Which pair of elements is most apt to form an ionic compound with each other?

The amount of energy needed for a reaction to take place is called activation energy. It determines the rate at which a chemical reaction occurs.

In a potential energy diagram, the amount of energy needed for the reaction to take place is called the activation energy. Activation energy is the minimum energy required for reactant molecules to collide with sufficient energy to form a high-energy activated complex or transition state. It determines the rate at which a chemical reaction occurs, with higher activation energy leading to slower reactions and lower activation energy leading to faster reactions.

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Answer : 24 atomic mass units (amu)

Explanation : Taking into consideration that the oxygen isotope has the molecular mass as 18 amu and the isotope of hydrogen will be tritium which will have the molecular mass as 3 amu.

The water molecule has two hydrogen and one oxygen; which means 2H and 1 O = [tex] H_{2}O[/tex]

So, to calculate the largest mass that a water molecule could have using other isotopes will be;

(2 X 3) + 18 = 24 amu.

So, the largest mass that a water molecule can have is 24 amu.

The largest mass that a water molecule could have using other isotopes = 24.0312

The elements in nature have several types of isotopes

Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

Atomic mass is the average atomic mass of all its isotopes

In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu

So the atomic mass obtained is the mass of the atom relative to the carbon atom

[tex]\large {\boxed{mass~average~atom~X~=~ \frac {mass\: isotope ~ 1 + mass ~ isotope ~ 2} {whole ~ atom ~ X}}[/tex]

An atomic mass unit = amu is a relative atomic mass of 1/12 the mass of an atom of carbon-12.

The 'amu' unit has now been replaced with a unit of 'u' only

for example, Carbon has 3 isotopes, namely ₆C¹², ₆C¹³, and ⁶C¹⁴

To determine the largest molecular mass for air, we use the largest isotope mass of Hydrogen and Oxygen as its constituent elements

Hydrogen (1 H) has 3 stable natural isotopes, namely ₁H¹, ₁H², and ₁H³

There are also unstable isotopes obtained through the synthesis in laboratories ₁H⁴, ₁H⁵, ₁H⁶ and ₁H7⁷

While Oxygen (₁₆O) has 3 natural isotopes ₈O¹⁶, ₈O¹⁷, and ₈O¹⁸

If we use the largest mass of isotopes, we use stable isotopes ₁H³ and ₈O¹⁸ to form 1 H₂O water molecule

₁H³ = 3,016 049 amu

and the atomic mass of ₈O¹⁸ = 17,999 amu

So that the mass of the water molecule

= 2. The atomic mass of ₁H³ + 1. atomic mass ₈O¹⁸

= 2. atomic mass of 3,0161 amu + 17,999 amu

= 24.0312

Water itself has no atomic mass, because water is a molecule

And we should be able to distinguish between the number of mass and atomic mass

The subatomic particle that has the least mass

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From a reaction equation, check the enthalpy change (ΔH). Negative ΔH represents an exothermic reaction (heat producing) while positive ΔH relates to endothermic reaction (heat absorbing). Other methods include reaction diagrams, bond energies, and Hess's Law.

An exothermic reaction, which produces heat, will have a negative ΔH, as the bonds in the products are stronger than the bonds in the reactants. This implies that energy is released during the reaction. For instance, when heat (q) is negative in a calorimetric determination, this suggests an exothermic process is taking place, and thermal energy is transferred from the system to its surroundings.

On the other hand, an endothermic reaction, which consumes heat, will have a positive ΔH, as the bonds in the products are weaker than the ones in the reactants. This indicates that energy is absorbed during the reaction. If heat (q) is positive in a calorimetric determination, it signifies that an endothermic process is happening, with thermal energy being transferred from the surroundings to the system.

Furthermore, reaction diagrams, bond energies, and Hess's Law can be applied to predict and calculate the enthalpy changes for reactions, thus determining whether they are exothermic or endothermic.

