What is the partial pressure of water vapor in an air sample when the total pressure is 1 atm, the partial pressure of nitrogen is 0.79 atm, the partial pressure of oxygen is 0.20 atm and the partial pressure of all other gases in air is 0.0044 atm?

Answer:  an aqueous potassium hydroxide solution

An alkaline fuel cell is a zero-emission device whose one major component is the electrolyte. An electrolyte, on the other hand,  is a solution that is able to conduct electricity. In an alkaline fuel cell. The electrolyte is an alkaline liquid, and potassium hydroxide also known as KOH is the only alkaline liquid among the choices. 

The formation of ionic compounds is favored when a metal, which loses electrons, combines with a nonmetal, which gains electrons. This process results in an ionic compound stabilized by ionic bonds between ions of opposite charges, a well-known example being sodium chloride (NaCl). It's also possible for an ionic compound to form between a metal atom and a halogen.

The conditions favoring the formation of ionic compounds typically involve a metal and a nonmetal. This is because metals, which have low ionization potential, tend to readily lose electrons and nonmetals, with high electron affinities, tend to gain electrons. In this process, such as the formation of sodium chloride (NaCl), the metal (sodium) loses an electron to form a cation (Na+), and the nonmetal (chlorine) gains an electron to form an anion (Cl-), resulting in an ionic compound. The compound is stabilized by ionic bonds, which are electrostatic attractions between ions of opposite charges.

The formation of ionic compounds ensures that both the metal and nonmetal achieve a stable electron configuration, often referred to as an octet. Additionally, ionic compounds can also form between a metal atom and a halogen as halogens are a group of nonmetals that are extremely electron-affinitive.

Lastly, it's important to note that not all combinations of metals and nonmetals produce ionic compounds. For instance, compounds that do not contain ions but consist of atoms bonded tightly together in molecules, usually form from two nonmetals and are called covalent compounds.

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[tex]\boxed{\text{4.79 g}}[/tex] of KBr is required to make 350 mL of 0.115 M KBr solution.

Further Explanation:

Concentration is defined as the amount or quantity of solute present in specific amount of solution. In order to evaluate concentration of various solutions, different concentration terms are used. Some of these are mentioned below.

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

The formula to calculate molarity of KBr solution is as follows:

  [tex]\text{Molarity of KBr solution}=\dfrac{\text{Moles of KBr}}{\text{Volume (L) of KBr solution}}[/tex]  ...... (1)

Rearrange equation (1) to calculate moles of KBr.

[tex]\begin{aligned}{\text{Moles of KBr}&=\left[{(\text{Molarity of KBr solution})\\&\text{ }\text{ }\text{ }(\text{Volume (L) of KBr solution})]\right]\end{aligned}}[/tex]        ...... (2)

Substitute 0.115 M for molarity of KBr solution and 350 mL for volume of KBr  solution in equation (2).

[tex]\begin{aligned}\text{Moles of KBr}&=(\text{0.115 M})(\text{350 mL)}\left(\dfrac{\text{10}^{-3}\text{ L}}{\text{1 mL}}\right)\\&=\text{0.04025 mol}\end{aligned}[/tex]

The formula to calculate moles of KBr is as follows:

[tex]\text{Moles of KBr}=\dfrac{\text{Mass of KBr}}{\text{Molar mass of KBr}}[/tex]                                       ...... (3)

Rearrange equation (3) for mass of KBr.

[tex]\text{Mass of KBr}=(\text{Moles of KBr})(\text{Molar mass of KBr})[/tex]               ...... (4)

Substitute 0.04025 mol for moles of KBr and 119.002 g/mol for molar mass of KBr in equation (4).

[tex]\begin{aligned}{\text{Mass of KBr}&=(\text{0.004025 mol})(\text{119.002 g/mol})\\&=\text{4.79 g}}\end{aligned}[/tex]

Therefore 4.79 g of KBr is required to make 350 mL of a 0.115 M KBr solution.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, KBr, 350 mL, 0.115 M, 4.79 g, mass, molar mass, concentration terms, concentration, solute, moles of solute, volume.

Option d: N and P

In a periodic table, elements within the same group have similar properties.

(a) K and He: Here, K (potassium) belongs to group 1 , alkali metals and He belongs to group 18, noble gas thus, they cannot have similar properties.

