Using the chemical equation: 4NO2(g) + O2(g) + 2H2O(l) = 4HNO3 (aq) : Calculate the theoretical yield in grams of nitric acid when 2.0 moles nitrogen dioxide react with 2.0 moles of oxygen gas.

The mass of NH3 produced from the complete reaction of 2.55g of H2 is calculated to be 14.45 g, based on the molar masses of H2 and NH3 and the stoichiometry of the given balanced chemical equation.

To calculate the mass of NH3 produced by the complete reaction of 2.55g of H2, we first need to determine the molar mass of H2. Knowing that the atomic mass of hydrogen is 1.0, we can say that the molar mass of H2 is 2.0 g/mol. Next, we need to find out how many moles of H2 are in 2.55 g:

Number of moles of H2 = mass (g) / molar mass (g/mol) = 2.55 g / 2.0 g/mol = 1.275 mol

Using the stoichiometry of the balanced equation (N2(g) + 3H2(g) → 2NH3(g)), we can see that 3 moles of H2 produce 2 moles of NH3. Therefore, we can set up a proportion to find the number of moles of NH3 that would be produced from 1.275 moles of H2:

(1.275 mol H2) * (2 mol NH3 / 3 mol H2) = 0.85 mol NH3

Now, to find the mass of NH3, we need to know its molar mass. The atomic mass of nitrogen is 14.0 and hydrogen is 1.0, so the molar mass of NH3 is 14.0 + (3 * 1.0) = 17.0 g/mol. Finally, we can calculate the mass of NH3 produced:

Mass of NH3 = number of moles * molar mass = 0.85 mol * 17.0 g/mol = 14.45 g

The molarity of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex] solution is [tex]\boxed{{\text{0}}{\text{.214 M}}}[/tex].

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

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

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

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex]solution is as follows:

[tex]{\text{Molarity of NaN}}{{\text{O}}_{\text{3}}}{\text{ solution}} = \frac{{{\text{Moles}}\;{\text{of}}\;{\text{NaN}}{{\text{O}}_{\text{3}}}}}{{{\text{Volume }}\left( {\text{L}} \right){\text{ of}}\;{\text{NaN}}{{\text{O}}_{\text{3}}}{\text{ solution}}}}[/tex]      …… (1)

The formula to calculate the moles of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex]is as follows:

[tex]{\text{Moles of NaN}}{{\text{O}}_{\text{3}}} = \frac{{{\text{Given mass of NaN}}{{\text{O}}_{\text{3}}}}}{{{\text{Molar mass of NaN}}{{\text{O}}_{\text{3}}}}}[/tex]                  …… (2)

The given mass of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex] is 45.4 g.

The molar mass of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex]is 84.99 g/mol.

Substitute these values in equation (2).

[tex]\begin{aligned}{\text{Moles of NaN}}{{\text{O}}_3}&=\left( {{\text{45}}{\text{.4 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{84}}{\text{.99 g}}}}} \right)\\&=0.5341\;{\text{mol}}\\\end{aligned}[/tex]

Substitute 0.5341 for the moles of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex]and 2.50 L for the volume of [tex]{\text{NaN}}{{\text{O}}_{\text{3}}}[/tex] solution in equation (1).

[tex]\begin{aligned}{\text{Molarity of NaN}}{{\text{O}}_{\text{3}}}{\text{ solution}}&=\frac{{{\text{0}}{\text{.5341 mol}}}}{{{\text{2}}{\text{.50 L}}}}\\&=0.21{\text{364 M}}\\&\approx{\text{0}}{\text{.214 M}} \\ \end{aligned}[/tex]

The molarity of the [tex]{\mathbf{NaN}}{{\mathbf{O}}_{\mathbf{3}}}[/tex]solution is 0.214 M.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity of NaNO3 solution, 2.50 L, volume of NaNO3 solution, moles of NaNO3, given mass, molar mass, 84.99 g/mol, 45.4 g, 0.214 M, NaNO3, molar mass, given mass.

Answer:

A) 0.225atm

Explanation:

P1V1 = P2V2

V1 = 450ml

P1 = 1.0atm

V2 = 2L = 2 X 100 = 2000ml

P2 =?

1.0 X 450 = P2 X 2000

P2 = (1.0 X 450)/2000

    = 0.225atm

Final answer:

The activation energy for the reverse reaction is calculated using the given activation energy for the forward reaction (100 kJ/mol) and the change in enthalpy of the reaction (-250 kJ/mol), resulting in an activation energy of 350 kJ/mol for the reverse reaction.

Explanation:

The question is about finding the activation energy for the reverse reaction based on the given activation energy and the change in enthalpy for the forward reaction. Using the provided data, Ea for the forward reaction is 100 kJ/mol and ΔH for the reaction is -250 kJ/mol.