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Constructing the ICE table for acid-dissociation  of sulfurous acid

 H₂SO₃ --> H⁺ + HSO₃⁻

I    0.163 M      0        0

C     - x           + x       + x

E   0.163 - x    + x       + x

Writing the acid dissociation expression  of sulfurous acid,

Kₐ = [H⁺ ][ HSO₃⁻]/ [H₂SO₃]

Plugging in the values we get,

\frac{x²}{(0.163 - x)} = 1.7 x 10⁻²

Since 1.7 x 10⁻² is small we can ignore x in the denominator,

\frac{x²}{0.163}  = 1.7 x 10⁻²

x = 0.052640 M  

pH = 1.28  

Thus the pH of a 0.163 M aqueous solution of sulfurous acid is 1.28.

The pH of a 0.163 M sulfurous acid solution is calculated by considering only the first dissociation step due to the significantly larger first acid-dissociation constant. The concentration of H+ ions is found by solving the equilibrium expression of dissociation, leading to the calculation of pH with the negative logarithm of the H+ concentration.

To calculate the pH of a 0.163 M aqueous solution of sulfurous acid, we must consider its acid-dissociation constants. Sulfurous acid is a diprotic acid with two dissociation steps. The first dissociation constant (Ka1) is significantly larger than the second (Ka2), which means that the first dissociation step will contribute the most to the hydronium ion concentration [H+] in solution. Due to the relatively large Ka1, we can approximate that the contribution from the second dissociation is insignificant for the calculation of pH in a 0.163 M H2SO3 solution.

The dissociation of H2SO3 can be represented as follows:

Since Ka1 > Ka2, we focus on Ka1 for the pH calculation. The dissociation can be described by the equilibrium expression:

Ka1 = [H+][HSO3-] / [H2SO3]

Assuming x represents the equilibrium concentration of H+ and HSO3-, the equation becomes:

Ka1 = (x)(x) / (0.163-x) ≈ (x2) / (0.163)

Solving for x yields the concentration of H+, and the pH is calculated using the equation pH = -log[H+].

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Final answer:

The contamination that occurs from a paint chip falling in soup can be physical but may also lead to chemical contamination if the paint contains lead, which poses serious health risks.

Explanation:

If a paint chip falls into soup, the type of contamination that occurs is called physical contamination, which involves foreign objects entering foodstuffs. However, if the paint is lead-based, this can also lead to chemical contamination, as lead is a toxic substance.

When paint peels and cracks, it creates lead dust that can be hazardous if ingested. Given the risks associated with lead, it is considered a significant public health concern, particularly because lead exposure can cause serious health issues.

The ingestion of lead can be particularly dangerous for children and can lead to numerous health problems, including cognitive impairments.

The mass of oxygen required to react with 7.6 g of ethanol is 5.3 g.

This problem is a stoichiometric problem. To solve problems like this we always begin with the balanced equation. Then, we use the stoichiometric ratios provided by the coefficients of the reactants and products.

1. Write the balanced chemical equation.

CH₃CH₂OH + O₂ → H₂O + CH₃COOH

2. Convert the mass of oxygen to moles.

[tex]7.6 \ g \ CH_3CH_2OH \times \frac{1 \ mol CH_3CH_2OH}{46.07 \ g} = 0.165 \ mol \ CH_3CH_2OH\\[/tex]

3. Determine the equivalent moles of oxygen that reacts with the given quantity of ethanol. The stoichiometric ratio for ethanol and oxygen indicated in the balanced chemical equation is 1:1.

[tex]moles \ O_2 = 0.165 \ mol \ CH_3CH_2OH \times \frac{1 \ mol O_2}{ 1 \ mol CH_3CH_2OH}\\\\moles \ O_2 = 0.165 \ mol \ O_2[/tex]

4. Convert the moles of O₂ to mass.

[tex]mass \ of \ O_2 = 0.165 \ mol \ O_2 \times \frac{32 \ g}{1 \ mol \ O_2 }\\\\\boxed {mass \ of \ O_2 \ = 5.28 \ g}[/tex]

The answer required should only have 2 significant digits. Therefore, the final answer is:

[tex]\boxed {\boxed {mass \ of \ O_2 \ = 5.3 \ g}}[/tex]

To calculate the mass of oxygen gas consumed by the reaction of 7.6g of ethanol, we need to convert the mass of ethanol to moles, and then use the balanced chemical equation to determine the moles of oxygen gas. Finally, we convert moles of oxygen gas to grams.