(b) Li and B: Here, Li (lithium) belongs to group 1, alkali metals and B (boron) belongs to group 13, it is the only non-metal in group 13. Thus, they cannot have similar properties.

(c) I and Ca: Here, I (iodine) belongs to group 17, halogen, they electronegative in nature and Ca (calcium) belongs to group 2, alkaline earth metal, they are electropositive in nature. Thus, they cannot have similar properties.

(d) N and P: Here, both N (nitrogen) and P (phosphorus) belongs to group 15, they both are non metals and have 5 electrons in their outermost shell thus, they both have similar properties.

Therefore, option (d) is correct.

Answer:

The electron configuration of a chlorine ion in a compound of BeCl2 is

1s²2s²2p⁶3s²3p⁶

Explanation:

Beryllium chloride (BeCl2) is an ionic compound where Be is an alkaline earth metal (group 2) and Cl is a halogen atom (group 17).

Alkaline earth metals have a charge of +2 while halogens exhibit a -1 charge. Hence in BeCl2 each Be atom exists as Be2+ cation and Cl exists as Cl- anion.

The atomic number of Cl = 17. The valence electron configuration for the neutral atom is:

Cl(Z=17) = 1s²2s²2p⁶3s²3p⁵

The chloride ion has an addition electron, therefore

Cl⁻(Z = 17+1 = 18) = 1s²2s²2p⁶3s²3p⁶

Final answer:

The volume at STP that 50.2 grams of CO2 will occupy is found by converting the mass to moles and then multiplying by the molar volume of a gas at STP, which is 22.4 liters per mole.

Explanation:

To find the volume at STP that 50.2 grams of CO2 (g) will occupy, we first need to convert the mass of CO2 to moles using the molar mass of CO2, which is approximately 44.01 g/mol. Next, we apply the concept that one mole of any gas at STP will occupy 22.4 liters. The calculation involves dividing the mass of CO2 by its molar mass to get the moles, and then multiplying the number of moles by 22.4 L/mol to find the volume.The steps are as follows:Calculate the number of moles: number of moles = mass (g) / molar mass (g/mol)

Calculate the volume at STP: volume (L) = number of moles x 22.4 L/mol

By following these steps, we can determine the volume of CO2 gas at STP conditions.

Answer:

0.668 g of barium sulfate

Explanation:

Given,

Balanced chemical equation: BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl.

Volume of BaCl₂ = 25.34 mL x [tex]\frac{1L}{1000 ml }[/tex]= 0.02534 L.

Molarity of BaCl₂ = 0.113 M

Molarilty = [tex]\frac{moles of solute}{L of the solution }[/tex]

Moles of solute = Molarilty x L of the solution

Moles of BaCl₂ = 0.113 M x 0.02534 L = 0.00286 mol.

From the balanced chemical equation there is a 1:1 molar ratio between BaCl₂ and BaSO₄

Therefore, moles of BaCl₂ = moles of BaSO₄

Moles of BaSO₄ = 0.00286 mol.

Mass of BaSO₄ = moles of BaSO₄ x Molar mass of BaSO₄

Mass of BaSO₄ = 0.00286 mol x 233.4 g/mol.

Mass of BaSO₄ = 0.668 g.

Answer:

0.015 for APEX

Explanation:

In the titration of a weak acid with a strong base, the pH after adding 20.00 mL of base can be determined by considering the concentration of the weak acid and the remaining base after the reaction.

To determine the pH after adding 20.00 mL of a 0.10 M strong base to a 40.00 mL sample of a 0.10 M weak acid with a Ka value of 1.8×10−5, we need to consider the reaction between the acid and the base.

Since the acid is weak, it does not dissociate completely. As a result, the pH can be calculated by considering the concentration of the weak acid and the remaining base after the reaction.

Using the titration curve as a reference, you can find the pH value corresponding to 20.00 mL of added base.

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An organic compound which contains only carbon and hydrogen atoms is called the hydrocarbon. The expanded structural formula for an alkane with three carbon atoms (Propane) is CH₃–CH₂–CH₃. The correct option is B.

The saturated hydrocarbons which consists of only single bonded carbon and hydrogen atoms without any other functional groups are known as alkanes. Their general formula is CₙH₂ₙ ₊ ₂ where 'n' represents the number of carbon atoms.

