3. Calculate the molarity of the HCI solution if 0.074 mol of HCl is dissolved in water to

make 2000 mL of the solution. Also calculate the pH of this solution. Hint: In case of

strong acids the [H3O+] is same as the molarity of acid solution.

The true statements are;

We define a redox reaction as one in which a specie is oxidized and another is reduced.

Now;

Eo cell = cell potential = -0.13 V - (+0.34 V) = -0.47 V

n =number of moles of electrons = 2 mole of electrons

K = equilibrium constant

ΔG = change in free energy

Eo cell = 0.0592/n log K

-0.47 =  0.0592/2 log K

log K =  -0.47  * 2/0.0592

K = 1.3 * 10^-16

ΔG = -nFEo cell

ΔG = -(2 * 96500 * -0.47)

ΔG = 90.7kJ

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The heat needed is (-) 14760 Joules because the ice is cooling.

The term heat refers to a form of energy that can be observed as a result of temperature difference between two bodies. We can recall that heat always flows from the hotter to the cooler body.

Given that the heat added to a substance can be calculated by the use of H = mcdT

Where;

H = heat

m = mass

c = specific heat capacity

T = temperature change

Hence;

H = 150 g * 2.03 Jg∘C * (-63 - (-15))

H =  (-) 14760 J

Hence the heat needed is (-) 14760 Joules because the ice is cooling.

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Answer:

.67 mole

Explanation:

Using ideal gas law

PV = n RT

P = pressure = 1 atm at STP

T = 273.15 K   at STP

V = 15 L

R = .082057  L-atm / K-mol

Subbing in those values:

1 (15) = n  (.082057) (273.15 )      results in n = .669 moles

Answer:

1.) 17.76 moles H₂O

2.) 323.1 g HCN

Explanation:

1.) An equation is balanced when there is an equal amount of each element on both sides of the reaction. The quantities can be modified by adding coefficients in front of the molecules.

The unbalanced equation:

Al₄C₃(s) + H₂O -----> Al(OH)₃(s) + CH₄(g)

Reactants: 4 aluminum, 3 carbon, 2 hydrogen, 1 oxygen

Products: 1 aluminum, 1 carbon, 7 hydrogen, 3 oxygen

The balanced equation:

Al₄C₃(s) + 12 H₂O -----> 4 Al(OH)₃(s) + 3 CH₄(g)

Reactants: 4 aluminum, 3 carbon, 24 hydrogen, 12 oxygen

Products: 4 aluminum, 3 carbon, 24 hydrogen, 12 oxygen

Now that the equation is balanced, you can use the relevant coefficients to construct a mole-to-mole ratio. This will allow you to convert between moles Al₄C₃ to moles H₂O.

 1.48 moles Al₄C₃           12 moles H₂O
----------------------------  x  ------------------------  =  17.76 moles H₂O
                                         1 mole Al₄C₃

2.) To find the mass of HCN, you need to (1) convert grams NH₃ to moles NH₃ (via molar mass), then (2) convert moles NH₃ to moles HCN (via mole-to-mole from equation coefficients), and then (3) convert moles HCN to grams HCN (via molar mass). It is important to arrange the coefficients in a way that allows for the cancellation of units.

Molar Mass (NH₃): 14.009 g/mol + 3(1.008 g/mol)

Molar Mass (NH₃): 17.033 g/mol

Molar Mass (HCN): 1.008 g/mol + 12.011 g/mol + 14.009 g/mol

Molar Mass (HCN): 27.028 g/mol

2 CH₄(g) + 3 O₂(g) + 2 NH₃(g)  ------> 2 HCN(g) + 6 H₂O(g)

203.6 g NH₃            1 mole            2 moles HCN           27.028 g
---------------------  x  ----------------  x  ------------------------  x  -----------------  =
                                17.033 g           2 moles NH₃             1 mole

=  323.1 g HCN

Answer:

0.645 L

Explanation:

To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.

