Hydrochloric acid is purchased in 12 molar concentrations. If I needed 200 ml of 2.0 molar HCl, how many milliliters of the concentrated solution would I need

Answer:

0.3 Coulomb charge flows through the toy car.

0.9 Joules of energy is lost by the battery

Explanation:

[tex]Current(I)=\frac{charge(Q)}{Time (T)}[/tex]

a)

Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)

milli Ampere = 0.001 Ampere

Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s

1 min = 60 seconds

Charge flows through the toy car be Q

[tex]0.50\times 0.001 A=\frac{Q}{600 s}[/tex]

[tex]Q=0.50 \times 0.001 A\times 600 s=0.3 C[/tex]

0.3 Coulomb charge flows through the toy car.

b)

[tex]Heat(H)=Voltage(V)\times Current(I)\times Time(T)[/tex]

Voltage of the battery = V = 3.0 V

Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)

Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s

[tex]H=V\times I\time T=3.0 V\times 0.50 \times 0.001 A\times 600 s[/tex]

H = 0.9 Joules

0.9 Joules of energy is lost by the battery

Explanation:

The chemiosmotic theory

Answer:

53.18 gL⁻¹

Explanation:

Given that:

[tex]Cu^{2+}_{(aq)}[/tex] [tex]+[/tex] [tex]2NH_{3(aq)}[/tex] [tex]------>[/tex]  [tex][Cu(NH_3)_2]^+_{(aq)}[/tex]      ------equation (1)

where;

Formation Constant  [tex](k_f) =[/tex] [tex]6.3*10^{10}[/tex]

However, the Dissociation of [tex]CuBr_{(s)[/tex] yields:

[tex]CuBr_{(s)}[/tex]      ⇄    [tex]Cu^{+}_{(aq)}[/tex]  [tex]+[/tex] [tex]Br^-_{(aq)}[/tex]      -------------- equation (2)

where;

the Solubility Constant [tex](k_{sp})[/tex]  [tex]= 6.3 *10^{-9[/tex]

From equation (1);

[tex](k_f) =[/tex] [tex]\frac{[[Cu(NH_3)_2]^+]}{[Cu^{2+}][NH_3]^{2}}[/tex]            ---------  equation (3)

From equation (2)

[tex](k_{sp})[/tex]  [tex]= [Cu^+][Br^-][/tex]           ---------  equation (4)

In [tex]NH_3[/tex], the net reaction for [tex]CuBr_{(s)[/tex] can be illustrated as:

[tex]CuBr_{(s)[/tex]   [tex]+[/tex] [tex]2NH_{3(aq)}[/tex]  ⇄  [tex][Cu(NH_3)_2]^+_{(aq)}[/tex]  [tex]+ Br^-_{(aq)}[/tex]

The equilibrium constant (K) can be written as :

[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]

If we multiply both the numerator and the denominator with  [tex][Cu^+][/tex] ; we have:

[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}*\frac{[Cu^+]}{[Cu^+]}[/tex]

[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+}{[NH_3]^2[Cu^+]}*{[Cu^+][Br^-]}[/tex]

[tex]K = k_f *k_{sp}[/tex]

[tex]K= (6.3*10^{10})*(6.3*10^{-9})[/tex]

[tex]K= 3.97*10^2[/tex]

[tex]K[/tex] ≅ [tex]4.0*10^2[/tex]

Now; we can re-write our equilibrium constant again as:

[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]

[tex]4.0*10^2 = \frac{(x)(x)}{(0.76-2x)^2}[/tex]

[tex]4.0*10^2 = \frac{(x)^2}{(0.76-2x)^2}[/tex]

[tex]4.0*10^2 = (\frac{(x)}{(0.76-2x)})^2[/tex]

By finding the square of both sides, we have

[tex]\sqrt {4.0*10^2} = \sqrt {(\frac{(x)}{(0.76-2x)})^2[/tex]

[tex]2.0*10 = \frac{x}{(0.76-2x)}[/tex]

[tex]20(0.76-2x) =x[/tex]

[tex]15.2 -40x=x[/tex]

[tex]15.2 = 40x +x[/tex]

