Calculate the mass of sodium chloride produced when 5.50 moles of sodium reacts in excess chlorine gas

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Rubbing alcohol is 70.0% isopropyl alcohol by volume. how many ml of isopropyl alcohol are in a full 1 pint (473 ml) container?

We have the following information:  

% V/V (percentage of volume per volume) = 70  

V1 (solute volume) = ? (in mL)  

V (volume of the solution) = 473 mL

We apply the data to the formula of volume percentage of the solute per volume of solution, we will have:  

[tex]\%\:v/v = \dfrac{v_1}{v} *100[/tex]

[tex]70 = \dfrac{v_1}{473}*100[/tex]

[tex]70*473 = 100*v_1[/tex]

[tex]33110 = 100\:v_1[/tex]

[tex]100\:v_1 = 33110[/tex]

[tex]v_1 = \dfrac{3311\diagup\!\!\!\!0}{10\diagup\!\!\!\!0}[/tex]

[tex]\boxed{\boxed{v_1 = 331.1\:mL\:of\:isopropyl\:alcohol}}\Longleftarrow(solute\:volume)\:\:\:\:\:\:\bf\purple{\checkmark}[/tex]

Answer:

331.1 ml of isopropyl alcohol

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[tex]\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}[/tex]

The concentration terms are molality, normality and mole fraction and percent by volume. Therefore, 331.1ml of isopropyl alcohol are in a full 1 pint (473 ml) container.

The percent by volume formula calculates the concentration of the solution by using the volume of the solute (the dissolved content in the solvent) and the volume of the solution. It is represented as (v/v%).

Mathematically,

percent by volume=(volume of solute÷ volume of solution)×100

% V/V (percentage of volume per volume) = 70  

V1 (solute volume) = ? (in mL)  

V (volume of the solution) = 473 mL

Substituting all the given values in the given formula, we get

70  =(volume of solute÷  473 )×100

volume of solute=331.1ml

Therefore, 331.1ml of isopropyl alcohol are in a full 1 pint (473 ml) container.

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The balanced nuclear equation for the beta decay of protactinium is 231Pa → 231U + β- + ν. In this equation, a protactinium-231 nucleus decays into a uranium-231 nucleus through the emission of a beta particle (β-) and an electron neutrino (ν).

The balanced nuclear equation for the beta decay of protactinium is:

231Pa → 231U + β- + ν

In this equation, a protactinium-231 nucleus decays into a uranium-231 nucleus through the emission of a beta particle (β-) and an electron neutrino (ν). The atomic number of the nucleus increases by one, from 91 to 92, due to the gain of a proton.

Final answer:

To find the mass of MgCl₂ produced from the reaction of milk of magnesia with stomach acid, the mass of Mg(OH)₂ is converted to moles, which is then used to calculate the moles and subsequently the mass of MgCl₂ formed, resulting in 2.97 grams.

Explanation:

The neutralization of stomach acid by milk of magnesia, which contains magnesium hydroxide (Mg(OH)2), can be represented by the following chemical reaction:

Mg(OH)₂(s) + 2 HCl(aq) → 2 H₂O(ℜ) + MgCl₂(aq)

To determine the mass of MgCl₂ produced when 1.82 g of Mg(OH)₂ reacts, we will follow these steps:

Step 1: Molar mass of Mg(OH)₂ = 24.31 (Mg) + 2(16.00 (O) + 1.01 (H)) = 58.33 g/mol

Step 2: Moles of Mg(OH)₂ = 1.82 g / 58.33 g/mol = 0.0312 mol

Step 3: From the reaction stoichiometry, 1 mol of Mg(OH)₂ produces 1 mol of MgCl₂.

So, moles of MgCl₂ produced = 0.0312 mol

Step 4: Molar mass of MgCl₂ = 24.31 (Mg) + 2(35.45 (Cl)) = 95.21 g/mol

Mass of MgCl₂ = 0.0312 mol * 95.21 g/mol = 2.97 g

Thus, 2.97 g of MgCl₂ will be produced.

the concentration of [tex]\( \text{OH}^- \)[/tex] ions in the solution is [tex]\( 1.0 \times 10^{-3} \, \text{mol/L} \)[/tex].

