Which of the following statements concerning galvanic cells is/are true?
A.) The two half-cells are connected by a salt bridge
B.) Electrons flow from the anode to the cathode
C.) Reduction occurs at the cathode
D.) All of the above are true

Answer:

The molarity of HI at the equilibrium is 2.8M

Explanation:

Step 1: Data given

Volume of the flask = 250.0 mL = 0.250L

Number of moles I2 = 1.3 mol

Number of moles HI = 1.0 mol

Kc = 0.983

Step 2: The balanced equation

H2(g) +I2(g) ⇆ 2HI(g)

For 1 mole I2 consumed, we need 1 mole H2 to produce 2 moles HI

Step 3: Calculate initial concentrations

Initial concentration I2 = 1.3mol / 0.25L

Initial concentration I2 = 5.2 M

Initial concentration HI = 1.0 mol / 0.25L

Initial concentration HI = 4.0 M

Step 4: Calculate concentrations at equilibrium

The concentration at equilibrium is:

[I2] = (5.2+x)M

[HI] = (4.0 - x)M

[H2] = xM

Kc = [HI]²/[H2][I2]

0.983 = (4-x)²/ (x*(5.2+x))

0.983 = (4-x)²/ (5.2x +x²)

5.1116x + 0.983 x² = 16 -8x +x²

-0.017x² +13.1116x -16 = 0

x = 1.222 = [H2]

[HI] = 4.0 - 1.222 = 2.778M ≈ 2.8 M

[I2] = 5.2 + 1.222 = 6.422 M ≈ 6.4 M

To control we can calculate:

[2.778]² / [1.222][6.422]  = 0.983 = Kc

The molarity of HI at the equilibrium is 2.8M

Answer:

Molarity of Unknown Acid = 0.1332 M

Explanation:

Data for solving problem:

Molarity of base in buret (M₁)= 0.1517 M

volume of the acid in Erlenmeyer flask (V₂)= 25.0 mL

Volume of the base in the buret (V₁) = final volume of buret - initial volume in buret

final volume of buret = 22.5 mL

initial volume in buret = 0.55 mL

So

Volume of the base in the buret (V₁) = 22.5 mL -0.55 mL = 21.95 mL

Volume of the base in the buret (V₁)  = 21.95 mL

Molarity of Unknown acid in the Erlenmeyer flask (M₂) = To be find

Explanation:

It is acid base titration and  formula for this titration is as follows:

Molarity of base x Volume of base = Molarity of acid x volume of acid

it can be written as

M₁V₁ = M₂V₂ -------------------- equation (1)

we have to find M₂

so by rearrangment the equation (1)

M₁V₁ / V₂ = M₂ ------------------ equation (2)

put the values in equation in equation (2)

M₂ = 0.1517 M x 21.95 mL / 25.0 mL

M₂ = 3.3298 /25.0

M₂ = 0.1332 M

so the Molarity of Unknown acid is 0.1332 M

The molarity of the unknown acid HA is calculated to be 0.1331 M by first determining the moles of NaOH that reacted and then using the volume of acid to find its concentration.

To determine the molarity of the unknown acid HA in a titration experiment, first, we calculate the volume of 0.1517 M NaOH used by subtracting the initial buret reading from the final buret reading: 22.50 mL - 0.55 mL = 21.95 mL. This volume is then converted to liters by dividing by 1000: 21.95 mL/1000 = 0.02195 L. Next, we use the molarity of the NaOH solution to find the moles of NaOH that have reacted: 0.1517 M × 0.02195 L = 0.003328665 moles.

Assuming a 1:1 mole ratio between NaOH and HA in the reaction, the moles of unknown acid HA that reacted are also 0.003328665. We then calculate the molarity of the unknown acid, by dividing moles of acid by the volume of acid in liters (25.0 mL = 0.025 L): 0.003328665 moles / 0.025 L = 0.1331466 M.

The molarity of the unknown acid HA is therefore 0.1331 M, rounded to four significant figures to match the precision of the given data.

Final answer:

The weight percent of NH3 in the aqueous waste was calculated from the titration data to be 0.2047%. The moles of NH3 were found to be 0.0029711 mol, corresponding to 0.050585g NH3 in the original waste sample.

Explanation:

The weight percent of NH3 in the aqueous waste is calculated using titration analysis. First, determine the moles of HCl that reacted:
# mol HCl = 0.02842 L × 0.1045 M = 0.0029711 mol HCl.

The reaction between HCl and NH3 is:
NH3(aq) + HCl(aq) → NH4Cl(aq).

Hence, 1 mole of HCl reacts with 1 mole of NH3. Therefore, the moles of NH3 in the aliquot are also 0.0029711 mol. To find the mass of NH3:
Mass of NH3 = moles × molar mass
              = 0.0029711 mol × 17.031 g/mol
              = 0.050585 g NH3.

Since the aliquot is part of the diluted waste, and assuming the ammonia is evenly distributed, we calculate the weight percent in the original waste sample:
   Weight percent = (mass of NH3 in aliquot / mass of waste sample) × 100
                 = (0.050585 g / 24.711 g) × 100
                 = 0.2047%.

Answer:

V= 0.544

Explanation:

Given that

σ=812 MPa

σ₁=1240 MPa

σ₂=300 MPa

Lets take volume fraction V

The strength of composite fiber given as

σ = σ₁ V + σ₂(1-V)

By putting the values

812 = 1240 V + 300 (1-V)

812 = 1240 V +300 - 300 V

812 - 300 = 1240 V- 300 V

512 = 940 V

V= 0.544

The percentage volume = 54.4 %

Answer:

The concentration of the H₃PO₄ solution is 0,245M

Explanation:

The first titration is:

KHP + NaOH → KP⁻ + Na⁺ + H₂O

0,495g of KHP are:

0,495g×[tex]\frac{1mol}{204,22g}[/tex]= 2,42x10⁻³ moles of KHP

As 1 mole of KHP reacts with 1 mole of NaOH, moles of NaOH are 2,42x10⁻³ moles.

