which stament is true about endothermic and exothermic reactions? 1. Energy is absorbed 2. energy is released in an endothermic reaction. 3. the products have more potential energy than the reactants in an exothermic reaction. 4. the products have more potential energy than the reactant in an endothermic reaction.

Le Chatelier's Principle tells us what happens to the equilibrium of a chemical system (reaction) when certain stresses are inflicted onto it.

When the temperature of a system is increased, the system moves away from the heat. For instance, for a forward exothermic reaction, it would move to the reactants side, favouring the endothermic reaction. For a forward endothermic reaction, it would however favour the forward reaction with an increase in heat.

The opposite occurs when heat is removed.

When the concentration of a reactant is increased, the equilibrium shifts to the right and favours the formation of products. The opposite occurs when the concentration of a product is increased, it shifts to the left.

Pressure and volume are inversely proportional, meaning an increase in pressure leads to a decrease in volume (and vice versa). When pressure is increased/volume is decreased, the system shifts in the direction of least moles/molecules. Count the sum of the coefficients on the reactants and products side to determine which side this is.

Again, the opposite occurs when pressure is decreased or volume is increased; the system shifts to the side with more moles.

Taking into account all the pieces of information mentioned above, here is what our answers should be to the given question:

A. Increasing [SO2]: shifts right

B. Removing O2: shifts left

C. Increasing temperature: shifts left

D. Decreasing pressure: shifts left

E. Add a catalyst: no effect

To calculate the molarity of the solution, we need to know the number of moles of BaI2 and the volume of the solution in liters.

First, let's calculate the number of moles of BaI2. We can use the formula:

Number of moles = Mass (in grams) / Molar mass

The molar mass of BaI2 can be calculated as follows:

Ba: atomic mass = 137.33 g/mol

I: atomic mass = 126.90 g/mol

2 x I = 2 x 126.90 g/mol = 253.80 g/mol

Total molar mass of BaI2 = 137.33 g/mol + 253.80 g/mol = 391.13 g/mol

Number of moles of BaI2 = 413 g / 391.13 g/mol ≈ 1.056 moles

Next, we need to convert the volume of the solution from milliliters to liters:

Volume of solution = 750 ml / 1000 = 0.75 L

Finally, we can calculate the molarity using the formula:

Molarity = Number of moles / Volume of solution

Molarity = 1.056 moles / 0.75 L ≈ 1.408 M

Therefore, the molarity of the BaI2 solution is approximately 1.408 M.

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Answer:

Given:

Mass of glucose (m) = 281.1 g

Molar mass of glucose (MM) = 180.2 g/mol

Volume of solution (V) = 325 mL

First, let's convert the volume of the solution from milliliters (mL) to liters (L):

V = 325 mL = 325/1000 L = 0.325 L

Next, we can calculate the mass of glucose in the solution using its molar mass and the given mass:

moles of glucose (n) = m / MM

n = 281.1 g / 180.2 g/mol

Now, we need to calculate the mass percent by volume:

mass percent by volume = (mass of glucose / mass of solution) x 100

mass of solution = mass of glucose

mass percent by volume = (mass of glucose / mass of solution) x 100

= (n x MM / V) x 100

Substituting the values:

mass percent by volume = ((281.1 g / 180.2 g/mol) x 180.2 g/mol) / 0.325 L) x 100

Calculating this expression will give us the mass percent by volume of glucose in the solution.

The molarity of the calcium ion (Ca²⁺) in the solution is 0.411 M and the molarity of the hydroxide ions (OH⁻) in the solution is 0.822 M.

The molarity of ions in a solution can be determined by considering the dissociation of the compound into its constituent ions. In the case of Ca(OH)₂, it dissociates into one calcium ion (Ca²⁺) and two hydroxide ions (OH⁻) per formula unit.

Since the solution is 0.411 M Ca(OH)₂, the molarity of the calcium ion (Ca²⁺) would also be 0.411 M because there is one calcium ion for every formula unit of Ca(OH)₂. The molarity of the hydroxide ions (OH⁻) would be twice that of the Ca²⁺ ion because there are two hydroxide ions per formula unit of Ca(OH)₂.