To find the concentration of [tex]\( \text{OH}^- \)[/tex] ions [tex](\( [\text{OH}^-] \))[/tex] in a solution given the concentration of [tex]\( \text{H}^+ \)[/tex] ions [tex](\( [H^+] \))[/tex], we can use the fact that in water at [tex]\( 25^\circ \text{C} \)[/tex] [tex](\( 298 \text{K} \))[/tex], the product of [tex]\( [\text{H}^+] \)[/tex] and [tex]\( [\text{OH}^-] \)[/tex] is constant and equal to [tex]\( 1.0 \times 10^{-14} \, \text{mol/L} \)[/tex]. This is known as the ion product of water [tex](\( K_w \)).[/tex]

[tex]\[ K_w = [\text{H}^+] \times [\text{OH}^-] \][/tex]

Given [tex]\( [\text{H}^+] = 1 \times 10^{-1} \, \text{mol/L} \)[/tex], we can rearrange the equation to solve for [tex]\( [\text{OH}^-] \)[/tex]:

[tex]\[ [\text{OH}^-] = \frac{K_w}{[\text{H}^+]} \][/tex]

Substitute the given values:

[tex]\[ [\text{OH}^-] = \frac{1.0 \times 10^{-14} \, \text{mol/L}}{1 \times 10^{-1} \, \text{mol/L}} \][/tex]

[tex]\[ [\text{OH}^-] = 1.0 \times 10^{-14} \, \text{mol/L} \times \frac{1}{1 \times 10^{-1}} \, \text{mol/L} \][/tex]

[tex]\[ [\text{OH}^-] = 1.0 \times 10^{-14} \times 10^{1} \, \text{mol/L} \][/tex]

[tex]\[ [\text{OH}^-] = 1.0 \times 10^{-3} \, \text{mol/L} \][/tex]

Therefore, the concentration of [tex]\( \text{OH}^- \)[/tex] ions in the solution is [tex]\( 1.0 \times 10^{-3} \, \text{mol/L} \)[/tex].

A is the wave with the longest wavelength. The graph clearly shows this since wave A has a larger wavelength than waves B and C. Wave A has the longest wavelength when compared to the other two waves.

C is the wave with the least amplitude. The graph shows this since wave C has a significantly lesser amplitude than waves A and B. Wave C has the least amplitude when compared to the other two waves. B is the most energetic wave.

The graph shows this because wave B has a larger amplitude than waves A and C. Wave B has the greatest amplitude when compared to the other two has the more energy in comparison to the two.

The longest wavelength wave has the lowest frequency and the lowest energy. A wave's wavelength is inversely related to its frequency, which implies that as the wavelength grows, so does the frequency. Because energy is proportional to frequency, the wave with the longest wavelength has the least energy.

The wave with the smallest amplitude has the least amount of energy. The largest displacement of a wave from its equilibrium position is referred to as its amplitude. The energy of a wave is exactly proportional to its amplitude, therefore as the amplitude falls, so does the energy.

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When 5.6 g of Copper (II) oxide are used, approximately 0.07035 moles of copper metal will be produced. This is calculated using the molar mass of CuO (79.55 g/mol) and the 1:1 stoichiometric ratio in the decomposition reaction of CuO to metallic Cu.

To determine how many moles of copper metal will be produced from 5.6 g of Copper (II) oxide, we need to use the molar mass of Copper (II) oxide (CuO). The molar mass of CuO is approximately 79.55 g/mol. By dividing the mass of Copper (II) oxide used by its molar mass, we can find the number of moles of CuO that we have.

Here is the calculation:

Number of moles of CuO = mass of CuO (g) / molar mass of CuO (g/mol)

Number of moles of CuO = 5.6 g / 79.55 g/mol

Number of moles of CuO = 0.07035 mol

Since Copper (II) oxide decomposes to copper and oxygen, and the stoichiometry of the decomposition reaction indicates a 1:1 mole ratio between CuO and metallic Cu, we can predict that the moles of copper metal produced would also be 0.07035 mol. Thus, 5.6 g of Copper (II) oxide would produce approximately 0.07035 mol of copper metal.