(Step 1)

Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol

Molar Mass (KOH): 56.104 g/mol

19.9 grams KOH              1 mole
--------------------------  x  -----------------------  =  0.355 moles KOH
                                     56.014 grams

(Step 2)

Molarity = moles / volume                            <----- Molarity ratio

0.550 M = 0.355 moles / volume                 <----- Insert values

(0.550 M) x volume = 0.355 moles              <----- Multiply both sides by volume

volume = 0.645 L                                          <----- Divide both sides by 0.550

Heating up a reaction does increase the number of collision only.

Hence, Option C is correct answer.

According to the collision theory, the reaction take place when the collision occur and molecules having sufficient energy or threshold energy for collision.

The collision of molecules depend on the kinetic energy of molecules. Greater the kinetic energy greater the number of molecules collide. Lesser the kinetic energy lesser will be collision.

With an increase in temperature, there is an increase in energy that can be converted into activation energy in a collision, and that will increase the reaction rate. A decrease in temperature would have the opposite effect. With an increase in temperature, there is an increase in the number of collision.

Thus from the above conclusion we can say that Heating up a reaction does increase the number of collision only.

Hence, Option C is correct answer.

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Answer:

1.72 M

Explanation:

To find the new molarity, you need to (1) convert both volumes from mL to L, then (2) calculate the moles (using the molarity equation), and then (3) calculate the new molarity (using the new volume and molarity equation). The final answer should have 3 sig figs.

(Step 1)

1,000 mL = 1 L

250.0 mL / 1,000 = 0.250 L

332.4 mL / 1,000 = 0.3324 L

(Step 2)

Molarity = moles / volume (L)

4.00 M = moles / 0.250 L

1.00 = moles

(Step 3)

New Volume = 0.3324 L + 0.250 L = 0.5824 L

Molarity = moles / volume

Molarity = 1.00 moles / 0.5824 L

Molarity = 1.72 M

Answer:

1.45 mol/kg

Explanation:

nilered videos

Answer:

1.47 m

Explanation:

The molality can be found using the following equation:

Molality = moles solute / mass (kg) solvent

In this case, NaCl is the solute and H₂O is the solvent. To find the molality, you need to (1) convert grams NaCl to moles (via the molar mass), then (2) convert grams H₂O to kilograms, and then (3) calculate the molality.

(Step 1)

Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol

Molar Mass (NaCl): 58.443 g/mol

 30 grams NaCl                 1 mole
--------------------------  x  ------------------------  =  0.513 moles NaCl
                                      58.443 grams

(Step 2)

1,000 grams = 1 kg

350 grams H₂O                   1 kg
--------------------------  x  -----------------------  =  0.35 kg H₂O
                                      1,000 grams

(Step 3)

Molality = moles solute / mass (kg) solvent

Molality = 0.513 moles NaCl / 0.35 kg H₂O

Molality = 1.47 m

Answer:

424 grams CaCl₂

Explanation:

To find the mass of CaCl₂, you need to (1) calculate the moles (using the molarity ratio) and then (2) convert moles to grams (using the molar mass). The final answer should have 3 sig figs to match the given value with the least amount of sig figs.

(Step 1)

Molarity (M) = moles / volume (L)

1.56 M = moles / 2.451 L

3.82 = moles

(Step 2)

Molar Mass (CaCl₂): 40.078 g/mol + 2(35.453 g/mol)

Molar Mass (CaCl₂): 110.984 g/mol

3.82 moles CaCl₂           110.984 g
----------------------------  x  -------------------  =  424 grams CaCl₂
                                          1 mole

The number of mole of solute particles are present in 1 mL (exact) of aqueous 0.0040 M Ba(OH)₂ is 0.000004 mole

Molarity is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

With the above formula, we can determine the number of mole present in the solution. Detail below:

Molarity = mole / Volume

Cross multiply

Mole = Molarity × volume

Mole of solute = 0.004 × 0.001

Mole of solute = 0.000004 mole

Thus, the mole of the solute in the solution is 0.000004 mole

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Answer:

kindly refer to the picture attached

Explanation:

Have a great day . keep smiling

Answer:

B.) 129.9 grams

Explanation:

To find the mass, you need to use the following equation:

Q = mcΔT

In this equation,

-----> Q = energy (J)

-----> m = mass (g)

-----> c = specific heat (J/g°C)

-----> ΔT = change in temperature (°C)

The specific heat of copper is 0.385 J/g°C. Knowing this, you can plug the given values into the equation and simplify to isolate "m".

