The value of ΔH° for the reaction below is -126 kJ. The amount of heat that is released by the reaction of 10.0 g of Na2O2 with water is ________ kJ. 2Na2O2 (s) + 2H2O (l) → 4NaOH (s) + O2 (g)

Answer: The balanced chemical equation is written below.

Explanation:

A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.

For the reaction of carbon dioxide with hydrogen gas, 9.85 kcal of energy is absorbed. So, this energy term will be written on the reactant side.

Thus, the balanced chemical equation for the reaction of carbon dioxide with hydrogen gas follows:

[tex]CO_2(g)+H_2(g)+9.85kcal\rightarrow CO(g)+H_2O(g)[/tex]

Hence, the balanced chemical equation is written above.

Answer: There will be no change in rate.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]A+B+C\rightarrow Products[/tex]

[tex]Rate=k[A]^x[B]^y[C]^z[/tex]

k= rate constant

x = order with respect to A  = 0

y = order with respect to B = [tex]\frac{1}{2}[/tex]

z=  order with respect to C = 2

Thus [tex]Rate=k[A]^0[B]^{\frac{1}{2}}[C]^2[/tex]

Given : when [A] is doubled and the other reactant concentrations are held constant.

Thus the new rate law is [tex]Rate'=k[2A]^0[B]^{\frac{1}{2}}[C]^2[/tex]

[tex]Rate'=k[2]^0[A]^0[B]^{\frac{1}{2}}[C]^2[/tex]

[tex]Rate'=k[A]^0[B]^{\frac{1}{2}}[C]^2[/tex] [tex](2^0=1)[/tex]

[tex]Rate'=Rate[/tex]

Thus the reaction rate would not change.

Doubling the concentration of A has no effect on the reaction rate because the reaction is zero order in A. The rate law for the reaction is rate = k[B]^1/2[C]², indicating independence from [A].

The question pertains to chemical kinetics, specifically how the reaction rate changes with respect to changes in reactant concentrations. Since the reaction is zero order with respect to A, doubling the concentration of A ([A]) will have no effect on the rate of the reaction. The rate law for this reaction could be represented as rate =k[B]^1/2[C]², which clearly shows that the reaction rate is independent of the concentration of A. Therefore, the reaction rate remains unchanged if the concentration of A is doubled while keeping B and C constant.

Answer:

Explanation:

The vapor pressure of a solution is a colligative property, which means that it is determined by the number of particles (molecules or ions) of solute present in a solution.

Raoult's law permits the calculations of the change of the vapor pressure of a solvent when a solute is added.

The equation is:

Where:

In the case of ionic solutes, you must take into account the number of ions that result from the ionization.

Calculating the molar fraction:

Calculating the change in the vapor pressure of the solution:

Vapor pressure of the solution:

Rounding to three significant figures (because 40.0g has three significant figures): 22.0 mmHg ← answer.

Raoult's law states that for a given solution, the partial pressure of each component is equal to the mole fraction in that solution. The vapor pressure of the solution is 22 mmHg.

The vapor pressure is defined as the force exerted by the vapors in the walls of a container. It is a colligative property, such that the amount of substance increased or decreased is dependent on the amount of solute present.

The equation can be represented as:

where,

Now, calculating the mole fraction, where:

Now, the vapor pressure of the solution can be calculated as:

[tex]\Delta[/tex] P = X[tex]_{\text{solute}}[/tex] x  P[tex]_{\text{solvent}}[/tex]

[tex]\Delta[/tex] P = 0.0755 × 23.76 mmHg = 1.78 mmHg

Hence, the vapor pressure of the solution:

P[tex]_{\text{solution}}[/tex] = [tex]\Delta[/tex] P + P[tex]_{\text{solvent}}[/tex]

P[tex]_{\text{solution}}[/tex] = 23.76 mmHg - 1.79 mm Hg = 21.97mmHg

Therefore, the vapor pressure of the solution is approximately 22 mmHg.

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Answer: Chlorine, amu 35.45

Calculate the relative atomic mass by multiplying the mass of each isotope by its percentage, then sum. The relative atomic mass is 35.5, which corresponds to the element chlorine.

The relative atomic mass of an element is the weighted average mass of the atoms in a naturally occurring sample of the element, taking into account the percentages of each isotope present. You seem to have two isotopes of the same element present: Isotope-35 and Isotope-37.

To calculate the relative atomic mass, you multiply the mass of each isotope by the percentage (in decimal form) in which it is found, and then sum those values. Let's see how this works:

Add these two values together (26.25 + 9.25) to find the relative atomic mass: 35.5. The element with a relative atomic mass of approximately 35.5 is chlorine (Cl).

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The first part of the answer is in the photo attached. I hope i can attach a second photo...

To find the change in pH, use the Henderson-Hasselbalch equation with the initial concentrations of NH3 and NH4+ to find the initial pH. Adjust concentrations after addition of HCl to find the final pH. Subtract initial pH from final to find change.

