A 6.53 g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces 1.73 L of carbon dioxide gas at 26 ∘C and 745 torr pressure.
a)Calculate the total number of moles of carbon dioxide that forms from these reactions.
Express your answer using three significant figures.
b) Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

Answer:

5 pumpkin pies could be made with 7.5 cups of sugar.

Explanation:

The conversion factor to solve the problem is:

[tex]7.5 cups of sugar\times\frac{1 pie}{1.5 cups of sugar} = 5 pies[/tex]

Answer:

1.877 x 10⁷ kg

Explanation:

To solve this problem we first use the equation:

Where Q is the heat needed to increase the temperature of a substance, m is the mass, c is the specific heat and ΔT is the difference in temperature.

For this problem:

We put the data in the equation and solve for Q:

So that's the energy required to heat 1000 kg of water, now we calculate the mass of wood using the equation:

Where b is the heating value of wood and m its mass:

Final answer:

A positive ion, or cation, has more protons than electrons, while a negative ion, or anion, has more electrons than protons, resulting in their net electrical charges.

Explanation:

The difference between a positive and negative ion lies in the balance between protons and electrons within an atom or molecule. When an atom has more protons than electrons, it becomes a positive ion or, as it's scientifically named, a cation. Conversely, an atom becomes a negative ion or anion if it gains extra electrons, surpassing the number of protons. This difference in the number of electrons leads to a net electrical charge, which determines whether the ion is positive or negative.

DID YOU KNOW? The names for these ions are pronounced "CAT-eye-ons" (cations) and "ANN-eye-ons" (anions), reflecting their respective charges.

Answer : The solubility of [tex]Al(OH)_3[/tex] is [tex]1.0\times 10^{-27}M[/tex]

Explanation :

First we have to calculate the pOH.

[tex]pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-12\\\\pOH=2[/tex]

Now we have to calculate the concentration of [tex]OH^-[/tex].

[tex]pOH=-\log [OH^-][/tex]

[tex]2=-\log [OH^-][/tex]

[tex][OH^-]=0.01[/tex]

The solubility equilibrium reaction will be:

[tex]Al(OH)_3\rightleftharpoons Al^{3+}+3OH^{-}[/tex]

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=[Al^{3+}][OH^{-}]^3[/tex]

Now put all the given values in this expression, we get:

[tex]1.0\times 10^{-33}=[Al^{3+}]\times (0.01)^3[/tex]

[tex][Al^{3+}]=1.0\times 10^{-27}M[/tex]

As, the solubility of [tex]Al(OH)_3[/tex] = [tex][Al^{3+}][/tex] = [tex]1.0\times 10^{-27}M[/tex]

Thus, the solubility of [tex]Al(OH)_3[/tex] is [tex]1.0\times 10^{-27}M[/tex]

Answer:

599.26 grams of potassium sulfate will be produced.

Explanation:

[tex]Cr_2(SO_4)_3(aq)+2K_3PO_4(aq)\rightarrow 2CrPO_4(s)+3K_2SO_4(aq)[/tex]

Moles of chromium (III) sulfate = [tex]\frac{450 g}{392 g/mol}=1.1480 mol[/tex]

According to reaction, 1 mole of chromium (III) sulfate gives 3 moles of potassium sulfate.

Then 1.1480 moles of chromium (III) sulfate will give:

[tex]\frac{3}{1}\times 1.1480 mol=3.4440 mol[/tex]

Mass of 3.4440 moles of potassium sulfate:

= 3.4440 mol × 174 g/mol = 599.26 g

599.26 grams of potassium sulfate will be produced.

The question seeks to determine the amount of heat released during the decomposition of 57.0 g of Fe2O3. After converting the given weight to moles, we find that 57.0 g corresponds to about 147 kJ. Therefore, the answer is E. the release of 147 kJ of heat.