To calculate the mass of oxygen gas consumed by the reaction of 7.6g of ethanol, we first need to convert the mass of ethanol to moles. Ethanol has a molar mass of 46.06 g/mol, so we divide 7.6g by the molar mass to get 0.165 moles of ethanol. The balanced chemical equation shows that for every 1 mole of ethanol, 3 moles of oxygen gas are consumed. Therefore, we multiply the moles of ethanol by the ratio of moles of oxygen gas to moles of ethanol (3/1) to get 0.495 moles of oxygen gas. Finally, we convert moles of oxygen gas to grams by multiplying by the molar mass of oxygen gas (32.00 g/mol), giving an answer of 15.8g of oxygen gas consumed.

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The empirical formula of phosphorus selenide, formed from a reaction between 45.2 mg of phosphorus and selenium resulting in 131.6 mg of the compound, is P4Se3.

To determine the empirical formula of phosphorus selenide, we have to find the ratio of the number of moles of phosphorus to the number of moles of selenium. First, convert the mass of each element to moles. The molar mass of phosphorus (P) is 30.97 grams per mole, and the molar mass of selenium (Se) is 78.97 grams per mole. Thus, we have 45.2 mg of P = 0.00146 mole and 86.4 mg of Se (= 131.6 mg - 45.2 mg) = 0.0011 mole.

Then, find the smallest whole number ratio of moles of P to Se, by dividing both by the smallest amount. The ratio of P:Se is 1.33:1, which is close to 4:3 when multiplied up.

So, the empirical formula of phosphorus selenide is P4 Se3.

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Determine the empirical formula of phosphorus selenide follows steps involving mass calculations and mole conversions, leading to the empirical formula of P₄Se₃.

To determine the empirical formula of phosphorus selenide, follow these steps:

Calculate the mass of selenium in the compound:

Mass of selenium = Mass of phosphorus selenide - Mass of phosphorus

Mass of selenium = 131.6 mg - 45.2 mg = 86.4 mg

Convert the masses to moles:

Moles of phosphorus (P):

45.2 mg P × (1 g / 1000 mg) × (1 mol P / 30.97 g P) = 0.00146 mol P

Moles of selenium (Se):

86.4 mg Se × (1 g / 1000 mg) × (1 mol Se / 78.96 g Se) = 0.00109 mol Se

Determine the simplest whole number ratio of moles:

Ratio of P to Se = 0.00146 / 0.00109 ≈ 1.34

The ratio of P to Se is approximately 1.34:1. This ratio is close to the simple fraction 4/3. Therefore, the ratio is adjusted to 4:3 to yield whole numbers.

Based on this ratio, the empirical formula of phosphorus selenide is P₄Se₃.

Answer:

The empirical formula of phosphorus selenide is P₄Se₃.

SO2 (5.00 g) and CO2 (5.00 g) are placed in a 750.0 ml container at 50.0 °C. the partial pressure of SO2 in the container was 2.762 atm.

To calculate the partial pressure, we can calulate the total pressure

[tex]Total pressure =  \frac{nRT}{V} SO2  + \frac{nRT}{V} CO2[/tex]

Where:

The amount of SO2 is n SO2 = mass/molar mass = m/M = [tex]\frac{5 }{32} + 16*2 = 0.078125[/tex] mol

The amount of CO2 is n CO2 = mass/molar mass = m/M = [tex]\frac{5}{14} + 16*2 = 0.1[/tex] mol

Total amount of gas is n(total) = [tex]n1 + n2 = 0.078 + 0.113 = 0.191[/tex]  mol

T = 50+273 = 323 K

[tex]V = \frac{750}{1000} = 0.75 [/tex]liters

Total Pressure  [tex]p*V = n*R*T[/tex]

Total Pressure [tex]= (0.078125*0.0821*\frac{323}{0.75} ) + (0.1*0.0821*\frac{323}{0.75}) = 2.7623 + 3.535 = 6.298[/tex] atm

Partial pressure = x SO2 * Total Pressure = [tex](\frac{no. of moles of SO2 }{total no. of moles} ) * Total pressure = (\frac{0.078125}{0.078125} +0.1) * 6.298 = 2.762 [/tex]atm

Grade:  9

Subject:  chemistry

Chapter:  pressure

Keywords:   the container, the partial pressure, SO2, CO2,  The mole fraction

The name of the element that is a member of the alkali metal family whose most stable ion contains 2 electrons is "Ba (barium).