A formula which denotes all the atoms and bonds present in a compound is defined as the expanded structural formula. But a condensed structural formula omits most of the bonds.

The formula of propane is C₃H₈ and it is a colourless gas.  Its expanded structural formula is CH₃–CH₂–CH₃ which contains three single bonds, 8 'H' atoms and 3 'C' atoms.

Thus the correct option is B.

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The expanded structural formula for an alkane with three carbon atoms is CH₃-CH₂-CH₃, which represents propane with the molecular formula C₃H₈.

The expanded structural formula for an alkane with three carbon atoms is CH₃-CH₂-CH₃. This compound is known as propane, which follows the general alkane formula CₙH₂ₙ+2. In this case, with three carbon atoms (n=3), the molecular formula becomes C₃H(₂x₃)₊₂, which simplifies to C₃H₈. Therefore, C₃H₈ is the correct molecular formula for propane, whereas C₃H₆ would be an alkene with a double bond between carbon atoms, not an alkane.

Answer: Esters for Plato

Explanation:

The ideal equation relates the temperature with the pressure and the volume of the gas. When the temperature is increased then the volume will be 4 cubic meters.

An ideal gas equation depicts the relation between the temperature to that of the volume and the pressure of the gas.

The formula is given as,

[tex]\rm \dfrac{P_{1}V_{1}}{P_{2}V_{2}} = \rm \dfrac{T_{1}}{T_{2}}[/tex]

Given,

Initial pressure = 100 kPa

Initial volume = 2 cubic meter

Initial temperature = 100 K

Final pressure = 200 kPa

Final volume = ?

Final temperature = 400 K

The final volume is calculated as:

[tex]\begin{aligned} \rm V_{2} &= \rm \dfrac{P_{1}V_{1}T_{2}}{T_{1}}\\\\&= \dfrac{100\times 2 \times 400}{200 \times 100}\\\\&= 4 \;\rm m^{3}\end{aligned}[/tex]

Therefore, 4 cubic meters is the volume of the final gas.

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The 3.0% (v/v) label indicates that for every 100 mL of solution, there are 3.0 mL of hydrogen peroxide. Thus, in a 400 mL bottle, there would be 12 mL of hydrogen peroxide.

The percentage given here is a volume/volume percentage, meaning that for every 100 mL of solution, there are 3.0 mL of hydrogen peroxide. To find out how much hydrogen peroxide is in a 400 mL bottle, you just need to scale this proportion up.

Set up an equation like this:

3.0 mL H₂O₂ / 100 mL solution = x mL H₂O₂ / 400 mL solution. Solving for x gives you:
x = (3.0 mL H₂O₂ / 100 mL solution) * 400 mL solution

So, there are 12 mL of hydrogen peroxide in the 400 mL bottle of solution.

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Final answer:

The molality of the solution is calculated by dividing the number of moles of glucose by the mass of water in kilograms, resulting in a solution with a molality of 0.544 m.

Explanation:

The question is asking to calculate the molality of a glucose solution. To find the molality, we need to know the mass of the solute (glucose) and the mass of the solvent (water) in kilograms.

Steps to Calculate Molality

First, calculate the number of moles of glucose (C6H12O6). Its molar mass is 180.16 g/mol.

Next, convert the mass of water from milliliters to kilograms. Since the density of water is 1.00 g/mL, we simply convert 150.0 mL to grams and then to kilograms.

Finally, use the equation for molality: molality (m) = moles of solute (mol) / mass of solvent (kg).

Let's perform the calculations:

Moles of glucose = 14.7 g / 180.16 g/mol = 0.0816 mol

Mass of water = 150.0 mL * 1.00 g/mL = 150.0 g = 0.150 kg

Molality of the solution = 0.0816 mol / 0.150 kg = 0.544 m

The empirical formula should be [tex]C_3H_7[/tex]

C: 83.7% = 83,7 g

H: 16.3% = 16.3 g

Now

[tex]C = 83.7 \div 12 = 6.975 mol\\\\H = 16.3 \div 1 = 16.3 mol[/tex]

The above does not represent the integers

So,

[tex]C = 6.975 \div 6.975 = 1\\\\H = 16.3 \div 6.975 = 2.3[/tex]

Therefore the above empirical formula should be used.

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