[tex]15.2 = 41x[/tex]

[tex]x = \frac{15.2}{41}[/tex]

[tex]x = 0.3707 M[/tex]

In gL⁻¹; the solubility of [tex]CuBr_{(s)[/tex] in 0.76 M [tex]NH_3[/tex] solution will be:

[tex]= \frac{0.3707 mole of CuBr}{1L}*\frac{143.45 g}{mole of CuBr}[/tex]

=  53.18 gL⁻¹

The problem involves finding the solubility of copper(I) bromide in a solution of ammonia. The solution entails expressing the formation of complex ions and dissolution of CuBr in terms of equilibrium expressions. Solving the system of expressions for solubility 's' can eventually give the solubility in g·L-1.

In this problem, we are looking for the solubility of copper(I) bromide (CuBr) in a solution with 0.76 M NH3. Since CuBr does not have high solubility in water, we are using ammonia which forms complex ions with copper, enhancing its solubility. We will utilize the given equilibrium constant (Kf) which helps us to understand how far the forward reaction (complex ion formation) proceeds.

To start, we'll write the reaction of Cu+ ion with NH3 and the corresponding Kf expression:

Cu+ (aq) + 2NH3 (aq) <=> Cu(NH3)2+ (aq); Kf = [Cu(NH3)2+] / [Cu+] [NH3]^2

Next, we need to find an expression for [Cu+] in terms of solubility and solve for solubility. Let's denote the solubility of CuBr as 's'. The dissolution of CuBr can be represented as:

CuBr (s) <=> Cu+ (aq) + Br- (aq)

We can see that for every mole of CuBr that dissolves, one mole of Cu+ ion is produced. Thus, [Cu+] = s.

Now using the Kf expression, we can solve the system of equations to find s, which effectively gives us the solubility of CuBr in 0.76 M NH3. After that, by taking the molecular weight of CuBr into account, we can finally express the solubility in g·L−1.

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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:

2Ag (s) + 2HNO₃ (aq)  2AgNO₃ (aq) + H₂ (g)

Explanation:

2Ag (s) + 2HNO₃ (aq)  2AgNO₃ (aq) + H₂ (g)

Ag is below H₂ in reactivity series. Therefore Ag does not spontaneously replace H₂ from any compound.

Answer:

The answer to the question is

2Ag (s) +2HNO₃ (aq) →2AgNO₃ (aq)+ H₂ (g)

Explanation:

The position of a cell on the reduction potential table determines if a reduction reaction will be spontaneous

The higher the positive reduction potential  the higher the spontaneity of the reduction reaction. that is if the E⁰ cell is positive the cell is a spontaneous voltaic cell, however if the E⁰ cell is negative the cell is a electrolytic non-spontaneous cell

On the standard potential table silver has a low positive potential of

Ag⁺ + e⁻ → Ag = 0.799

It is an oxidizing agent tending to scarcely release electrons, therefore for the reaction to take place, there has to be some external supply of electromotive force.

Answer:

12 molecules of carbon dioxide

24 molecules of water

Explanation:

Given parameters:

Number of molecules of methane = 12 molecules

Unknown:

Number of molecules of carbon dioxide = ?

Number of molecules of water = ?

Solution:

The given amount of methane is the limiting reagent in this reaction. By this, we can ascertain the amount of products that will be formed and the extent of the reaction.

We first write the balanced chemical equation for this reaction;

             CH₄     +    2O₂    →     CO₂   +     2H₂O

From balanced equation;

                 1 mole of methane produced 1 mole of carbon dioxide

  12 molecules of methane will produce 12 molecules of carbon dioxide

Also;

             1 mole of methane produced 2 moles of water;

12 molecules of methane will produce (12 x 2)molecules = 24 molecules of water

When 12 molecules of methane react with excess oxygen, 12 molecules of carbon dioxide and 24 molecules of water are produced, following a 1:2:1:2 ratio of methane to oxygen to carbon dioxide to water.