To find the activation energy for the reverse reaction, we can use the concept that the sum of the activation energies for the forward and reverse reactions is equal to the difference in energy between the products and reactants. This relationship is derived from the potential energy diagram of a chemical reaction.

The activation energy for the reverse reaction can be calculated using the equation:
Ea(reverse) = Ea(forward) + ΔH

Substituting the given values:

Ea(reverse) = 100 kJ/mol - (-250 kJ/mol)

Ea(reverse) = 100 kJ/mol + 250 kJ/mol

Ea(reverse) = 350 kJ/mol

Therefore, the activation energy for the reverse reaction is 350 kJ/mol.

The speed of a wave can be determined by multiplying its frequency by its wavelength. In this case, a wave with a frequency of 14 hertz and a wavelength of 3 meters will travel at a speed of 42 meters per second.

This is a question related to the physics of wave motion. The speed of a wave can be calculated using the formula:

Speed = Frequency x Wavelength

. Given the frequency of the wave is 14 hertz and the wavelength is 3 meters, you can plug these values into the formula. Therefore, the speed of the wave would be:

14 Hertz x 3 meters = 42 meters per second

. Hence, the wave will travel at a speed of 42 meters per second.

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The correct formula for the compound formed between the iron(III) ion and the oxide ion is Fe2O3.

The compound formed between the iron(III) ion and the oxide ion is called iron(III) oxide. The formula for this compound can be determined by balancing the charges of the ions. The iron(III) ion has a charge of +3 and the oxide ion has a charge of -2. To balance the charges, we need two iron(III) ions for every three oxide ions. Therefore, the correct formula for the compound is Fe2O3.

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Answer: The given statement is false.

Explanation:

Oxidizing agents are defined as the agents that helps in the oxidation of other substance and itself gets reduced. These agents undergo reduction reactions and reduction reaction is the reaction in which an atom gains electrons.

Reducing agents are defined as the agents that helps in the reduction of other substance and itself gets oxidized. These agents undergo oxidation reactions and oxidation reaction is the reaction in which an atom looses electrons.

So, a strong oxidizing agent will gain electrons easily

Hence, the given statement is false.

Final answer:

The subject of the sentence is 'horses,' as it is the main noun that the sentence is about, while 'Arabian,' 'Appaloosa,' and 'Morgan' are adjectives. Furthermore, horses can be classified as mammals in the animal kingdom. Option C

Explanation:

The subject in the sentence 'Arabian, Appaloosa, and Morgan horses will be at the county horse show this week.' is horses. When identifying the subject of a sentence, you are looking for the main noun or noun phrase that the sentence is about.

In this case, 'Arabian,' 'Appaloosa,' and 'Morgan' serve as adjectives describing the kinds of horses that will be present at the show. Therefore, the correct answer is C. horses.

As for classification, it's easy enough to classify the horse in the animal kingdom. That's one level of classification. But horses also belong to other groups; one important group is the mammals. These animals all have fur and nurse their young, which are key characteristics of mammals. Option C

Answer: a liquid changing to a gas

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An endothermic reaction is one that absorbs heat. The dissolving of sodium chloride or sugar in water is usually endothermic since these processes typically absorb heat. The transition of a liquid to a gas is also an endothermic process as it requires heat input.

An endothermic reaction is a process that absorbs heat from the surroundings. When sodium chloride dissolves in water, it can be an endothermic process as the solution usually gets cooler, indicating that heat is absorbed from the surroundings to break the ionic bonds and dissolve the salt. The dissolving of sugar in water generally is also considered slightly endothermic for similar reasons. A liquid changing to a gas, such as water boiling, is an endothermic process as it requires heat to overcome the intermolecular forces in the liquid. Finally, although dissolving strong hydrochloric acid in water is also an interaction with water, it is typically an exothermic process, where heat is released.

The value of [H+] for the solution with [OH−] = 4.0×10−8 is calculated using the formula [H+] = Kw / [OH−], yielding a hydronium ion concentration of 2.5×10−7 M.

To find the value of the hydronium ion concentration ([H+]) for a solution with a given hydroxide ion concentration ([OH−]), you can use the ion product constant for water (Kw), which is always 1.0 × 10−14 M2 at 25°C. The formula is [H+] = Kw / [OH−]. When the [OH−] is 4.0 × 10−8, we can calculate the [H+] as follows:

Therefore, the hydronium ion concentration of the solution is 2.5 × 10−7 M.

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The concentration of Na+ ions in the final solution is determined by calculating the total moles of Na+ in the final solution and dividing by the total volume of the solution. In this example, it is calculated to be 1.33 M.

