A scientist is at the hospital visiting a friend's young daughter, who has been

In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,

Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - -

At t =equilibrium (x-s) s 2s

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$K_{sp}=s\times {2s}^2$

$K_{sp}=4s^3$

Given s = 7.4×10⁻³ M

So, Ksp is:

$K_{sp}=4\times (7.4\times 10^{-3})^3$

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - 0.20

At t =equilibrium (x-s') s' 0.20+2s'

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2$

Solving for s', we get

s' = 4.0×10⁻⁵ M

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

3 0

2 years ago

50cm3 of sodium hydroxide solution was titrated against a solution of sulfuric acid. The concentration of the sodium hydroxide s

miskamm [114]

Answer:

49 g/L is the concentration of the acid

Explanation:

Firstly, we proceed to write the equation of reaction.

2NaOH + H2SO4 ——-> Na2SO4 + 2H2O

We can see that 1 mole of the base reacted with two moles of the acid.

kindly note that dm^3 is same as liter

Firstly, we need to get the concentration of the reacted sulphuric acid in g/L

we use the simple titration equation below;

CaVa/CbVb = Na/Nb

From the question;

Ca = ?

Va = 25 cm^3

Cb = 20 g/L

we convert this to concentration in mol/L

Mathematically, that is concentration in g/L divided by molar mass in g/mole

molar mass of NaOH = 40 g/mol

so we have; 20g/L / 40 = 0.5 mol/L

Vb = 50 cm^3

Na = 1

Nb = 2

Where C represents concentrations, V volumes and N , number of moles

Now, substitute the values;

Ca * 25/0.5 * 50 = 1/2

25Ca/25 = 0.5

So Ca = 0.5 mol/L

Now to get the concentration of H2SO4 in g/L

What we do is to multiply the concentration in mol/L by molar mass in g/mol

That would be 0.5 * 98 = 49 g/L

4 0

2 years ago

What is the oxidation number of pt in k2ptcl6?

padilas [110]

Your compound is $K_{2}PtCl_{6}$ .

Remember that the oxidation numbers in a neutral compound must add up to zero. Cl has an oxidation number of -1 because it is a halogen K has an oxidation number of +1 because it is an alkali metal, which exhibits an oxidation state of +1 in compounds.

Since you have 6 atoms of Cl, you have -1(6) = -6 for the Cl. Since you 2 atoms of K, you have +1(2) = +2 for the K. The oxidation number of Pt must make all the oxidation numbers add up to zero:
+2 + (-6) + oxidation number of Pt = 0
-4 + oxidation number of Pt = 0
Oxidation number of Pt = 4

4 0

2 years ago

A vessel contains Ar(g) at a high pressure. Which of the following statements best helps to explain why the measured pressure is

Luda [366]

Answer:

4. The combined volume of the Ar atoms is too large to be negligible compared with the total volume of the container.

Explanation:

Deviations from ideality are due to intermolecular forces and to the nonzero volume of the molecules themselves. At infinite volume, the volume of the molecules themselves is negligible compared with the infinite volume the gas occupies.

However, the volume occupied by the gas molecules must be taken into account. Each molecule does occupy a finite, although small, intrinsic volume.

The non-zero volume of the molecules implies that instead of moving in a given volume V they are limited to doing so in a smaller volume. Thus, the molecules will be closer to each other and repulsive forces will dominate, resulting in greater pressure than the one calculated with the ideal gas law, that means, without considering the volume occupied by the molecules.

5 0

2 years ago

Please Help Me!!

Cerrena [4.2K]

Answer:

1. Galvanic oxidation. Example is the corrosion of aluminium wires when in contact with copper wires under wet conditions.

2. Rainwater or Damp/moist air

3. Chromium-plated steel screws or stainless steel screws or galvanized steel screws

Explanation:

1. Galvanic oxidation or corrosion occurs when two different metals with different electrode potentials are brought into contact with each other by means of an electrolyte (usually a aqueous solution), such that a redox reaction occurs leading to one metal with the more negative electrode potential (the anode) becoming oxidized, while the other less negative potential (the cathode) is reduced.

In order for galvanic corrosion to occur, three elements are required.

i. Two metals with different corrosion potentials (anode and cathode)

ii. Direct metal-to-metal electrical contact

iii. A conductive electrolyte solution (e.g. water) must connect the two metals on a regular basis.

For example oxidation (corrosion) of aluminium wires when in contact with copper wire under wet conditions.

2. The most likely electrolyte will be rainwater containing dissoved solutes (if the panel is in an exposed part of the house) or damp/moist air.

3. From the table, the most likely screw will be chromium-plated steel screws or stainless steel (made of iron and nickel) screws or galvanized steel (zinc-plated) screws.

All these possible screw components have a more negative electrode potential than copper. Thus they will serve as the anode in a galvanic oxidation with copper.

3 0

1 year ago