The question asks us to calculate the change in pH when 5.00 ml of 0.100 M HCl is added to 75.0 ml of a buffer solution that is 0.100 M in NH3 and 0.100 M in NH4Cl. This is a problem related to acid-base chemistry. We have a strong acid (HCl) being added to a buffer solution composed of NH3 (a weak base) and NH4Cl (the conjugate acid of the base).

First, it's important to use the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), where pKa is the -log of the Ka, [A-] is the concentration of base (NH3 in this case) and [HA] is the concentration of conjugate acid (NH4+ in this case). Plug the initial concentrations into the equation to find initial pH. Then, calculate the moles of HCl being added and adjust the concentrations of NH4+ and NH3 accordingly. Using the new concentrations in Henderson-Hasselbalch equation will give the final pH. The difference between the initial and final pH is the change in pH.

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Answer: The mass of water that can be formed are 33.48 g.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]   ....(1)

Given mass of ammonia gas = 21.1 g

Molar mass of ammonia gas = 17 g/mol

Putting values in above equation, we get:  

[tex]\text{Moles of ammonia}=\frac{21.1g}{17g/mol}=1.24mol[/tex]

Given mass of oxygen gas = 73.9 g

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:  

[tex]\text{Moles of oxygen gas}=\frac{73.9g}{32g/mol}=2.30mol[/tex]

For the given chemical equation:

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(l)[/tex]

By Stoichiometry of the reaction:

4 moles of ammonia gas reacts with 5 moles of oxygen gas.

So, 1.24 moles of ammonia gas will react with = [tex]\frac{5}{4}\times 1.24=1.55moles[/tex] of oxygen gas.

As, given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

So, ammonia gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the above reaction:

4 moles of ammonia gas is producing 6 moles of water.

So, 1.24 moles of ammonia gas will produce = [tex]\frac{6}{4}\times 1.24=1.86moles[/tex] of water.

Now, calculating the mass of water by using equation 1, we get:

Moles of water = 1.86 moles

Molar mass of water = 18 g/mol

Putting all the values in equation 1, we get:

[tex]1.86mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=33.48g[/tex]

Hence, the mass of water that can be formed are 33.48 g

The equilibrium constant for the phosgene formation reaction, given the partial pressures of carbon monoxide, chlorine gas, and phosgene at equilibrium, is approximately 0.204.

The student has asked to calculate the equilibrium constant (Kp) for the formation of phosgene (COCl₂) from carbon monoxide (CO) and chlorine gas (Cl₂) at a given temperature. The equilibrium constant for a reaction can be determined using the partial pressures of the gases at equilibrium. The equation for Kp in this reaction is Kp = PCOCl₂ / (PCO * PCl₂). Given the partial pressures at equilibrium (PCO = 0.830 atm, PCl₂ = 1.30 atm, PCOCl₂ = 0.220 atm), we can calculate Kp as follows:

Kp = 0.220 atm / (0.830 atm * 1.30 atm) = 0.220 / 1.079 = 0.204

Therefore, the equilibrium constant, Kp, for the reaction under the given conditions is approximately 0.204.

Answer:

see explanation...

Explanation:

Given 125ml standard flask

Measured volume to top = 157ml

Measured volume with stopper = 140ml

------------------------------------------------------------------------------------------

Based on the wording of the question, the volume of the flask is ...

= 157ml without stopper*, or

= 140ml with stopper.

If interest is in volume occupied by H₂(g) above 5ml of rxn solution ...

=> Volume without stopper* = 157ml - 5ml = 152ml above rxn solution

=> Volume with stopper = 140ml - 5ml = 135ml (Assuming stopper doesn't pop out of container). :-)

*Assuming gas fills available volume without escaping container.

Answer:

Volume of flask without stopper=157 mL-5mL=152 mL

Volume of flask with stopper=140-5=135 mL

Answer:

The rate of the reaction is [tex]6.24\times 10^{-6} M/s[/tex].

Explanation:

[tex]NH4^+(aq) + NO_2^-(aq)\rightarrow N_2(g) + 2H_2O(l)[/tex]

Concentration of [tex][NH_4^{+}]=0.26 M[/tex]

Concentration of [tex][NO_2^{-}]=0.080 M[/tex]

Rate constant of the reaction = k= [tex]3.0\times 10^{-4} M^{-1} s^{-1}[/tex]

[tex]R= k[NH_{4}^+][NO_{2}^-][/tex]

[tex]R=3.0\times 10^{-4} M^{-1} s^{-1}\times 0.26 M\times 0.080 M[/tex]

[tex]R=6.24\times 10^{-6} M/s[/tex]

The rate of the reaction is [tex]6.24\times 10^{-6} M/s[/tex].

Answer:

Rate of Reaction = 6.24 x 10–6 M/s

Explanation:

Rate of reaction = k[NH4+][NO2–]

Concentration of [NH4+] = 0.26 M

Concentration of [NO2–] = 0.080 M

k= 3.0 × 10–4/ M · s

Rate of Reaction = (3.0 × 10–4/ M · s)( 0.26 M)(0.080 M)

Answer: The [tex]\Delta H_f[/tex] for [tex]C_3H_5N_3O_9[/tex] in the reaction is -365.5 kJ/mol.