In the alveoli of the lungs, CO2 and O2 are exchanged with blood in the surrounding capillaries via diffusion. This passive process occurs due to the concentration gradients of the gases and is aided by the large surface area of the alveoli and the thin walls of both the alveoli and capillaries.

In the alveoli and lung capillaries, CO2 and O2 are exchanged by means of diffusion. This process is known as pulmonary gas exchange and is crucial for respiration. Oxygen diffuses from the alveoli into the blood because there is a higher concentration of oxygen in the inhaled air than in the blood of the capillaries. Conversely, carbon dioxide, which is more concentrated in the blood, diffuses from the capillaries into the alveoli to be exhaled. This exchange occurs passively and requires no energy due to the concentration gradients between the gases in the alveoli and the blood.

The efficiency of this gas exchange is supported by the large surface area provided by the vast number of alveoli in the lungs and the thin-walled nature of the alveoli and capillaries. Constant blood flow and regular breathing maintain a steep concentration gradient for oxygen and carbon dioxide, facilitating the diffusion process.

An Arrhenius base is a substance that dissociates in water to form hydroxide ions (OH⁻). 
In this example lithium hydroxide is an Arrhenius base:

LiOH(aq) → Li⁺(aq) + OH⁻(aq).

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons (H⁺). 
For example hydrochloric acid: HCl(aq) → H⁺(aq) + Cl⁻(aq).

4) red litmus turn blue whe base is drop on it and blue litmus turn red when acid is drop on it.

5) bases have pH greater than 7, acids have pH less than 7.

The properties that would possess by both acids and bases is they conduct electricity.

Acids are sour in taste and turn blue litmus paper to red.

They react with metals to liberate hydrogen.

Bases are solution that accept electron pair bonds.

NaOH is a base and HCl is an acid.

Thus, the correct option is 3, they conduct electricity.

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To achieve the desired 40% antifreeze solution, you should drain 70 liters of the 20% solution and replace it with 70 liters of 100% antifreeze.

To get a 40% antifreeze solution in the radiator, we need to drain a certain amount of the 20% solution and replace it with 100% antifreeze. Let's break down the problem step by step:

1. Calculate the amount of antifreeze in the original 70-liter 20% solution:
- The original solution is 20% antifreeze, so 20% of 70 liters is (20/100) * 70 = 14 liters of antifreeze.

2. Determine how much antifreeze we need in the final solution:
- We want a 40% antifreeze solution, so the final solution should have (40/100) * (70 liters) = 28 liters of antifreeze.

3. Find the amount of solution to drain:
- The difference between the desired and current amount of antifreeze is 28 liters - 14 liters = 14 liters.
- Since the original solution is 20% antifreeze, we can equate 14 liters to 20% of the total solution, and solve for the total amount of solution.
- Let's represent the total solution as "x" liters. So, (20/100) * x = 14 liters.
- We can solve this equation by dividing both sides by 20/100: x = 14 liters / (20/100) = 70 liters.

4. Calculate the amount of solution to drain and replace with 100% antifreeze:
- We need to replace 70 liters of the 20% solution with 100% antifreeze.
- So, we should drain and replace 70 liters of the original solution with 70 liters of 100% antifreeze.

Therefore, to achieve the desired 40% antifreeze solution, you should drain 70 liters of the 20% solution and replace it with 70 liters of 100% antifreeze.

Final answer:

To determine how many liters of a 20% antifreeze solution should be drained and replaced with 100% antifreeze to make a 40% solution in a 70-liter system, we set up and solve an equation based on the volume and concentration of antifreeze before and after draining and adding pure antifreeze.

Explanation:

Let's denote the amount of solution to be drained and replaced with 100% antifreeze as 'x' liters. The amount of antifreeze in the original solution is 20% of 70 liters, or 0.2 × 70 = 14 liters. When 'x' liters of this solution are removed, 'x' liters of pure antifreeze are added, so the new volume of antifreeze in the solution will be (14 - 0.2x + x) liters. The total volume of the solution remains 70 liters, and we want it to be a 40% antifreeze solution. So, we set up the equation:

Solving for 'x', we find the amount of solution to be replaced with 100% antifreeze to obtain the 40% solution desired.