The question pertains to the heat change associated with the decomposition of Iron (III) oxide (Fe2O3). Given that Fe2O3 molecular weight is ~159.69 g/mol and the fact that every 2 mol of Fe2O3 releases 824.2 kJ of heat, we can calculate the energy associated with 57 grams of Fe2O3 as follows:

First, deduce the number of moles of Fe2O3 in 57 grams: 57g / 159.69 g/mol = 0.357 mol of Fe2O3

Since 2 mol of Fe2O3 correspond to 824.2 kJ, we can conclude that 0.357 mol corresponds to: (824.2 kJ / 2) x 0.357 = 147 kJ

Considering the question described the decomposition process as 'results in the release of', it suggests the heat change is exothermic or heat-releasing. Hence, the correct answer should be: E. the release of 147 kJ of heat.

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Answer:

158.0 pm

Explanation:

In this case, the bond length of the HI molecule is equal to the sum of the atomic radii of its components. Which is to say:

Bond length HI = Atomic Radius H + Atomic Radius I

Bond Length = 25.0 pm + 133 pm

Bond Length = 158.0 pm

Final answer:

The bond length of the HI molecule can be predicted by summing the atomic radii of hydrogen (25.0 pm) and iodine (133 pm), resulting in an estimated bond length of 158 pm.

Explanation:

The question seeks to predict the bond length of the HI molecule based on the atomic radii of hydrogen (H) and iodine (I). Bond length is crucial for understanding molecular structure and is reflective of the optimal distance between two bonded atoms where the potential energy of the system is at its lowest. Given that the atomic radius of H is 25.0 pm and that of I is 133 pm, the bond length of the HI molecule can be estimated by summing these radii.

To predict the bond length of HI, we simply add the atomic radii of H and I:

Atomic radius of H = 25.0 pm

Atomic radius of I = 133 pm

Estimated bond length of HI = (25.0 pm + 133 pm) = 158 pm

This approach assumes that the bond length is approximately the sum of the individual atomic radii, which is often a reasonable approximation for such predictions.

Answer:

2. ΔG is zero, ΔH is positive, and ΔS is positive

Explanation:

When the ice is being converted to water ate 0ºC and 1 atm, there is an equilibrium between the solid and the liquid. At the equilibrium point, ΔG (the free energy) is zero. It is negative for spontaneous reactions and positive for nonspontaneous reactions.

For the phase change happens, the ice must absorb heat from the surroundings, so it's an endothermic reaction, and because of that ΔH (the enthalpy) must be positive. It is negative for exothermic reactions.

In the liquid state, the molecules have more energy and the randomness is higher than the solid-state. The entropy (S) is the measure of the randomness, so if it's increasing, ΔS must be positive.

Final answer:

The conversion of ice to water at 0°C and 1 atm has a negative ΔG, negative ΔH, and positive ΔS.

Explanation:

For the conversion of ice to water at 0°C and 1 atm:

ΔG is negative, ΔH is negative, and ΔS is positive.

ΔG is zero, ΔH is positive, and ΔS is positive.

ΔG is positive, ΔH is negative, and ΔS is positive.

Option 1 is the correct answer. When ice converts to water at 0°C and 1 atm, the Gibbs free energy change (ΔG) is negative, indicating that the process is spontaneous. The enthalpy change (ΔH) is negative, as heat is absorbed from the surroundings, and the entropy change (ΔS) is positive, as there is an increase in disorder.

Answer: Mineral A changed because Molecular energy transferred is equal or greater than than its Activation energy

Mineral B didn't change because Molecular energy transferred is less than its Activation energy.

Explanation:

The molecules of Mineral A has been disturbed by the addition of energy causing a change and the entropy is increased. The Molecules of Mineral B has not been disturbed because the residual energy has not been overcome and therefore yielding no visible change

The plausible reason for the change in mineral A will be the lower activation energy than the transferred energy.

The addition of energy will result in the change in the minerals if the activation energy is exceeded.

The minerals found in mining A and B have transferred the energy. The minimum amount of energy required to do the transition in the atoms in the activation energy.

The possible reason for the change in the mineral A will be the lower activation energy for the mineral A. The mineral B does not change because the activation energy of mineral B is higher as compared to mineral A.

