If δh°rxn and δs°rxn are both positive values, what drives the spontaneous (favored) reaction and in what direction at standard conditions?

Answer: The balanced chemical equation is written above.

Explanation:

Balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side. These equations follow law of conservation of mass.

When ferrous sulfate reacts with lead (II) nitrate, leads to the production of iron (II) nitrate and lead (II) sulfate.

The chemical equation for the above reaction follows:

[tex]FeSO_4(aq.)+Pb(NO_3)_2(aq.)\rightarrow Fe(NO_3)_2(aq.)+PbSO_4(s)[/tex]

By Stoichiometry of the reaction:

1 mole of aqueous solution of ferrous sulfate reacts with 1 mole of aqueous solution of lead (II) nitrate to produce 1 mole of aqueous solution of iron (II) nitrate and 1 mole of solid lead (II) sulfate.

Hence, the balanced chemical equation is written above.

In the metric system, a millimeter measures length. The word "milli" is derived from the Latin word "mille," which means one thousandth. A millimeter is therefore one-thousandth of a meter. A millimeter is represented by the letter "mm." Here 261 nm is 0.000261 millimeters.

A nanometer is a length measurement that is one billionth of a meter. The nanometer is denoted by the sign "nm" in the international system of units, sometimes known as SI units. One nanometer can be represented as 1 x 10⁻⁹ meters in scientific notation.

Here,

1 mm = 1000000 nm

261 nm = 0.000261 mm

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0.000261 millimeters

The answer is D.

10 mm2

The molality of solution formed by dissolving 34.8 g of LiI in 500 mL of water is [tex]\boxed{0.520{\text{ m}}}[/tex].

Further Explanation:

Concentration terms are used to determine concentration of various components present in any mixture. Some of these are mentioned below.

1. Molarity (M)

2. Mole fraction (X)

3. Molality (m)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

7. Parts per billion (ppb)

Molality is defined as moles of solute that are present per kilograms of solvent. It is represented by m and its unit is mol/kg. The expression for molality of solution is as follows:

[tex]{\text{Molarity of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Mass }}\left( {{\text{kg}}} \right){\text{ of solvent}}}}[/tex]  

Given information:

Mass of LiI: 34.8 g

Volume of water: 500.0 mL

To calculate:

Molality of LiI solution

How to proceed:

Step1: Moles of LiI that are present in 34.8 g of LiI has to be determined. This is done with the help of equation (1).

The formula to calculate the moles of LiI is as follows:

[tex]{\text{Moles of LiI}} = \dfrac{{{\text{Mass of LiI}}}}{{{\text{Molar mass of LiI}}}}[/tex]                                            …… (1)

The mass of LiI is 34.8 g.

The molar mass of LiI is 133.841 g/mol.

Substitute these values in equation (1).

[tex]\begin{aligned}{\text{Moles of LiI}} &= \frac{{{\text{34}}{\text{.8 g}}}}{{{\text{133}}{\text{.841 g/mol}}}} \\&= 0.26{\text{ mol}} \\\end{aligned}[/tex]  

Step 2: Mass of water is to be evaluated.

At standard temperature and pressure conditions, one liter of water can be considered equivalent to 1 kilograms of water.

Therefore mass of water can be calculated as follows:

[tex]\begin{aligned}{\text{Mass of water}} &= \left( {500.0{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)\left( {\frac{{1{\text{ kg}}}}{{1{\text{ L}}}}} \right) \\&= 0.5{\text{ kg}} \\\end{aligned}[/tex]  

Step 3: Molality of LiI solution is to be calculated. This is done with the help of equation (2).

The formula to calculate molality of LiI solution is as follows:

[tex]{\text{Molality of LiI solution}} = \dfrac{{{\text{Moles of LiI}}}}{{{\text{Mass of water}}}}[/tex]                                                …… (2)

Substitute 0.26mol for moles of LiI and 0.5 kg for mass of water in equation (2).

