Balance the following equation with the smallest whole number coefficients. Choose the answer that is the sum of the coefficients in the balanced equation. Do not forget coefficients of "one."
Cr2(SO4)3 + RbOH Cr(OH)3 + Rb2SO4

(a) 10
(b) 12
(c) 13
(d) 14
(e) 15

Answer:

B. It allows scientists to predict things about objects that are too small to see.

Explanation:

Hello,

In science, we're always looking for the study of tinier things as long as we're interested about what the matter is made of. Several atomic models such as Bohr's, Rutherford's, Dalton's and the actual one have been proposed in order for us to understand how the protons, neutrons and electrons are assembled into the atom because they are too small to see (about 0.5 fm) thus, with that better understanding, we can research and develop novel applications for new materials based on the micromolecular behavior of the atom.

Best regards.

Answer:

Molar concentration of ammonia gas is 0.0987 M.

Explanation:

Concentration of hydrogen gas = [tex][H_2]=0.15 M[/tex]

Concentration of nitrogen gas = [tex][N_2]=0.017 M[/tex]

Concentration of ammonia gas = [tex][NH_3]=x[/tex]

Equilibrium constant of the reaction = [tex]K_c=1.7\times 10^2[/tex]

[tex]K_c=\frac{[NH_3]^2}{[H_2]^3\times [N_2]}[/tex]

[tex]1.7\times 10^2=\frac{x^2}{(0.15 M)^3\times (0.017 M)}[/tex]

[tex]9.5737\times 10^{-3} M^2=x^2[/tex]

x = 0.0987 M

Molar concentration of ammonia gas is 0.0987 M.

The molar concentration of NH₃ is approximately 0.0987 M.

To find the molar concentration of NH₃, we use the equilibrium constant expression.

Step-by-Step Solution:

Write the equilibrium constant expression:

Kc = [NH₃]₂ / ([H₂]₃ × [N₂])

Insert the known values into the expression:

1.7 × 10² = [NH₃]₂ / (0.15)³ × 0.017

Solve for [NH₃]:

First, compute the denominator:

(0.15)³ = 0.003375 and 0.003375 × 0.017 = 5.7375 × 10⁻⁵

Now solve for [NH₃]₂:

[NH₃]₂ = 1.7 × 10² × 5.7375 × 10⁻⁵

[NH₃]₂ = 0.00975475

Finally, take the square root of both sides to find [NH₃]:

[NH₃] = √0.00975475

[NH₃] ≈ 0.0987 M

Therefore, the molar concentration of NH₃ is approximately 0.0987 M.

Final answer:

To identify the beaker showing HA behaving as a weak acid, we need to look for the beaker in which the concentration of H3O+ and A- is relatively low, indicating partial dissociation.

Explanation:

In the given question, we are asked to identify the beaker that shows the acid HA behaving as a weak acid in water.

A weak acid is one that only partially dissociates in water, creating a small amount of hydronium ions (H3O+) and the conjugate base (A-). Strong acids, on the other hand, completely dissociate into hydronium ions and the conjugate base.

To identify the beaker showing HA behaving as a weak acid, we need to look for the beaker in which the concentration of H3O+ and A- is relatively low, indicating partial dissociation.

The number of molecules of carbon dioxide present in 243.6 g of carbon dioxide is 3.33 × 10²⁴ molecules

From the question,

We are to determine the number of molecules of carbon dioxide that are in 243.6g of carbon dioxide

First, we will determine the number of moles of carbon dioxide present

Using the formula,

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

Molar mass of carbon dioxide = 44.01 g/mol

Number of moles of carbon dioxide = [tex]\frac{243.6}{44.01}[/tex]

Number of moles of carbon dioxide = 5.5351 moles

Now, for the number of molecules,

Using the formula,

Number of molecules = Number of moles × Avogadro's constant

Number of molecules = 5.5351 × 6.022 × 10²³

Number of molecules of carbon dioxide = 3.33 × 10²⁴ molecules

Hence, the number of molecules of carbon dioxide present in 243.6 g of carbon dioxide is 3.33 × 10²⁴ molecules

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Final answer:

When determining the percent acetic acid in vinegar, measuring the mass of vinegar is preferred over volume because mass is not affected by temperature and concentration changes. The mass can be used to calculate moles of acetic acid, which is then used to find molarity or mole fraction.