The molarity of the hydroxide ions = 2 × 0.411 M = 0.822 M.

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Answer:

[tex]\huge\boxed{\sf CNH_4}[/tex]

Explanation:

Given compound us,

C₂N₂H₈

= 2 : 2 : 8

= 2 ÷ 2 : 2 ÷ 2 : 8 ÷ 2

= 1 : 1 : 4

So, we can write the formula as:

[tex]\rule[225]{225}{2}[/tex]

Metamorphic rocks can be harder, less porous, and have crystals that can be lined, describing some of the ways in which metamorphic rocks differ from sedimentary rocks.

There are two different types of rocks: sedimentary rocks and metamorphic rocks. Igneous or sedimentary pre-existing rocks undergo changes under extreme heat and pressure to form metamorphic rocks. This process results in the recrystallization of minerals, leading to the formation of a new rock with distinct physical and chemical characteristics.

Therefore, the correct option is B.

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Answer:

Exothermic Reaction

Product  = Calcium hydroxide + hydrogen

Explanation:

1. The mass of chlorine formed can be obtain as follow:

2AlCl₃ -> 2Al + 3Cl₂

From the balanced equation above,

267 g of AlCl₃ decomposed to produce 213 g of Cl₂

Therefore,

10 g of AlCl₃ will decompose to produce = (10 × 213) / 267 = 7.98 g of Cl₂

Thus, the mass of chlorine formed is 7.98 g

2. The mass of aluminium formed can be obtain as follow:

2AlCl₃ -> 2Al + 3Cl₂

From the balanced equation above,

267 g of AlCl₃ decomposed to produce 54 g of Al

Therefore,

10 g of AlCl₃ will decompose to produce = (10 × 54) / 267 = 2.02 g of Al

Thus, the mass of aluminum formed is 2.02 g

The percentage yield of aluminum formed can be obtain as follow:

Percentage yield = (Actual /Theoretical) × 100

Percentage yield of aluminum = (1.2 / 2.02) × 100

Percentage yield of aluminum = 59.4%

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The addition of a catalyst to this reaction would cause a change in "I" indicated energy differences.

If a catalyst is added to a reaction, it typically affects the activation energy (Ea) of the reaction. The activation energy is the energy barrier that needs to be overcome for the reaction to proceed.

In the context of the energy diagram for the reaction X + Y -> Z, the addition of a catalyst would primarily affect the energy difference related to the activation energy. Let's consider the options:

It is generally expected that the addition of a catalyst would primarily affect the activation energy (Ea) of the reaction, which is typically associated with the energy difference labeled as "I" on energy diagrams.

Therefore, the answer is: I only: The addition of a catalyst would cause a change in the energy difference labeled as "I" on the energy diagram.

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Answer: A : 3

Explanation: 18 CO2 / 6 CO2 = 3 C6H12O6

Answer:

A = 3.

Explanation:

Here is how:

To determine the number of molecules of C6H12O6 (glucose) needed to produce 18 molecules of CO2, we need to consider the balanced chemical equation for the complete combustion of glucose:

C6H12O6 + 6O2 -> 6CO2 + 6H2O

From the balanced equation, we can see that 1 molecule of glucose (C6H12O6) produces 6 molecules of CO2. Therefore, we can set up a proportion to find the number of glucose molecules needed:

1 molecule of glucose produces 6 molecules of CO2

x molecules of glucose produce 18 molecules of CO2

Using the proportion:

1/6 = x/18

To solve for x, we can cross-multiply:

6x = 18

Dividing both sides by 6:

x = 3

Therefore, 3 molecules of C6H12O6 are needed to produce 18 molecules of CO2.

Answer:

Phosphorus(P) and Oxygen(O)=Covalent bond

Chlorine(Cl) and Sodium(Na) = Ionic bond

Silver (Ag) and Silver (Ag)= Metallic bond

The molar mass of the gas is approximately 20.72 g/mol.

