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2 years ago

Determine the concentrations of hg22 and cl– in a saturated solution of hg2cl2 in

Chemistry

1 answer:

lutik1710 [3]2 years ago

8 0

To solve this, we need the solubility constant for Hg2Cl2, which is:

Ksp = 1.4 x 10^-18

The ionic formula would be:

Hg2Cl2(s) ---> Hg2 2+(aq) + 2Cl-(aq)

Therefore the total amounts of ions produced = x M Hg2 2+ and 2x M Cl-

The formula for Ksp in this case is:

Ksp = [Hg2 2+] [Cl-]^2

1.4 x 10^-18 = (x) * (2x)^2

1.4 x 10^-18 = 4x^3 
x = 7.0 x 10^-7 M = [Hg2 2+]

2x = 14 x 10^-7 M = [Cl-]

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What volume would 0.435 moles of hydrogen gas, h2, occupy at stp?

faust18 [17]

The conversion factor for volume at STP is $\frac {1mol}{22.4L}$ or $\frac {22.4L}{1mol}$ . Since we want volume, we would use $\frac {22.4L}{1mol}$ . We conclude with the following calculations:

$0.435mol H_{2} * \frac {22.4LH_{2}}{1molH_{2}} = 9.744L$

The answer is 9.744L H2

3 0

2 years ago

Read 2 more answers

The dissociation of calcium carbonate has an equilibrium constant of Kp = 1.16 at 1073 K . CaCO3(s) ⇄ CaO(s) + CO2(g) If you pla

tino4ka555 [31]

Answer:

<h3> $Pressure of CO_2 in the container=1.6 atm$ </h3>

Explanation:

First balance the chemical equation:

$CaCO_3(s)$ ⇄ $CaO(s) + CO_2(g)$

two components are solid so these two will not exert any kind of pressure in the container so at equilibrium only CO2 will apply pressure on the container

Therefore only partial pressure of CO2 will be taken for the calculation of equilibrium pressure constant i.e. Kp

$K_p=[CO_2]$

$[CO_2]=p$

$K_p=p$

$p=K_p = 1.16atm$

$Pressure of CO_2 in the container=1.6 atm$

8 0

2 years ago

The mole fraction of A (XA ) in the vapor phase of a mixture of two liquids is 0.24 and the sum of the partial pressures is 740

Yakvenalex [24]

Answer:

Mole fraction of B is 0.76

Explanation:

The pressure of a mixture of gases using vapor pressure and mole fraction of each gas is:

P = P°aₓXa + P°bₓXb + ... + P°nₓXn

Where P is pressure of the system, P° is vapor pressure of pure gas and X is its mole fraction.

In the problem, the pressure of the mixture of two gases is 740torr, mol fraction of A is 0.24 and vapor pressure of B is 800torr, that is:

740torr = P°aₓ0.24 + 800torrₓXb

Also, the sum of mole fractions for each of the compounds in the mixture is 1, that is:

1 = Xa + Xb

As Xa = 0.24

1 = 0.24 + Xb

1-0.24 = Xb

0.76 = Xb

Mole fraction of B is 0.76

4 0

2 years ago

If water’s density is 1.0 g/mL, then would the perfume be more or less dense than water? Would the perfume float on top or sink

Alex17521 [72]

Answer:

usually the perfumes are made of aromatic hydrocarbons invloving

cetone, ethanol, benzaldehyde, formaldehyde, limonene, methylene chloride, camphor, ethyl acetate, linalool and benzyl alcohol. which have density lower than the water hence they will float on the top of the water.

Hope this helps you

Explanation:

5 0

2 years ago

Using the following thermochemical data: 2Y(s) + 6HF(g) → 2YF3(s) + 3H2(g) ΔH° = –1811.0 kJ/mol 2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2

Luba_88 [7]

Answer:

ΔH° = 182.4 kJ/mol

Explanation:

The ΔH wanted is for the reaction :

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g)

This is a Hess Law problem where e will have to algebraically manipulate the first and second equations , add them together, and arrive at the desired equation above.

Notice if we reverse the first equation and divide it by 2 and add to the the second only divided by two, we will arrive to the desired equation:

2YF3(s) + 3H2(g) → 2Y(s) + 6HF(g) ΔH° = 1811.0 kJ/mol (change sign)

dividing by two :

YF3(s) + 3/2H2(g) → Y(s) + 3HF(g) ΔH° = 905.5 kJ/mol Eq 1

2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2(g) ΔH° = –1446.2 kJ/mol

dividing this one by two,

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

Now adding 1 and 2

YF3(s) + 3/2H2(g) → Y(s) + 3HF(g) ΔH° = 905.5 kJ/mol Eq 1

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

________________________________________________________

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g). ΔH° = 905.5 + (-723.1) kJ/mol

ΔH° = 182.4 kJ/mol

Notice how the Y(s) and H2 cancel nicely and the coefficients are the right ones.

8 0

2 years ago