A pure substance made up entirely of atoms with almost the identical count of protons in respective nuclei was known as an element.

It is known that Ba element can form +2 oxidation state very easily which belongs to the group 1.

Therefore, the name of the element that is a member of the alkali metal family whose most stable ion contains 2 electrons is "Ba (barium).

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To complete the chart, we need to determine the number of valence electrons for each element. Starting from Period 1, we place electrons in the subshells according to the periodic table.

Start at Period 1 of Figure 2.8.2. Place two electrons in the 1s subshell (1s²). Proceed to Period 2 (left to right direction). Place the next two electrons in the 2s subshell (2s²) and the next six electrons in the 2p subshell (2pº).

Answer:

Option A = 17.03 g/mol

Explanation:

Given data:

Atomic mass of nitrogen = 14.01 g/mol

Atomic mass of hydrogen = 1.008 g/mol

Molecular mass of NH₃ = ?

Explanation:

Molecular mass of NH₃ = (14.01 g/mol × 1) + (1.008 g/mol × 3)

Molecular mass of NH₃ = 14.01 g/mol + 3.024 g/mol

Molecular mass of NH₃ =  17.03 g/mol

Ammonia consist of hydrogen and nitrogen both are nonmetals that's why ammonia is an covalent compound.

The half-cell reaction with the most positive standard half-cell potential (E°) has the greatest tendency to occur at the cathode since it contains the strongest oxidizing agent.

In an electrochemical cell, the half-cell reaction that has the greatest tendency to occur at the cathode is the one with a more positive standard half-cell potential (E°). The cathode is the site of reduction in an electrochemical cell. The half-cell with the higher E° has the highest tendency because it has the strongest oxidizing agent (reactant species in the half-cell reaction). This is because according to the electrochemical series, the stronger the oxidizing agent, the more positive the E°.

For instance, if you compare the silver(I)/silver(0) half-reaction to the copper(II)/copper(0) half-reaction, the silver one would predominate at the cathode since its entry for standard potential is above the copper one, thus it has a more positive E°, and the reaction is predicted to be spontaneous (E°cathode > E°anode and so E°cell > 0).

A useful established reference for cell potential measurements is the standard hydrogen electrode (SHE), which has an assigned potential of exactly 0 V. This helps in comparing different half-cell reactions and determining which is more likely to occur at the cathode.

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Final answer:

The concentration of hydroxide ion in a solution of methylamine is approximately 9.07 x 10^-7 M.

Explanation:

The concentration of hydroxide ion in a solution of methylamine can be found by using the relation:

Kw = [H+][OH-]

Given that the Kb for methylamine is 4.4 x 10-4, we can calculate the concentration of hydroxide ion using the formula:

[OH-] = sqrt(Kw/Kb) = sqrt(1.0 x 10-14/4.4 x 10-4)

Substituting the values and solving, we find that the concentration of hydroxide ion in the solution is approximately 9.07 x 10-7 M.

The pH of acid is between 0-7 on pH scale while for base pH range is from 7-14. Therefore,  the pH of solution is  1.84. pH is a unitless quantity.

pH is a measurement of amount of hydronium ion H₃O⁺ in a given sample. More the value of hydronium ion concentration, more will be the solution acidic.

On subtracting pH from 14, we get pOH which measures the concentration of hydroxide ion in a given solution. pH depend on the temperature. At room temperature pH scale is between 0 to 14. pH of neutral solution is 7.

Ka for  Co (H2O)[tex]_6[/tex]³⁺ is 1.0 ✕ 10⁻⁵

Ka = 1.0 ✕ 10⁻⁵

x²/[0.3-x]x =1.0 ✕ 10⁻⁵

Substituting all the given values in the above equation,  we get

[H+]= 0.014

The formula for calculating the pH of solution is given as

pH = -log [H+]

pH = 1.84

Therefore,  the pH of solution is 1.84.