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Answer:

a and B are correct

Explanation:

The energy of an electron associated with one shell around the nucleus is quantized. And is given by the formula

[tex]E= -\frac{13.6 eV}{n^2}[/tex].

where n= shell number 1,2,3...

It is true that energy of an electron cab be measured and Each electron around an atom has a discrete measure of energy associated with it. Moreover, quantity of electron energy remains constant in a shell and changes only when the electron changes its shell.

hence C is incorrect.

Answer: B: Each electron around an atom has a discrete measure of energy.

Answer:

Part A: 708 mmHg

Part B: 0.01 mol O2

Explanation:

Total pressure in a gas mixture = Sum of each partial pressure in the mixture so:

Total pressure = 730mmHg

The mixture has 2 compounds, the O2 and the vapor of water.

730mmHg - 22mmHg = 708 mmHg. Now that we have Pp O2, let's apply the Ideal Gas Law to find the mols

P.V = n . R . T

First of all, covert the mmHg in atm

760 mmHg ____ 1 atm

708 mmHg _____ 708/760 = 0.93 atm

and convert 262mL in L

262/1000 = 0.262L

0.93 atm . 0.262L = n . 0.082 . 297K

(0.93 . 0.262)/ (0.082. 297) = n

0.01 mol = n

The partial pressure of the O2 gas in the reaction of KClO3 decomposing is 708 mmHg, and the number of moles of O2 gas is approximately 0.011 moles.

The reaction of potassium chlorate (KClO3) decomposing into solid potassium chloride (KCl) and oxygen (O2) gas when heated is related to the concepts of partial pressure and moles in chemistry.

Part A: The partial pressure of a gas is the pressure contributed by that individual gas in a mixture of gases. If the total pressure is 730 mmHg, and the vapor pressure of water at 24°C is 22 mmHg, the partial pressure of the O2 (oxygen gas) would be the total pressure minus the vapor pressure of water. So, the partial pressure of the O2 gas is 730 mmHg - 22 mmHg = 708 mmHg.

Part B: To find the moles of the O2 gas, the ideal gas law could be used which links pressure, volume, temperature and number of moles in a gas. The law states that (Pressure x Volume) = (number of moles x Ideal Gas Constant x Temperature in Kelvin). We know the volume, partial pressure and absolute temperature, but an appropriate conversion of units is required. Therefore, adjusting to standard SI units, we have the volume 262 mL converted to 0.262 L, the partial pressure 708 mmHg converted to 0.93 atmospheres (1 atmosphere = 760 mmHg), and the temperature 24°C converted to 297 K (Kelvin = Celsius + 273.15). Short calculation later we get around 0.011 moles of O2 gas.

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Answer:

ii) They emit various tyoes of radiation

Explanation:

They fact that they emit radiation makes them ideal to use as a tracer because that radiation can be detected and followed. This means that you can know where the isotope is going throught.

Cause of that, radioisotopes are very helpful in medical applications such as tracing blood veins.

Answer :  The partial pressure of [tex]H_2[/tex] is, 726.2 mmHg

Solution :

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

[tex]P_T=p_{H_2}+p_{H_2O}[/tex]

where,

[tex]P_T[/tex] = total partial pressure = barometric pressure = 746 mmHg

[tex]P_{H_2}[/tex] = partial pressure of hydrogen gas = ?

[tex]P_{H_2O}[/tex] = partial pressure of water vapor = 19.8 mmHg  (assume)

Now put all the given values is expression, we get the partial pressure of the hydrogen gas.

[tex]746mmHg=p_{H_2}+19.8mmHg[/tex]

[tex]p_{H_2}=726.2mmHg[/tex]

Therefore, the partial pressure of [tex]H_2[/tex] is, 726.2 mmHg

Answer:

a)  cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

b)  0.068 V.

