Why do diastereomers have different chemical properties?

The reasonable transmittance values when a white light shines through \(CuSO_{4}\) solution is A. Red light: %T = 10% Green light: %T = 90%

If a solution appears purple then, B. It is transmitting purple light.

For a solution of CuSO_{4}, reasonable transmittance values would be:

A. Red light: %T = 10% Green light: %T = 90%

This is because the solution absorbs most of the red light and transmits most of the green light. Also, higher the absorbance of light, lower will be the % transmittance.

If a solution appears purple: B. It is transmitting purple light.

This is because the solution is reflecting and transmitting the purple light, which is why we perceive it as purple.

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To change a supersaturated lemonade solution to an unsaturated state, you need to dilute the solution by adding more water and mixing it well.

1. Identify the supersaturated solution: A supersaturated lemonade solution is one in which there is an excess of solute (such as sugar) dissolved in the solvent (such as water) beyond its saturation point.
2. Dilute the solution: To convert the supersaturated solution to an unsaturated state, you can add more solvent (in this case, water) to the solution. This will decrease the concentration of the solute (sugar) in the solution
3. Mix the solution thoroughly: After adding the solvent, mix the solution well to ensure that the solute and solvent are evenly distributed. This will result in an unsaturated solution where the solute concentration is below its saturation point.

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Answer:

what are you supposed to answer???

Explanation:

Total, 41.46 grams of sulfuric acid (H₂SO₄) would be required to react completely with 7.6 grams of aluminum (Al) in the given reaction.

Balanced chemical equation for the reaction between aluminum (Al) and sulfuric acid (H₂SO₄) is;

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

From the balanced chemical equation, we can see that 2 moles of aluminum (Al) react with 3 moles of sulfuric acid (H₂SO₄) to produce 1 mole of aluminum sulfate (Al₂(SO₄)₃) and 3 moles of hydrogen gas (H₂).

First, we need to calculate the molar mass of aluminum (Al) and sulfuric acid (H₂SO₄) from the periodic table;

Molar mass of Al = 26.98 g/mol

Molar mass of H₂SO₄ = 2 x (1.01 g/mol) + 32.07 g/mol + 4 x (16.00 g/mol) = 98.08 g/mol

Now, we can use the given mass of aluminum (7.6 g) and its molar mass (26.98 g/mol) to calculate the moles of aluminum;

moles of Al = mass of Al / molar mass of Al = 7.6 g / 26.98 g/mol ≈ 0.2817 moles

Since 2 moles of aluminum (Al) react with 3 moles of sulfuric acid (H₂SO₄), the number of moles of sulfuric acid (H₂SO₄) required is;

moles of H₂SO₄ = (3/2) x moles of Al = (3/2) x 0.2817 moles ≈ 0.4226 moles

Finally, we can calculate the mass of sulfuric acid (H₂SO₄) required using its molar mass (98.08 g/mol);

mass of H₂SO₄ = moles of H₂SO₄ x molar mass of H₂SO₄ = 0.4226 moles x 98.08 g/mol

≈ 41.46 g

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The main difference between a catalyzed and uncatalyzed reaction is that the activation energy of the catalyzed reaction will be lower. Option A is correct.

The main difference between a catalyzed and an uncatalyzed reaction is that a catalyst lowers the activation energy required for the reaction to occur, which increases the rate of the reaction. This means that in a catalyzed reaction, the reactants can more easily overcome the energy barrier required for the reaction to proceed, and therefore the activation energy is lower.

Option b) is not always true and depends on the specific reaction and the type of catalyst being used. A catalyst can make a reaction more favorable in terms of its rate, but not necessarily in terms of its free energy change.

Option c) and d) are not necessarily true and do not describe the main difference between catalyzed and uncatalyzed reactions. The enthalpy and entropy changes of a reaction depend on the specific reaction and the conditions in which it occurs, and the presence of a catalyst does not necessarily make the reaction more favorable in terms of enthalpy or entropy.

Hence, A. the activation energy of the catalyzed reaction is lower is the correct option.