If 12 molecules of methane (CH4) react with plenty of oxygen (O2), the balanced chemical equation for the combustion of methane shows a 1:2:1:2 ratio. This means for every one methane molecule, two oxygen molecules are consumed, and one carbon dioxide (CO2) molecule and two water (H2O) molecules are produced. Therefore, 12 molecules of methane reacting with excess oxygen will produce 12 molecules of carbon dioxide and 24 molecules of water.

This equation states that for every 1 molecule of methane and 2 molecules of oxygen that react, we will produce 1 molecule of carbon dioxide and 2 molecules of water.

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

Answer:

The limiting reactant can be described as the substance that is depleted first and stops

Explanation:

Imagine you have hydrogen and oxygen to produce water.

The reaction is:  2H₂(g) + O₂(g) → 2H₂O(g)

You have 1 mol of each reactant. As you see ratio is 2:1, so the limiting reactant is the hydrogen.

You know by stoichiometry, that 2 moles of H₂ need 1 mol of O₂ to react

If I have 1 mol of H₂, I will need the half of moles of O₂, so 0.5 moles. It is ok  because I have 1 mole, as I need the half, then half a mole will remain unreacted. This is what is called excess reagent,

If I make to react 1 mol of oxgen I need 2 moles of H₂. As I have 1 mol, of course I will need 2 moles but the thing is I have 1 mol.

This is the limiting reactant. I do not have enough of reactant so the reaction will happen until I complete to use it, that's why we can say that is depleted first and stops.

In a chemical reaction, when you have data of both reactants you can determine the limiting. Otherwise the excersise must tell that one ractant is in excess, to work with the limiting. Limiting reactant is the first step to work with the reaction, all the operations must be done by it. You do not use the reagent in excess

Answer : The value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]

Explanation :

For the given chemical reaction:

[tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]

The expression of [tex]K_p[/tex] for above reaction follows:

[tex]K_p=(P_{NH_3})^2\times P_{CO_2}[/tex]        ........(1)

                      [tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]

Initial:                                                      0               0

At eqm:                                                   2x              x

As we are given that:

Total pressure of gas at equilibrium = 0.116 atm

2x + x = 0.116 atm

3x = 0.116 atm

x = 0.0387 atm

Putting values in expression 1, we get:

[tex]K_p=(2x)^2\times (x)[/tex]

[tex]K_p=(2\times 0.0387)^2\times (0.0387)[/tex]

[tex]K_p=2.32\times 10^{-4}[/tex]

Thus, the value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]

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

To find the amount of NaCl needed to increase the boiling temperature of water by 1.500°C, one must understand molal boiling point elevation and van't Hoff factor. Using the formula and given constants, we calculate the grams of NaCl required through the steps of determining molality, calculating the moles of NaCl, and then converting to grams.

To calculate the amount of NaCl needed to raise the boiling temperature of water by 1.500°C, we first understand the concept of molal boiling point elevation and van't Hoff factor. Given that the Kb for water is 0.5100°C/m, we use the formula ΔTb = i*Kb*m, where ΔTb is the boiling point elevation, i is the van't Hoff factor for NaCl (which is 2 because NaCl disassociates into Na+ and Cl- ions), Kb is the ebullioscopic constant for the solvent (water), and m is the molality of the solution.

To find the molality (m), we rearrange the equation as m = ΔTb / (i*Kb). Substituting the known values, we get m = 1.500 / (2*0.5100) = 1.4706 mol/kg. Knowing the molality and the mass of the solvent (water), we can then calculate the moles of NaCl required, which is molality * mass of solvent in kg. Finally, converting moles of NaCl to grams using its molar mass (58.44 g/mol), gives us the total grams of NaCl needed. This step-by-step process provides a clear link between the theoretical concepts and their practical application.

Approximately 85.78 grams of [tex]NaCl[/tex] would need to be added to 1001 grams of water to increase the boiling temperature of the solution by [tex]1.500 \°C[/tex].

