He hydroxide ion concentration in household ammonia is 3.2 × 10−3 m at 25 °c. what is the concentration of hydronium ions in the solution

The average rate of reaction for CuCl₂ is found by dividing the concentration change (0.345 M) by the time interval (30.0 s), resulting in an average reaction rate of 0.0115 M/s for both the disappearance of CuCl₂ and the formation of Cu.

The student is asking about calculating the average rate of reaction using changes in the concentration of a reactant over a given time interval. The average rate of reaction can be calculated by dividing the change in concentration of a reactant by the time period over which the change occurred. In this case, the concentration of CuCl₂ drops from 1.000 M to 0.655 M over 30.0 seconds.

To find the average rate at which CuCl₂ has reacted, we can use the formula:

The average rate of reaction for the disappearance of CuCl₂ is 0.0115 M/s. Since the reaction stoichiometry shows 3 moles of CuCl₂ produces 3 moles of Cu, the average rate of formation of Cu is also 0.0115 M/s.

When 0.83g of lithium nitride reacts with water, 1.7g of lithium hydroxide is produced according to the balanced chemical equation and stoichiometry.

To determine the mass of lithium hydroxide (LiOH) produced, we first need to understand the balanced chemical equation of the reaction.
Lithium nitride (Li₃N) reacts with water (H₂O) to form lithium hydroxide and ammonia (NH₃), which can be represented as:

Li₃N + 3H₂O → 3LiOH + NH₃

From the equation, we can see that 1 mole of lithium nitride reacts with 3 moles of water to produce 3 moles of lithium hydroxide.
Next, we need to convert the mass of lithium nitride to moles using the molar mass of lithium nitride, which is approximately 34.83 g/mol.

0.83 g Li₃N * (1 mol/34.83 g) = 0.0238 mol

Since the molar ratio of Li₃N to LiOH is 1:3, we multiply this mole by 3 to get the moles of lithium hydroxide produced.

Moles of LiOH = 0.0238 mol * 3 = 0.0714 mol

Finally, we convert the moles of LiOH to grams using the molar mass of LiOH, which is 23.95 g/mol.

Mass of LiOH = 0.0714 mol * 23.95 g/mol = 1.7 g

So, when 0.83g of lithium nitride reacts with water, 1.7g of lithium hydroxide is produced.

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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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1. Calculation of volume of gas: https://brainly.com/question/3636135

2. Determine how many moles of water produce: https://brainly.com/question/1405182

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.

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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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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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.

Final answer:

The empirical formula of the compound is CH2O.

Explanation:

To determine the empirical formula of a substance, we need to find the ratio of the number of atoms of each element in the compound.

Step 1: Convert the given masses of each element to moles. Using the molar mass of each element (C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol), we have 5.28 g C = 0.44 mol C, 0.887 g H = 0.877 mol H, and 3.52 g O = 0.22 mol O.

Step 2: Divide each element's mole value by the smallest mole value to get the simplest ratio. In this case, the smallest mole value is 0.22 mol O. Dividing the other mole values by 0.22 gives us 0.44 mol C, 0.877 mol H, and 1 mol O.

The empirical formula of the compound is CH2O.

Answer:

water carbon dioxide

Explanation:

Final answer:

To determine the equilibrium pressure of NO, set up an ICE table using the values of the equilibrium constant (Kc) and initial partial pressures. Assume the increase in NO pressure is 'x' and the decrease in N2O and O2 pressures is '2x' and 'x', respectively. Calculate the equilibrium pressures by adding the initial pressures to their corresponding changes.

Explanation:

To determine the equilibrium pressure of NO, we need to set up an ICE table and use the values of the equilibrium constant (Kc) and initial partial pressures of the three gases. Let's assume the increase in the pressure of NO is 'x'. Since the stoichiometric coefficient of NO is 2, the decrease in the pressures of N2O and O2 will be '2x' and 'x', respectively. The equilibrium partial pressures of the three gases can be calculated by adding the initial pressures to their corresponding changes.

Initial pressures:

Change in pressures:

Equilibrium pressures:

Since the equilibrium constant (Kc) is given as 1.71 × 10-1, which represents the ratio of the equilibrium concentrations of products to reactants, we can set up the following equation:

Kc = [NO]2 / ([N2O] * [O2])

Substituting the equilibrium pressures into this equation:

1.71 × 10-1 = ([1.96 × 10-3 + x]2 / ([1.96 × 10-3 - 2x] * [1.96 × 10-3 - x]))

Now, solve this equation to find the value of 'x', which represents the equilibrium increase in the pressure of NO.

Once 'x' is determined, substitute it back into the equations for the equilibrium pressures to calculate the final equilibrium pressures of N2O, O2, and NO.

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.

Answer:

1.1 x 10-3 g

Explanation:

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

Answer:

data that adds up to 100%

Explanation:

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

Answer:

0.8 mph

Explanation:

divide distance by time

31.2/ 36 = 0.866..

The reaction tends to completion hence K should be in the order of K>103.

The equilibrium constant is a number that indicates the extent to which reactants are converted to products in a chemical reaction. A high value of equilibrium constant indicates that the system contains mostly products and few reactants at equilibrium. A low equilibrium constant indicates that the concentration of reactants is greater than the concentration of products at equilibrium.

For the reaction;  2H2(g) + O2(g) ⇌ 2H2O(g), we know the reaction to be explosive and tend to completion since it is a combustion reaction. This means that the reactants are mostly converted to products and the equilibrium constant will be large. Hence K should be in the order of K>103.

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Hydrogen is used as a rocket fuel because it is very light and reacts explosively and completely with oxygen. For the combustion reaction 2H2(g)+O2(g)⇌2H2O(g) what is the likely magnitude of the equilibrium constant K?

1. K<10−3

2. 10−3

3. K=0

4. K>103

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory:

Acetylsalicylic acid when dissolved in water donates its proton to form conjugate base and water gains the proton to form conjugate acid.The net ionic equation is given as:

[tex]HC_9H_7O_4(0+H_2O\rightarrow (C_9H_7O_4)^{-}+H_3O^+[/tex]

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:

[tex]C_6H_{15}O_6[/tex]

Explanation:

Hello,

In this case, one computes the molar mass for the given empirical formula as follows:

[tex]M_{empirical}=12*2+1*5+16*2=61g/mol[/tex]

In such a way, one computes how many times the molecular formula's molecular mass is contained into the empirical formula's molecular mass in order to determine the whole number relating them as shown below:

[tex]\frac{183.2g/mol}{61g/mol}=3[/tex]

Finally, the molecular formula is three times the empirical formula, hence:

[tex]C_6H_{15}O_6[/tex]

Best regards.