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To determine the concentrations of the HCl and NaOH solutions, we can use the concept of titration.

To determine the concentrations of the HCl and NaOH solutions, we can use the concept of titration. In the first titration, 27.5 ml of NaOH was used to titrate 100.0 ml of HCl. From this, we can determine the concentration of HCl using the balanced chemical equation:

HCl + NaOH -> NaCl + H2O

By using the given information and the equation, we can determine that the concentration of HCl is 0.500 M. Similarly, in the second titration, 18.4 ml of NaOH was used to titrate 50.0 ml of 0.0782 M H2SO4. By using the balanced chemical equation:

NaOH + H2SO4 -> NaHSO4 + H2O

Therefore, the concentrations of the HCl and NaOH solutions are 0.500 M and 0.168 M, respectively.

12 moles of ammonia  can be made.

In chemistry, a mole, sometimes spelled mole, is a common scientific measurement unit for significant amounts of very small objects like atoms, molecules, or other predetermined particles.

Given that

1 mole N₂  required 2 mole NH₃

6 mole N₂ required  x mole NH₃

x=2×6/1= 12 mole.

Thus,12 moles of ammonia  can be made.

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Final answer:

The volume occupied by 15.6 g of H2O(g) at 36.2 degrees Celsius and 1.25 atm can be calculated using the Ideal Gas Law. The volume is approximately 18.13 liters.

Explanation:

To find the volume occupied by 15.6 g of H2O(g) at 36.2 degrees Celsius and 1.25 atm, we can use the Ideal Gas Law, which is PV = nRT. Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, we convert the mass of water to moles:

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

Now we can plug these values into the Ideal Gas Law:

So, the volume occupied by 15.6 g of H2O(g) at 36.2 degrees Celsius and 1.25 atm is approximately 18.13 liters.

A neutral salt is produced from the reaction between a strong acid and a strong base, as these completely dissociate in water and the protons from the acid neutralize the hydroxide ions from the base.

The reaction that will produce a neutral salt is the neutralization reaction between a strong acid and a strong base. When equal amounts of a strong acid like hydrochloric acid (HCl) are mixed with a strong base such as sodium hydroxide (NaOH), the products are a salt (NaCl in this case) and water (H2O), and they do not exhibit characteristics of either an acid or a base.

This is because strong acids and strong bases completely dissociate in water, giving a neutral solution as the protons (H+) from the acid neutralize the hydroxide ions (OH−) from the base, resulting in the formation of water.

The general form of a compound can be written in the form of:

[X(a) Y(b)] ^ c

Where a and b are subscripts, and c is the superscript of the whole formula.

The relationship that we can derived here between the constants is:

(valence of X) * a + (valence of Y) * b = c

Since the formula obviously has no superscript, therefore c = 0. We also know in chemistry class that the valence of Cl is -1, therefore:

valence of X * 1 + (-1) * 3 = 0

valence of X - 3 = 0

valence of X = 3

This is an example of combustion reaction, a substance is reacted with oxygen to form products. The balanced reaction for the burning of FeS2 is:

2 FeS2 + 5.5 O2 ---> 4 SO2 + Fe2O3

Therefore based on stoichiometric ratio of the reaction, for every 2 moles of FeS2, 4 moles of SO2 is produced.

First, lets calculate for the number of moles of FeS2 supplied: (molar mass of FeS2 = 119.965 g/mol)

n FeS2 = 202.33 g / (119.965 g/mol)

n FeS2 = 1.6866 mol

Calculate for the number of moles of SO2 produced using the ratio:

n SO2 = 1.6866 mol (4 / 2) = 3.373 mol SO2

Converting to mass: (molar mass SO2 = 64.066 g/mol)

m SO2 = 3.373 mol * 64.066 g/mol

m SO2 = 216.10 g         (ANSWER)

Final answer:

When 202.33 g of iron pyrite (FeS2) is burned, it reacts with oxygen to produce approximately 216.10 g of sulfur dioxide (SO2), which is a harmful air pollutant.