Explanation:

In the analysis of vinegar, it is often more accurate to measure the mass of the vinegar sample rather than the volume because mass is not affected by temperature or concentration variations as volume can be. When determining the percent acetic acid in vinegar, we can use the sample's mass to calculate the number of moles of acetic acid present. Let's say we have a sample where the concentration of acetic acid was previously found to be 0.839 Molarity (M). If we find that a certain volume of vinegar contains 75.6 g of acetic acid, we can use the molarity and the mass of acetic acid to determine the volume of the vinegar solution.

To calculate the mole fraction of acetic acid in the solution, the masses of both acetic acid and water in the sample are required. Using an example from LibreTexts, if 100.0 g of vinegar contains 3.78 g of acetic acid, then there are 96.2 g of water in the solution. From the masses, we determine the moles of acetic acid and water and then divide the number of moles of acetic acid by the total number of moles of substances in the sample to get the mole fraction.

Answer:

[tex]S_2O_3[/tex]

Explanation:

Hello,

Nomenclature, is a powerful tool in chemistry to name chemical compounds to successfully distinguish among the thousands and thousands of existing molecular compounds. In rule to stipulate a chemical name y is via specifying the amount of the composing elements into the molecule, thus, the prefix di, accounts for two atoms and the prefix tri for three (this is related with the valence electrons the element has), in such a way, disulfur trioxide is represented with two sulfur atoms and three oxygen atoms, this is:

[tex]S_2O_3[/tex]

Which would be a novel molecule since the sulfur has been reported to have +2,+4 and +6 as the positive oxidation states which are possible to dovetail with oxygen as it is negative.

Kind regards.

To find the number of moles of silver atoms in the silver ring, divide the number of particles by Avogadro's number.

To find the number of moles of silver atoms in the silver ring, we can use Avogadro's number. Avogadro's number tells us that 1 mole of any substance contains 6.022 x 10^23 particles. In this case, we are given that the silver ring contains 5.15 x 10^22 silver atoms. We can use the following equation:

Number of moles = Number of particles / Avogadro's number

Substituting the given values, we get:

Number of moles = 5.15 x 10^22 silver atoms / (6.022 x 10^23 atoms/mol) ≈ 0.0855 mol ≈ 0.086 mol (rounded to two significant figures)

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Answer: A.

Explanation:  it makes some inexpensive materials look more appealing

There are approximately 7.81x 10¹³ molecules of ghrelin in 1 liter of blood.

A fasting individual has a very low ghrelin content in their blood, around 1.3 x 10²⁰ moles per litre. Avogadro's number—the number of molecules per mole—can be used to compute ghrelin molecules per litre of blood.

We compute the number of molecules using the formula:

Concentration (mol/L) × Volume (L) × Avogadro's number = Number of molecules.

Change the values:

Number of molecules = [tex]1.3 *10^{-10}\ mol/L * 1 L * 6.022 *10^{23}\ mol/mol.[/tex]

In 1 litre of blood, roughly 7.81 x 10¹³ ghrelin molecules are produced. This exceedingly low value shows how sensitive biological systems are to even tiny quantities of appetite-regulating signalling chemicals like ghrelin.

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To calculate the number of ghrelin molecules in 1 L of blood with a concentration of 1.3 × 10^-10 M, multiply this concentration by Avogadro's number, resulting in approximately 78.3 × 10^13 molecules.

The concentration of the appetite-regulating hormone ghrelin is about 1.3 × 10^-10 M in the blood of a fasting person. To find out how many molecules of ghrelin are in 1 L of blood, we can use Avogadro's number, which is 6.022 × 10^23 molecules/mol. Multiplying the molarity of ghrelin by Avogadro's number gives us the total number of ghrelin molecules per liter of blood.

To calculate:

Therefore, there are approximately 78.3 × 10^13 molecules of ghrelin in 1 liter of blood.

To find the maximum mass of S8 produced, we first convert mass of the reactants to moles, then use stoichiometry to find the amount of S8 each can produce. The reactant that produces the least S8 is the limiting reactant. Finally, we convert the moles of S8 to grams.

The calculation for maximum mass of S8 that can be produced in the reaction 8SO2 + 16H2S = 3S8 + 16H2O, first requires understanding of how to use stoichiometry and limiting reactants concept. We start by converting the given mass of the reactants (87.0 g) to moles using their molar mass. For SO2, it's 64 g/mol and for H2S, it's 34 g/mol. We get 1.36 mol of SO2 and 2.56 mol of H2S.