Explanation:

Enthalpy change of a reaction is defined as the difference in enthalpy of all the products and the reactants each multiplied with their respective number of moles. The equation that is used to calculate the enthalpy change of a reaction is:

[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}][/tex]

For the given chemical reaction:

[tex]4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+10H_2O(g)+6N_2(g)+O_2(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H_{rxn}=[(12\times \Delta H_f_{(CO_2)})+(10\times \Delta H_f_{(H_2O)})+(6\times \Delta H_f_{(N_2)})+(1\times \Delta H_f_{(O_2)})]-[(4\times \Delta H_f_{(C_3H_5N_3O_9)})][/tex]

We are given:

[tex]\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(N_2)}=-0kJ/mol\\\Delta H_f_{(CO_2)}=-393.5kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-5678kJ[/tex]

Putting values in above equation, we get:

[tex]-5678=[(12\times (-393.5))+(10\times (-241.8))+(6\times (0))+(1\times (0))]-[(4\times \Delta H_f_{(C_3H_5N_3O_9)})]\\\\\Delta H_f_{(C_3H_5N_3O_9)}=-365.5kJ/mol[/tex]

Hence, the [tex]\Delta H_f[/tex] for [tex]C_3H_5N_3O_9[/tex] in the reaction is -365.5 kJ/mol.

The standard enthalpy of formation (ΔfHo) for nitroglycerin which decomposes by the given equation can be calculated using the formula for enthalpy change and the provided enthalpies of formation for CO2(g) and H2O(g). Nitrogen and Oxygen are in their standard states so their enthalpies of formation are zero. The calculated ΔfHo for Nitroglycerin is about -364.6 kJ/mol.

To calculate the standard enthalpy of formation (ΔfHo) for nitroglycerin, we first need to know that the formation reactions are those that produce 1 mol of a substance from its elements in their standard states. Here, the combustion reaction for nitroglycerin given is: 4 C3H5N3O9 (l) → 12 CO2 (g) + 10 H2O (g) + 6 N2 (g) + O2 (g). And the provided ΔrxnHo = −5678 kJ.

Also, the standard enthalpy of formations for CO2(g) and H2O(g) are given as -393.5 kJ/mol and -241.8 kJ/mol, respectively. But, the enthalpy of formation for elements in their standard state (here N2 and O2) is zero.

Using the equation: ΔrxnHo = Σ ΔfHo(products) - Σ ΔfHo(reactants), and the knowledge of stoichiometry we get: -5678 kJ = [12(-393.5 kJ/mol) + 10(-241.8 kJ/mol) + 6 * 0 + 1 * 0] - [4 * ΔfHo (C3H5N3O9)]

By calculating we get the ΔfHo for Nitroglycerin = -364.6 kJ/mol (approx)

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Answer:

molar mass HA = 87.8 g/mole

Explanation:

Given 0.140g HA + 14.5ml(0.110M NaOH) => NaA + H₂O

Rxn is a 1:1 rxn ration => moles HA neutralized = moles NaOH used

=> 0.140g/molar mass of HA = (0.110M)(0.0145L)

=> molar mass of HA = (0.140g)/[(0.110moles/L)(0.0145L)] = 87.8 g/mole

Answer: The molar mass of monoprotic acid is 87.72 g/mol

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of monoprotic acid

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=1\\M_1=?M\\V_1=35.0mL\\n_2=1\\M_2=0.110M\\V_2=14.5mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 35.0=1\times 0.110\times 14.5\\\\x=\frac{1\times 0.110\times 14.5}{1\times 35.0}=0.0456M[/tex]

To calculate the molecular mass of acid, we use the equation used to calculate the molarity of solution:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

We are given:

Molarity of solution = 0.0456 M

Given mass of acid = 0.140 g

Volume of solution = 35.0 mL

Putting values in above equation, we get:

[tex]0.0456M=\frac{0.140\times 1000}{\text{Molar mass of acid}\times 35}\\\\\text{Molar mass of acid}=\frac{0.140\times 1000}{0.0456\times 35}=87.72g/mol[/tex]

Hence, the molar mass of monoprotic acid is 87.72 g/mol

The mass of PbSO₄ that could be produced is 13.74 g

From the question,

We are to determine the mass of PbSO₄ that could be produced from the reaction

The given balanced chemical equation for the reaction is

Na₂SO₄(aq) + Pb(NO₃)₂(aq) ⟶ 2NaNO₃(aq) + PbSO₄(s)

This means

1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂ to produce 2 moles of NaNO₃ and 1 mole of PbSO₄

Now, we will determine the number of moles of each reactant present

Volume = 100.0 mL = 0.1 L

Concentration = 0.453 M  

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles of Na₂SO₄ = 0.453 × 0.1

Number of moles of Na₂SO₄ = 0.0453 mol

Volume = 100.0 mL = 0.1 L

Concentration = 0.907 M

∴ Number of moles of Pb(NO₃)₂ = 0.907 × 0.1

Number of moles of Pb(NO₃)₂ = 0.0907 mol

Since, 1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂

Then,

0.0453 mole of Na₂SO₄ will react with 0.0453 mole of Pb(NO₃)₂

Now, from the balanced chemical equation

1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂ to produce 1 mole of PbSO₄