Explanation:

If iron pyrite (FeS2) is not removed from coal before burning, it can react with oxygen to form sulfur dioxide (SO2), which is a harmful air pollutant. To determine how much SO2 is produced from 202.33 g of FeS2 we can use stoichiometry. The balanced chemical equation for the reaction is:

4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2

From the equation, we see that 4 moles of FeS2 produce 8 moles of SO2. First, we calculate the moles of FeS2 in 202.33 g.

Given that the molar mass of FeS2 is approximately 119.98 g/mol (55.845 g/mol for Fe and 32.065 g/mol for S, multiplied by 2 because there are two sulfurs), we perform the following calculation:

202.33 g FeS2 × (1 mol FeS2 / 119.98 g FeS2) = 1.686 moles FeS2

Applying the stoichiometry from the balanced chemical equation:

1.686 moles FeS2 × (8 moles SO2 / 4 moles FeS2) = 3.372 moles SO2

Finally, we convert moles of SO2 to grams using its molar mass (approximately 64.066 g/mol):

3.372 moles SO2 × 64.066 g/mol = 216.10 g SO2

So, 202.33 g of FeS2 would produce approximately 216.10 g of SO2 when burned.

The balanced equation for the reaction between sodium carbonate and nickel(II) chloride is Na2CO3(aq) + NiCl2(aq) -> NiCO3(s) + 2NaCl(aq). This means one molecule of sodium carbonate reacts with one molecule of nickel(II) chloride to produce a molecule of nickel(II) carbonate and two molecules of sodium chloride.

The chemical reaction between aqueous sodium carbonate (Na2CO3) and aqueous nickel(II) chloride (NiCl2) can be represented and balanced as follows:

This balanced molecular equation indicates that one molecule of aqueous sodium carbonate reacts with one molecule of aqueous nickel(II) chloride to produce one molecule of solid nickel(II) carbonate and two molecules of aqueous sodium chloride.

In terms of phases, the sodium carbonate and nickel(II) chloride start as aqueous (dissolved in water) compounds, while the produced nickel(II) carbonate is solid, and the sodium chloride is still aqueous.

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To remedy overly acidic soil, use lime to raise pH, add organic matter like compost, and monitor pH levels regularly for optimal plant growth..

To remedy soil that is overly acidic (low pH), gardeners can take several steps to raise the pH to a more neutral or slightly acidic range, which is typically better for most plants.

Here’s how you can do it:

1. Test the Soil pH: Use a soil pH testing kit or send a sample to a lab to determine the current pH level accurately. This will help you gauge how much adjustment is needed.

2. Add Lime: Lime is the most common amendment used to raise soil pH. There are two main types:

Follow the application rates recommended based on your soil test results. Lime should be worked into the soil thoroughly for best results.

3. Use Wood Ash: Wood ash from hardwoods (like oak or maple) can also raise soil pH because it contains potassium carbonate. However, its use should be limited and monitored due to its high alkalinity and potential to raise pH too much if over-applied.

4. Add Organic Matter: Incorporating organic matter like compost, well-rotted manure, or peat moss can help buffer pH levels and improve soil structure over time. While organic matter alone won't drastically change pH, it can make the soil more hospitable to plants that prefer slightly acidic conditions.

5. Avoid Aluminum Sulfate and Ammonium-Based Fertilizers: These can lower pH levels, so they should be avoided in soils that are already acidic.

6. Monitor and Retest: After making amendments, retest the soil periodically (every few months or at least annually) to monitor pH levels and make adjustments as necessary.

7. Consider Plant Preferences: Some plants prefer acidic soils (e.g., blueberries, azaleas), so make sure the pH adjustments align with the needs of the plants you intend to grow.

By carefully applying these methods, gardeners can gradually raise the pH of overly acidic soil to create a more balanced environment for healthy plant growth.

We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows, 

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

12.0 M x V1 = 3.50 M x 75.0 mL

V1 = 21.88 mL

Therefore, we need about 21.88 mL of the 12.0 M of hydrochloric acid solution to make 75.0 mL of the 3.50 M hydrochloric acid solution.