[tex]\begin{aligned}{\text{Molality of LiI solution}} &= \frac{{{\text{0}}{\text{.26 mol}}}}{{{\text{0}}{\text{.5 kg}}}} \\&= 0.520{\text{ m}} \\\end{aligned}[/tex]  

Hence, molality of given solution comes out to be 0.520 m.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: concentration, concentration terms, solutions, molarity, molality, LiI, moles, mass, molar mass, 0.520 m, 0.5 kg, 0.26 mol.

equation is

pH=14+log(.2)

There are 3 atoms in total. Two come from Hydrogen hence the H2. The third comes from the single Oxygen (O). Hope this helps!

Three atoms make up the water molecule (H2O) as a whole.

According to the context, the term may or may not include ions that meet this requirement. A molecule is a collection of two or more atoms held together by attractive forces known as chemical bonds.

The same atoms can combine in various ratios to create various molecules. For instance, water (H2O) is made up of two hydrogen atoms and one oxygen atom, whereas hydrogen peroxide (H2O2) is made up of two hydrogen atoms and two oxygen atoms.

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When potassium hypochlorite is dissolved in water, it dissociates into potassium ions and hypochlorite ions. Thus, the net ionic equation for this process is: KClO(s) -> K+(aq) + ClO-(aq).

The net ionic equation represents the actual reaction happening in solution, excluding the spectator ions. When potassium hypochlorite (KClO) is dissolved in water, it dissociates completely into potassium ions (K+) and hypochlorite ions (ClO-).

So, the complete ionic equation is: KClO(s) -> K+(aq) + ClO-(aq).

The net ionic equation is the same as the complete ionic equation because there are no spectator ions in this case. Thus, the net ionic equation for the equilibrium that is established when potassium hypochlorite is dissolved in water is: KClO(s) -> K+(aq) + ClO-(aq)

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The net ionic equation for the equilibrium established when potassium hypochlorite dissolves in water involves only the hypochlorite ion reacting with water to form hypochlorous acid and hydroxide ion.

When potassium hypochlorite (KClO) is dissolved in water, it dissociates into potassium (K+) ions and hypochlorite (ClO-) ions.

Potassium is a spectator ion and does not participate in the equilibrium reaction that is established in the solution. Therefore, the net ionic equation for the equilibrium when potassium hypochlorite is dissolved in water only involves the hypochlorite ion and water.

The equation is as follows:

ClO-(aq) + H2O(l) ---> HClO(aq) + OH-(aq)

This equation represents the equilibrium between the hypochlorite ion and the hypochlorous acid and hydroxide ion in aqueous solution.

In the formation of ammonia, different initial amounts of reactants result in different Kc values at equilibrium. While the temperature is constant, varying concentrations of reactants and products lead to different equilibrium constants Kc.

The chemical equation is given by 2NH3(g) = N2(g) + 3H2(g). In the first scenario, you start with 1.30 mol of N2 and 1.65 mol of H2 in a 1.00 L container, with 0.100 mol of NH3 at equilibrium. This implies a decrease in N2 and H2 by x amount and an increase in NH3 by 2x. Given the stoichiometric relationships, we can calculate that the equilibrium concentrations of N2 and H2 are 1.20 M and 1.30 M respectively.

Subsequently, we can calculate Kc using the equilibrium concentrations, Kc = [NH3]^2/(N2][H2]^3) = (0.10)^2/(1.20 * (1.55)^3) = 422 M^-2.

In the second scenario, the same calculations result in a Kc = 20.5 M^-2. The discrepancy in the Kc values arises due to the differences in the initial amounts of reactants and the products formed. Although the temperature is constant and the reaction is the same, the amount of reactants used and the products formed vary which impacts the Kc value.

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Final answer:

The presence of an insoluble impurity such as sand in a compound causes a lower observed melting point and a broader melting range, known as melting point depression. The impurity disrupts the crystal lattice, lowering the energy required for the substance to melt, and physically hinders a uniform melting process.

Explanation:

Effect of Insoluble Impurities on Melting Point

The presence of an insoluble impurity, such as sand, in a compound can alter its melting point. Typically, the observed melting point of a compound with impurities is lowered and the range over which melting occurs is broadened. This phenomenon is known as melting point depression, which is akin to freezing point depression. The introduction of an insoluble impurity disrupts the orderly crystal lattice of a pure substance, thereby requiring less energy to break the intermolecular forces among the molecules when heat is applied. Consequently, the substance starts to melt at a lower temperature.