Then, by using the stoichiometric coefficients present in the balanced equation, we'll find the amount of S8 that can be produced by each reactant. The limiting reactant is the one which produces least amount of product. In this case, it's SO2. Finally, we convert the moles of S8 to grams using its molar mass (256 g/mol).

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Final answer:

The ratio of the frequency of effective collisions for a reaction at higher to lower temperature can be found using the Arrhenius equation, which illustrates that the reaction rate increases with temperature due to more molecules having sufficient energy to overcome the activation energy barrier.

Explanation:

The question revolves around the concept known as the Arrhenius equation, which describes how the rate of a chemical reaction is affected by changes in temperature and activation energy. The ratio 'f' represents the frequency of effective collisions leading to a reaction. The Arrhenius equation suggests that as temperature increases, the rate constant of the reaction also increases because a greater fraction of molecules will have the necessary energy to overcome the activation energy barrier.

To calculate the ratio of the frequency of effective collisions at two different temperatures for a reaction with an activation energy (Ea) of 205 kJ/mol, we use the Arrhenius equation:

K = Ae-(Ea/RT)

Where K is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant (8.314 J/mol K), and T is the temperature in kelvins. By taking the ratio of the rate constants at 505 K and 485 K (K2/K1), we can find the ratio of 'f' at the higher temperature to the lower temperature.

The analysis of the change in reaction mechanism with temperature typically shows that, as seen with a reaction with a lower activation energy of 54 kJ/mol, a temperature increase, even by 10 degrees Celsius, can significantly impact the reaction rate, often doubling it. While the exact ratio of 'f' would require computation, the concept remains that the reaction rate will increase with temperature due to the exponential factor in the Arrhenius equation.

Correctly balancing a chemical equation involves adding coefficients to ensure the number of atoms for each element is equal on both sides. As the given equation has a typo, a general method for balancing equations and an example with sodium and chlorine was described.

The question asks about balancing a particular chemical equation to find the coefficient of NaI. However, the given equation seems to have a typographical error and is not complete. In the context of this question, let's address a general approach on how to balance a chemical equation:

Write down the unbalanced equation.

Determine the number of atoms of each element in the reactants and products.

Adjust coefficients (not subscripts) to get the same number of atoms of each element on both sides.

Repeat steps until all elements are balanced.

Check to ensure that the total charge is the same on both sides if the equation includes ionic species.

For example, the equation for the reaction of sodium (Na) with chlorine gas (Cl2) to form sodium chloride (NaCl) is balanced by placing a '2' in front of Na and NaCl:

2 Na (s) + Cl₂ (g) → 2 NaCl (s)

In this balanced equation, we see that the number of sodium and chlorine atoms are equal on both sides, confirming that the equation is correctly balanced.

oxygen (O) and chlorine (Cl)

As the electronegativity of an element increases It is more likely to make an ionic bond. e.g. the metallic elements. On the moderate values of electronegativity, elements tend to make covalent bond.

Oxygen and chlorine will make covalent bond because the electronegativity difference between these two elements is 0,5 so there bond will be covalent in nature.

Final answer:

To calculate the pH at different points in the titration of HCl with NaOH, you need to consider the moles of reactants and products and use the Henderson-Hasselbalch equation.

Explanation:

In a titration of 25.00 mL of 0.150 M HCl with 0.250 M NaOH, the pH can be calculated at different points:

(a) When neutralization is 50% complete, we can assume that half of the HCl has reacted with NaOH. This means that the moles of HCl neutralized is half of the initial moles present. Use this information to calculate the moles of NaOH consumed and the remaining HCl. From there, you can use the Henderson-Hasselbalch equation to calculate the pH.

(b) When 1.00 mL of NaOH is added beyond the equivalence point, you can assume that all of the HCl has been neutralized and there is excess NaOH. Calculate the moles of NaOH consumed based on the volume added and use it to determine the concentration of NaOH remaining. Then, use the concentration of NaOH and the hydroxide ion concentration to calculate the pH.

In the ionic compound KmNO₄, the name of the anion is permanganate. Therefore, option D is correct.

Ionic compounds are compounds that are composed of ions held together by electrostatic forces called ionic bonds. These compounds are typically formed between a metal cation and a nonmetal anion.

In an ionic compound, the metal cation donates one or more electrons to the nonmetal anion. It results in the formation of positively and negatively charged ions. The attraction between these opposite charges leads to the formation of a stable crystal lattice structure.