Explanation:

A) Cu2+ + 2e- euilibrium cu (s)

 Hg2Cl2 + 2e- equilibrium 2Hg (l) + 1cl-

Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

B) To calculate the cell voltage

E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+

putting values we get

 = 0.339V + (90.05916V/2)log(0.100) = 0.309V

 E_cell = E Cu2+/Cu - E SCE = 0.309 V - 0.241 V = 0.068V.

The half-reaction for the copper electrode in a copper sulfate solution is Cu(s) -> Cu2+(aq) + 2e-. However, you cannot calculate the cell voltage without additional information about the Calomel electrode's standard reduction potential.

The subject of this question is electrochemistry, which involves redox reactions and measurements of electrode potential.

(a) The half-reaction for the Cu electrode in your scenario, when the Cu wire is dipped into the CuSO4 solution, can be represented as follows: Cu (s) -> Cu2+ (aq) + 2e-. This reaction shows that copper metal (Copper in zero oxidation state) is being oxidized to Copper(II)(in +2 oxidation state) ions by losing 2 electrons.

For the cell voltage, we do not have sufficient information to calculate. To perform this calculation, we would also need the standard reduction potential for the Calomel electrode, or some other point of comparison.

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Answer:

1. C has been oxidized in the reaction.

2. Mn has been reduced in the reaction.

3/4 The reducing agent is the C and the oxidizing agent is the Mn.

The reaction in an acidic solution is:

16H⁺  +  2MnO₄⁻  + 5C₂O₄⁻²   →  10CO₂  +  2Mn²⁺  +  8H₂O

10 moles of electrons are been transferred.

Explanation:

MnO₄⁻ (aq) + H₂C₂O₄ (aq)  →  Mn²⁺ (aq) + CO₂ (g)

In the permanganate, Mn acts with +7 as oxidation number, and in product side, we have Mn2+. It has decrease the oxidation number, so it has been reduced. This is the oxidizing agent.

In the oxalic acid, carbon has +3 as oxidation number, and in CO2, we see that acts with +4 so, it has increase it. This element is the reducing agent and has been oxidated.

8H⁺  +  MnO₄⁻  +  5e⁻  →  Mn²⁺  +  4H₂O

It is completed on the opposite side where there are oxygen, with as many water as there are oxygen, and on the opposite side complete with protons to balance the H

C₂O₄⁻²   →  2CO₂  +  2e⁻  

In oxalate, the carbon that acted with +3 gained an electron to oxidize to +4, but since there are 2 carbons, it gained 2 electrons. The oxygen is balanced by adding a 2 in stoichiometry

The halfs reaction have to be multipplied .2 (reduction) and .5 (oxidation) to balance the electrons in the main equation.

(8H⁺  +  MnO₄⁻  +  5e⁻  →  Mn²⁺  +  4H₂O) . 2

(C₂O₄⁻²   →  2CO₂  +  2e⁻ ) . 5

16H⁺  +  2MnO₄⁻  +  10e⁻  + 5C₂O₄⁻²   →  10CO₂  +  10e⁻  +  2Mn²⁺  +  8H₂O

16H⁺  +  2MnO₄⁻  + 5C₂O₄⁻²   →  10CO₂  +  2Mn²⁺  +  8H₂O

Answer:

The strong acids are fully ionized in aqueous solution, and they contains higher concentration of hydrogen ions. Strong acids are lower pH in nature. Some examples of strong acids are:

1) Hydrochloric acid.

2) Nitric acid.

3) Sulfuric acid.

The weak acids are not fully ionized, means they are partially ionized in aqueous solution, and they contains lower concentration of hydrogen ions. Weak acids are higher pH in nature than strong acid. Some examples of weak acids are:

1) Ethanoic acid.

2) Acetic acid.

3) Nitrous acid.

SiO₂(s) is not an example of a molecular solid

A molecular solid is a of solid in which its molecules are held together by weak intermolecular forces such as van der Waals forces.

Molecular solids are soft, often volatile, have low density, have low melting temperatures, and are electrical insulators.