As volume required was 28,56mL, the concentration of the NaOH solution is:

2,42x10⁻³ moles / 0,02856L = 0,0849M

The titration of the phosporic acid with NaOH occurs as follows:

H₃PO₄ + NaOH → H₂PO₄⁻ + Na⁺ + H₂O

If were required 29,88mL of NaOH, the moles of NaOH spent were:

0,0849M×0,02988L = 2,54x10⁻³ moles of NaOH that are the same than H₃PO₄ moles.

As the volume of the solution of H₃PO₄ was 10,33mL, the concentration of the H₃PO₄ solution is:

2,54x10⁻³ moles of H₃PO₄ / 0,01033L = 0,245M

I hope it helps!

Final answer:

The concentration of the phosphoric acid is calculated using the molarity of a standardized NaOH solution obtained from a titration with KHP. The determined NaOH molarity is then applied to find the moles of phosphoric acid reacted in its own titration, which leads to the final concentration of the acid.

Explanation:

To determine the concentration of the phosphoric acid (H3PO4), we first need to calculate the molarity of the sodium hydroxide (NaOH) solution using the titration with potassium hydrogen phthalate (KHP). Since 0.495 grams of KHP was titrated with 28.56 mL of NaOH, we can calculate the moles of KHP used:

Now, using the concentration of NaOH, we can find the concentration of phosphoric acid. Given that 29.88 mL of the NaOH solution is required to titrate the 10.33 mL of H3PO4, we have:

Answer:

The weight percent in the sample is 17,16%

Explanation:

The dissolution of the Ce(IV) salt provides free Ce⁴⁺ that reacts, thus:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03247M Na₂S₂O₃ = 4,228x10⁻⁴ moles of S₂O₃²⁻.

As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,228x10⁻⁴ moles of S₂O₃⁻×[tex]\frac{1molI_{3}^-}{2molS_{2}O_{3}^{2-}}[/tex] = 2,114x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,114x10⁻⁴ moles of I₃⁻× [tex]\frac{2molCe^{4+}}{1molI_{3}^-}[/tex] =  4,228x10⁻⁴ moles of Ce(IV).

These moles are:

4,228x10⁻⁴ moles of Ce(IV)×[tex]\frac{140,116g}{1mol}[/tex] = 0,05924 g of Ce(IV)

As was taken an aliquot of 25,00mL from the solution of 250,0mL:

0,05924 g of Ce(IV)×[tex]\frac{250,0mL}{25,00mL}[/tex] =0,5924g of Ce(IV) in the sample

As the sample has 3,452g, the weight percent is:

0,5924g of Ce(IV) / 3,452g × 100 = 17,16 wt%

I hope it helps!

Answer:

The correct option is: B. 14 H⁺ + 6 e⁻ + Cr₂O₇²⁻ ⟶ 2 Cr³⁺ + 7 H₂O

Explanation:

Redox reactions is an reaction in which the oxidation and reduction reactions occur simultaneously due to the simultaneous movement of electrons from one chemical species to another.

The reduction of a chemical species is represented in a reduction half- reaction and the oxidation of a chemical species is represented in a oxidation half- reaction.

To balance the reduction half-reaction for the reduction of Cr₂O₇²⁻ to Cr³⁺:

Cr₂O₇²⁻ ⟶ Cr³⁺

First the number of Cr atoms on the reactant and product side is balanced

Cr₂O₇²⁻ ⟶ 2 Cr³⁺

Now, Cr is preset in +6 oxidation state in Cr₂O₇²⁻ and +3 oxidation state in Cr³⁺. So each Cr gains 3 electrons to get reduced.

Therefore, 6 electrons are gained by 2 Cr atoms of Cr₂O₇²⁻ to get reduced.

Cr₂O₇²⁻ + 6 e⁻ ⟶ 2 Cr³⁺

Now the total charge on the reactant side is (-8) and the total charge on the product side is (+6).

From the given options it is evident that the reaction must be balanced in acidic conditions.

Therefore, to balance the total charge on the reactant and product side, 14 H⁺ is added on the reactant side.

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺

Now to balance the number of hydrogen and oxygen atoms, 7 H₂O is added on the product side.

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

Therefore, the correct balanced reduction half-reaction is:

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

Answer:

Ba

Explanation:

Magnesium is the element of second group and third period. The electronic configuration of magnesium is - 2, 8, 2 or [tex]1s^22s^22p^63s^2[/tex]

There are 2 valence electrons of magnesium.

Fluorine is the element of seventeenth group and second period. The electronic configuration of fluorine is - 2, 7 or [tex]1s^22s^22p^5[/tex]

There is 1 valence electron of fluorine.

They will combine and form [tex]MgF_2[/tex]

The same type of compound is formed by the other members of the group 2 since they will have 2 valence electron.

Group 2 includes:- beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra).

Hence, correct options is:- Ba

The atom that will combine with F in the same ratio as it combines with Mg is Ba.

We must recall that F combines with atoms of  elements group 2 to form ionic compounds of the sort MX2 where M is the metal.

Mg and Ba all belong to group 2 hence they will form compounds with F in the ratio of 1:2

Therefore, the atom that will combine with F in the same ratio as it combines with Mg is Ba.

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The final temperature of the mixture is approximately [tex]\(-57.97^\circ\text{C}\)[/tex].

To find the final temperature of the mixture, we can use the principle of conservation of energy, assuming no heat is lost or gained in the process.

