}{3}dt } =[/tex] [tex]e^{\frac{1}{3}t[/tex]

so we have:

[tex]e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A[/tex] [tex]= 0.8e^{\frac{1}{3}t[/tex]

[tex]\frac{d}{dt}[e^{\frac{1}{3}t}A][/tex] [tex]= 0.8e^{\frac{1}{3}t[/tex]

[tex]Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C[/tex]

∴ [tex]A(t) = 2.4 +Ce^{-\frac{1}{3}t[/tex]

Since A(0) = 12

Then;

[tex]12 =2.4 + Ce^{-\frac{1}{3}}(0)[/tex]

[tex]C= 12-2.4[/tex]

[tex]C =9.6[/tex]

Hence;

[tex]A(t) = 2.4 +9.6e^{-\frac{t}{3}}[/tex]

[tex]A(0) = 2.4 +9.6e^{-\frac{10}{3}}[/tex]

[tex]A(t) = 2.74[/tex]

∴ the concentration at 10 minutes is ;

=  [tex]\frac{2.74}{6000}*100[/tex]%

= 0.0456667 %

= 0.046% to three decimal places

Answer: C. The Second Law of Thermodynamics.

Explanation:

The law of gravity is related to the force interaction due to gravitation between two bodies (i.e. a person and planet Earth, a star and a planet).

The law of diminishing return is not used in Physics, but in Economics and describes the diminishing of marginal returns in time.

The first law of Thermodynamics analyses the energy interactions of a system in a quantitative manner and it is a generalized form of the Principle of Energy Conservation. It is oriented to the analyses of energy inflows and outflows.

The second law of Thermodynamics analyses the energy interactions of a system in a qualitative manner and it is based on thermodynamic property named Entropy, which may helpful to measure irreversibilities associated with system and irreversibility generation as well, provided that a comparison with another equivalent system exists.

Irreversibilities depends on the characteristics of system and nature of energy sources. Empirically, it is known that heat offers a lower quality than electricity in order to get a available work. That is to say, there is a lower of obtaining available work from heat than from electricity.

Hence, the statement is a consequence of using the second law of Thermodynamics and, therefore, the correct answer is C.

Answer:  32.9 Liters

Explanation:

To calculate the moles, we use the equation:

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

1. moles of [tex]FeS_2=\frac{96.7g}{119.99g/mol}=0.806mol[/tex]

2. moles of [tex]O_2[/tex]

[tex]PV=nRT[/tex]

P = pressure of the gas = 1.20  atm

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]398K[/tex]

[tex]1.20\times 55.0=n\times 0.0821\times 398[/tex]

[tex]n=2.02[/tex]

[tex]4FeS_2(s)+11O_2(g)\rightarrow 2Fe_2O_3(s)+8SO_2(g)[/tex]

According to stoichiometry:

11 moles of oxygen reacts with 4 moles of [tex]FeS_2[/tex]

Thus 2.02 moles of oxygen reacts with =[tex]\frac{4}{11}\times 2.02=0.73[/tex] moles of [tex]FeS_2[/tex]

Thus oxygen acts as limiting reagent and [tex]FeS_2[/tex] is excess reagent.

As 11 moles of oxygen gives = 8 moles of [tex]SO_2[/tex]

2.02 moles of oxygen gives =[tex]\frac{8}{11}\times 2.02=1.47[/tex] moles of [tex]SO_2[/tex]

[tex]PV=nRT[/tex]

P = pressure of the gas = 1  atm (at STP)

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]273K[/tex]   (at STP)

[tex]1\times V=1.47\times 0.0821\times 273[/tex]

[tex]V=32.9L[/tex]

Thus volume of [tex]SO_2[/tex] (at STP) formed from the reaction is 32.9 L

The volume of SO₂ formed from the reaction at STP is 32.92 L

From the question,

We are to determine the volume of SO₂ formed from the reaction

The given balanced chemical equation for the reaction is

4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(g)

This means,

4 moles of FeS₂ reacts with 11 moles of oxygen to produce 2 moles of Fe₂O₃ and 8 moles of SO₂

First, we will determine the number of moles of each reactant present

Mass = 96.7 g

From the formula

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

Molar mass of FeS₂ = 119.99 g/mol

∴ Number of moles of FeS₂ present =[tex]\frac{96.7}{119.99}[/tex]

Number of moles of FeS₂ present = 0.8059 mole

Using the formula

PV = nRT

[tex]n =\frac{PV}{RT}[/tex]

Putting the given parameters into the formula, we get

[tex]n = \frac{1.2 \times 55.0}{0.08206 \times 398}[/tex]

[tex]n = \frac{66}{32.65988}[/tex]

n = 2.0208 moles

Since,

4 moles of FeS₂ reacts with 11 moles of oxygen to produce 8 moles of SO₂

Then,

[tex]\frac{2.0208 \times 4}{11}[/tex] moles of FeS₂ will react with 2.0208 moles of oxygen to produce [tex]\frac{2.0208 \times 4}{11} \times 2[/tex]  moles of SO₂