Example of molecular solid include CO₂(s) (dry ice), iodine (I₂(s)), C₂₅H₅₂(s) (paraffin wax), H₂O(s)

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Answer:

The balloon will occupy the volume of 2.91 L at 62.0 °C .

Explanation:

Using Charle's law  

[tex]\frac {V_1}{T_1}=\frac {V_2}{T_2}[/tex]

Given ,  

V₁ = 2.64 L

V₂ = ?

T₁ = 31.0 °C

T₂ = 62.0 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (31.0 + 273.15) K = 304.15 K  

T₂ = (62 + 273.15) K = 335.15 K  

Using above equation as:

[tex]\frac{2.64}{304.15}=\frac{V_2}{335.15}[/tex]

[tex]V_2=\frac{2.64\cdot \:335.15}{304.15}[/tex]

New volume = 2.91 L

The balloon will occupy the volume of 2.91 L at 62.0 °C .

The correct electron configurations are:

A) O2−: [He]2s²2p⁶.

D) Si: [Ne]3s²3p².

E) Ni: [Ar]4s²3d⁸.

The correct electron configurations from the provided list are as follows:

A) The electron configuration of O2− (the oxygen ion with a charge of -2) is [He]2s²2p⁶. This configuration accounts for the addition of two electrons to the neutral oxygen atom's electron configuration of 1s²2s²2p⁴, resulting in a total of eight electrons.

D) The electron configuration of Si (silicon) is [Ne]3s²3p². Silicon, with an atomic number of 14, has 14 electrons. The electron configuration follows the Aufbau principle, filling the 1s, 2s, 2p, and 3s subshells.

E) The electron configuration of Ni (nickel) is [Ar]4s²3d⁸. Nickel, with an atomic number of 28, has 28 electrons. This configuration represents the filling of the 1s, 2s, 2p, 3s, 3p, 4s, and 3d subshells.

Option B and C are not correct:

B) The electron configuration of Cs (cesium) is [Xe]6s¹. Cesium, with an atomic number of 55, has only one valence electron in the 6s orbital.

C) The electron configuration of Ag+ (the silver ion with a charge of +1) is [Kr]5s²4d⁹. This configuration arises from the removal of one electron from the neutral silver atom's electron configuration, [Kr]5s²4d¹⁰, leaving it with 47 electrons.

Understanding electron configurations is essential in chemistry, as they determine the chemical properties and reactivity of elements. These configurations are based on the filling of electron orbitals following specific rules and principles, such as the Aufbau principle, Pauli exclusion principle, and Hund's rule.

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Answer:

Sodium, potassium, calcium , magnesium etc

Explanation:

To determine the species which is being oxidized or reduced in a reaction, we make use of oxidation number. While oxidation involves electron loss, reduction involves electron gain. The specie which gain electrons is being reduced and thus experience a decrease in oxidation number and thus said to be the oxidizing agent. The species which lose electrons experience and increase in oxidation number and thus is the reducing agent.

There are some elements however which never get oxidized or reduced in the course of a chemical reaction. What we mean by this is that they neither get oxidized nor reduced but maintain the same original oxidation number.

Examples of these kind of elements include the groups 1 and 2 metals.

Answer:

0.008

Explanation:

From the balanced equation, 3 moles of oxygen gas produced 2 moles of the product. Then, 0.012 moles of oxygen gas will produces 0.012 x 2/3 = 0.008

When 0.012 mol of O2 completely reacts with excess aluminum according to the balanced chemical equation, 0.008 mol of Al2O3 is produced, rounded to two significant figures.

To calculate the number of moles of Al2O3 produced from the complete reaction of 0.012 mol of O2, we use stoichiometry based on the balanced chemical equation:

4Al(s) + 3O2(g) → 2Al2O3(s).

From the equation, 3 moles of oxygen gas (O2) produce 2 moles of aluminum oxide (Al2O3). To find the moles of Al2O3 produced, use the mole ratio:

(mol Al2O3) = (mol O2) × (2 mol Al2O3 / 3 mol O2)

Substituting the given value:

(mol Al2O3) = (0.012 mol O2) × (2 mol Al2O3 / 3 mol O2) = 0.008 mol Al2O3

So, when 0.012 mol of O2 completely reacts, 0.008 mol of Al2O3 is produced, assuming excess aluminum is present. This answer is rounded to two significant figures.