The subject of the question deals with determining the concentration of sodium ions (Na+) in a mixed solution of sodium sulfate (Na₂SO₄) and sodium chloride (NaCl). The concentration of Na+ ions in the solution can be calculated using the molarity (M) relationships of the two solutions.

First, we determine the amount of Na+ contributed from each solution. The Na₂SO₄ solution will contribute 2 moles of Na+ for each mole of Na₂SO₄, and the NaCl solution will contribute 1 mole of Na+  per mole of NaCl.

In Na₂SO₄, mols = Molarity * Volume (L) = 0.800 M * 0.100 L = 0.080 moles. Each mole of Na₂SO₄ gives 2 moles of Na+, hence total moles of Na+ from Na₂SO₄ is 2 * 0.080 = 0.160 moles.

In NaCl, mols = Molarity * Volume (L) = 1.20 M * 0.200 L = 0.240 moles. Total moles of Na+ from NaCl is 0.240 moles. The total Na+ in the solution is the sum of the Na+ from each, so total moles of Na+ = 0.160 + 0.240 = 0.400 moles.

Finally, molarity of Na+ in the final solution is total moles of Na+ divided by total volume (in Liters). Since it is given that volumes are additive, total volume = 0.100 + 0.200 = 0.300 L. Therefore, molarity of Na+ (M) = 0.400 moles / 0.300 L = 1.33 M. So, the concentration of Na+ in the final solution is 1.33 M.

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c₀(CH₃NH₂) = 0,57 M.

c(CH₃NH₃⁺) = c(OH⁻) = x.

c(NH₂OH) = 0,57 M - x.

Kb = c(CH₃NH₃⁺) · c(OH⁻) / c(CH₃NH₂).

0,00044 = x² /  (0,57 M - x). 

Solve quadratic equation: x = c(OH⁻) = 0,0156 mol/L.

pOH = -log(0,0156 mol/L.) = 1,80.

pH = 14 - 1,80 = 12,2.

The pH value is 12.2

Given:

Question:

The pH value of methylamine

The Process:

Methylamine  is a weak base. When a weak base reacts with water, it produces its conjugate acid and hydroxide ions.

[tex]\boxed{ \ CH_3NH_2_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3NH_3_{(aq)} + OH^-_{(aq)} \ }[/tex]

Let's prepare the equilibrium system to calculate the concentration of hydroxide ions. In chemical equilibrium, the liquid phase has no effect.

[tex]\boxed{ \ K_b = \frac{ [CH_3NH_3] [OH^-] }{ [CH_3NH_2] } \ }[/tex]

Here Kb acts as Kc or equilibrium constant.

[tex]\boxed{ \ 4.4 \times 10^{-4} = \frac{ x \cdot x }{ 0.57 - x } \ }[/tex]

[tex]\boxed{ \ 4.4 \times 10^{-4} = \frac{x^2}{0.57 - x} \ }[/tex]

[tex]\boxed{ \ 2.508 \times 10^{-4} - 4.4 \times 10^{-4}x = x^2 \ }[/tex]

[tex]\boxed{ \ x^2 + 4.4 \times 10^{-4}x - 2.508 \times 10^{-4} = 0 \ }[/tex]

The solution is obtained through the formula of quadratic equations, i.e., [tex]\boxed{ \ x = [OH^-] = 0.0156 \ M \ }[/tex]

Next, we calculated the pOH value followed by the pH value.

[tex]\boxed{ \ pOH = -log [OH^-] \ }[/tex]

[tex]\boxed{ \ pOH = -log [0.0156] \ }[/tex]

We get [tex]\boxed{ \ pOH = 1.81 \ }[/tex]

[tex]\boxed{ \ pH + pOH = 14 \ }[/tex]

[tex]\boxed{ \ pH = 14 - pOH \ }[/tex]

[tex]\boxed{ \ pH = 14 - 1.81 \ }[/tex]

Thus [tex]\boxed{\boxed{ \ pH = 12.19 \ rounded \ to \ 12.2 \ }}[/tex]

- - - - - - -

Quick Steps

0.57 M methylamine (CH₃NH₂)

[tex]K_b = 4.4 \times 10^{-4}[/tex]

We immediately use the formula to calculate the concentration of hydroxide ions for weak bases.

[tex]\boxed{\boxed{ \ [OH^-] = \sqrt{K_b \times base \ concentration} \ }}[/tex]

[tex]\boxed{ \ [OH^-] = \sqrt{4.4 \times 10^{-4} \times 0.57} \ }[/tex]

[tex]\boxed{ \ [OH^-] = 0.0158 \ M \ }[/tex]

Like the steps above, we calculated the pOH value followed by the pH value.