To solve this problem, we will use the concept of boiling point elevation, which states that the boiling point of a solvent is increased by an amount proportional to the molal concentration of the solute. The proportionality constant is known as the ebullioscopic constant (Kb). The formula to calculate the boiling point elevation is:

[tex]\[ \Delta T_b = i \cdot K_b \cdot m \][/tex]

 where:

- [tex]\( \Delta T_b \)[/tex] is the increase in boiling point temperature,

- [tex]\( i \)[/tex] is the van 't Hoff factor (the number of moles of particles in solution per mole of solute; for [tex]NaCl[/tex], [tex]\( i = 2 \)[/tex] because it dissociates into two ions, [tex]Na^+ and Cl^-)[/tex],

- [tex]\( K_b \)[/tex] is the ebullioscopic constant for the solvent (given as [tex]0.5100 \°C[/tex]/m for water),

- [tex]\( m \)[/tex] is the molality of the solution (moles of solute per kilogram of solvent).

Given:

- [tex]\( \Delta T_b = 1.5000 °C \)[/tex]

- [tex]\( K_b = 0.5100 °C/m \)[/tex]

- [tex]\( i = 2 \)[/tex] for [tex]NaCl[/tex]

- Mass of water (solvent) = 1001 g (which is approximately 1 kg, since 1000 g = 1 kg)

First, we will rearrange the formula to solve for the molality (m):

[tex]\[ m = \frac{\Delta T_b}{i \cdot K_b} \][/tex]

Substitute the given values:

[tex]\[ m = \frac{1.5000 °C}{2 \cdot 0.5100 °C/m} \][/tex]

[tex]\[ m = \frac{1.5000 °C}{1.0200 °C/m} \][/tex]

[tex]\[ m \approx 1.4695 m \][/tex]

Now that we have the molality, we can calculate the number of moles of [tex]NaCl[/tex] needed to achieve this molality in 1 kg of water:

[tex]\[ \text{moles of NaCl} = m \cdot \text{mass of solvent in kg} \][/tex]

[tex]\[ \text{moles of NaCl} = 1.4695 \text{ m} \cdot 1 \text{ kg} \][/tex]

[tex]\[ \text{moles of NaCl} \approx 1.4695 \][/tex]

The molar mass of [tex]NaCl[/tex] is approximately 58.44 g/mol. To find the mass of [tex]NaCl[/tex] needed, we multiply the number of moles by the molar mass:

[tex]\[ \text{mass of NaCl} = \text{moles of NaCl} \cdot \text{molar mass of NaCl} \][/tex]

[tex]\[ \text{mass of NaCl} \approx 1.4695 \text{ mol} \cdot 58.44 \text{ g/mol} \][/tex]

[tex]\[ \text{mass of NaCl} \approx 85.7797 \text{ g} \][/tex]

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:

We can for 5.93 grams potassium fluoride

Explanation:

Step 1: Data given

Mass of potassium = 4.00 grams

Mass of fluorine = 3.00 grams

Molar mass potassium = 39.10 g/mol

Molar mass fluorine gas =38.00 g/mol

Step 2: The balanced equation

2K (s) + F2 (g) → 2KF (s)

Step 3: Calculate moles potassium

Moles potassium = 4.00 grams / 39.10 g/mol

Moles potassium = 0.102 moles

Step 4: Calculate moles F2

Moles F2 = 3.00 grams / 38.00 g/mol

Moles F2 = 0.0789 moles

Step 5: Calculate limiting reactant

Potassium is the limiting reactant. There will react 0.102 moles

Fluorine gas is in excess. There will react 0.102/ 2 = 0.051 moles

There will remain 0.0789 - 0.051 = 0.0279 moles

Step 6: Calculate moles potassium fluoride

For 2 moles potassium we need 1 mol fluorine to produce 2 moles potassium fluoride

For 0.102 moles K we need 0.102 moles KF

Step 7: Calculate mass KF

Mass KF = moles KF * molar mass KF

Mass KF = 0.102 moles * 58.10 g/mol

Mass KF = 5.93 grams

Final answer:

By calculating the moles of potassium and fluorine used in the reaction from their given masses, and considering the stoichiometry of the balanced equation, it's determined that 4.59 grams of potassium fluoride can be produced.