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Answer:

827 mL

Explanation:

To answer this question we use the definition of Molarity:

Molarity = mol / L

[Cl⁻] = mol Cl⁻ / L

Now we calculate the moles of Cl⁻ present in 42.0 g of MgCl₂⋅6H₂O:

Molar mass of MgCl₂⋅6H₂O = 24.3 + 2*35.45 + 6*18 = 203.2 g/mol

moles of Cl⁻ = 42.0 g MgCl₂⋅6H₂O ÷ 203.2 g/mol * [tex]\frac{2molCl^{-}}{1molMgCl_{2}.6H_{2}O}[/tex] = 0.4134 mol Cl⁻

Finally we use the definition of molarity to calculate the volume:

0.500 M = 0.4134 mol Cl⁻ / xL

xL = 0.827 L =  827 mL

Answer:

The equilibrium concentration of [PCl₅] = 0.24M

Explanation:

This is the reaction.

PCl₅(g) → PCl₃(g) +Cl₂(g)

Now let's make, the expression for Kc

Remember that concentrations must be in M

Kc = ( [PCl₃] . [Cl₂] ) / [PCl₅]

4.2x10⁻² = [0.1] . [0.1] / [PCl₅]

[PCl₅] =  [0.1] . [0.1] / 4.2x10⁻²

[PCl₅] = 0.24M

Using the given equilibrium constant (Kc) and the concentrations of PCl3 and Cl2 at equilibrium, the equilibrium concentration of PCl5 is determined to be 0.24 M.

This question refers to the concept of equilibrium in chemistry, specifically the application of the equilibrium constant (Kc) in determining the concentrations of reactants and products in chemical reactions.

The given chemical reaction is: PCl₅(g)  PCl(g) + Cl₂(g). The equilibrium constant expression for the reaction is Kc = [PCl₃][Cl₂] / [PCl₅]. Here, [PCl₃], [Cl₂], and [PCl₅] represent molar concentrations of PCl₃, Cl₂, and PCl₅, respectively, at equilibrium.

We are given that Kc = 4.2 x 10-2,, [PCl₃]= 0.10M, [Cl₂]= 0.10M. Substituting these values into the equilibrium constant expression, we can solve for [PCl₅], giving:

[PCl₅]= [PCl₃][Cl₂] / Kc = (0.10)(0.10) / (4.2 x 10-2) = 0.24 M.

So, the equilibrium concentration of PCl₅ in this reaction is 0.24 M, thus the correct option is (e) 0.24 M.

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Statement a is true because of Avogadro's Law. Statement b is also true since molar mass of O2 is greater than H2, so 5.0g of O2 contains fewer molecules than H2. Statement c is false as the number of molecules in given volume and temperature is constant.

The subject of this question is chemistry, specifically the concept of Avogadro's law. Avogadro's law states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.

Statement a is TRUE. Given equal volumes and temperatures, the flasks do contain the same number of molecules. This principle is referred to as Avogadro's Law.

Statement b is TRUE. The molar mass of O2 is indeed greater than H2, meaning that 5.0 g of O2 has fewer molecules than 5.0 g of H2, contrary to what statement b suggests.

Statement c is FALSE. Equal volume flasks at the same temperature will have the same number of molecules, regardless of pressure. Which specific gas is involved does not change this fact.

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The correct options are as follows:

a) False. Because the molar mass of O2 is greater than the molar mass of H2, 5.0 g of O2 will contain fewer molecules than 5.0 g of H2.

b) True. Because the molar mass of O2 is greater than the molar mass of H2, 5.0 g of O2 will contain fewer molecules than 5.0 g of H2.

c) False. Regardless of the pressure, if the volumes and temperatures are the same, the number of molecules will be the same due to Avogadro's Law.