Then,

0.0453 mole of Na₂SO₄ will react with 0.0453 mole of Pb(NO₃)₂ to produce 0.0453 mole of PbSO₄

∴ The number of moles of PbSO₄ produced is 0.0453 mole

Now, for the mass of PbSO₄ produced

From the formula

Mass = Number of moles × Molar mass

Molar mass of PbSO₄ = 303.26 g/mol

∴ Mass of PbSO₄ produced = 0.0453 × 303.26

Mass of PbSO₄ produced = 13.737678 g

Mass of PbSO₄ produced ≅ 13.74 g

Hence, the mass of PbSO₄ that could be produced is 13.74 g

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The question involves the stoichiometric calculation during a chemical reaction. In the reaction,  Pb(NO₃)₂ and Na₂SO₄ react to produce  PbSO₄. Given the volumes and molarities, we determined Na₂SO₄ to be the limiting reactant. Therefore, approximately 13.72 g of  PbSO₄ can be produced.

This question is trying to understand a stoichiometric calculation involved in a chemical reaction. We start by determining the limiting reactant. To do this, divide the number of moles of each reactant by the stoichiometric coefficient in the balanced chemical equation. The reactant that returns the smallest number is the limiting reactant.

In this case, the reaction uses one mole of Pb(NO₃)₂ and one mole of Na₂SO₄ to produce PbSO₄, therefore the molar ratio is 1:1. Pb(NO₃)₂ has a higher molarity, hence it will not be the limiting factor. Therefore, Na₂SO₄ is the limiting reactant. Considering this, we calculate that (0.453 moles/L × 0.1 L = 0.0453 moles of Na₂SO₄) are present.

To find the mass of PbSO₄ that can be produced, we use the molar mass of PbSO₄ (303.26 g/mol)—based on the atomic masses of lead, sulfur, and oxygen. From this we can calculate the mass as follows: 0.0453 moles × 303.26 g/mol = 13.72398 g of PbSO₄.

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Final answer:

The ideal gas pressure in the container if all the liquid acetone evaporated is calculated by converting the sample mass to moles, ensuring all units are consistent, and applying the Ideal Gas Law. The result is an ideal gas pressure of approximately 0.064 atm.

Explanation:

To calculate the ideal gas pressure in the container if all of the liquid acetone evaporated, we need to determine the amount of acetone in moles and use the Ideal Gas Law. Given that the vapor pressure of acetone is 100 mm Hg at 281 K, we will first convert the 6.06E-2 g of acetone into moles using acetone's molar mass.

The molar mass of acetone (CH3)2CO is approximately 58.08 g/mol. By dividing the mass of the acetone sample by its molar mass, we obtain:

Number of moles = mass/molar mass = 6.06E-2 g / 58.08 g/mol ≈ 1.04E-3 mol

Next, we apply the Ideal Gas Law which is PV=nRT. To use this law, we must ensure all units are consistent. The vapor pressure must be in atmospheres, temperature in Kelvin, and volume in liters. The conversion from mm Hg to atm is 1 atm = 760 mm Hg, so 100 mm Hg = 100/760 atm. Also, convert the volume from mL to L by dividing by 1000.

Using these conversions and the Ideal Gas Law:

P = (nRT)/V = (1.04E-3 mol * 0.0821 L atm/(K mol) * 281 K) / (360 mL * 1 L/1000 mL) ≈ 0.064 atm

This is the ideal gas pressure if all of the acetone vaporizes at 281 K in the 360 mL container.

To determine the number of moles of HNO3 formed from 8.44 moles of NO2, we need to use the mole ratio from the balanced chemical equation.

To determine the number of moles of HNO3 formed from 8.44 moles of NO2, we need to use the mole ratio from the balanced chemical equation. The ratio of HNO3 to NO2 is 2:3, which means for every 3 moles of NO2, 2 moles of HNO3 are formed. Therefore, we can calculate:

(8.44 mol NO2) / (3 mol NO2) * (2 mol HNO3) = 5.63 mol HNO3

So, 5.63 moles of HNO3 are formed from 8.44 moles of NO2.

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To find the moles of HNO3 formed, we can use the stoichiometric coefficients and set up a proportion. The answer is 5.63 moles HNO3.

To determine the number of moles of HNO3 formed from 8.44 moles of NO2, we can use the stoichiometric coefficients in the balanced equation. From the equation, we see that 3 moles of NO2 produce 2 moles of HNO3. Therefore, if we have 8.44 moles of NO2, we can set up the following proportion:

(3 moles NO2) / (2 moles HNO3) = (8.44 moles NO2) / (x moles HNO3)

Solving for x, we find that x = (2 moles HNO3) * (8.44 moles NO2) / (3 moles NO2) = 5.63 moles HNO3.