[tex]\boxed{ \ pOH = -log [OH^-] \ }[/tex]

[tex]\boxed{ \ pOH = -log [0.0158] \ }[/tex]

[tex]\boxed{ \ pOH = 1.8 \ }[/tex]

[tex]\boxed{ \ pH = 14 - pOH \ }[/tex]

[tex]\boxed{ \ pH = 14 - 1.8 \ }[/tex]

Thus [tex]\boxed{\boxed{ \ pH = 12.2 \ }}[/tex]

Keywords: determine, the pH, 0.57 M, methylamine, CH₃NH₂, CH₃NH₃, OH⁻, Kb, Kc, equilibrium constant, weak base

Final answer:

The percent ionization of HA in a 0.10 M solution, with a Ka of 3.6×10⁻⁷, is approximately 0.19%.

Explanation:

The percent ionization of a weak acid can be calculated using its Ka value and the initial concentration of the acid. For HA, with a Ka of 3.6×10⁻⁷ and an initial concentration of 0.10 M, the percent ionization is determined as follows: First, set up the reaction as HA → H+ + A-. The equilibrium expression is Ka = [H+][A-] / [HA]. Assuming x is the amount ionized, we have Ka = x² / (0.10 - x). Solving this equation for x, we approximate that x is small compared to the initial concentration, so 0.10 - x is nearly 0.10. Therefore, Ka ≈ x² / 0.10 M. Solving for x gives x = sqrt(Ka × 0.10 M), and the percent ionization = (x / 0.10 M) × 100%. Substituting in the given values, we get percent ionization ≈ sqrt(3.6×10⁻⁷ × 0.10 M) / 0.10 M × 100% = 0.19%.

Answer : The number of moles of nitrogen present in nitrous oxide is 3.32 moles.

Explanation : Given,

Mass of nitrous oxide = 73.0 g

Molar mass of nitrous oxide = 44 g/mole

Now we have to calculate the moles of [tex]N_2O[/tex].

Formula used :

[tex]\text{ Moles of }N_2O=\frac{\text{ Mass of }N_2O}{\text{ Molar mass of }N_2O}[/tex]

[tex]\text{ Moles of }N_2O=\frac{73.0g}{44g/mole}=1.66moles[/tex]

Now we have to calculate the moles of nitrogen in nitrous oxide.

In [tex]N_2O[/tex] molecule, there are 2 moles of nitrogen atoms and 1 mole of oxygen atom.

As, 1 mole of [tex]N_2O[/tex] contains 2 moles of nitrogen

So, 1.66 moles of [tex]N_2O[/tex] contains [tex]1.66\times 2=3.32[/tex] moles of nitrogen.

Therefore, the number of moles of nitrogen present in nitrous oxide is 3.32 moles.

Answer: Option (A) is the correct answer.

Explanation:

When a gas is heated then its molecules gain more amount of kinetic energy. And, as kinetic energy is the energy obtained by an object due to its motion.

Therefore, with increase in kinetic energy of molecules of the gas there will occur more number of collisions. Hence, the gas will move more rapidly from one place to another.

Therefore, upon heating of a gas the energy absorbed by the gas will get converted into kinetic energy due to which gas move much more rapidly.

Potential energy is the energy obtained by an object due to its position and not because of its movement.

Thus, we can conclude that when a gas is heated all of the absorbed energy is converted to kinetic energy.

Final answer:

Carbon monoxide and carbon dioxide both contain carbon and oxygen atoms. CO has one oxygen atom with a triple bond to carbon, whereas CO2 has two oxygen atoms, each with a double bond to carbon. CO is toxic while CO2 contributes to global climate change.

Explanation:

The chemical composition of carbon monoxide (CO) is similar to that of carbon dioxide (CO2) in that both compounds consist of carbon (C) and oxygen (O) atoms. However, the key difference is in the number of oxygen atoms. CO has one oxygen atom, while CO2 has two oxygen atoms. The molecular structures of both compounds reveal these differences. CO has a triple bond between the carbon and the oxygen atom, which includes two covalent bonds and one dative covalent bond. In contrast, the CO2 molecule has a linear structure with a double bond to each oxygen atom, forming an O=C=O configuration.

In terms of Lewis structures, CO's Lewis structure consists of a carbon atom triple-bonded to an oxygen atom with a lone pair on the oxygen, while CO2's Lewis structure displays the carbon atom with two double bonds, each connected to an oxygen atom with two sets of lone pairs.

Both CO and CO2 are important in context as they have significant environmental and health impacts. Carbon monoxide is a toxic gas with the potential to bind to hemoglobin, making it a competitive inhibitor for oxygen transport in the bloodstream. Carbon dioxide, while non-toxic at normal concentrations, is a significant greenhouse gas contributing to global climate change.