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--The given question is incomplete, the complete question is

"The main difference between a catalyzed and an uncatalyzed reaction is that a) the activation energy of the catalyzed reaction is lower b) the catalyzed reaction has a more favorable free energy change c) the catalyzed reaction has a more favorable enthalpy change d) the catalyzed reaction has a more favorable entropy change."--

The acceleration can be defined as the change in velocity with respect to time. The velocity of the particle is 1200 m/s.

The movement of the particles is described with the rate of change in the position of the object with respect to time. The change in position, velocity, and acceleration is given by the Newton's kinematics equations.

The velocity of particle with respect to acceleration can be given as:

\(v=u+at\)

Where, the initial position rest has the initial velocity, \(u=0\rm \;m/s\)

The acceleration of the particle, \(a= 8. 0\;\rm m/s^2\)

The time for the final velocity, \(t=150\;\rm sec\)

Substituting the values in the equation for final velocity, v:

\(v= 0+ 8. 0\;\times\;150\;\rm m/s\\\textit v=1200\;m/s\)

The velocity of the particle after 150 seconds is 1200 m/s.

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Answer:

protons- 10

electrons- 10

neutrons- 10

Explanation:

no of protons/ electrons = atomic number

no of neutrons = atomic mass - atomic number

Answer: By the very laws of the universe, matter cannot be created or destroyed, the Big Bang cannot have happened by its own power. There was a creator involved.

Answer:

observation

Explanation: you observe something then create a question and a hypotesis

The first step in the Scientific Method is to make objective observations.

Hope this helps!

To calculate the flow rate of 2 L 05 NS (M) to infuse in 20 hr with a drop factor of 16 gtt/mL, we need to follow a few steps. Here's how to calculate the flow rate in gtt/min:First, we need to convert 2 L 05 NS to mL. 1 L = 1000 mL, so 2 L = 2000 mL. Since we have 50 mL left, we can add it up to get a total volume of 2050 mL.

Next, we need to calculate the total time in minutes, since the flow rate is in gtt/min. 20 hours = 20 × 60 = 1200 minutes.The formula for calculating the flow rate is: Flow rate (gtt/min) = Volume (mL) ÷ Time (min) ÷ Drop factor (gtt/mL)Now we can substitute the given values into the formula:

Flow rate (gtt/min) = 2050 mL ÷ 1200 min ÷ 16 gtt/mLFlow rate (gtt/min) = 0.10677 ≈ 0.11 gtt/min (rounded to 2 decimal places)Therefore, the flow rate of 2 L 05 NS (M) to infuse in 20 hr with a drop factor of 16 gtt/mL is approximately 0.11 gtt/min.

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Consider the reaction A(g) +B(g) → C(g) +D(g) for which AH° = +85.0 kJ and AS--66.0 J/K. You may assume that AH° and Asº do not change with temperature. This reaction is: nonspontaneous at low l but spontaneous at high T. The correct option is (c).

Considering the reaction A(g) + B(g) → C(g) + D(g) with ΔH° = +85.0 kJ and ΔS° = -66.0 J/K, we can use the Gibbs free energy equation to determine the spontaneity of the reaction at different temperatures. The Gibbs free energy equation is:
ΔG° = ΔH° - TΔS°
In this case, ΔH° is positive, and
ΔS° is negative.

To determine the spontaneity, we can analyze the signs of ΔG° at different temperatures:
a) At low T, the TΔS° term will be small and negative, making ΔG° positive, which indicates a nonspontaneous reaction.
b) At high T, the TΔS° term will be large and negative, potentially making ΔG° negative, which indicates a spontaneous reaction.

Thus, the reaction is nonspontaneous at low temperatures but spontaneous at high temperatures, which corresponds to option (c).

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The number of moles of NaClO that were added to the water are calculated to be 6.23 × \(10^{-11}\) mol.

In this case, we know that the pH of the solution is 10.50. Using the formula of pH, we can find the concentration of hydrogen ions.

The pH of a solution can be calculated using the following formula:

pH = -log[\(H^+\)]

where [\(H^+\)] is the concentration of hydrogen ions in the solution.