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Answer:

[HBr] = 4.7M at equilibrium

Explanation:

Answer:

K in KClO2 = +1

Cl in KClO2 = +3

O in KClO2 = -2

K in KCl = +1

Cl in KCl = -1

O in O2 = 0

Chlorine is going from +3 to -1 so it is being reduced

Oxygen is going from -2 to 0 so it is being oxidized

Explanation:

Potassium is a constant +1

Chlorine could be -1, +1, +3, +5, or +7

Oxygen can be either -2 or +2

Every reactant has to equal zero if there is not a given charge.

Each element can only have certain charges, for example in the previous answer K = +5. Potassium can only be the charge of +1

Chlorine was reduced and oxygen was oxidized in the reaction as shown.

The oxidation number of an atom in a compound is the charge that the atom appears to have as determined by a certain set of rules.

On the left hand side, the oxidation numbers of the elements are;

K = +1

Cl = +3

O = -2

On the right hand side;

K = +1

Cl = -1

O = zero

The element that was oxidized is oxygen. Its oxidation number increased from -2 to zero. The element that was reduced is chlorine. Its oxidation number decreased from +3 to -1.

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Answer:

0.116 g.

Explanation:

mass of C = (no. of moles of C)(atomic mass of C) = (0.117 mol)(12.01 g/mol) = 1.405 g.

mass of H = (no. of moles of H)(atomic mass of H) = (0.233 mol)(1.01 g/mol) = 0.235 g.

∴ mass of O = mass of unknown sample - (mass of C + mass of H) = 3.50 g - (1.405 g + 0.235) = 1.86 g.

∴ no. of moles of O = (mass of O)/(atomic mass of O) = (1.86 g)/(16.0 g/mol) = 0.116 g.

Answer: The empirical formula of the ether will be [tex]C_4H_{10}O[/tex]

Explanation:

The chemical equation for the combustion of ether follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=3.565g[/tex]

Mass of [tex]H_2O=1.824g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 3.565 g of carbon dioxide, [tex]\frac{12}{44}\times 3.565=0.972g[/tex] of carbon will be contained.

In 18g of water, 2 g of hydrogen is contained.

So, in 1.824 g of water, [tex]\frac{2}{18}\times 1.824=0.202g[/tex] of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.972g}{12g/mole}=0.081moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.202g}{1g/mole}=0.202moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.327g}{16g/mole}=0.0204moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0204 moles.

For Carbon = [tex]\frac{0.081}{0.0204}=3.97\approx 4[/tex]

For Hydrogen  = [tex]\frac{0.202}{0.0204}=9.9\approx 10[/tex]

For Oxygen  = [tex]\frac{0.0204}{0.0204}=1[/tex]

The ratio of C : H : O = 4 : 10 : 1

Hence, the empirical formula for the given compound is [tex]C_4H_{10}O_1=C_4H_{10}O[/tex]

Answer:

T₂ = 33.2⁰C

Explanation:

Needed in problem text is the temperature of the acid and base solutions before reaction and the accepted (published) molar heat of neutralization (strong acids) for HCl by NaOH. => Assuming solution temperature is 25⁰C* for both acid and base solutions before reaction and the molar heat of neutralization for HCl by NaOH is 55.7Kj/mole,**  then …

* Standard Thermodynamic conditions => 25⁰C (298K) & 1.00 Atm.

**(https://chemdemos.uoregon.edu/demos/Heat-of-Neutralization-HClaq-NaOHaq)  

NaOH + HCl => NaCl + H₂O

=> 50ml(1.05M NaOH) + 25ml(1.86M HCl)  

=> 0.05(1.05)mole NaOH + 0.025(1.86)mole HCl

=> 0.0525mole NaOH + 0.0465mole HCl

=> (0.0525 – 0.0465)mole NaOH excess + 0.0465mole NaCl + H₂O + Heat

=> 0.0060mole NaOH in excess + 0.0465mole NaCl + H₂O + Heat

Note => NaOH neutralizes 0.0465mole HCl (Limiting Reactant) and produces 0.0465mole NaCl  & H₂O + Heat of Neutralization.

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Heat flow (Q) = Heat received by solvent water from the NaOH + HCl reaction  

=> Q = mcΔT = mc(T₂ - T₁) = specific heat produced by 0.0465mole HCl

=> Q(m) = Molar Heat of Neutralization = Q/mole = mcΔT/n

- m = mass of solvent water receiving heat = (50ml + 25ml)1g/ml = 75g

- c = specific heat of water = 4.184j/g⁰C

- T₂ - T₁ = T₂ - 25⁰C

- Q(m) = 55,700 joules/mole (published heat of neutralization)

- n = moles of HCl neutralized = 0.0465mole HCl

=>Q(m) = mcΔT/n = 55,700j/mole = (75g)(4.184j/g⁰C)(T₂ - 25⁰C) /0.0465mole

Solving for T₂ => T₂ = 33.2⁰C

Final answer:

The final temperature for the neutralization of HCl by NaOH cannot be determined without specific values for initial temperatures, heat of neutralization, specific heat capacity, and density. For a similar reaction with provided values, the final temperature can be calculated based on the produced heat and the specific heat capacity.