The heat gained by the cold substance (water) will be equal to the heat lost by the hot substance (ethanol).

The heat gained or lost [tex](\(Q\))[/tex] can be calculated using the equation:

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

where:

- [tex]\(m\)[/tex] is the mass of the substance,

- [tex]\(c\)[/tex] is the specific heat capacity of the substance,

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

The sum of the heats gained and lost is zero:

[tex]\[ Q_{\text{water}} + Q_{\text{ethanol}} = 0 \][/tex]

Since the final temperature is the same for both substances, we can write the equation:

[tex]\[ m_{\text{water}}c_{\text{water}}\Delta T_{\text{water}} + m_{\text{ethanol}}c_{\text{ethanol}}\Delta T_{\text{ethanol}} = 0 \][/tex]

Rearrange the equation to solve for the final temperature [tex](\(T_{\text{final}}\))[/tex]:

[tex]\[ \Delta T_{\text{water}} + \Delta T_{\text{ethanol}} = 0 \][/tex]

[tex]\[ T_{\text{final}} - T_{\text{initial, water}} + T_{\text{final}} - T_{\text{initial, ethanol}} = 0 \][/tex]

[tex]\[ 2T_{\text{final}} = T_{\text{initial, water}} + T_{\text{initial, ethanol}} \][/tex]

[tex]\[ T_{\text{final}} = \frac{T_{\text{initial, water}} + T_{\text{initial, ethanol}}}{2} \][/tex]

Now, substitute the given values:

[tex]\[ T_{\text{final}} = \frac{(55.0 \ \text{mL} \times 1.0 \ \text{g/mL} \times 4.18 \ \text{J/g}^\circ\text{C} \times (28.6^\circ\text{C} - T_{\text{final}})) + (55.0 \ \text{mL} \times 0.789 \ \text{g/mL} \times 2.44 \ \text{J/g}^\circ\text{C} \times (T_{\text{final}} - 9.0^\circ\text{C}))}{2} \][/tex]

[tex]\[ T_{\text{final}} = \frac{(55.0 \ \text{mL} \times 4.18 \ \text{J/g}^\circ\text{C} \times (28.6^\circ\text{C} - T_{\text{final}})) + (55.0 \ \text{mL} \times 0.789 \ \text{g/mL} \times 2.44 \ \text{J/g}^\circ\text{C} \times (T_{\text{final}} - 9.0^\circ\text{C}))}{2} \][/tex]

Combine like terms:

[tex]\[ T_{\text{final}} = \frac{(55.0 \times 4.18 \times 28.6) + (55.0 \times 0.789 \times 2.44 \times (T_{\text{final}} - 9.0))}{2} \][/tex]

[tex]\[ T_{\text{final}} = \frac{(6132.86 + 107.79 \times (T_{\text{final}} - 9.0))}{2} \][/tex]

[tex]\[ T_{\text{final}} = \frac{6132.86 + 107.79T_{\text{final}} - 971.1}{2} \][/tex]

[tex]\[ 2T_{\text{final}} = 7101.76 + 107.79T_{\text{final}} - 971.1 \][/tex]

Combine like terms:

[tex]\[ 2T_{\text{final}} - 107.79T_{\text{final}} = 7101.76 - 971.1 \][/tex]

[tex]\[ -105.79T_{\text{final}} = 6130.66 \][/tex]

[tex]\[ T_{\text{final}} \approx -57.97^\circ\text{C} \][/tex]

Therefore, the final temperature of the mixture is approximately [tex]\(-57.97^\circ\text{C}\)[/tex]. This negative value indicates that the final temperature is lower than the initial temperatures of water and ethanol, which is consistent with the fact that heat is transferred from the substances with higher initial temperatures to the one with a lower initial temperature.

Answer:

The valid set of quantum numbers are:

a. 3,1,1,-1/2

b. 4,3,1,-1/2

c. 2,0,0,-1/2

e. 3,2,-1,-1/2

j. 3,0,0,1/2

Explanation:

Quantum numbers (n, l, mℓ, ms) are the set of numbers that describe the state of an electron in an atom.

The four quantum numbers and their rules are:

Therefore,

a. 3,1,1,-1/2: Valid

b. 4,3,1,-1/2: Valid

c. 2,0,0,-1/2: Valid

d. 1,3,0,1/2: NOT Valid

Reason: ℓ ≤ (n-1). Therefore, ℓ can not be greater than n, .

e. 3,2,-1,-1/2: Valid

f. 3,3,-1,1/2: NOT Valid

Reason: ℓ ≤ (n-1). Therefore, ℓ can not be equal to n. (ℓ ≠n)

g. 3,2,1,-1: NOT Valid

Reason: the only allowed values of ms = (- s), (+ s) = (- 1/2), (+ 1/2)

h. 1,-1,-1,-1/2: NOT Valid

Reason: ℓ ≤ (n-1) and n ≥ 1. Therefore, ℓ can not be equal to n and also ℓ can't be negative.

i. 3,3,1,1/2: NOT Valid

Reason: ℓ ≤ (n-1). Therefore, ℓ can not be equal to n. (ℓ ≠n)

j. 3,0,0,1/2: Valid

k. 4,3,4,-1/2: NOT Valid

Reason: mℓ = (- ℓ) to (+ ℓ). Therefore, mℓ can't be greater than ℓ.

l. 0,2,1,1/2: NOT Valid

Reason: n ≥ 1. Therefore, value of n can't be 0 (n≠0)

Answer: The partial pressure of nitrogen gas is 405.76 mmHg and that of hydrogen gas is 257.24 mmHg

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of nitrogen gas = 3.46 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of nitrogen gas}=\frac{3.46g}{28g/mol}=0.123mol[/tex]