For example sodium chloride is ionic compound and strong electrolyte and dissociates in water on hydrated sodium cations and chlorine anions:

NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
In water, oxygen is negative and attract the positive ions of sodium, hydrogen is positive and attract the negative ions of chlorine.

Answer: If the molecule is symmetric, the effect of polarity will cancel out

Explanation: Apex

Answer is: molarity of sodium hydroxide solution is c. 4 M.
m(NaOH) = 80.0 g.
n(NaOH) = m(NaOH) ÷ M(NaOH).
n(NaOH) = 80 g ÷ 40 g/mol.
n(NaOH) = 2 mol.
V(NaOH) = 500 mL ÷ 1000 mL/L = 0.5 L.
c(NaOH) = n(NaOH) ÷ V(NaOH).
c(NaOH) = 2 mol ÷ 0.5 L.
c(NaOH) = 4 mol/L.

n - amount of substance.

The molarity of the solution indicates the concentration of the solute dissolved in the solution. The molarity of the 500 mL solution is 4M. Thus, option c is correct.

Molarity is the ratio of the moles of the solute that were dissolved in a solvent to make a liter of solution. The molar concentration is calculated as:

Molarity (M) = moles ÷ volume

Given,

The mass of sodium hydroxide = 80 gm

Volume of solution = 500 mL

Molar mass of sodium hydroxide = 39.997 g/mol

Moles of sodium hydroxide is calculated as:

Moles = mass ÷ molar mass

= 80 ÷ 40

= 2 moles

Molarity is calculated as:

M = n ÷ V

= 2 moles ÷ 0.5 L

= 4 M

Therefore, option c. the molarity of sodium hydroxide solution is 4 M.

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To produce 3.5 moles of aluminum oxide (Al₂O₃), you need 188.86 grams of aluminum (Al) in the presence of excess oxygen (O₂).

This is determined using the balanced chemical equation and molar masses.

To determine how many grams of aluminum (Al) are required to produce 3.5 moles of aluminum oxide (Al₂O₃) in the presence of excess oxygen (O₂), you need to use stoichiometry and the molar masses of the compounds involved.

The balanced chemical equation for the formation of aluminum oxide is:

4Al + 3O₂ → 2Al₂O₃

From the equation, 4 moles of aluminum (Al) produce 2 moles of aluminum oxide (Al₂O₃). This means that 2 moles of Al₂O₃ are produced from 4 moles of Al.

To find out how many moles of Al are needed for 3.5 moles of Al₂O₃, set up the proportion:

(4 moles Al / 2 moles Al₂O₃) = (x moles Al / 3.5 moles Al₂O₃)

Solving for x:

x = (4/2) * 3.5 = 7 moles of Al

The molar mass of Al is approximately 26.98 g/mol. To find the mass of 7 moles of Al:

7 moles Al * 26.98 g/mol = 188.86 g Al

Thus, 188.86 grams of aluminum are required to produce 3.5 moles of aluminum oxide in the presence of excess oxygen.

Final answer:

The equation for the base ionization constant of PO4^3- is K_base = K_1 * K_2 * K_3.

Explanation:

The base ionization constant of the PO4^3- ion can be determined from the ionization reactions it undergoes.

The ionization reactions for the PO4^3- ion are:

HPO4^2- (aq) + H2O(l) → H2PO4^-(aq) + OH^-(aq), with a base ionization constant of K1.

H2PO4^-(aq) + H2O(l) → H3O^+(aq) + HPO4^2-(aq), with a base ionization constant of K2.

HPO4^2-(aq) + H2O(l) → H3O^+(aq) + PO4^3-(aq), with a base ionization constant of K3.

The base ionization constant for the PO4^3- ion is the product of the base ionization constants of each ionization reaction:

Kbase = K1 * K2 * K3.