Explanation:

The question pertains to the heat of neutralization of HCl(aq) by NaOH(aq). To find out the final temperature of the solution, we would typically calculate the amount of heat produced during the reaction and use it together with the specific heat capacity of the solution. However, the details required for calculation (initial temperatures, heat of neutralization value, specific heat capacity, density) are not provided in the question. In an example provided in the reference, when 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH, both at 22.0 °C in a coffee cup calorimeter, the mixture reached 28.9 °C. The exothermic reaction between HCl and NaOH produces NaCl and H₂O, and the heat produced can be calculated using the change in temperature, the mass of the solution, and the specific heat capacity.

Answer: The mass of nitrogen gas produced will be 45.64 grams.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]    .....(1)

Given mass of [tex]N_2O_4=50.0g[/tex]

Molar mass of [tex]N_2O_4=92.02g/mol[/tex]

Putting values in equation 1, we get:

[tex]\text{Moles of }N_2O_4=\frac{50g}{92.02g/mol}=0.543mol[/tex]

Given mass of [tex]N_2H_4=45.0g[/tex]

Molar mass of [tex]N_2O_4=32.05g/mol[/tex]

Putting values in  equation 1, we get:

[tex]\text{Moles of }N_2H_4=\frac{45g}{32.05g/mol}=1.40mol[/tex]

For the given chemical reaction:

[tex]N_2O_4(l)+2N_2H_4(l)\rightarrow 3N_2(g)+4H_2O(g)[/tex]

By stoichiometry of the reaction:

1 mole of [tex]N_2O_4[/tex] reacts with 2 moles of [tex]N_2H_4[/tex]

So, 0.543 moles of [tex]N_2O_4[/tex] will react with = [tex]\frac{2}{1}\times 0.543=1.086moles[/tex] of [tex]N_2H_4[/tex]

As, the given amount of [tex]N_2H_4[/tex] is more than the required amount. Thus, it is considered as an excess reagent.

Hence, [tex]N_2O_4[/tex] is the limiting reagent.

By Stoichiometry of the reaction:

1 mole of [tex]N_2O_4[/tex] produces 3 moles of nitrogen gas

So, 0.543 moles of [tex]N_2O_4[/tex] will produce = [tex]\frac{3}{1}\times 0.543=1.629moles[/tex] of nitrogen gas.

Now, calculating the mass of nitrogen gas from equation 1, we get:

Molar mass of nitrogen gas = 28.02 g/mol

Moles of nitrogen gas = 1.629 moles

Putting values in equation 1, we get:

[tex]1.629mol=\frac{\text{Mass of nitrogen gas}}{28.02g/mol}\\\\\text{Mass of nitrogen gas}=45.64g[/tex]

Hence, the mass of nitrogen gas produced will be 45.64 grams.

The total heat energy required to convert 25 grams of ice at -4.00°C to steam at 110.0 °C can be calculated in several steps involving warming of ice to 0°C, melting of ice, heating water to 100°C, vaporization of water, and heating of steam to the final temperature. Each step requires the use of specific heat values, molar heats of fusion and vaporization, and the mass of the starting ice chunk. The total energy calculated would be the sum of these steps, yields 75579 Joules.

The question is about calculating the total energy, or enthalpy change, needed to convert ice at -4.00°C to steam at 110.0°C. We need to take into account the warming of ice, the melting of ice, the heating of water, the vaporization of water, and the heating of steam. This process involves several steps and requires using the principle of conservation of energy and heat transfer equations.

First, the ice is warmed to 0°C : ΔH1 = mcΔT = (25.0 grams) (2.09 J/g-K)(4 K) = 209 J

Then, the ice is melted: ΔH2 = n(ΔHfus) = (25.0 grams/18.015 g/mol)(6.01 kJ/mol)= 8.33 kJ = 8330 J

The water is heated to 100°C: ΔH3 = mcΔT = (25.0 grams) (4.18 J/g-K)(100 K) = 10450 J

The water is vaporized: ΔH4 = n(ΔHvap) = (25.0 grams/18.015 g/mol)(40.67 kJ/mol) = 56.13 kJ = 56130 J

Finally, the steam is heated to 110°C: ΔH5 = mcΔT = (25.0 grams) (1.84 J/g-K)(10 K) = 460 J

The total energy change is the sum of all these changes: ΔHtotal = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5 = 209J + 8330J + 10450J + 56130J + 460J = 75579 J

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Answer:

VP(solution) = 186.3 Torr

Explanation:

Given 24g Lactose (C₁₂H₂₂O₁₁) IN 200g H₂O @338K (65°C) & 187.5Torr

VP(soln) = VP(solvent) - [mole fraction of solute(X)·VP(solvent)]  => Raoult's Law

VP(H₂O)@65°C&187.5Torr = 187.5Torr

moles Lactose = (24g/342.3g/mol) = 0.0701mole Lactose

moles Water = (200g/18g/mol) = 11.11mole Water

Total moles = (11.11 + 0.0701)mole = 11.181mole

mole fraction Lac = n(lac)/[n(lac) + n(H₂O)] = (0.0701/11.181) = 6.27 x 10⁻³

VP(solution) = 187.5Torr - (6.27 x 10⁻³)187.5Torr = 186.3Torr

Final answer:

To find the vapor pressure of water above a lactose solution, calculate the mole fraction of water in the solution using moles of lactose and water, then apply Raoult’s Law by multiplying the mole fraction by the vapor pressure of pure water at 338 K, resulting in a vapor pressure of 186.3 torr.