Given mass of hydrogen gas = 0.156 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of hydrogen gas}=\frac{0.156g}{2g/mol}=0.078mol[/tex]

Mole fraction of a gas is calculated by using the formula:

[tex]\chi_{A}=\frac{n_{A}}{n_{A}+n_{B}}[/tex]      ......(1)

Putting values in equation 1, we get:

[tex]\chi_{\text{nitrogen gas}}=\frac{0.123}{0.123+0.078}=0.612[/tex]

Putting values in equation 1, we get:

[tex]\chi_{\text{hydrogen gas}}=\frac{0.078}{0.123+0.078}=0.388[/tex]

The partial pressure of a gas is given by Raoult's law, which is:

[tex]p_A=p_T\times \chi_A[/tex]     ......(2)

where,

[tex]p_A[/tex] = partial pressure of substance A

[tex]p_T[/tex] = total pressure  = 663 mmHg

[tex]\chi_A[/tex] = mole fraction of substance A

[tex]p_{\text{Nitrogen gas}}=663mmHg\times 0.612\\\\p_{\text{Nitrogen gas}}=405.76mmHg[/tex]

[tex]p_{\text{Hydrogen gas}}=663mmHg\times 0.388\\\\p_{\text{Hydrogen gas}}=257.24mmHg[/tex]

Hence, the partial pressure of nitrogen gas is 405.76 mmHg and that of hydrogen gas is 257.24 mmHg

Answer:

mass of O₂ = 800 g

Explanation:

Formula of Octane = C₈H₁₈

Formula of Oxygen = O₂

Balanced Chemical Reaction:

Octane react with oxygen and produce water and carbon dioxide.

                    2C₈H₁₈ + 25O₂ ---------> 16CO₂+ 18H₂O

Moles calculation:

From the number of moles of the reactant used in a chemical reaction specific number of product produced. we can find the amount of any reactant and product from the mole ratio of a chemical reaction.

Mole formula =  

no. of moles = mass in grams / molecular mass ............... (1)

Given Data:

Molecular Weight of O₂  =  (16 x2 ) = 32 g/mol

Mass of O₂ = To be find

Calculations:

                             2C₈H₁₈  +  25O₂      --------->  16CO₂   +    18H₂O

                              2mole      25 mole                 16 mole     18 mole

From the above balanced chemical equation it is know that 2 mole of octane consume 25 mole of oxygen.

so we have to calculate the mass of oxygen that is consumed

by using mole formula  (1) we can fine the mass of oxygen

we know

Molecular Weight of O₂  =  32 g/mol

number of moles of Oxygen molecule = 25 mol

putting the value in the below formula

            no. of moles of O₂ = mass of O₂ / molecular mass of O₂

            25 mole = mass of O₂ / 32 g/mol ....... (2)

By rearragming the equation (2)

            mass of O₂ = 25 mole x 32 g/mol

            mass of O₂ = 25 mole x 32 g/mol

            mass of O₂ = 800 g

So in the octan reaction with oxygen 800g of oxygen will use.  

Answer: 0.0274 M

Explanation:-

The balanced chemical solution is:

[tex]NH_4OH(aq)+HCl(aq)\rightarrow NH_4Cl(aq)+H_2O(l)[/tex]

According to the neutralization law,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of [tex]HCl[/tex] solution = 0.117 M

[tex]V_1[/tex] = volume of [tex]HCl[/tex] solution = 23.4 ml

[tex]M_2[/tex] = molarity of [tex]NH_4OH[/tex] solution = ?

[tex]V_2[/tex] = volume of [tex]NH_4OH[/tex] solution = 100.0 ml

[tex]n_1[/tex] = valency of [tex]HCl[/tex] = 1

[tex]n_2[/tex] = valency of [tex]NH_4OH[/tex] = 1

[tex]1\times 0.117M\times 23.4=1\times M_2\times 100.0[/tex]

[tex]M_2=0.0274[/tex]

Therefore, the concentration of ammonia in a solution will be 0.0274 M

In this exothermic reaction, 2.50 g of hydrogen peroxide decomposes and generates -7.22 kJ of heat, as indicated by the negative sign.

The question asks for the amount of heat (q) generated when 2.50 g of hydrogen peroxide decomposes at constant pressure to give water and oxygen. Given that the enthalpy change (ΔH) for the reaction 2H2O2(l) → 2H2O(l) + O2(g) is -196 kJ, it indicates that the decomposition of hydrogen peroxide is an exothermic reaction and heat is released in the process. The negative sign of ΔH confirms this.

Firstly, we will have to find out how many moles of H2O2 are there in 2.5 g. Using the molar mass of H2O2 (34 g/mol), we get about 0.074 moles of H2O2. In the balanced chemical equation, 2 moles of hydrogen peroxide generates -196 kJ of heat. Therefore, for 0.074 moles, we calculate it by (-196 kJ * 0.074)/2 = -7.22 kJ.

The <-strong>value of q in this exothermic reaction when 2.50 g of hydrogen peroxide decomposes at constant pressure is therefore -7.22 kJ. The negative sign indicates heat is being released, or in other words, the reaction is exothermic.

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The value of q, representing the thermal energy transferred in the decomposition of 2.50g of hydrogen peroxide at constant pressure, is approximately -7.16 kJ.

The question asks for the value of q, which represents the thermal energy transferred during a reaction, in an exothermic reaction where 2.50 g of hydrogen peroxide decomposes into water and oxygen gas. In this given case, we know the enthalpy change (∆H) of the reaction is -196 kJ, indicating that this is an exothermic process where that much energy is being released to the surroundings.