To determine the molar mass of the gas, we can use the ideal gas law equation:

PV = nRT

where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = ideal gas constant (0.0821 L·atm/(mol·K))

T = temperature (in Kelvin)

First, let's convert the given values to the appropriate units:

Pressure = 432 torr = 432/760 atm

Volume = 333 mL = 333/1000 L

Temperature = 23°C = 23 + 273.15 K

Substituting the values into the ideal gas law equation:

(432/760) atm * (333/1000) L = n * 0.0821 L·atm/(mol·K) * (23 + 273.15) K

Simplifying the equation:

0.191 atm * 0.333 L = n * 0.0821 L·atm/(mol·K) * 296.15 K

Solving for the number of moles (n):

n = (0.191 atm * 0.333 L) / (0.0821 L·atm/(mol·K) * 296.15 K)

n ≈ 0.01012 moles

Finally, we can calculate the molar mass using the formula:

Molar mass = mass of the gas sample / moles of gas

Molar mass = 0.210 g / 0.01012 moles

Molar mass ≈ 20.72 g/mol

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Answer:  505 grams K3PO4 x (3 x 222 grams HCI)/ (3 x K3PO4) = 555.5 grams KCl

Explanation:

Answer:

C. The sun is closer to Earth than other stars.

Explanation:

The sun appears larger and brighter than other stars because it is much closer to Earth. The sun is the closest star to Earth, at a distance of about 93 million miles. Other stars are much farther away, so they appear smaller and less bright in the sky.

For the following questions:

2. The periodic table was first arranged by increasing atomic mass. False

The periodic table was first arranged by increasing atomic number. This was done by Dmitri Mendeleev in 1869.

3. The short configuration of Hf is [Xe] 6s2 4f14 5s1. True

The short configuration of Hf is [Xe] 6s2 4f14 5s1. This is because Hf has 72 electrons, and the electron configuration of Xe is [Kr] 5s2 4d10 5p6. So, the electron configuration of Hf can be written as [Xe] 6s2 4f14 5s1.

4. 150 LX 4.0 moles would equate to a molarity of 0.0266 mol/L. False

Molarity is defined as the moles of solute per liter of solution. So, to calculate the molarity of a solution, we need to divide the moles of solute by the volume of solution in liters. In this case, we have 150 L of solution and 4.0 moles of solute. So, the molarity of the solution is 4.0 moles / 150 L = 0.0266 mol/L.

5. A block 1.35 m x 2.467 m = 3.3 m². False

The area of a rectangle is calculated by multiplying the length by the width. So, the area of a block that is 1.35 m long and 2.467 m wide is 1.35 m x 2.467 m = 3.319 m².

6. The density of an unknown solid weighs 3.00 g in 5.0 mL = 0.60 g/mL. True

Density is defined as mass per unit volume. So, to calculate the density of a substance, we need to divide the mass of the substance by the volume of the substance. In this case, we have a solid that weighs 3.00 g and has a volume of 5.0 mL. So, the density of the solid is 3.00 g / 5.0 mL = 0.60 g/mL.

7. 2 moles of helium would occupy 50 L of a balloon filled with it at STP. True

At STP, one mole of any gas occupies 22.4 L. So, two moles of helium would occupy 2 x 22.4 L = 44.8 L.

8. The empirical formula of a compound is half of the molecular? False

The empirical formula of a compound is the simplest whole-number ratio of the atoms in the compound. The molecular formula of a compound is the actual number of atoms in the compound. So, the empirical formula of a compound is not necessarily half of the molecular formula.

9. Multiple compounds can have the same empirical formulas? True

Multiple compounds can have the same empirical formulas. For example, the empirical formula of methane, ethane, and propane are all CH₃. However, the molecular formulas of methane, ethane, and propane are CH₄, C₂H₆, and C₃H₈, respectively.

10. Stoichiometric calculations can only be achieved by converting to moles? False

Stoichiometric calculations can be achieved by converting to moles, but they can also be achieved by using other units, such as grams or liters.