Let's analyze each statement:

a) This statement is false. The volume of a gas is directly proportional to the number of moles of gas at a constant temperature and pressure (Avogadro's Law). Since the molar mass of O2 (approximately 32 g/mol) is much greater than that of H2 (approximately 2 g/mol), 5.0 g of O2 will represent fewer moles than 5.0 g of H2. Consequently, the flask containing O2 will have fewer molecules than the flask containing H2, despite the volumes being equal.

b) This statement is true. As explained above, because O2 has a higher molar mass than H2, 5.0 g of O2 will contain fewer moles and hence fewer molecules than 5.0 g of H2.

c) This statement is false. According to Avogadro's Law, equal volumes of gases at the same temperature and pressure contain the same number of molecules. Therefore, the pressure does not affect the number of molecules if the volume and temperature are constant. The number of molecules in each flask will be the same because both flasks are at the same temperature and have the same volume. The pressure within each flask will adjust according to the number of moles of gas present (as per the ideal gas law, PV = nRT), but this does not change the number of molecules.

To calculate the number of moles for each gas:

 For O2:

Number of moles of O2 = mass of O2 / molar mass of O2

Number of moles of O2 = 5.0 g / 32 g/mol ≈ 0.156 mol

For H2:

Number of moles of H2 = mass of H2 / molar mass of H2

Number of moles of H2 = 5.0 g / 2 g/mol = 2.5 mol

Clearly, the number of moles of H2 is greater than the number of moles of O2 for the same mass, which confirms that the number of molecules in the H2 flask is greater than in the O2 flask.

Answer:

1.20 V

Explanation:

[tex]Pb(s) + Br_2(l)\rightarrow Pb^{2+}(aq) + 2Br^-(aq)[/tex]

Here Pb undergoes oxidation by loss of electrons, thus act as anode. Bromine undergoes reduction by gain of electrons and thus act as cathode.

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

Given,

[tex]Pb^{2+}(aq) + 2 e^-\rightarrow Pb(s)[/tex]

[tex]E^0_{[Pb^{2+}/Pb]}= -0.13\ V[/tex]

[tex]Br_2(l) + 2 e^-\rightarrow 2 Br(aq)[/tex]

[tex]E^0_{[Br_2/Br^{-}]}=+1.07\ V[/tex]

[tex]E^0=E^0_{[Br_2/Br^{-}]}- E^0_{[Pb^{2+}/Pb]}[/tex]

[tex]E^0=+1.07- (-0.13)\ V=1.20\ V[/tex]

The cell potential for the given reaction is calculated by subtracting the anode's potential from the cathode's potential since the reaction at the anode is an oxidation and the cathode involves a reduction. The standard cell potential for this reaction is 1.20 V, and this positive value indicates the reaction is spontaneous under standard conditions.

The standard cell potential is calculated by adding the standard reduction potential of the cathode to the standard reduction potential of the anode. According to the reaction, Pb(s) is being oxidized to Pb2+(aq), thus acting as the anode: E° = -0.13 V. Meanwhile, Br2(l) is being reduced to 2Br-(aq), acting as the cathode: E° = +1.07 V. When calculating the overall cell potential, we subtract the anode's potential from the cathode's potential, because the reaction at the anode is oxidation (loss of electrons), while the cathode involves a reduction (gain of electrons).

Therefore, the standard cell potential, E°cell, is calculated as E°cathode - E°anode = 1.07 V - (-0.13 V) = 1.20 V. This value is positive, indicating that the reaction is spontaneous under standard conditions as per the cells' relative oxidizing strength.

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Answer:

-41,9kJ/mol NaOH

Explanation:

For the solution process:

NaOH(s) →Na⁺(aq) + OH⁻(aq)

The released heat is:

Q = -C×m×ΔT

Where Q is the released heat, C is specific heat of the solution (4,18J/g°C), m is the mass of water (100,0g) and ΔT is (32,0°C-23,9°C)

Replacing:

Q = -3385,8J

This heat is released per 3,23g of NaOH. Now, the heat released (ΔH) per mole of NaOH is:

[tex]\frac{-3385,8J}{3,23gNaOH} *\frac{40g}{1mol}[/tex]= -41929J/molNaOH ≡

-41,9kJ/mol NaOH

I hope it helps!