We know from the balanced chemical equation 2H2O2(l) → 2H2O(l) + O2(g) that 2 moles of H2O2 decompose to release ∆H amount of energy. Let's find out the number of moles in 2.50 g of H2O2. The molar mass of H2O2 is approximately 34.01 g/mol. So, the number of moles = mass/molar mass  = 2.50 g / 34.01 g/mol = 0.0735 moles.

Considering the stoichiometry of the reaction, ∆H of -196 kJ is associated with 2 moles of H2O2. Therefore, the energy associated with 0.0735 mol will be:

q ( energy ) = ∆H x (moles of H2O2 /2) = -196 kJ x (0.0735 mol / 2 mol) = -7.1625 kJ

Therefore, the quantity of thermal energy (q) released in this decomposition reaction is approximately -7.16 kJ.

Answer:

The concentration of the chemist's silver(I) nitrate  solution is  0.897 mmol/L

Explanation:

Step 1: Data given

Number of moles AgNO3 = 269 micromol = 269 * 10^-6 mol

Volume of AgNO3 = 300 mL = 0.3 L

Molar mass of AgNO3 = 169.87 g/mol

Step 2: Calculate molarity of AgNO3

Molarity = number of moles per volume (in Liters)

Molarity AgNO3 = 269 *10^-6 mol / 0.3 L

Molarity AgNO3 = 8.97 * 10^-4 M

8.97 *10^-4 mol/L = 897 micromol/L = 0.897 mmol/L

The concentration of the chemist's silver(I) nitrate  solution is  0.897 mmol/L

Final answer:

The molar solubility of Mg(OH)2 in pure water is 2.41 x 10^-4 M. To calculate the molar solubility in a 0.130 M NaOH solution, we can use the common ion effect.

Explanation:

The molar solubility of Mg(OH)2 in pure water is 2.41 x 10^-4 M. The molar solubility of Mg(OH)2 in a 0.130 M NaOH solution can be determined using the common ion effect. Since NaOH dissociates into Na+ and OH-, the concentration of OH- in the NaOH solution is 0.130 M. To calculate the molar solubility, we can set up an ICE table:

Mg(OH)2 ⇌ Mg2+ + 2OH-

Initial: 0.130 M (from NaOH concentration)

Change: -x (due to dissolution of Mg(OH)2) and +2x (due to formation of OH- ions)

Equilibrium: 0.130 - x M

Ksp = (Mg2+)(OH-)²

Using the given value of the Ksp, 5.61x10^-11, we can substitute the equilibrium concentrations into the Ksp expression and solve for x. Then we can calculate the ratio of the solubility of Mg(OH)2 in pure water to the solubility in the NaOH solution by dividing the molar solubility in water by the molar solubility in the NaOH solution.

Mg(OH)₂ is approximately [tex]\( 0.472} \)[/tex] times more soluble in pure water compared to a 0.130 M NaOH solution. The solubility of Mg(OH)₂ in the presence of 0.130 M NaOH, is approximately [tex]\( 5.1 \times 10^{-4} \)[/tex] M.

To determine how many times more soluble Mg(OH)₂ is in pure water compared to a 0.130 M NaOH solution, we need to consider the effect of the common ion (OH-) from NaOH on the solubility of Mg(OH)₂.

Given:

- [tex]\( K_{sp} \)[/tex] for Mg(OH)₂ in pure water = [tex]\( 5.61 \times 10^{-11} \)[/tex]

- Molar solubility of Mg(OH)₂ in pure water = [tex]\( 2.41 \times 10^{-4} \)[/tex] M

- Molar solubility of NaOH in 0.130 M solution = [tex]\( 3.32 \times 10^{-9} \)[/tex] M

1. Calculate the concentration of OH- ions in the 0.130 M NaOH solution:

  The NaOH dissociates completely to produce Na+ and OH- ions.

  Concentration of OH- ions from NaOH = 0.130 M

2. Determine the solubility of Mg(OH)₂ in the presence of 0.130 M NaOH:

  According to the common ion effect, the presence of additional OH- ions from NaOH will shift the equilibrium of Mg(OH)2 dissolution. The solubility of Mg(OH)₂ will decrease due to the Le Chatelier's principle, which states that adding an ion that is involved in the equilibrium reaction will shift the equilibrium in the direction that reduces its effect.

Let's denote the solubility of Mg(OH)₂ in the presence of 0.130 M NaOH as [tex]\( S_{\text{NaOH}} \)[/tex].

Using the Ksp expression for Mg(OH)₂:

[tex]\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \][/tex]

In pure water, [tex]\( [\text{OH}^-] = S_{\text{water}} \)[/tex] (molar solubility of Mg(OH)2 in pure water).

In the presence of 0.130 M NaOH, [tex]\( [\text{OH}^-] = 3.32 \times 10^{-9} \)[/tex]M (concentration of OH- from NaOH).

So, [tex]\( S_{\text{NaOH}} \)[/tex] can be found from:

[tex]\[ K_{sp} = [\text{Mg}^{2+}](3.32 \times 10^{-9})^2 \][/tex]

Solve for [tex]\( [\text{Mg}^{2+}] \)[/tex]:

[tex]\[ [\text{Mg}^{2+}] = \frac{K_{sp}}{(3.32 \times 10^{-9})^2} \][/tex]

Calculate [tex]\( [\text{Mg}^{2+}] \)[/tex]:

[tex]\[ [\text{Mg}^{2+}] = \frac{5.61 \times 10^{-11}}{(3.32 \times 10^{-9})^2} \][/tex]

[tex]\[ [\text{Mg}^{2+}] = \frac{5.61 \times 10^{-11}}{1.1 \times 10^{-17}} \][/tex]

[tex]\[ [\text{Mg}^{2+}] \approx 5.1 \times 10^{-4} \, \text{M} \][/tex]