Explanation:

To calculate the vapor pressure of water above a solution prepared by adding 24 g of lactose (C12H22O11) to 200 g of water at 338 K, we use Raoult’s Law. Raoult's Law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution. First, calculate the mole fraction of water in the solution:

Moles of lactose = mass (g) / molar mass (g/mol). For lactose, molar mass = 342.3 g/mol, so moles of lactose = 24 g / 342.3 g/mol = 0.0701 mol.

Moles of water = mass (g) / molar mass (g/mol). For water, molar mass = 18.015 g/mol, so moles of water = 200 g / 18.015 g/mol = 11.10 mol.

Total moles = moles of lactose + moles of water = 0.0701 mol + 11.10 mol = 11.1701 mol.

Mole fraction of water = moles of water / total moles = 11.10 / 11.1701 = 0.9937.

Finally, calculate the vapor pressure of water: Vapor pressure = mole fraction of water × vapor pressure of pure water = 0.9937 × 187.5 torr = 186.3 torr.

Answer: The mass of magnesium consumed will be 14.731 g.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]  .....(1)

Given mass of nitrogen gas = 5.65 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of }N_2=\frac{5.65g}{28g/mol}=0.202mol[/tex]

For the given chemical equation:

[tex]3Mg+N_2\rightarrow Mg_3N_2[/tex]

By Stoichiometry of the reaction:

1 mole of nitrogen gas reacts with 3 moles of magnesium.

So, 0.202 moles of nitrogen gas will react with = [tex]\frac{3}{1}\times 0.202=0.606mol[/tex] of magnesium.

Now, calculating the mass of magnesium by using equation 1, we get:

Moles of magnesium = 0.606 moles

Molar mass of magnesium = 24.31 g/mol

Putting values in equation 1, we get:

[tex]0.606mol=\frac{\text{Mass of magneisum}}{24.31g/mol}\\\\\text{Mass of magnesium}=14.731g[/tex]

Hence, the mass of magnesium consumed will be 14.731 g.

Answer: The correct answer is Option c.

Explanation:

Weak acid is defined as the acid which does not get completely dissociated into its ions when dissolved in water. For Example: [tex]CH_3COOH,H_2CO_3[/tex] etc..

Strong acid is defined as the acid which gets completely dissociated into its ions when dissolved in water. For Example: [tex]HCl,HNO_3[/tex] etc..

Weak base is defined as the base which does not get completely dissociated into its ions when dissolved in water. For Example: [tex]NH_3,NH_4OH[/tex] etc..

Strong base is defined as the base which gets completely dissociated into its ions when dissolved in water. For Example: [tex]NaOH,KOH[/tex] etc..

From the above information, it is clearly visible that the correct answer is Option c.

The correct statement is that C. NH3 is a weak base, and HCl is a strong acid.

When a weak base, such as ammonia (NH3), reacts with a strong acid, like hydrochloric acid (HCl), a chemical reaction occurs. The acid donates protons (H+) to the base, forming water and the conjugate acid of the weak base. For example, in the reaction between NH3 and HCl, [tex]NH3 + HCl -- NH4+ + Cl-,[/tex]  ammonium chloride is formed. The resulting solution is acidic due to the presence of excess H+ ions.

Thus, the correct statement among the options is c. NH3 is a weak base, and HCl is a strong acid. Ammonia (NH3) is a weak base because it donates only a partial amount of its lone pair of electrons. Hydrochloric acid (HCl) is a strong acid because it completely ionizes in water to produce a high concentration of hydrogen ions.

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Answer:

60.00% is the volume percent the ethylene glycol in the antifreeze mixture.

Explanation:

Volume of the ethylene glycol in the mixture = 3.00 gal

Volume of the water in the mixture = 2.00 gal

Total volume of the mixture =3.00 gal + 2.00 gal = 5.00 gal

Volume percent the ethylene glycol :

[tex]\frac{\text{Volume of solute}}{\text{Volume of solution or mixture}}\times 100[/tex]

[tex]\%=\frac{3.00 gal}{5.00 gal}\times 100=60.00\%[/tex]

60.00% is the volume percent the ethylene glycol in the antifreeze mixture.

Answer: The molarity of hydroiodic acid in the titration is 0.485 M.

Explanation:

To calculate the molarity of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HI[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ca(OH)_2[/tex]

We are given:

[tex]n_1=1\\M_1=?M\\V_1=20.3mL\\n_2=2\\M_2=0.186M\\V_2=26.5mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 20.3=2\times 0.186\times 26.5\\\\M_1=0.485M[/tex]

Hence, the molarity of hydroiodic acid is 0.485M.