When a compound undergoes melting, impurities like sand do not integrate into the crystal lattice. Instead, they remain as separate entities. As the majority component (for instance, a pure chemical) begins to melt, it may form small pools of liquid, which do not contain the impurity. Since the sand is insoluble, it doesn't contribute to the solution phase and hence doesn't affect the liquid's composition or properties directly. However, the presence of the sand broadens the range over which the compound melts because it can physically hinder the melting process. This leads to a melting range rather than a sharp melting point.

In practice, the presence of impurities can be identified by a melting point that is lower than expected for the pure compound and by a broader range of temperatures over which melting occurs, which is indicative of a less pure substance. Such a broad range is due to the impure solid melting at various temperatures, influenced by the amount and distribution of the impurity.

Answer:

The reaction will proceed to the right (favoring the products).

Explanation:

Let's consider the following isomerization reaction.

cis-2-butene ⇌ trans-2-butene

To predict in what direction will shift to reach equilibrium, we have to calculate the reaction quotient (Qp).

[tex]Qp=\frac{p(trans-2-butene)}{p(cis-2-butene)} =\frac{1.00}{1.00} =1.00[/tex]

Since Qp (1.00) < Kp (3.40), the reaction will proceed to the right, so that the pressure of the product increases, the pressure of the reactant decreases, and Qp reaches the value of Kp.

0.300 M IKI represents the concentration which is in molarity of a potassium iodide solution. This means that for every liter of solution there are 0.300 moles of potassium iodide. Knowing that molarity is a ratio of solute to solution.

By using a conversion factor:

100 ml x (1L / 1000 mL) x (0.300 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g

Therefore, in the first conversion by simply converting the unit of volume to liter, Molarity is in L where the volume is in liters. The next step is converted in moles from volume by using molarity as a conversion factor which is similar to how density can be used to convert between volume and mass. After converting to moles it is simply used as molar mass of Kl which is obtained from periodic table to convert from mole to grams.

In order to get the grams of IKI to create a 100 mL solution of 0.600 M IKI, use the same formula as above:

100 ml x (1L / 1000 mL) x (0.600 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g

Final answer:

To obtain a 100 mL solution of 0.300 M KI, it would take approximately 4.98 grams of KI. To create a 100 mL solution of 0.600 M KI, it would take approximately 9.96 grams of KI.

Explanation:

To determine how many grams of KI are needed to create a 100 mL solution of 0.300 M KI, we can use the equation:

Molarity (M) = moles of solute / volume of solution (L)

First, convert the volume from mL to L: 100 mL = 0.100 L

Then, rearrange the equation to solve for moles of solute:

moles of solute = Molarity x volume of solution

moles of solute = 0.300 M x 0.100 L

moles of solute = 0.030 mol KI

Finally, use the molar mass of KI (166 g/mol) to convert moles to grams:

grams of solute = moles of solute x molar mass

grams of solute = 0.030 mol KI x 166 g/mol

grams of solute = 4.98 g KI

Therefore, it would take approximately 4.98 grams of KI to obtain a 100 mL solution of 0.300 M KI.

Similarly, to create a 100 mL solution of 0.600 M KI, the calculation would be:

moles of solute = 0.600 M x 0.100 L

moles of solute = 0.060 mol KI

grams of solute = 0.060 mol KI x 166 g/mol

grams of solute = 9.96 g KI

It would take approximately 9.96 grams of KI to create a 100 mL solution of 0.600 M KI.

The root-mean square speed of the CO₂ molecules is 339 m/s.

The given parameters;

The molecular mass of the CO₂ = (12 + 16x2) = 44 g = 0.044 kg

The average mass of the CO₂ molecule is calculated as;

[tex]average \ mass \ = \frac{molecular \ mass}{Avogadro's \ number} = \frac{0.044}{6.02 \times 10^{23}} = 7.31 \times 10^{-26} \ kg[/tex]

The root-mean square speed is calculated by applying kinetic energy equation;

[tex]E = \frac{1}{2} mv^2\\\\v^2 = \frac{2E}{m} \\\\v= \sqrt{\frac{2E}{m}} \\\\v = \sqrt{\frac{2(4.2 \times 10^{-21})}{7.31\times 10^{-26}}}\\\\v = 338.99 \ m/s \ \approx 339 \ m/s[/tex]

Thus, the root-mean square speed is 339 m/s.