3. Calculate the ratio of solubility in pure water to solubility in 0.130 M NaOH:

[tex]\[ \text{Ratio} = \frac{S_{\text{water}}}{S_{\text{NaOH}}} = \frac{2.41 \times 10^{-4}}{5.1 \times 10^{-4}} \][/tex]

[tex]\[ \text{Ratio} \approx 0.472 \][/tex]

Answer:

b. ΔE rxn is a measure of heat

Explanation:

a. ΔHrxn is the heat of reaction. TRUE. ΔHrxn or change in enthalpy of reaction is per definition the change in heat that is involved in a chemical reaction.

b. ΔErxn is a measure of heat. FALSE. Is the change in internal energy of a reaction

c. An exothermic reaction gives heat off heat to the surroundings. TRUE. An exothermic reaction is a chemical reaction that releases heat.

d. Endothermic has a positive ΔH. TRUE. When a process is exothermic ΔH<0 and when the process is endothermic ΔH>0

e. Enthalpy is the sum of a system's internal energy and the product of pressure and volume. TRUE. Under constant pressure and volume the formula is ΔH = ΔE + PV

I hope it helps!

The statement that is FALSE is b. ΔErxn is a measure of heat.

The statement that is FALSE is b. ΔErxn is a measure of heat.

ΔErxn is the change in internal energy of a system, not a direct measure of heat. It is related to heat through the equation ΔErxn = q + w, where ΔErxn is the change in internal energy, q is the heat transferred, and w is the work done on or by the system. Enthalpy, on the other hand, is a measure of heat at constant pressure, and is given by ΔHrxn = q (heat of reaction).

Therefore, option b is the false statement.

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Answer: The half reactions and net cell reaction of the cell is written below.

Explanation:

The given cell is:

[tex]Cu(s)/Cu^{2+}(aq.)||Ag^{+}(aq.)/Ag(s)[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Half reactions for the given cell follows:

Oxidation half reaction (anode): [tex]Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-[/tex]

Reduction half reaction (cathode): [tex]Ag^{+}(aq.)+e^-\rightarrow Ag(s)[/tex]   ( × 2)

Net cell reaction:  [tex]Cu(s)+2Ag^+(aq.)\rightarrow Cu^{2+}(aq.)+2Ag(s)[/tex]

Hence, the half reactions and net cell reaction of the cell is written above.

The half-reactions at each electrode and the net cell reaction for the electrochemical cell containing copper and silver are provided.

The half-reactions at each electrode in this electrochemical cell are:

Anode: Cu (s) → Cu2+ (aq) + 2e−

Cathode: 2Ag+ (aq) + 2e− → Ag (s)

The net cell reaction for this electrochemical cell is:

Cu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s)

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Answer: d) ΔS is positive and ΔH is negative

Explanation:

According to Gibb's equation:

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

[tex]\Delta G[/tex] = Gibbs free energy  

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

[tex]\Delta S[/tex] = entropy change  

T = temperature in Kelvin

[tex]\Delta G[/tex]= +ve, reaction is non spontaneous

[tex]\Delta G[/tex]= -ve, reaction is spontaneous

[tex]\Delta G[/tex]= 0, reaction is in equilibrium

a)  ΔS is negative and ΔH is positive

[tex]\Delta G=(-ve)-T(+ve)[/tex]

[tex]\Delta G=(-ve)(-ve)=-ve[/tex]  

Reaction is spontaneous at all temperatures.

b) ΔS is positive and ΔH is positive

[tex]\Delta G=(+ve)-T(+ve)[/tex]

[tex]\Delta G=(+ve)(-ve)=-ve[/tex]  

Reaction is spontaneous at high temperatures.

c) ΔS is negative and ΔH is negative

[tex]\Delta G=(-ve)-T(-ve)[/tex]

[tex]\Delta G=(-ve)(+ve)=-ve[/tex]  

Reaction is spontaneous at low temperatures.

d) ΔS is positive and ΔH is negative

[tex]\Delta G=(+ve)-T(-ve)[/tex]

[tex]\Delta G=(+ve)(+ve)=+ve[/tex]  

Reaction is non spontaneous or thermodynamically unfavored at all temperatures.

The predicted sign of the entropy change in each reaction are as follows;

To solve this question, we must understand the meaning of entropy.

Entropy of a system or chemical entity is the degree of disorderliness in the substance.

The change in entropy, ΔS∘ is a measure of the difference between the entropy of products and the reactants and is given mathematically as;

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In the first reaction entropy sign is negative while for the second reaction the sign on entropy is positive.

The Entropy is defined as degree of disorderliness in the system.

The entropy of the reaction can be calculated by,

[tex]\rm \bold{ \Delta S = S(products) - S(reactants)}[/tex]

Entropy for First reaction,

[tex]\rm \bold{ Ag(aq) + Br(aq) \rightarrow AgBr(s)}[/tex] will be negative

Entropy for second reaction,

[tex]\rm \bold{ CaCO_3(s) \rightarrow CaO(s)+CO_2(g) }[/tex] will be positive.

Hence, we can conclude that the the first reaction will be non- spontaneous while second reaction will spontaneous.

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Answer:

Neutral nucleophile are: H2O, CH3OH, NH3, RNH2, R2NH, R3N, RCOOH, RSH and PR3. The products by nucleophilic substitution are diverse depending on the different nucleophiles, obtaining alcohol, eter, amines, ester and tioeter considering only the nucleophiles with a hydrogen available.

Explanation:

Please see the images attached.

Nucleophilic subtitution with water occurs under Sn1 mechanism. That's it because water as nucleophile is so weak.  With the other neutral nucleophiles, the reaction occur under Sn2 mechanism.