Final answer:

The molarity of the hydroiodic acid solution is calculated using the stoichiometry of its reaction with calcium hydroxide and the volumes and molarity of the titrant, resulting in a molarity of 0.485 M for the hydroiodic acid.

Explanation:

To find the molarity of the hydroiodic acid solution, we need to first understand the reaction occurring between hydroiodic acid (HI) and calcium hydroxide (Ca(OH)₂). The balanced chemical equation for this reaction is: 2HI(aq) + Ca(OH)₂(aq) → CaI₂(aq) + 2H₂O(l). This equation indicates that two moles of HI react with one mole of Ca(OH)₂.

Using the titration information provided: 26.5 mL of 0.186 M Ca(OH)₂ were required to neutralize 20.3 mL of HI. From this, we can calculate the moles of Ca(OH)₂ used (Moles = Molarity × Volume in L), which is (0.186 M) × (0.0265 L) = 0.004929 moles of Ca(OH)₂. Given the stoichiometry of the reaction, there are twice as many moles of HI, so 0.009858 moles of HI were neutralized.

To find the molarity of the hydroiodic acid solution, we use the formula Molarity = Moles/Volume in L. Here, it is 0.009858 moles / 0.0203 L = 0.485 M HI.

Answer : The mass of [tex]CO_2[/tex] produced will be, 37.488 grams.

Explanation : Given,

Mass of [tex]C_4H_{10}[/tex] = 12.4 g

Mass of [tex]O_2[/tex] = 56.9 g

Molar mass of [tex]C_4H_{10}[/tex] = 58 g/mole

Molar mass of [tex]O_2[/tex] = 32 g/mole

Molar mass of [tex]CO_2[/tex] = 44 g/mole

First we have to calculate the moles of [tex]C_4H_{10}[/tex] and [tex]O_2[/tex].

[tex]\text{Moles of }C_4H_{10}=\frac{\text{Mass of }C_4H_{10}}{\text{Molar mass of }C_4H_{10}}=\frac{12.4g}{58g/mole}=0.213moles[/tex]

[tex]\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{56.9g}{32g/mole}=1.778moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]

From the balanced reaction we conclude that

As, 2 moles of [tex]C_4H_{10}[/tex] react with 13 mole of [tex]O_2[/tex]

So, 0.213 moles of [tex]C_4H_{10}[/tex] react with [tex]\frac{13}{2}\times 0.213=1.385[/tex] moles of [tex]O_2[/tex]

From this we conclude that, [tex]O_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]C_4H_{10}[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]CO_2[/tex].

As, 2 moles of [tex]C_4H_{10}[/tex] react to give 8 moles of [tex]CO_2[/tex]

So, 0.213 moles of [tex]C_4H_{10}[/tex] react to give [tex]\frac{8}{2}\times 0.213=0.852[/tex] moles of [tex]CO_2[/tex]

Now we have to calculate the mass of [tex]CO_2[/tex].

[tex]\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2[/tex]

[tex]\text{Mass of }CO_2=(0.852mole)\times (44g/mole)=37.488g[/tex]

Therefore, the mass of [tex]CO_2[/tex] produced will be, 37.488 grams.

Answer: From the given gases, the greatest rate of effusion is of [tex]CH_4[/tex]

Explanation:

Rate of effusion of a gas is determined by a law known as Graham's Law.

This law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]

It is visible that molar mass is inversely related to rate of effusion. So, the gas having lowest molar mass will have the highest rate of effusion.

For the given gases:

Molar mass of [tex]NH_3=17g/mol[/tex]

Molar mass of [tex]HCl=36.5g/mol[/tex]

Molar mass of [tex]CH_4=16g/mol[/tex]

Molar mass of [tex]Ar=40g/mol[/tex]

Molar mass of [tex]HBr=81g/mol[/tex]

The molar mass of methane gas is the lowest. Thus, it will have the greatest rate of effusion.

Hence, the greatest rate of effusion is of [tex]CH_4[/tex]

Answer:

CH4

CH4

Explanation:

To synthesize the given ether, you would need ethyl bromide as the alkyl halide and ethoxide as the nucleophile.

To synthesize the ether CH3CH2OCH2CH2CHCH3CH3, you would need an alkyl halide and a nucleophile. One possible set of reactants that could be used is ethyl bromide (CH3CH2Br) as the alkyl halide and ethoxide (CH3CH2O-) as the nucleophile. The reaction can be represented as:

CH3CH2Br + CH3CH2O- → CH3CH2OCH2CH2CHCH3CH3 + Br-

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C=[tex]p_o= 187.54 mmHg[/tex]

Vapor pressure of the solution at 65 °C= [tex]p_s[/tex]

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water=[tex]n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol[/tex]

Moles of ethylene glycol=[tex]n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol[/tex]

[tex]\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}[/tex]

[tex]\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}[/tex]

[tex]p_s=173.83 mmHg[/tex]

173.83 mmHg is the vapor pressure of a ethylene glycol solution.