Plugging in the values.

10.50 = -log[\(H^+\)]

\([H^+]=10^{10.50}\)

\([H^+] =3.162 \times 10^{-11} mol/L\)

Now, let us find concentration of NaClO.

Since sodium hypochlorite is a salt, it dissociates in water to form ions.

The balanced equation for the dissociation of sodium hypochlorite is:

NaClO(s) → \(Na^+(aq) + ClO^-(aq)\)

The concentration of sodium hypochlorite can be calculated from the concentration of hypochlorite ions using the stoichiometry of the reaction.

Since there is a 1:1 ratio between NaClO and , the concentration of NaClO is also \(3.162 \times 10^{-11} mol/L\).

To find the number of moles of NaClO added to the solution, we need to multiply the concentration by the volume:

moles NaClO = concentration x volume

moles NaClO = \(3.162 \times 10^{-11} mol/L \times 2L\)

moles NaClO = 6.23 × \(10^{-11}\) mol.

Therefore, the number of moles of NaClO that were added to the water is 6.23 × \(10^{-11}\) mol..

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The life cycle of a star the size of our sun starts with a nebula, a huge cloud of dust and gas. The gravity of the nebula compresses the cloud until the atoms fuse together, creating a star. This stage is known as pre-birth.

Next, the star enters a main sequence phase, where it shines brightly for millions of years, powered by nuclear fusion.

Eventually the star will slowly run out of hydrogen and the fusion process will become less and less efficient. The star will expand and become what is known as a red giant. This marks the end of the main sequence phase.

After this point, the star expels its outer layers, leaving behind a small, hot core called a white dwarf. It will continue to cool down until it becomes a black dwarf.

If the star is at least 5 times bigger than our sun, it will enter a different life cycle. After the red giant phase, the star will undergo a supernova explosion, which will spew dust and gas into the surrounding areas. These remainders will later form new stars, planets, and other cosmic bodies. What's left of the star will become a neutron star or a black hole.

The size of the atoms a and the number of holes in the atoms can be used to decode the element.

A model is a miniature depiction of reality. Molecular models often consists of boxes that contain balls which are used to represent elements. In common parlance, the color is used to show the type of element.

Apart from the color, the size of the atoms a and the number of holes in the atoms can be used to decode the element.

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The amount of Silicon left after 300 years is 75g

It is given that the initial amount of Si is 100 times decay is 300 years and the half-life of Silicon is 710 years.

The radioactive decay formula is given by,

A = A₀ x 2^(-t/h);

where;

A is the resulting amount after t time, Ao is the initial amt (t=0),t is the time of decay, and h is the half-life of the substance.

On substituting the values from the given we get,

A = 100x2^(-300/710)

A = 100 x 0.746112347

A = 74.6112347 grams left after 300 yrs

Therefore, the number of grams of silicon left after 300 years is 74.6112347g. This value could be rounded off to 75 grams as in the whole number

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3990.86 moles of NH₃ (g) have been produced after the reaction has proceeded to completion.

The first step in solving this problem is to determine the limiting reactant. This is the reactant that will be completely used up in the reaction, thereby limiting the amount of product that can be formed. To do this, we need to calculate the number of moles of each reactant present in the vessel.

To find the number of moles of N₂ (g) present, we use the formula:

moles = mass / molar mass

The molar mass of N₂ is 28.02 g/mol. Therefore, the number of moles of N₂ present in the vessel is:

moles of N₂ = 56 kg / (28.02 g/mol) = 1995.43 mol

To find the number of moles of H₂ (g) present, we use the formula:

moles = mass / molar mass

The molar mass of H₂ is 2.02 g/mol. Therefore, the number of moles of H₂ present in the vessel is:

moles of H₂ = 9.0 kg / (2.02 g/mol) = 4455.45 mol

Now we need to determine which reactant is the limiting reactant. To do this, we compare the number of moles of each reactant to the stoichiometric ratio given in the balanced chemical equation. From the equation, we see that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.