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Percent yield or yield is mathematically defined as:

Yield = Actual amount / Theoretical amount

So to solve the yield, let us first calculate the theoretical amount of POCl3 produced. The balanced chemical reaction for this is:

6 PCl5 + P4O10  --->  10 POCl3

Since P4O10 is stated to be supplied in large amount, then PCl5 becomes the limiting reactant.

So we calculate for POCl3 based on PCl5. To do this let us convert the amount into moles: (molar mass PCl5 = 208.24 g/mol)

n PCl5 = 42.66 grams / (208.24 g/mol)

n PCl5 = 0.205 mol

Now based on the stoichiometric ratio of the reaction:

n POCl3 = 0.205 mol (10 POCl3 / 6 PCl5)
n POCl3 = 0.3414 mol POCl3

Converting to mass (molar mass POCl3 = 153.33 g/mol)

m POCl3 = 0.3414 mol (153.33 g/mol)

m POCl3 = 52.35 g

Calculating for yield:

Yield = 47.22 g/ 52.35 g

Yield = 0.902

%Yield = 90.2 %                      (ANSWER)

The percent yield of POCl₃ is calculated by dividing the actual yield (47.22 grams) by the theoretical yield (52.36 grams), then multiplying by 100 to obtain a percent yield of 90.18%.

To calculate the percent yield of POCl₃, we first need the balanced chemical equation for the reaction between PCl₅ and P₄O₁₀. The equation is:

6 PCl₅ + P₄O₁₀ → 10 POCl₃

Next, we will convert the mass of PCl₅ to moles:

Molar mass of PCl₅ = 208.24 g/mol

Moles of PCl₅ = 42.66 g / 208.24 g/mol = 0.2049 moles

Using the stoichiometry of the balanced equation, we calculate the theoretical yield of POCl₃:

For every mole of PCl₅, (10/6) moles of POCl₃ are produced. Therefore, the moles of POCl₃ produced from 0.2049 moles of PCl₅ are:

Moles of POCl₃ = 0.2049 moles PCl₅ × (10/6) = 0.3415 moles

Molar mass of POCl₃ = 153.33 g/mol

Theoretical yield of POCl₃ = 0.3415 moles × 153.33 g/mol = 52.36 grams

Now, we can calculate the percent yield:

Percent yield = (Actual yield / Theoretical yield) × 100

Percent yield = (47.22 g / 52.36 g) × 100 = 90.18%

Therefore, the percent yield of POCl₃ is 90.18%.

The cell reaction that occurs is as follows:

 [tex]\boxed{{\text{Mg + C}}{{\text{u}}^{{\text{2 + }}}} \rightleftarrows {\text{M}}{{\text{g}}^{2 + }}{\text{ + Cu}}}[/tex]

The line notation of cell is as follows:

[tex]\left. {{\text{Mg}}\left( {\text{s}} \right)} \right|{\text{M}}{{\text{g}}^{2 + }}\left( {{a_{{\text{M}}{{\text{g}}^{2 + }}}}} \right)\left\| {{\text{C}}{{\text{u}}^{2 + }}\left( {{a_{{\text{C}}{{\text{u}}^{2 + }}}}} \right)\left| {{\text{Cu}}\left( {\text{s}} \right)} \right.} \right.[/tex]

The standard emf value of the cell is [tex]\boxed{2.7{\text{ V}}}[/tex].

Further Explanation:

Redox reaction:

It is a type of chemical reaction in which the oxidation states of atoms are changed. In this reaction, both reduction and oxidation are carried out at the same time. Such reactions are characterized by the transfer of electrons between the species involved in the reaction.

The general representation of a redox reaction is,

 [tex]{\text{X}} + {\text{Y}} \to {{\text{X}}^ + }+{{\text{Y}}^ - }[/tex]

The oxidation half-reaction can be written as:

[tex]{\text{X}} \to {{\text{X}}^ + } + {e^ - }[/tex]

The reduction half-reaction can be written as:

[tex]{\text{Y}} + {e^ - } \to {{\text{Y}}^ - }[/tex]  

Here, X is getting oxidized and its oxidation state changes from  to +1 whereas B is getting reduced and its oxidation state changes from 0 to -1.

The element which has higher oxidation potential is oxidized at anode and the element with the less oxidation potential is reduced at cathode in the cell.

The standard oxidation potential for [tex]{\text{Mg/M}}{{\text{g}}^{2 + }}[/tex] is  [tex]+ 2.363{\text{ V}}[/tex].

The standard oxidation potential for [tex]{\text{Cu/C}}{{\text{u}}^{2 + }}[/tex] is [tex]- 0.337{\text{ V}}[/tex].

