All categories

2 years ago

explain what you would do expect caesium astatide to look like .will it be soluble in water ?explain your reasoning

Chemistry

1 answer:

son4ous [18]2 years ago

3 0

Answer:

it will not be soluble in water Becoz it can only be

separated by passing it through silver nitrate solution

Explanation:

i hope you understand

You might be interested in

Sea water's density can be calculated as a function of the compressibility, B, where p = po exp[(p - Patm)/B]. Calculate the pre

vaieri [72.5K]

Answer:

The pressure 10,000 m below the surface of the sea is 137.14 MPa.

The density 10,000 m below the surface of the sea is 2039 kg/m3

Explanation:

P0 and ρ0 are the pressure and density at the sea level (atmosferic condition). As the depth of the sea increases, both the pressure and the density increase.

We can relate presure and density as:

$\frac{dP}{dy}=\rho*g=\rho_0*g*e^{(P-P_0)/\beta\\\\$

Rearranging

$\frac{dP}{e^{(P-P_0)/\beta}}= \rho_0*g*dy\\\\\int\limits^{P}_{P_0} {e^{-(P-P_0)/\beta}}dP =\int\limits^y_0 {\rho_0*g*dy}\\\\(-\beta*e^{-(P-P_0)/\beta})-(\beta*e^0)=\rho_0*g*(y-0)\\\\-\beta*(e^{-(P-P_0)/\beta}-1)=\rho_0*g*y\\\\e^{-(P-P_0)/\beta}=1-\frac{\rho_0*g*y}{\beta}\\\\-\frac{P-P_0}{\beta} =ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\$

With this equation, we can calculate P at 10,000 m below the surface:

$P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-200MPa*ln(1-\frac{1027kg/m^3*9.81m/s^2*10,000m}{200MPa})\\\\P-P_0=-200MPa*ln(1-\frac{1027*9.81*10,000Pa}{200*10^6Pa})\\\\P-P_0=-200MPa*ln(1-0.5037)\\\\P-P_0=-200MPa*(-0.6857)=137.14MPa$

The density at 10,000 m below the surface of the sea is

$\rho=\rho_0*e^{(P-P_0)/\beta}\\\rho=1027kg/m^3*e^{(137.14/200)}=1027*e^{0.686}kg/m^3\\\rho=1027*1.985 kg/m^3\\\rho=2039\,kg/m^3$

4 0

2 years ago

Phosphoric acid has a pka of 2.1. at what ph will 75% of phosphoric acid be in the conjugate base form?

Colt1911 [192]

Answer:

pH is 2.58

Explanation:

The pH of the buffer made from the mixture of phosphoric acid (H₃PO₄) and its conjugate base form (H₂PO₄⁻) follows the Henderson-Hasselbalch equation:

pH = pka + log [H₂PO₄⁻] /[H₃PO₄]

If 75% of the buffer is in the conjugate base form (H₂PO₄⁻), 25% will be as H₃PO₄. Replacing to find pH:

pH = 2.1 + log [75%] /[25%]

pH = 2.58

pH is 2.58

8 0

2 years ago

Read 2 more answers

(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW

Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = 2.3x10⁻³M InF₃. Thus:

[In⁺³] = 2.3x10⁻³M InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M In⁺³

[F⁻] = 2.3x10⁻³M InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M In⁺³] × [7.0x10⁻³M F⁻]³ = 7.9x10⁻¹⁰

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, solubility is 6.31x10⁻⁶M

3 0

2 years ago

Barium oxalate is used as a colorant to produce the green color in fireworks. Imagine that you have been assigned to prepare bar

JulsSmile [24]

Answer:

2.1

Explanation:

Calculation of moles of $Ba(OH)_2.8H_2O$

Mass of copper = 5.3 g

Molar mass of copper = 315.46 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{5.3\ g}{315.46\ g/mol}$

Moles of $Ba(OH)_2.8H_2O$ = 0.0168 moles

According to the reaction,

$Ba(OH)_2.8H_2O + H_2C_2O_4.2H_2O\rightarrow BaC_2O_4 + 12 H_2O$

1 mole of $Ba(OH)_2.8H_2O$ react with 1 mole of $H_2C_2O_4.2H_2O$

0.0168 moles of $Ba(OH)_2.8H_2O$ react with 0.0168 moles of $H_2C_2O_4.2H_2O$

Moles of $H_2C_2O_4.2H_2O$ = 0.0168 moles

Molar mass of $H_2C_2O_4.2H_2O$ = 126.07 g/mol

Thus,

Mass = Moles * Molar mass = 0.0168 moles * 126.07 g/mol = 2.1 g

Answer - 2.1

5 0

2 years ago

The Sun is a main sequence star. During the last stage of its life cycle, it will become a white dwarf. It will shrink in size,

dangina [55]

The one that is tiny and white.

3 0

1 year ago

Read 2 more answers

explain what you would do expect caesium astatide to look like .will it be soluble in water ?explain your reasoning​

explain what you would do expect caesium astatide to look like .will it be soluble in water ?explain your reasoning