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The mass of [tex]O_2[/tex] needed to produce 8.65 x [tex]10^{23[/tex] atoms of silver is 22.96 grams.

In the given balanced equation, it is stated that 2 moles of [tex]Ag_2O[/tex] produce 4 moles of Ag and 1 mole of [tex]O_2[/tex].

First, let's calculate the number of moles of Ag:

8.65 x [tex]10^{23[/tex] atoms of Ag / (6.022 x [tex]10^{23[/tex] atoms/mol) = 1.435 moles of Ag

Since 2 moles of  [tex]Ag_2O[/tex] produce 1 mole of  [tex]O_2[/tex], the number of moles of  [tex]O_2[/tex] needed is half of the moles of  [tex]Ag_2O[/tex].

Number of moles of  [tex]O_2[/tex]= 1.435 moles of  [tex]Ag_2O[/tex] / 2 = 0.7175 moles

Now, we'll use the molar mass of  [tex]O_2[/tex] to find the mass:

Molar mass = 32.00 g/mol

Mass of  [tex]O_2[/tex] = number of moles × molar mass

                   = 0.7175 moles × 32.00 g/mol = 22.96 g

Therefore, the mass of  [tex]O_2[/tex] needed to produce 8.65 x [tex]10^{23[/tex] atoms of silver is 22.96 grams.

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The chromium ions with a +3 oxidation state are reduced to chromium atoms with an oxidation state of 0.The reduction of Cr^3+ to Cr in this chemical equation is an example of a reduction reaction.

The chemical equation Cr^3+ + 3e^- = Cr represents the reduction of chromium ions (Cr^3+) to elemental chromium (Cr). In this reaction, the chromium ions gain three electrons to form neutral chromium atoms. Reduction reactions involve the gain of electrons and a decrease in the oxidation state of an element.

During the reduction process, the chromium ions are undergoing a change in their electronic configuration, gaining three electrons to achieve a stable configuration. This reduction reaction typically occurs in the presence of a reducing agent that donates electrons, allowing the chromium ions to be reduced. By gaining three electrons, the chromium ions are reduced to their elemental form, which has a neutral charge and an oxidation state of 0.

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The limiting reactant will be the one that produces fewer moles of BaSO4. The excess reactant will be the one that has moles left over after the reaction.

To determine the grams of BaSO4 produced and the limiting reactant, we need to compare the stoichiometry of the balanced chemical equation for the reaction between Ba(NO3)2 and Na2SO4, which is:

Ba(NO3)2 + Na2SO4 → BaSO4 + 2NaNO3

First, calculate the number of moles for each reactant:

Moles of Ba(NO3)2 = 200.0 g / molar mass of Ba(NO3)2

Moles of Na2SO4 = 100.0 g / molar mass of Na2SO4

Then, calculate the moles of BaSO4 formed by comparing the stoichiometric coefficients:

Moles of BaSO4 formed = Moles of Ba(NO3)2 (according to the stoichiometry ratio)

Next, calculate the grams of BaSO4 formed:

Grams of BaSO4 formed = Moles of BaSO4 formed × molar mass of BaSO4

To identify the limiting reactant, compare the moles of BaSO4 formed from each reactant. The reactant that produces fewer moles of BaSO4 is the limiting reactant.

To determine the excess reactant remaining, calculate the moles of excess reactant and then convert it to grams.

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Based on Table G, the mass of KCl that must be dissolved in 200 grams of H₂O at 10 °C to make a saturated solution is 60 g.

Based on Table G, the solubility of KCl at 10°C is given as 30 g/100 g water.

To calculate the mass of KCl that can be dissolved in 200 grams of water at 10°C, we can set up a proportion:

(30 g KCl / 100 g water) = (x g KCl / 200 g water)

Cross-multiplying and solving for x, we get:

x g KCl = (30 g KCl / 100 g water) * (200 g water)

x g KCl = 60 g KCl

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Answer:

[tex]\large \textsf{If the Keq of a reaction is 4$\times$10$^{-7}$, then:}\\\\\large \textsf{$\implies$ the equilibrium lies slightly to the left.}[/tex]

The position or extent of a chemical equilibrium can be expressed quantitatively using the equilibrium constant (Keq). If the value of Keq is large, then the equilibrium lies to the right (the product side). If the value of Keq is small, then the equilibrium lies to the left (the reactant side).