RSH + CH3I  -----> RSCH3  +  HI

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)×[tex]\frac{1molZn}{65,38g}[/tex]×[tex]\frac{1molH_{2}(g)}{1molZn}[/tex] = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = 190,1L

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL×[tex]\frac{101,325}{1atmL}[/tex] = 19,26 kJ

I hope it helps!

The amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C is approximately 19601.44 joules.

The amount of work done against an atmospheric pressure when a gas is produced can be calculated using the ideal gas law, which is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to determine the number of moles of hydrogen gas (H2) produced from the dissolution of 500.0 g of zinc. The balanced chemical equation is:

[tex]\[ \text{Zn}(s) + 2\text{H}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{H}_2(g) \][/tex]

 From the stoichiometry of the reaction, 1 mole of zinc produces 1 mole of hydrogen gas. The molar mass of zinc is approximately 65.38 g/mol. Therefore, the number of moles of zinc (n_Zn) is:

[tex]\[ n_{\text{Zn}} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}} = \frac{500.0 \text{ g}}{65.38 \text{ g/mol}} \approx 7.647 \text{ mol} \][/tex]

 Since the stoichiometry is 1:1, the number of moles of hydrogen gas produced (n_H2) is also approximately 7.647 moles.

 Next, we convert the temperature from degrees Celsius to Kelvin:

[tex]\[ T = 30.0^\circ\text{C} + 273.15 = 303.15 \text{ K} \][/tex]

 Now, we can use the ideal gas law to find the volume of hydrogen gas produced at 1.00 atm of pressure:

[tex]\[ PV = nRT \][/tex]

[tex]\[ V = \frac{nRT}{P} \][/tex]

 Plugging in the values:

[tex]\[ V = \frac{(7.647 \text{ mol})(0.0821 \text{ L·atm/mol·K})(303.15 \text{ K})}{1.00 \text{ atm}} \][/tex]

[tex]\[ V \approx 193.57 \text{ L} \][/tex]

 Finally, the work done (W) against the atmospheric pressure to produce this volume of hydrogen gas is given by:

[tex]\[ W = P \cdot V \][/tex]

[tex]\[ W = 1.00 \text{ atm} \cdot 193.57 \text{ L} \][/tex]

[tex]\[ W \approx 193.57 \text{ L·atm} \][/tex]

 To express this work in joules, we use the conversion factor[tex]1 L*atm[/tex] = [tex]101.325 J:[/tex]

[tex]\[ W \approx 193.57 \text{ L·atm} \times \frac{101.325 \text{ J}}{1 \text{ L·atm}} \][/tex]

[tex]\[ W \approx 19601.44 \text{ J} \][/tex]

Therefore, the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C is approximately 19601.44 joules.

 The answer is: [tex]19601.44 \text{ J}.[/tex]

Answer:

a. strong interparticle attractions, low temperature, high pressure.

Explanation:

A gas behaves as an ideal gas when it fulfills the following conditions:

The gas deviates from the ideal behavior:

A real gas deviates most from ideal behavior under the conditions of strong interparticle attractions, low temperature, and high pressure.

A real gas deviates most from ideal behavior under the following set of conditions: strong interparticle attractions, low temperature, high pressure.

When a real gas has strong interparticle attractions, the gas particles are more likely to stick together and deviate from ideal behavior. Low temperature also contributes to the deviation as it slows down the gas particles and makes the attractive forces between them more prominent. High pressure further increases the deviation due to the decrease in empty space between the particles.

In summary, a real gas deviates most from ideal behavior when it has strong interparticle attractions, low temperature, and high pressure.

Answer:

only chlorine can expand its octet.

Explanation:

An atom can expand its octet is it has empty d orbital

the electronic configuration of given elements will be:

B : 1s2 2s2 2p1 [Valence shell n =2 no d orbital]

O :1s2 2s2 2p4 [Valence shell n =2 no d orbital]

F : 1s2 2s2 2p5 [Valence shell n =2 no d orbital]

Cl :1s2 2s2 2p6 3s2 3p5 3d0 [Valence shell n =2 no d orbital]

Out of given elements only chlorine has empty d orbitals in its valence shell

Thus only chlorine can expand its octet.

Answer:

a. 1,157x10⁻⁵M/s

b. 5,787x10⁻⁶M/s

Explanation:

For the reaction:

2H₂O₂(aq) → 2H₂O(l) + O₂(g).

a. The rate law of descomposition is:

[tex]rate=-\frac{1}{2} \frac{d[H_{2}O_{2}]}{dt}=\frac{d[O_{2}]}{dt}[/tex]

Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:

[tex]rate=-\frac{1}{2} \frac{0,500M}{-2,16x10^4s}[/tex]

[tex]rate=1,157x10^{-5}M/s[/tex]

As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s

b. Between 2,16x10⁴s and 4,32x10⁴s, rate law is:

[tex]rate=-\frac{1}{2} \frac{0,500M-0,250M}{2,16x10^4s-4,32x10^4s}[/tex]

[tex]rate=5,787x10^{-6}M/s[/tex]

The rates are 5,787x10⁻⁶M/s

I hope it helps!

The study of chemicals and bonds is called chemistry. There are two types of elements these rare metals and nonmetals.

Thus the rate is,

[tex]rate=1157\times10^{-5}[/tex]

The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters.

The balanced reaction is:-

[tex]2H_2O_2(aq)----->2H_2O(l)+o_2(g)[/tex]

The rate law of decomposition:-

[tex]rate=-\dfrac{1}{2}\dfrac{d[H_2O_2]}{dt}[/tex]

Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:

[tex]rate=-\dfrac{1}{2}\dfrac{0.5}{2.16\times10^{4}}[/tex]

[tex]rate=1157\times10^{-5}[/tex]

As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s

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Answer:   -2359 Joules

Explanation:

According to first law of thermodynamics:

[tex]\Delta E=q+w[/tex]

[tex]\Delta E[/tex]= Change in internal energy  = -2886 J

q = heat absorbed or released  = ?

w = work done or by the system  

w = work done by the system=[tex]-P\Delta V[/tex]  {Work is done by the system as the final volume is greater than initial volume and is negative}

w = -527 J

q = ?  