Q = mcΔT                                                            <----- Equation

5000 J = m(0.385 J/g°C)(200 °C - 100 °C)        <----- Insert values

5000 J = m(0.385 J/g°C)(100)                            <----- Subtract

5000 J = m(38.5)                                                <----- Multiply 0.385 and 100

129.9 = m                                                             <----- Divide both sides by 38.5

Answer:

13.0 L

Explanation:

To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (kPa)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (8.314 kPa*L/mol*K)

-----> T = temperature (K)

After converting the temperature from Celsius to Kelvin, you can plug the given values into the equation and simplify to find "V". The final answer should have 3 sig figs to match the sig figs of the given values.

P = 855 kPa                                  R = 8.314 kPa*L/mol*K

V = ? L                                           T = 26.0 °C + 273.15 = 299.15 K

n = 4.47 moles

PV = nRT

(855 kPa)V = (4.47 moles)(8.314 kPa*L/mol*K)(299.15 K)

(855 kPa)V = 11117.48496

V =  13.0 L

Taking into account the Henry's Law, 1.297 g carbon dioxide will dissolve at 790 mm Hg in 1.01 L water.

A change in pressure does not appreciably influence the solubility of solids or liquids or liquids in liquids; however, that of gases in solvents increases when the partial pressure of the gases increases. The solubility of a gas depends on pressure and temperature.

In this way, Henry's Law describes the effect of pressure on the solubility of gases. This law states that the solubility of a gas in contact with the surface of a liquid at a given temperature is directly proportional to the partial pressure of said gas on the liquid.

Mathematically, Henry's law is expressed as:

C=k×P

Where:

At 2 different partial pressure values, the Henry's law is expressed as:

[tex]\frac{C1}{C2} =\frac{P1}{P2}[/tex]

In this case, you know:

Replacing in Henry's Law:

[tex]\frac{1.228\frac{g}{L} }{C2} =\frac{755 mmHg}{790mmHg}[/tex]

Solving:

[tex]1.228\frac{g}{L} =\frac{755 mmHg}{790mmHg}xC2[/tex]

C2= [tex]\frac{1.228\frac{g}{L} }{\frac{755 mmHg}{790mmHg}}[/tex]

C2= 1.285 [tex]\frac{g}{L}[/tex]

Then, the concentration of carbon dioxide at 790 mmHg is 1.285 g/L. But you have 1.01 L of water. So the amount of gas dissolved can be calculated as 1.01 L×1.285 g/L = 1.297 g

Finally, 1.297 g carbon dioxide will dissolve at 790 mm Hg in 1.01 L water.

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Explanation:

Answer:

d

Explanation:

The + charge of the proton and the - charge of the electron create a electrostic force of attraction ( opposites attract)

Answer:

Having a specific set of possible values

Answer:

is a solution whose pH is above 7or is a basic solution

The three disadvantages of hard water are:

1. Hard water is unsuitable for washing because soap does not readily create a lather in it.

2. Soap may react and produce, wasting the soap.

3. Due to the development of calcium and magnesium carbonates, tea kettles will get furred.

Water that contains a lot of minerals is referred to as hard water. In limestone, chalk, or gypsum deposits, which are mostly composed of calcium and magnesium carbonates, bicarbonates, and sulphates, hard water is created as a result of percolation. A moderate health advantage from drinking hard water is possible. In industrial settings where water hardness is regulated to prevent expensive breakdowns in boilers, cooling towers, and other water-handling equipment, it can present serious issues. Hard water is frequently detected in home settings by the build-up of lime scale in kettles and water heaters as well as by the absence of foam when soap is agitated in water. Wherever there is a worry about water hardness, water softening is frequently employed to lessen the negative impacts of hard water.

write three disadvantages of hard water.​

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The property of water is that It exists in all three states on Earth.

The major residences to be able to be mentioned in this text are its appeal to polar molecules, its high specific warmth, the excessive warmness of vaporization, the decrease density of ice, and its high polarity.