In terms of sizing, a small value of Keq usually ranges from 10⁻¹⁰ to 10⁻⁵⁰ and beyond. A large value of Keq usually ranges from 10¹⁰ and onwards.

∴ for a Keq of 4×10⁻⁷, we say that the equilibrium lies slightly to the left.

The ideal gas law and stoichiometry must be used to calculate the volume of carbon dioxide gas produced by the breakdown of 4.09 g of calcium carbonate at STP (Standard Temperature and Pressure).

Use the molar mass of calcium carbonate (CaCO3) to determine how many moles it contains. CaCO3 has a molar mass of 100.09 g/mol.

CaCO3 mass divided by its molar mass equals the number of moles of CaCO3: 4.09 g/100.09 g/mol.

The number of moles of carbon dioxide (CO2) generated may be calculated using the stoichiometric ratio from the balancing equation. By using the equation:

A unit of CaCO3 and CO2 is produced.

CO2 moles equal the same number of moles of CaCO3.

Use the ideal gas law to translate the volume of carbon dioxide into moles.

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Diatomic: Composed of two atoms. Polar: A bond with a negative end and a positive end. Nonpolar: A bond in which neither atom takes more than its share of electrons. Metallic: A type of bond that allows valence electrons to move freely among ions. Electronegativity: Determines what type of bond will form.

The ability of an atom or functional group to draw electrons to itself is known as electronegativity in chemistry.

Diatomic molecules consist only of two atoms, whether they are from the same or distinct chemical elements.

Since charges fluctuate, a momentary dipole moment occurs in a so-called nonpolar molecule at any given time if the charge arrangement is spherically symmetric when averaged across time.

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Answer:

An atom contains three basic particles namely protons, neutrons and electrons. The nucleus of the atom contains protons and neutrons where protons are positively charged and neutrons are neutral. The electrons are located at the outermost regions called the electron shell.

Taking into account definition of theoretical yield, Cl₂ is the limiting reagent and the theoretical yield is 100.31 grams of AlCl₃ if 60 g Al react with 80 g of Cl₂ and produce aluminum chloride

In first place, the balanced reaction is:

2 Al + 3 Cl₂ → 2 AlCl₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction.

To determine the limiting reagent, it is possible to use a rule of three as follows: if by stoichiometry 54 grams of Al reacts with 212.7 grams of Cl₂, 60 grams of Al reacts with how much mass of Cl₂?

mass of Cl₂= (60 grams of Al× 212.7 grams of Cl₂)÷54 grams of Al

mass of Cl₂= 236.33 grams

But 236.33 grams of Cl₂ are not available, 80 grams are available. Since you have less mass than you need to react with 60 grams of Al, Cl₂ will be the limiting reagent.

The theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 212.7 grams of Cl₂ form 266.7 grams of AlCl₃, 80 grams of Cl₂ form how much mass of AlCl₃?

mass of AlCl₃= (80 grams of Cl₂×266.7 grams of AlCl₃)÷212.7 grams of Cl₂

mass of AlCl₃= 100.31 grams

Finally, the theoretical yield is 100.31 grams of AlCl₃.

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The concentration of ethanol in units of volume/volume percent is 29.49%.

To calculate the concentration of ethanol in units of volume/volume percent, we need to determine the volume of ethanol relative to the total volume of the solution.

Total volume of the solution = volume of water + volume of ethanol

Total volume = 569 mL + 238 mL

Total volume = 807 mL

To express the concentration as volume/volume percent, we can calculate the ratio of the volume of ethanol to the total volume of the solution and multiply by 100 to obtain a percentage.