Therefore, the maximum number of moles of NH₃ that can be produced is:

max moles of NH₃ = (1995.43 mol N₂) / 1 * (2 mol NH₃ / 1 mol N₂) = 3990.86 mol NH₃

max moles of NH₃ = (4455.45 mol H₂) / 3 * (2 mol NH₃ / 1 mol H₂) = 2970.30 mol NH₃

From this calculation, we see that the maximum number of moles of NH₃ that can be produced is limited by the amount of N₂ present in the vessel. Therefore, N₂ is the limiting reactant.

To find the actual number of moles of NH₃ produced, we use the stoichiometric ratio from the balanced chemical equation. We know that 1 mole of N₂ reacts with 2 moles of NH₃. Therefore, the number of moles of NH₃ produced is:

moles of NH₃ = (1995.43 mol N2) / 1 * (2 mol NH₃ / 1 mol N₂) = 3990.86 mol NH₃

Therefore, 3990.86 moles of NH₃ (g) have been produced after the reaction has proceeded to completion.

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When the energy of light emitted from an atom is 2.3x10 to the negative 15th joules, the wavelength of the light is 8.618 × 10^-8 m.

its wavelength can be calculated using the formula:

wavelength = hc/E

Let's discuss the terms used in the formula:

λ = wavelength = Planck's constant

= 6.626 × 10^-34 J.sc

= speed of light

= 2.998 × 10^8 m/s

E = Energy

Now, substitute the values in the formula:

wavelength = hc/E

wavelength = (6.626 × 10^-34 J.s) × (2.998 × 10^8 m/s) / (2.3 × 10^-15 J)

wavelength = 8.618 × 10^-8 m

Therefore, the wavelength of the light is 8.618 × 10^-8 m.

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Answer:

C. A linear, nonpolar molecule

Explanation:

Molecules which are alike usually have the same degree of pull which results in them sharing electrons. This sharing of electrons is known as the molecules exhibiting Covalent bonding between them.

The equal pull also results in the cancelling out of electrons and favoring non polar bonds due to the absence of free electrons which would have been able to interact with H2O in a polar binding system.

Answer:

C

Explanation:

APEX

Answer:

I believe that would be a liquid.

Explanation:

Explanation:

You can easily approximate the number of atoms from the mass of a sample because the mass number of an isotope approximately equals the mass of 1 mole of atoms in grams. It's a historical convention that would be too expensive or uncomfortable to change now.

Answer:

That's why chemists used large number of moles

This is your answer ☺️. If I'm right so,

Please mark me as brainliest. thanks!!!

An empirical formula presents the atoms of all the elements with the most simplified whole number ratios. In other words, divide each subscript by their greatest common factor.

The greatest common factor of 2 and 4 is 2.

2/2 = 1

4/2 = 2

So, the empirical formula for tetrachloroethene is CCl₂.

8. The number of molecules in 745 grams of C₁₄H₁₈N₂O₅ is 1.53×10²⁴ molecules

9. The mass of 5×10²⁵ molecules of Fe₂(SO₃)₃ is 29201.26 g

10. The mass of 6.30 moles of Ba₃(PO₄)₂ is 3786.3 g

The number of molecules present in 745 grams of C₁₄H₁₈N₂O₅ can be obtained as follow:

From Avogadro's hypothesis,

1 mole of C₁₄H₁₈N₂O₅ = 6.022×10²³ molecules

But

1 mole of C₁₄H₁₈N₂O₅ = 294 g

Therefore, we can say that

294 g of C₁₄H₁₈N₂O₅ = 6.022×10²³ molecules

Thus,

745 g of C₁₄H₁₈N₂O₅ = (745 × 6.022×10²³) / 294

745 g of C₁₄H₁₈N₂O₅ = 1.53×10²⁴ molecules

Thus, the number of molecules is 1.53×10²⁴ molecules

The mass of 5×10²⁵ molecules of Fe₂(SO₃)₃ can be obtained as follow:

6.022×10²³ molecules = 1 mole of Fe₂(SO₃)₃

But

1 mole of Fe₂(SO₃)₃ = 351.7

Therefore, we can say that

6.022×10²³ molecules = 351.7 g of Fe₂(SO₃)₃

Thus,

5×10²⁵ molecules = (5×10²⁵ × 351.7) / 6.022×10²³

5×10²⁵ molecules = 29201.26 g of Fe₂(SO₃)₃

Thus, we can conclude that the mass is 29201.26 g

The mass of 6.30 moles of Ba₃(PO₄)₂ can be obtained as follow:

Mole = mass / molar mass

6.3 = Mass of Ba₃(PO₄)₂ / 601

Cross multiply

Mass of Ba₃(PO₄)₂= 6.3 × 601

Mass of Ba₃(PO₄)₂= 3786.3 g

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An adjustable-volume micropipette with a range of 0.5-10 μL would be the most accurate for dispensing 125 μL of solution.

What is the micropipette ?

A micropipette is a precision instrument used to accurately measure and transfer very small volumes of liquid, typically between 0.5 µL and 10 mL. It is commonly used in laboratories to prepare samples for chemical analysis and in medical applications to dispense precise amounts of medication. The micropipette is composed of a plunger, a tip, and a cylinder. The plunger is used to draw liquid into the tip, and the cylinder is used to release the liquid. The micropipette is usually operated using a thumbwheel or a push button that controls the plunger. The tip of the micropipette is designed to fit a range of different sized microtubes, allowing for accurate and repeatable transfer of liquids. The micropipette can be calibrated for accuracy, making it an invaluable tool for laboratories that need precise measurements of liquids.

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When drawing the Lewis structure of a molecule, start by determining the total number of available valence electrons based on each element's group number. Then, use the total number of electrons needed for each element to be stable, generally based on its charge, to determine the ionic charge by finding the difference between the number of needed and available electrons divided by two.

For example, for a neutral oxygen atom in Group 6A or 16, it has six valence electrons. To achieve a stable octet, it needs two more electrons, which makes its ionic charge -2. Similarly, a nitrogen atom in Group 5A or 15 has five valence electrons, and it needs three more electrons to achieve a stable octet, which makes its ionic charge -3.

Once you have determined the ionic charges for each element in the molecule, you can start constructing the Lewis structure by placing the atoms in a way that satisfies the octet rule, where each atom (except hydrogen) has eight electrons in its outermost shell

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The p-value is greater than the significance level of 0.05, we do not reject the null hypothesis and conclude that all proportions are equal.

Firstly, let us conduct a Chi-square test of independence of categorical variables based on the information given above. We have three different cases of hypothesis testing that we have to solve one by one.

Case 1: HD​:pSun ​=rhoMan ​=pTue ​=pWed ​=pThu ​=pFri ​=pSat ​=71​
Ha​ : Not all proportions are equal.

Test Statistic
For this hypothesis, we need to compute the test statistic that is given as:
\($$\chi^2=\sum_{i=1}^{k}\frac{(O_i-E_i)^2}{E_i}$$\)  where k is the number of groups/categories. Since we have 7 days of the week, \(k = 7. $O_i$ and $E_i$\) are the observed and expected frequencies respectively.  
Here, we have equal proportions of 71 for each day of the week.
Therefore, the expected frequencies are also equal to 71.
\($$E_i = 71, i=1,2,3,4,5,6,7.$$\)

We also have to use the given information to compute the observed frequencies,
\($O_i$.$$O_1 = 90, O_2 = 99, O_3 = 122, O_4 = 123, O_5 = 130, O_6 = 160, O_7 = 126$$\)

Therefore, the test statistic can be computed as \($$\chi^2=\frac{(90-71)^2}{71} + \frac{(99-71)^2}{71} + \frac{(122-71)^2}{71} + \frac{(123-71)^2}{71} + \frac{(130-71)^2}{71} + \frac{(160-71)^2}{71} + \frac{(126-71)^2}{71}$$$$\chi^2=180.14\)

Now we have to find the p-value of this test. Since the number of degrees of freedom is k - 1 = 7 - 1 = 6, the p-value can be found using the chi-square distribution table with 6 degrees of freedom at the 0.05 significance level. The p-value is 0.000014. ConclusionSince the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that not all proportions are equal.