Answer:

Specific heat of gold is 0,133J/g°C

Explanation:

In this problem, the heat of the gold is transferred to water and the calorimeter, that means:

[tex]q_{Lost By Metal} = q_{GainedWater} + q_{Gained Calorimeter}[/tex]

The Q lost by metal is:

Q = C×m×ΔT, Where m is mass (61,68g), ΔT is (99,01°C-23,98°C = 75,03°C) and C is sepecific heat of gold

The Q gained by water is:

Q =  C×m×ΔT, Where m is mass (79,34g), ΔT is (23,98°C-22,14°C = 1,84°C) and C is sepecific heat of water (4,184J/g°C)

The Q gained by calorimeter is:

Q =Cc×ΔT Where Cc is calorimeter constant (1,80J/°C), and ΔT is (23,98°C-22,14°C = 1,84°C)

Replacing:

C×61,68g×75,03°C = 4,184J/g°C×79,34g×1,84°C + 1,80J/°C×1,84°C

4628g°C×C = 610,8J + 3,3J

4628g°C×C = 614,1J

C = 0,133J/g°C

I hope it helps

The specific heat of gold can be calculated using the equation representing the transfer of heat in the experiment. The sum of the heat absorbed by the water and the calorimeter (each calculated using respective mass, specific heat, and temperature change) equals the heat lost by the gold.

The specific heat of a substance is typically calculated using the formula q = mcΔT, where q is the heat absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature. In this experiment, heat is transferred from the gold to the water and the calorimeter, so we need to set up an equation with the sum of the heat absorbed by the water and the calorimeter equal to the heat lost by the gold.

For the water, we use the specific heat value of water as 4.18 J/g°C, and for the calorimeter, we use the given calorimeter constant. Setting up the equation and solving for the specific heat of gold gives:

(61.68g)(c)[99.01°C-23.98°C] = (79.34g)(4.18 J/g°C)[23.98°C-22.14°C] + (1.80 J/°C)(23.98°C-22.14°C)

Solving the above equation will yield the specific heat of gold.

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Answer:

69%

Explanation:

Say the hydrocarbon is C2H4, the equation of reaction would be;

C2H4 + H2 ----------> C2H6

The hydrocarbon, C2H4 is the limiting reagent.

From the question, the hydrocarbon C2H4 flows into a catalytic reactor at 26.2 atm and 250°C with a flow rate of 1100 L/min

Using, PV=nRT--------------------(1).

n= PV/RT to find the number of moles.

n= 26.2 atm × 1100 L/ 0.0821 L. atm/mol per kelvin × 523 K.

=28,820 atm.L/ 42.9383 L.atm/mol.

= 671.2 mole.

One mole of C2H4 produced one mole of C2H6.

Mass of C2H6 = 30 × 671.2

= 20,136 g = 20.136 kg.

Percent yield = actual yield/ theoretical yield × 100% -------(2)

Actual yield= 13.9 kg, theoretical yield = 20.136 kg

Substitute the parameters into equation (2). We have;

Percent yield = 13.9 kg/ 20.136× 100

Percent yield= 0.69 × 100

Percent yield= 69%

Note: P= pressure, V= volume, T= temperature and n = number of moles

Answer:

The temperature will change and become 2/3 of its original.

Explanation:

Using Ideal gas equation for same mole of gas as

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]

Given ,  

The volume of the sample gets reduced 1/3 of the original. So,

V₂ = 1/3V₁

The pressure of the sample is doubled of the original. So,

P₂ = 2P₁

Using above equation as:

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{2\times P_1}\times {\frac{1}{3}\times V_1}}{T_2}[/tex]

[tex]T_2=\frac{2}{3}\times T_1[/tex]

The temperature will change and become 2/3 of its original.