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The molecular formula of the compound given the percentages and conditions is determined to be N₂O₄ by converting mass percentages to moles, using the ideal gas law to find molar mass, and then refining the empirical formula.

Finding the Molecular Formula:

To determine the molecular formula of the compound with 30.45% nitrogen (N) and 69.55% oxygen (O) by mass, we will follow several steps:

The molecular formula of the compound is N₂O₄.

Answer:

Carbon and carbon tetrahydride (methane).

Explanation:

Hello,

In this case, it is not necessary to compute the particles for all the given masses, it is enough by knowing each substance's moles by knowing their molar mass and subsequently proof they equals 1 mole as long as 1 mole equals 6.022x10²³ which is the Avogadro's number; in such a way, the molar masses are:

[tex]M_C=12.01g/mol\\M_{SiO_2}=60.08g/mol\\M_{O_3}=48g/mol\\M_{CH_4}=16.04g/mol[/tex]

Therefore, the two cases are carbon and carbon tetrahydride or methane as shown below:

[tex]ParticlesC=12.01gC*\frac{1molC}{12.01gC}*\frac{6.022x10^{23}particlesC}{1molC} =6.022x10^{23}particlesC\\ParticlesCH_4=12.01gCH_4*\frac{1molCH_4}{12.01gCH_4}*\frac{6.022x10^{23}particlesCH_4}{1molCH_4}=6.022x10^{23}particlesCH_4[/tex]

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Final answer:

Nothing happens when a piece of copper is placed in 1M hydrochloric acid because copper is not reactive enough to displace hydrogen from the acid and no observable chemical reaction occurs.

Explanation:

When a piece of copper is placed in 1M hydrochloric acid (HCl), the correct option would be d. nothing happens. Copper is a less reactive metal and does not react with hydrochloric acid to produce hydrogen gas or any other substances. Reactions such as those with zinc in hydrochloric acid form hydrogen gas because zinc is a more active metal and can displace hydrogen from the acid. However, copper does not have sufficient reactivity to do this, and so when copper is placed into hydrochloric acid, there is no observable chemical reaction.

The United States Food and Drug Administration (FDA) publishes the food code, providing guidelines and regulations for safe food handling and production.

The agency that publishes the food code is the United States Food and Drug Administration (FDA). This agency issues codes that regulate the production, handling, and sale of food in the United States. Details about harmful microorganisms found in food and related aspects are also released by the FDA, as shown in their 'Bad Bug Book'.

The FDA's food codes serve as a model and guide for strengthening food safety by detailing specific procedures and practices required to prevent foodborne illnesses. Different from USDA, which conducts research and provides data on food and nutrition, the FDA directly oversees the food safety and integrity. Thus, when it comes to food code, the FDA should be the top-priority source.

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The mass of [tex]\(\text{As-76}\)[/tex] remaining after 538 minutes is approximately [tex]\( 270.38 \text{ g} \).[/tex]

To determine the remaining mass of [tex]\(\text{As-76}\)[/tex] after 538 minutes given its half-life of 26.0 hours, we can use the concept of radioactive decay.

First, convert the given time from minutes to hours:

[tex]\[ 538 \text{ minutes} \times \frac{1 \text{ hour}}{60 \text{ minutes}} = 8.97 \text{ hours} \][/tex]

Next, we use the formula for radioactive decay:

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

Plug in the values:

Calculate the fraction of the substance remaining after 8.97 hours:

[tex]\[ N(t) = 344 \left( \frac{1}{2} \right)^{\frac{8.97}{26.0}} \][/tex]

First, compute the exponent:

[tex]\[ \frac{8.97}{26.0} \approx 0.344 \][/tex]

Now calculate the remaining mass:

[tex]\[ N(t) = 344 \left( \frac{1}{2} \right)^{0.344} \][/tex]

[tex]\[ N(t) = 344 \times 0.786 \][/tex]

[tex]\[ N(t) \approx 270.38 \text{ g} \][/tex]

So, the mass of [tex]\(\text{As-76}\)[/tex] remaining after 538 minutes is approximately [tex]\( 270.38 \text{ g} \).[/tex]

To calculate the pH of a 0.060 M carbonic acid solution, first, consider the first dissociation of the H2CO3. Calculate [H+] by taking the square root of Ka1 x [H2CO3]. Finally, express [H+] in terms of pH. The calculated pH is 3.69.