The primary homes of water are its polarity, concord, adhesion, floor anxiety, excessive unique warmth, and evaporative cooling. A water molecule is barely charged on each ends. this is because oxygen is more electronegative than hydrogen.

Water is good sized functionality to dissolve a ramification of molecules has earned it the designation of “standard solvent,” and it's far this capacity that makes water such an invaluable life-sustaining pressure. On a biological level, water's role as a solvent facilitates cells transport and use substances like oxygen or vitamins.

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The equation for the autoionization of water is-

NH3 + H2O → NH4+ + OH− (H2O acts as an acid)

HCl + H2O → H3O+ + Cl− (H2O acts as a base)

Autoionization of water, the autoionization constant Kw, and the relationship between [H⁺] and [OH⁻] in aqueous solutions.

Water can undergo autoionization to form and ions.

The expression for the autoionization constant is

[tex]The expression for the autoionization constant isK_\text{w}=[\text{H}_3\text{O}^+][\text{OH}^-]\quad\quad\text{(Eq. 1)}K w​ =[H 3​ O + ][OH − ][/tex]

In this way the autoionization of water,

NH3 + H2O → NH4+ + OH− (H2O acts as an acid)

HCl + H2O → H3O+ + Cl− (H2O acts as a base) is given.

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B. At the equivalence point of a titration of the [H+] concentration is equal to 7.

The equivalence point of a titration is a point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution.

At the equivalence point in an acid-base titration, moles of base equals moles of acid and the solution only contains salt and water.

At the equivalence point, equal amounts of H+ and OH- ions combines as shown below;

H⁺ + OH⁻  → H₂O

The pH of resulting solution is 7.0 (neutral).

Thus, the pH at the equivalence point for this titration will always be 7.0.

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Answer:

5.5 grams

Explanation:

To find how much of the isotope remains, you need to use the half-life equation:

[tex]N(t) = N_0(\frac{1}{2})^\frac{t}{h}[/tex]

In this equation,

-----> N(t) = remaining mass (g)

-----> N₀ = initial mass (g)

-----> t = time (hrs)

-----> h = half-life (hrs)

You can find the remaining mass by plugging the given values into the equation and solving.

N(t) = ? g                          t = 30 hrs

N₀ = 22 g                         h = 15 hrs

[tex]N(t) = N_0(\frac{1}{2})^\frac{t}{h}[/tex]                                         <----- Half-life equation

[tex]N(30) = (22)(\frac{1}{2})^\frac{30}{15}[/tex]                                    <----- Insert values

[tex]N(30) = (22)(\frac{1}{2})^2[/tex]                                      <----- Simplify exponent

[tex]N(30) = (22)(0.25)[/tex]                                   <----- Solve [tex](\frac{1}{2} )^2[/tex]

[tex]N(30) = 5.5[/tex]                                              <----- Multiply 22 and 0.25

Answer:

Silver

Explanation:

What mass of copper was discharge at the cathode in a cell

Answer:

C

Explanation:

For every action there is an equal but opposite reaction..... the forces are the same

The acid dissociation constant(Ka) is 0.0095

The reaction for this dissociation of acid is

HSO4 ⇄H+ + SO4 -2

The dissociation constant can be determined from the following expression

[tex]Ka= \frac{[H+] [SO4]-2}{[HSO4}[/tex]  

[H+] = 10-PH

       = 10-1.35

[H+]  = 0.0447

[H+] = 0.0447 mol / L

From equation, [H+] = [SO4-2] = 0.0447 mol / L

[SO4-2] = 0.0447 mol / L

From the values of [H+], [SO4-2] and [HSO4] Ka can be calculated as follows,

Ka = 0.0447 * 0.0447 / 0.200

     = 0.0019 / 0.200

     = 0.0095

Hence the value of the acid dissociation constant (Ka) for the given reaction is  0.0095

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Answer:

D.) reducing agent

Explanation:

Oxidized chemicals gain electrons. In order to gain these electrons, another chemical must lose electrons. So, the chemical (which will be oxidized) acts as a reducing agent, causing the other chemical to be reduced and lose electrons.