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1 year ago

The molar solubility of x2s in pure water us 0.0395

1 answer:

7 0

The Ksp shows the extent of dissolution of a solute in solution. The Ksp of the solution is 2.5 * 10^-4.

The Ksp is an equilibrium constant that is usd to show the extent of solubility of a solute in solution.

The equation for the solution of X2S is;

X2S(s) ⇄2X^+(aq) + S^2-(aq)

Ksp = [2s]^2 [s]

Since s = 0.0395 M

Ksp = 4s^3

Ksp = 4[0.0395 ]^3

Ksp = 2.5 * 10^-4

Learn more about Ksp: brainly.com/question/27132799

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Given the following, determine ΔG°f at 298 K for SnO. Sn(s) + SnO2(s) → 2SnO(s) ; ΔG° = 12.0 kJ at 298K

Virty [35]

Answer:

The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.

Explanation:

$Sn(s) + SnO_2(s)\rightarrow 2SnO(s), \Delta G_{f}^{o} = 12.0 kJ$

$\Delta G_{f,SnO_2}^{o}= -515.8kJ/mol$

$\Delta G_{f,Sn}^{o}= 0 kJ/mol$

$\Delta G_{f,SnO}^{o}=?$

$\Delta G_{f}^{o}=\sum ((\Delta G_{f}^{o})_ {products})-\sum ((\Delta G_{f}^{o})_ {reactants})$

$12.0 kJ=(2 mol\times \Delta G_{f,SnO}^{o})-(1mol\times 0 kJ/mol+1 mol\times -515.8 kJ/mol)$

$\Delta G_{f,SnO}^{o}=-251.9 kJ/mol$

The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.

3 0

2 years ago

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kaheart [24]

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1 year ago

Mg(s) + ½O2(g) → MgO(s) + 146 kcal/mole H2(g) + ½O2(g) → H2O(g), ΔH = -57.82 kcal/mole What type of reaction is represented by t

adell [148]

Answer is: both reactions are exothermic.

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3 years ago

15.5 g of an unknown metal at 165.0°C is dropped into 150.0mL of H2O at 23.0°C in a coffee cup calorimeter. The metal and H2O re

tino4ka555 [31]

Answer:

Specific heat capacity of metal is 2.09 j/g.°C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of metal = 15.5 g

Initial temperature = 165.0°C

Initial temperature of water = 23.0°C

Final temperature = 30.0°C

Specific heat capacity of metal = ?

Specific heat capacity of water = 4.184 J/g°C

Volume of water = 150.0 mL or 150.0 g

Solution:

Formula:

- Qm = +Qw

Now we will put the values in formula.

-15.5 g × c × [ 30.0°C - 165.0°C] = 150 g × 4.184 J/g°C × [ 30.0°C - 23.0°C]

15.5 g × c × 135°C = 4393.2 j

2092.5 g.°C × c = 4393.2 j

c = 4393.2 j/2092.5 g.°C

c = 2.09 j/g.°C

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2 years ago

Which statement best describes an array of 10 elements where the elements are of a class type?

ladessa [460]

The array will be created with the elements equal to null.

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2 years ago