To calculate the pH of a 0.060 M carbonic acid solution, we have to consider the stepwise dissociation of the carbonic acid, denoted by its dissociation constants Ka1 and Ka2. To see how this works, let's look at the first dissociation of H2CO3:

Ka1 = [H+][HCO3-]/[H2CO3]

Given, [H2CO3] = 0.060 M and Ka1 = 4.3 × 10-7, we can solve for [H+]. According to the ICE (Initial, Change, Equilibrium) approach, we approximate [H+] by the square root of Ka1 x [H2CO3]. Therefore, [H+] = √(4.3 × 10^-7 × 0.060) = 2.04 x 10^-4. We may ignore the second dissociation (Ka2) as it contributes negligibly to [H+].

Finally, we express [H+] in terms of pH. pH is the negative log to the base 10 of [H+], hence, pH = -log[H+] = -log(2.04 x 10^-4) = 3.69

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The midpoint of AB is (-5,1), if A(-3,8) and B(-7,-6).

The midpoint of coordinates is found by measuring the distance between two endpoints and dividing the obtained result by 2. Apart from this, one more method is there which is to add the two X-coordinates of the endpoints and divide them by 2. The same concept is applied to the Y-coordinates as well.

According to the question,

A = (-3,8) and B = (-7,-6).

The formula is as follows:

M(x,y) = ((x₁+x₂)/2 , (y₁+y₂)/2).

Putting the above values in the mentioned formula, we get:

M(x,y) = ((-3-7)/2 , (8-6)/2).

M(x,y) = (-5 , 1).

Therefore, the midpoint of AB is (-5,1), if A(-3,8) and B(-7,-6).

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Note: M1 = 50.00 ml

          V1 = 0.150 M

         M2 = 0.100 M

Asked: V2?

Answer: First, first realize the reaction:

               NaOH + CH3COOH = Na (+) + CH3COO (-) + H2O

              Second, enter all known numbers into the molarity formula:

                M = n / V

                 n = M * V

                M1 * V1 = M2 * V2

                V2 = M1 * V1 / M2

                V2 = 50.00 ml * 0.150 M / 0.100 M

                V2 = 75 ml

So, the NaOH needed to neutralize 50.00 ml of a solution of 0.1150 m of acetic acid (ch 3 co) is 75 ml.

In chemistry, molarity (abbreviated M) is one measure of the concentration of the solution. The molarity of a solution expresses the number of moles of a substance per liter of solution. For example, 1.0 liter of solution contains 0.5 mol of compound X, so this solution is called a 0.5 molar (0.5 M) solution. Generally, the concentration of aqueous aqueous solutions is expressed in molar units. The advantage of using molar units is the ease of calculation in stoichiometry because the concentration is expressed in moles (proportional to the actual number of particles). The disadvantage of using this unit is inaccuracy in volume measurement. Also, the volume of a liquid changes with temperature, so the molarity of the solution can change without adding or reducing any substances. Also, in a solution that is not very thin, the molar volume of the substance itself is a function of concentration, so the molarity-concentration relationship is not linear.

A neutralization reaction is a reaction where acids and bases react in aqueous solution to produce salt and water. The liquid sodium chloride that is produced in a reaction is called salt. Salt is an ionic compound consisting of cations from bases and anions from acids. Salt is an ionic compound that is not an acid or a base.

Strong-base Strong Acid Reaction

When the same amount of strong acid such as hydrochloric acid is mixed with a strong base such as sodium hydroxide, the result is a neutral solution. The reaction product does not have the characteristics of either acid or base.

Reactions Involving Weak Acids or Weak Bases

Reactions where at least one component is weak generally do not produce a neutral solution.

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Grade: College

Subject: Chemistry

keywords: molarity