Since [tex]{\text{Mg}}[/tex] has higher oxidation potential thus the oxidation of [tex]{\text{Mg}}[/tex] takes place at anode and reduction of [tex]{\text{C}}{{\text{u}}^{2 + }}[/tex] takes place at cathode.

The oxidation half-reaction of [tex]{\text{Mg/M}}{{\text{g}}^{2 + }}[/tex] can be written as:

[tex]{\text{Mg}} \to {\text{M}}{{\text{g}}^{2 + }} + 2{e^ - }[/tex]     ......(1)

The reduction half-reaction [tex]{\text{Cu/C}}{{\text{u}}^{2 + }}[/tex] can be written as:

[tex]{\text{C}}{{\text{u}}^{2 + }} + 2{e^ - } \to {\text{Cu}}[/tex]     ......(2)

Add reaction (1) and (2) and eliminate common terms to determine the net reaction for the given cell.

[tex]{\text{Mg + C}}{{\text{u}}^{{\text{2 + }}}} \rightleftarrows {\text{M}}{{\text{g}}^{2 + }}{\text{ + Cu}}[/tex]           ......(3)

The expression of the cell is written as follows:

[tex]\left. {{\text{Mg}}\left( {\text{s}} \right)} \right|{\text{M}}{{\text{g}}^{2 + }}\left( {{a_{{\text{M}}{{\text{g}}^{2 + }}}}} \right)\left\| {{\text{C}}{{\text{u}}^{2 + }}\left( {{a_{{\text{C}}{{\text{u}}^{2 + }}}}} \right)\left| {{\text{Cu}}\left( {\text{s}} \right)} \right.} \right.[/tex]                             ......(4)

The expression to calculate the standard emf of the cell is as follows:

[tex]E_{{\text{cell}}}^0 = E_{{\text{anode}}}^0 - E_{{\text{cathode}}}^0[/tex]                  ......(5)

Substitute [tex]+ 2.363{\text{ V}}[/tex] for [tex]E_{{\text{anode}}}^0[/tex] and [tex]- 0.337{\text{ V}}[/tex] for [tex]E_{{\text{cathode}}}^0[/tex] in equation (5).

[tex]\begin{aligned}E_{{\text{cell}}}^0&=\left({ + 2.363{\text{ V}}}\right)-\left( { - 0.337{\text{ V}}} \right)\\&= 2.7{\text{ V}}\\\end{aligned}[/tex]        

Learn more:

1. Which occurs during the redox reaction? https://brainly.com/question/1616320

2. Oxidation and reduction reaction: https://brainly.com/question/2973661

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Electrochemistry

Keywords: Mg/M2+, Cu/Cu2+, half-cell reaction, standard emf of the cell, oxidation state, reduction, oxidation, redox reaction, transfer of electrons, reducing agents, oxidizing agents.

To calculate the standard emf of a cell with Mg/Mg2+ and Cu/Cu2+ at 25 °C, subtract the anode potential (-2.37 V) from the cathode potential (+0.34 V) to get +2.71 V, indicating a spontaneous reaction. The line notation for the cell is Mg(s) | Mg2+(aq) || Cu2+(aq) | Cu(s).

To calculate the standard emf of a cell using the Mg/Mg2+ and Cu/Cu2+ half-cell reactions at 25 °C, we use the standard reduction potentials from the electrochemical series. The standard reduction potential for the Mg2+ to Mg half-cell is -2.37 V and for the Cu2+ to Cu half-cell is +0.34 V. Because oxidation occurs at the anode and reduction at the cathode, the Mg/Mg2+ cell will act as the anode and the Cu/Cu2+ cell as the cathode.

The overall cell reaction, under standard conditions, is derived by combining the half-reactions:

Mg(s) -> Mg2+(aq) + 2e- (Oxidation)

Cu2+(aq) + 2e- -> Cu(s) (Reduction)

When combined, the overall reaction is:

Mg(s) + Cu2+(aq) -> Mg2+(aq) + Cu(s)

The standard cell potential (Ecell) is calculated by taking the difference between the standard reduction potentials of the cathode and anode:

Ecell = Ecathode(red) - Eanode(ox) = 0.34 V - (-2.37 V) = +2.71 V

The positive value of Ecell indicates that the reaction is spontaneous under standard conditions. The line notation for the cell is:

Mg(s) | Mg2+(aq) || Cu2+(aq) | Cu(s)