That is,

0.7348 moles of FeS₂ will react with 2.0208 moles of oxygen to produce 1.4697 moles of SO₂

∴ Number of moles of SO₂ formed is 1.4697 moles

Now for the volume of SO₂ formed at STP

Since

1 mole of a gas has a volume of 22.4 L at STP

Then

1.4697 moles of the SO₂ will have a volume of 1.4697 × 22.4 L

1.4697 × 22.4 L = 32.92L

Hence, the volume of SO₂ formed from the reaction at STP is 32.92 L

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Answer:

The answer to your question is empirical formula = CH

                                                    molecular formula = C₆H₆

Explanation:

Data

Carbon  92.2%

Hydrogen  7.76%

Formula mass = 78.1 g

Process

1.- Express the percents as grams

Carbon            92.2 g

Hydrogen          7.76 g

2.- Convert the grams to moles

Carbon                 12 g ---------------- 1 mol

                            92.2 g -------------  x

                            x = (92.2 x 1)/12

                            x = 7.68 moles

Hydrogen             1 g ------------------ 1 mol

                            7.76 g -------------  x

                            x = 7.76 moles

3.- Divide by the lowest number of moles

Carbon         7.68 / 7.68 = 1

Hydrogen     7.76 / 7.68 = 1.01

4.- Write the empirical formula

                                      CH

5.- Calculate the molecular weight of the empirical formula

CH = 12 + 1 = 13

6.- Divide the molecular weight by the molecular weight of the empirical formula

                78.1 / 13 = 6

7.- Write the molecular formula

               6(CH) = C₆H₆        

The empirical formula of the compound is CH. The molecular formula of the compound is [tex]\rm \bold C_6H_6[/tex]

A molecular formula has been the one that represents the exact number of atoms in a compound. The empirical formula has been able to represent the whole number ratio of atoms present in a compound.

The given compound has 92.2% Carbon and 7.76% Hydrogen. The weight will be:

Carbon = 92.2 g

Hydrogen = 7.76 g

From the weight, the moles of elements in the compound can be calculated.

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Moles of carbon = [tex]\rm \dfrac{92.2}{12}[/tex]

Moles of carbon = 7.76 moles

Moles of hydrogen = weight of hydrogen

Moles of hydrogen = 7.76 moles

For the empirical formula, divide the moles of each element to the nearest mole number;

Carbon = [tex]\rm \dfrac{7.76}{7.76}[/tex]

Carbon = 1

Hydrogen = [tex]\rm \dfrac{7.76}{7.76}[/tex]

Hydrogen = 1

Thus, the empirical formula of the compound will be CH.

For finding the molecular formula, the molecular weight from the empirical formula has been divided with the molecular weight of the compound to find the number of atoms.

Molecular weight of empirical formula = weight of carbon + weight of hydrogen

Molecular weight of empirical formula = 12 + 1

Molecular weight of empirical formula = 13 grams

Molecular mass of compound = 78.1 grams.

Number of atoms of each element = [tex]\rm \dfrac{78.1}{13}[/tex]

Number of atoms of each element = 6

Thus, the molecular formula of the compound is [tex]\rm C_6H_6[/tex].

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Answer:

0.07 M

Explanation:

We have to start with the reaction between the Calcium hydroxide and the Hydroiodic acid:

[tex]Ca(OH)_2~+~HF~->~CaF_2~+~H_2O[/tex]

Then we have to balance the reaction:

[tex]Ca(OH)_2~+~2HF~->~CaF_2~+~2H_2O[/tex]

We have to keep in mind that we have to do use the volume in "L" units (14.7 mL= 0.0147L). When we plug the values into the equation we will get:

[tex]M=\frac{#mol}{L}[/tex]

[tex]0.164~M=\frac{#mol}{0.0147}[/tex]

[tex]#mol=~0.0024[/tex]

Now, we have to use the information from the balanced reaction to finding the moles of Calcium hydroxide. When we check the reaction we found a molar ratio of 1:2 ( 1 mol of [tex]Ca(OH)_2[/tex]: 2 mol of  [tex]HF[/tex]). With this molar ratio, we can find the moles of [tex]HF[/tex].

[tex]0.0024~mol~HF\frac{1~mol~Ca(OH)_2}{2~mol~HF}[/tex]

[tex]0.0012~mol~Ca(OH)_2[/tex]

Finally, to find the molarity we have to divide the moles of [tex]Ca(OH)_2[/tex] and the volume of [tex]Ca(OH)_2[/tex] in liters (17.2 mL=0.0172L) so:

[tex]M=\frac{0.012}{0.0172}=0.07[/tex]

The molarity of [tex]Ca(OH)_2[/tex] is 0.07M.