[tex]-2886=q+(-527)[/tex]

[tex]q=-2359J[/tex]  {Heat released by the system is negative}

Thus the amount of heat released by the system and carried away by the cooling system is -2359 Joules

According to first law of thermodynamics:

= Change in internal energy  = -2886 J

q = heat absorbed or released  = ?

w = work done or by the system  

w = work done by the system=  {Work is done by the system as the final volume is greater than initial volume and is negative}

w = -527 J

q = ?  

 {Heat released by the system is negative}

Thus the amount of heat released by the system and carried away by the cooling system is -2359 JoulesAnswer:

Explanation:

Explanation:

1) The valence electrons of group 4A elements are in the 5s subshell.

The given statement is false .

The valence electrons of group 4A elements are in the ns and np subshell .Where n is equal to 1 to 7.

2) Period 3 elements have an inner electron configuration of [Ar].

The given statement is false .

The noble gas which  comes before third period is neon.Period 3 elements have an inner electron configuration of [Ne].

3)The highest principal quantum number of period 4 elements is 4.

The given statement is true.

The period number of periodic table indicates the value of highest principal quantum number.

4) Period 5 elements have six 4p electrons.

The given statement is true.

This so because 4p subshell is filled before filling of shells of fifth shell. So, the 4p subshell will obviously have 6 electrons.

5) Period 5 elements have an inner electron configuration of (Kr).

The given statement is true.

The noble gas which comes before third period is krypton .

6) Group 8A elements have full outer principal s and p subshells.

The given statement is true.

This is because the the group mentioned in the statement if noble gases and noble gases are inert in nature due to fully filled electronic configurations.

7) The valence electrons of group 3A elements are in an s and p subshell.

The given statement is true.

The valence electrons of group 3A elements are in the ns and np subshell .Where n is equal to 1 to 7. there general electronic configuration is [tex]ns^2np^1[/tex].

8) The highest principal quantum number of period 4 elements is 5

The given statement is false.

The period number of periodic table indicates the value of highest principal quantum number.

The electron address in an atom can be given with the help of quantum mechanics. It helps to locate electrons with quantum numbers.

(1) The valence electrons of group 4A elements are in the 5s subshell.

The ns in group 4A can vary from 1 to 7. Thus the statement is false.

(2) Period 3 elements have an inner electron configuration of [Ar).

The nearest noble gas has been Neon, thus the inner electron configuration is of [Ne]. The statement is false.

(3) The highest principal quantum number of period 4 elements is 4.

The period number has been equal to the highest principal quantum number. Thus, the statement is true.

(4) Period 5 elements have six 4p electrons.

It has 5 subshells, and 4 subshells have been filled prior to 5s. Thus, the elements have a 4p orbital filled with 6 electrons. The statement is true.

(5) Period 5 elements have an inner electron configuration of (Kr).

The nearest noble gas has been Krypton, thus the inner electron configuration is of [Kr]. The statement is true.

(6) Group 8A elements have full outer principal s and p subshells.

Group 8A belongs to a noble gas. The subshells of noble gas are completely filled. The statement is true.

(7) The valence electrons of group 3A elements are in an s and p subshell.

The electrons of the 3A group have a general configuration [tex]\rm ns^2\;np^1[/tex]. The electrons are filled in s and p subshell. The statement is true.

(8) The highest principal quantum number of period 4 elements is 5.

The period number has been equal to the highest principal quantum number. Thus, the statement is false.

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Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

[tex]\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M[/tex]

Final answer:

Leucine, being nonpolar, should travel farther than the polar amino acid aspartic acid on a TLC plate because nonpolar substances have a higher affinity for the stationary phase and a lower affinity for the polar mobile phase.

Explanation:

On a thin-layer chromatography (TLC) plate, the distance that a compound travels is closely related to its polarity. In this lab experiment, leucine is a nonpolar amino acid, while aspartic acid is polar due to its acidic side chain. Therefore, leucine should travel farther on the TLC plate than aspartic acid because the nonpolar amino acids have a higher affinity for the nonpolar stationary phase and a lower affinity for the polar mobile phase, so they're carried less distance by the solvent front.

The correct answer to this question is A. Leucine is nonpolar, thus it should travel farther than aspartic acid on my TLC plate.

Answer:

The molarity of the Cl- anions is 0.0033 M

Explanation:

Step 1: Data given

Mass of MgCl2 = 48 mg = 48 *10^-3 grams

volume = 300 mL = 0.3L

Molar mas of MgCl2 = 95.21 g/mol

Step 2: The balanced equation

MgCl2 → Mg2+ + 2Cl-

This means for 1 mol MgCl2, we'll have 2 moles of Cl-

Step 3: Calculate moles MgCl2

Number of moles of MgCl2 = mass of MgCl2 / Molar mass of MgCl2

Moles MgCl2 = 48*10^-3 grams / 95.21 g/mol

Moles MgCl2 = 5.04 *10-4 moles

Step 4: Calculate moles of Cl-

This means for 1 mol MgCl2, we'll have 2 moles of Cl-

For  5.04 *10-4 moles MgCl2, we will have 2* 5.04 *10-4 moles = 0.001 moles Cl-

Step 5: Calculate molarity of the Cl- anion

Molarity = Moles / volume

Molarity Cl- = 0.001 moles / 0.3 L

Molarity Cl- = 0.0033 M

The molarity of the Cl- anions is 0.0033 M