Explanation:

The question asks how many grams of potassium fluoride can form when 4.00 grams of potassium react with 3.00 grams of fluorine gas according to the balanced chemical equation 2K (s) + F2 (g) → 2KF (s). To solve this, we first find the molar mass of potassium (K) and fluorine (F2), which are approximately 39.10 g/mol and 38.00 g/mol, respectively. Then, we calculate the moles of K and F2 available by dividing the given masses by their respective molar masses. With potassium: 4.00 g / 39.10 g/mol = 0.102 moles of K; with fluorine: 3.00 g / 38.00 g/mol = 0.079 moles of F2. As the equation shows a 2:1 ratio, fluorine limits the reaction. Therefore, we can form 0.079 moles of KF. The molar mass of KF is approximately 58.10 g/mol (39.10 g/mol for K + 19.00 g/mol for F). Thus, the mass of KF that can form is 0.079 moles × 58.10 g/mol = 4.59 grams.

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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20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration? The pKa of CH3CH2NH3+ is 10.75.

Answer:

Bromocresol green, color change from pH = 4.0 to  5.6

Explanation:

The equation for the reaction is as follows:

[tex]C_2H_5NH_{2(aq)[/tex]     +     [tex]H^+_{(aq)[/tex]      ⇄        [tex]C_2H_5NH_{3(aq)}^+[/tex]

Given that concentration of [tex]C_2H_5NH_{2(aq)[/tex] = 10%

i.e 10 g of [tex]C_2H_5NH_{2(aq)[/tex] in 100 ml solution

molar mass = 45.08 g/mol

number of moles = [tex]\frac{10}{45.08}[/tex]

= 0.222 mol

Molarity of [tex]C_2H_5NH_{2(aq)[/tex] = 0.222 × [tex]\frac{1000}{100}mL[/tex]

= 2.22 M

However, number of moles of [tex]C_2H_5NH_{2(aq)[/tex] in 20 mL can be determined as:

number of moles of [tex]C_2H_5NH_{2(aq)[/tex] = 20 mL × 2.22 M

= [tex]44*10^{-3} mole[/tex]

Concentration of [tex]C_2H_5NH_{2(aq)[/tex] = [tex]\frac{44*10^{-3}*1000}{20}[/tex]

= 2.22 M

Similarly, The pKa Value of [tex]C_2H_5NH_{2(aq)[/tex]  is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

Finally, the pH value at equivalence point is:

pH= [tex]\frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C][/tex]

pH = [tex]\frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22][/tex]

pH = 5.21

∴

The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6.

For titration of the weak base ethylamine with a strong acid like HCl, the pH at the equivalence point will be less than 7, so an indicator that changes color in an acidic pH range such as methyl orange would be most suitable.

The appropriate indicator in this case would depend on the equivalence point of the titration. Ethylamine, CH3CH2NH2, is a weak base, and it's being titrated with HCl, a strong acid. The reaction between them generates water, and ethylammonium chloride which functions as a weak acid. This setup indicates that the pH at the equivalence point will be less than 7.

Because the pH at the equivalence point will be in the acidic range (below 7), we should select an indicator which changes color around this pH. A good choice would be methyl orange, which changes color from red to yellow across a pH range of approximately 3.1 to 4.4.

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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:

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:

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.

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:

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:

10.4psi

Explanation:

Since the problem involves pressure and volume at constant temperature, Boyle's law is the appropriate formula to use.

Boyle's law States that at a constant temperature, the volume of a given mass of a bass is inversely proportional to its pressure.

Using Boyle's law

P1v1 = p2v2

P1 is 18.1psi

V1 is 2.77L

P2 =?

V2 is 4.81

Temperature is constant

P2 =P1v1/ v2

= 18.1 x 2.77 / 4.82

= 10.4psi

Answer:

C. the use of hydrogen gas as an electron donor.

Explanation:

Hydrogenotrophy is the convertion of hydrogen gas to other compounds as part of its metabolism.