Concentration of ethanol = (volume of ethanol / total volume of solution) x 100

Concentration of ethanol = (238 mL / 807 mL) x 100

Concentration of ethanol = 0.2949 x 100

Concentration of ethanol = 29.49%

Therefore, the concentration of ethanol in the solution is approximately 29.49%.

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The balanced chemical equation is CaS + AlC → A + CaC

To balance the chemical equation CaS + AlC → A + CaC, we need to ensure that the same number of atoms of each element is present on both sides of the equation. Here's the step-by-step process to balance the equation:

Begin by counting the number of atoms of each element on both sides of the equation.

Left side (reactants):

Calcium (Ca): 1

Sulfur (S): 1

Aluminum (Al): 1

Carbon (C): 1

Right side (products):

A: 1

Calcium (Ca): 1

Carbon (C): 1

Sulfur (S): 0

Start by balancing the elements that appear in the fewest compounds. In this case, we can balance sulfur (S) first. Since there is only one sulfur atom on the left side and none on the right side, we need to add a coefficient of 1 in front of A on the right side to balance the sulfur.

CaS + AlC → 1A + CaC

Next, balance calcium (Ca) by adding a coefficient of 1 in front of CaS on the left side.

1CaS + AlC → 1A + CaC

Now, balance aluminum (Al) by adding a coefficient of 1 in front of AlC on the left side.

1CaS + 1AlC → 1A + CaC

Finally, balance carbon (C) by adding a coefficient of 1 in front of CaC on the right side.

1CaS + 1AlC → 1A + 1CaC

The balanced chemical equation is:

CaS + AlC → A + CaC

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To calculate the pH of the solution obtained by mixing HCl and NaOH, we need to consider the neutralization reaction between the two compounds. The reaction between HCl (hydrochloric acid) and NaOH (sodium hydroxide) produces water (H₂O) and forms a salt (NaCl).

Given:

Volume of HCl solution (V₁) = 40 cm³

Concentration of HCl solution (C₁) = 0.2 M

Volume of NaOH solution (V₂) = 30 cm³

Concentration of NaOH solution (C₂) = 0.1 M

1. Determine the moles of HCl and NaOH used:

Moles of HCl = Concentration (C₁) × Volume (V₁)

Moles of HCl = 0.2 M × 0.04 L (converting cm³ to L)

Moles of HCl = 0.008 mol

Moles of NaOH = Concentration (C₂) × Volume (V₂)

Moles of NaOH = 0.1 M × 0.03 L (converting cm³ to L)

Moles of NaOH = 0.003 mol

2. Determine the limiting reagent:

The stoichiometry of the reaction between HCl and NaOH is 1:1, meaning that they react in a 1:1 ratio. Whichever reactant is present in a smaller amount will be the limiting reagent.

In this case, NaOH is present in a smaller amount (0.003 mol), which means it will be fully consumed during the reaction.

3. Determine the excess reagent and its remaining moles:

Since NaOH is the limiting reagent, we need to find the remaining moles of HCl.

Moles of HCl remaining = Moles of HCl initially - Moles of NaOH reacted

Moles of HCl remaining = 0.008 mol - 0.003 mol

Moles of HCl remaining = 0.005 mol

4. Calculate the concentration of HCl in the resulting solution:

Volume of resulting solution = Volume of HCl solution + Volume of NaOH solution

Volume of resulting solution = 0.04 L + 0.03 L

Volume of resulting solution = 0.07 L

Concentration of HCl in the resulting solution = Moles of HCl remaining / Volume of resulting solution

Concentration of HCl in the resulting solution = 0.005 mol / 0.07 L

Concentration of HCl in the resulting solution ≈ 0.071 M

5. Calculate the pH of the resulting solution:

pH = -log[H⁺]

pH = -log(0.071)

Using logarithm properties, we can determine the pH value:

pH ≈ -log(0.071)

pH ≈ -(-1.147)

pH ≈ 1.147

Therefore, the pH of the solution obtained by mixing 40 cm³ of 0.2 M HCl and 30 cm³ of 0.1 M NaOH is approximately 1.147.