What is the volume of 0.200 mol of an ideal gas at 200. kPa and 400. K?

Suppose 231.8 mgmg of PbCl2PbCl2 was added to 15.0 mLmL of water in a flask, and the solution was allowed to reach equilibrium a

vovangra [49]

Answer:

ksp = 2.2 x ⁻⁴

Explanation:

The equilibrium here is:

PbCl₂ (s) ⇄ Pb²⁺ + 2 Cl⁻

we can recognize it as a product solubilty equilibrium once we are told that some undissolved PbCl₂ remained.

The equilibrium constant, Ksp is given by the equation

Ksp = [Pb²⁺][Cl⁻]²

where [Pb²⁺] and [Cl⁻]² are the concentrations (M) of Pb²⁺ and Cl⁻ in solution.

we have the mass of solid PbCl₂ placed in solution, so we can determine the number of moles it represents, and if we substract the moles of undissolved PbCl₂ we will know the moles of Pb²⁺ and Cl⁻ which went into solution.

From there we can calculate the molarity (M= moles/L solution) and finally plug the values into our expression for Ksp to answer this question.

molar mas PbCl₂ = 278.1 g/mol

1 milligram = 1 x 10⁻³ g

mol PbCl₂ initially = 231.8 x 10⁻³ g / 278.1 mol = 8.3 x 10⁻⁴ mol

Volume solution = 15 mL x 1L / 1000 mL = 0.015 L

mol undissolved PbCl₂ = 74 x 10⁻³ g / 278.1 g/mol = 2.7 x 10⁻⁴ mol

mol PbCl dissolved = 8.3 x 10⁻⁴ mol - 2.7 x 10⁻⁴ mol = 5.7 x 10⁻⁴ mol

Concentration of Pb²⁺ in solution = 5.7 x 10⁻⁴ mol / 0.015 L = 3.8 x 10⁻² M

Concentration of Cl⁻ in solution = 2 x 3.8 x 10⁻² M = 7.6 x 10⁻² M

(Note from the formula we we get 2 mol Cl⁻ per mol PbCl₂)

Plugging these values into the expression for Ksp we have

Ksp = 3.8 x 10⁻² x (7.6 x 10⁻²)² = 2.2 x 10⁻⁴

8 0

3 years ago

How many moles of sulfuric acid (H2SO4) are needed to react completely with 6.8 moles of lithium hydroxide (LiOH)? (4 points) 2L

Keith_Richards [23]

Answer:

3.4 mol H2SO4

Explanation:

Took the test

3 0

3 years ago

If one molecule of glucose (C6H12O6) reacts completely with three molecules of oxygen (O2), what is the TOTAL number of atoms th

USPshnik [31]

Answer;

30 atoms

Explanation;

When glucose, C6H12O6, reacts with oxygen it produces carbon dioxide and water. If a molecule of glucose reacts completely with three molecules of oxygen, then 30 atoms would be produced because there are 24 atoms in glucose and 6 atoms in oxygen molecules.

The equation for the reaction would be;

C6H12O6 + 6O2(g) → 6CO2(g) + 6 H2O(l)

7 0

3 years ago

Read 2 more answers

A quantity of NaCl was dissolved in 100 cm of solution to obtain

soldi70 [24.7K]

Answer:

2.93g

Explanation:first, let us calculate the number of mole of NaCl present in the solution. This is illustrated below:

Molarity = 0.5M

Volume = 100cm^3 = 100/1000 = 0.1L

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole of NaCl = 0.5 x 0.1 = 0.05mole

Now we can obtain the mass of NaCl as follows:

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mole of NaCl = 0.05mol

Mass of NaCl =?

Mass = number of mole x molar Mass

Mass of NaCl = 0.05 x 58.5

Mass of NaCl = 2.93g

7 0

4 years ago

Bauxite must go through two processes to produce aluminum metal. The yield of the Bayer process, which extracts aluminum oxide f

Mashcka [7]

The yield of aluminium obtained from 1 m^3 of bauxite is 419810 g

<h3>What is a Percent yield</h3>

A percent yield of a substance measures the amount of the substance actually obtained as a percentage ratio of expected yield.

Percent yield = actual yield / expected yield × 100%

<h3>How to calculate the mass of aluminium obtained from bauxite </h3>

From the data given:

40 % of the bauxite is converted to aluminium oxide.

Volume of bauxite = 1 m^3

40 % of 1 m^3 = 0.4 m^3

volume of aluminium oxide = 0.4 m^3

density of aluminium oxide = 3965 kg/m^3

Using mass = density × volume

mass of aluminium oxide = 0.4 × 3965 kg

mass of aluminium oxide = 1586 kg

Formula of aluminium oxide is Al203

molar mass of aluminium oxide = 102 g

percentage mass of aluminium in one mole of aluminium oxide = mass of aluminium / mass of aluminium oxide × 100 %

Percentage mass of aluminium in aluminium oxide = 54/102 × 100

Percentage mass of aluminium in aluminium oxide = 52.94 %

Expected mass of aluminium from aluminium oxide = 52.94 × 1586

Expected mass of aluminium = 839.62 kg

Actual yield = 40 % × 839.62

Actual yield of aluminium = 419.81 kg

mass of aluminium in grams = 419810 g

Therefore, mass of aluminium obtained from 1 m^3 of bauxite is 419810 g

Learn more about percent yield at: brainly.com/question/8638404

3 0

2 years ago