2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone

Explanation:

To convert grams into moles

[tex]moles = grams \times \frac{1 mole}{grams}[/tex]

We have 5.00 g acetone

[tex]moles = 5 \times \frac{1}{58.1}[/tex]

[tex]moles = 0.0861[/tex]

Heat liberated = moles [tex]\times[/tex] heat of vapourization

                         =0.0861 mol x 30.3 kJ/mol

                        = 2.61 kJ

Therefore, 2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone

Answer:Covalent

Explanation:

Covalent bonds are formed by sharing of electrons between two atoms. The shared electrons are normally situated between the nuclei of the both atoms. However, atoms of some elements posses an usual ability to attract the shared electrons of a bond towards itself. We say that such atoms have a high electronegativity. High electronegativity leads to the existence of polar covalent bonds since the shared electron pair is now closer to one of the bonding atoms than the other. The atom to which the electron pair is closer becomes partially negative while the other becomes partially positive. This is only possible in a covalent bond where electrons are shared between bonding atoms. The charge separation is known as a dipole.

Note that ionic bonds involve a complete transfer of electrons leading to the formation of ions and is not applicable here.

Answer:

The answer to the question

The size of a population increases if the number of individuals added to the population is equal to the number of individuals leaving the population, is this true?

is No

Explanation:

The size of a population increases if the number of individuals added to the population is more than the number of individuals leaving the population

In biology, a population is the total number of organism of a particular species or group living together at a given geographical location. It is the number of inhabitants of a place.

Population growth is the increase in the number count of the specie or group in a population.

Answer:

nitrogen monoxide: NO

nitrogen dioxide: NO₂

Explanation:

Nitrogen monoxide is composed by 1 atom of O (prefix "mono-") and 1 atom of N. Nitrogen dioxide is composed by 2 atoms of O (prefix "di-") and 1 atom of N. As the oxigen atom in oxides has the valency -2 (it shares 2 electrons), the nitrogen has valency +2 in NO and +4 in NO₂.

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.

Answer:

Explanation:

The detailed steps and step by step calculations is as shown in the attached file.

To calculate the percent yield of ZnS, use the given masses of Zn and S8 and compare to the theoretical yield. For the oxides formed, determine the moles of remaining Zn and S8 using the balanced chemical equation.

To calculate the percent yield of ZnS, we need to compare the actual yield (104.4 g) to the theoretical yield. The theoretical yield can be calculated using the stoichiometry of the reaction. The balanced chemical equation is: Zn + S8 → ZnS. The molar ratio between Zn and ZnS is 1:1, which means that the moles of ZnS formed will be equal to the moles of Zn used. Convert the given masses of Zn and S8 to moles using their molar masses, then compare the moles to calculate the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction. Once you have determined the limiting reactant and calculated the theoretical yield, you can calculate the percent yield using the formula: (actual yield / theoretical yield) x 100.

For part (b), if all the remaining reactants combine with oxygen, we need to determine the remaining moles of Zn and S8 after the reaction with ZnS is complete. Convert these moles to masses using their molar masses, and then use the balanced chemical equation to determine the moles of each oxide formed. The balanced chemical equation for the reaction of zinc with oxygen is: 2Zn + O2 → 2ZnO. The molar ratio between Zn and ZnO is 2:2, so the moles of ZnO formed will be equal to the moles of Zn remaining. The balanced chemical equation for the reaction of sulfur with oxygen is: S8 + 8O2 → 8SO2. The molar ratio between S8 and SO2 is 1:8, so the moles of SO2 formed will be 8 times the moles of S8 remaining. Convert the moles of each oxide formed to masses using their molar masses.

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1.72 atm of hydrogen gas is created when 2.29 atm of phosphine breaks down at 953 K for 70.5 seconds.

Explanation:

The first step in this calculation is to use the first-order decay equation to find the remaining moles of phosphine. For a first-order reaction with a half-life of 35.0 s, after 70.5 s (or two half-lives), 25% of the phosphine would remain. To find the remaining moles of phosphine, we multiply 2.29 atm by 25%, yielding 0.57 atm.

From the balanced equation for the decomposition of phosphine (4PH3 -> P4 + 6H2), we see that 1 mole of phosphine yields 3/2 moles of H2. Thus, the number of moles of H2 produced is 75% of the original moles of PH3 (1.72 atm).

The sum of the partial pressures of phosphine and hydrogen is 0.57 + 1.72 = 2.29 atm, which is the final pressure in the 2.20 L vessel after 70.5 s.

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