The total number of accidents is \($$90+99+122+123+130+160+126=850$$\)

The percentage of accidents occurring on each day of the week can be found as follows:
\($$Sunday: $$\frac{90}{850}\times 100 = 10.59\%$$Monday: $$\frac{99}{850}\times 100 = 11.65\%$$Tuesday: $$\frac{122}{850}\times 100 = 14.35\%$$Wednesday: $$\frac{123}{850}\times 100 = 14.47\%$$Thursday: $$\frac{130}{850}\times 100 = 15.29\%$$Friday: $$\frac{160}{850}\times 100 = 18.82\%$$Saturday: $$\frac{126}{850}\times 100 = 14.82\%$$\)

From the above percentages, we can see that Friday has the highest percentage of traffic accidents.

Case 2:
HD​: Not all proportions are equal.

Ha​:pSun ​=pMon ​=pTue ​=rhoWed ​=pThu ​=rhoFri ​=rhoSat ​=71

Test Statistic

\($$E_1 = 78.57, E_2 = 86.57, E_3 = 106.86, E_4 = 107.43, E_5 = 113.57, E_6 = 139.43, E_7 = 109.14$$\)

We already know the observed frequencies,
\($$O_1 = 90, O_2 = 99, O_3 = 122, O_4 = 123, O_5 = 130, O_6 = 160, O_7 = 126.$$\)

The test statistic can be computed as:

\($$\chi^2=\frac{(90-78.57)^2}{78.57} + \frac{(99-86.57)^2}{86.57} + \frac{(122-106.86)^2}{106.86}+ \frac{(123-107.43)^2}{107.43} + \frac{(130-113.57)^2}{113.57} + \frac{(160-139.43)^2}{139.43} + \frac{(126-109.14)^2}{109.14} $$$$ \implies \chi^2=34.98$$\)
The p-value is 0.000001.
Conclusion- Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that not all proportions are equal.

Case 3:

All proportions are equal.
Test Statistic
The expected frequency for each group is
\(E = \frac{850}{7} = 121.43\)
We already know the observed frequencies,
\($$O_1 = 90, O_2 = 99, O_3 = 122, O_4 = 123, O_5 = 130, O_6 = 160, O_7 = 126.$$\)

The test statistic is,

\($$\chi^2=\frac{(90-121.43)^2}{121.43} + \frac{(99-121.43)^2}{121.43} + \frac{(122-121.43)^2}{121.43} + \frac{(123-121.43)^2}{121.43} + \frac{(130-121.43)^2}{121.43} + \frac{(160-121.43)^2}{121.43} + \frac{(126-121.43)^2}{121.43}} \\\implies \chi^2=9.17$$\)

The p-value is 0.1664.

Since the p-value is greater than the significance level of 0.05, we do not reject the null hypothesis and conclude that all proportions are equal.

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If more \(H_{2}\) and \(N_{2}\) are added to the reaction 3\(H_{2}\) + N2 → 2\(NH_{3}\) after it reaches equilibrium, two events will occur Shift in Equilibrium and Increased Yield of \(NH_{3}\)

1. Shift in Equilibrium: According to Le Chatelier's principle, when additional reactants are added, the equilibrium will shift in the forward direction to consume the added reactants and establish a new equilibrium. In this case, more \(NH_{3}\) will be produced to counteract the increase in \(H_{2}\) and \(N_{2}\).

2. Increased Yield of \(NH_{3}\): The shift in equilibrium towards the forward reaction will result in an increased yield of \(NH_{3}\). As more \(H_{2}\) and \(N_{2}\) are added, the reaction will favor the production of \(NH_{3}\) to maintain equilibrium. This will lead to an increase in the concentration of \(NH_{3}\) compared to the initial equilibrium state.

It is important to note that the equilibrium position will ultimately depend on factors such as the concentrations of \(H_{2}\), \(N_{2}\), and \(NH_{3}\), as well as the temperature and pressure of the system. By adding more reactants, the equilibrium will adjust to achieve a new balance, favoring the formation of more \(NH_{3}\).

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