(H E L P)

The value of the equilibrium constant for the following chemical equation is Kr 40 at 25°C. Calculate the solubility of Al(OH)s(

dedylja [7]

Answer:

0,040 M

Explanation:

The global reaction of the problem is:

Al(OH) (s) + OH⁻ ⇄ Al(OH)₂⁻(aq) K= 40

The equation of equilibrium is:

K = $\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}$

The concentration of OH⁻ is:

pOH = 14 - pH = <em>3</em>

pOH = -log [OH⁻]

[OH⁻] = 1x10⁻³

Thus:

40 = $\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}$

<em>0,04M = $\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}$ </em>

This means that 0,04 M are the number of moles that the solvent can dissolve in 1L, in other words, solubility.

I hope it helps!

7 0

3 years ago

An 8.5g Ice Cube is placed into 255g water.

Troyanec [42]

LAw of conservation of Energy is an important concept to solve this problem. The energy released is equal to the energy absorbed.
Ice undergoes latent heat. MEaning there is a change in phase but not temperature and the energy is solved by (enthalpy of fusion)(mass) = 333.5J/g)(8.5g). =2834.75J.
This is equal to the energy released by the water. The energy is computed by (mass)(specific heat of water)(temperature change) = (255g)(4.16J/gK)(T)

Final equation is:
2834.75 = 255(4.16)(T)
T = 2.67K

7 0

3 years ago

Read 2 more answers

If 252 grams of iron are reacted with 321 grams of chlorine gas, what is the mass of the excess reactant leftover after the reac

mario62 [17]

Answer:

Iron is in excess.

1) The mass of the iron remaining = 83.38 grams

2) Ethane is in excess. There will remain 90.06 grams ethane

Explanation:

Step 1: Data given

Mass of iron = 252 grams

Mass of Cl2 = 321 grams

Molar mass of Fe = 55.845

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Fe(s)+3Cl2(g)⟶2FeCl3(s)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe = 252.0 grams / 55.845 g/mol = 4.512 moles

Moles Cl2 = 321.0 grams / 70.90 g/mol = 4.528 moles

Step 4: Calculate the limiting reactant

For 2 moles Fe we need 3 moles Cl2 to produce 2 moles Fecl3

Cl2 is the limiting reactant. It will completely be consumed (4.528 moles).

Fe is in excess. There will 4.528 * 2/3 = 3.019 moles be consumed

There will remain 4.512 - 3.019 = 1.493 moles of Fe

The mass of the iron remaining = 1.493 * 55.845 g/mol =83.38 grams

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If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?

Step 1: Data given

Mass of ethane = 152.0 grams

mass of O2 =231.0 grams

Molar mass of ethane = 30.07 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles ethane = 152.0 grams / 30.07 g/mol = 5.055 moles

Moles O2 = 231.0 grams / 32.0 g/mol = 7.22 moles

Step 4: Calculate limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed (7.22 moles).

Ethane is in excess. There will react 7.22 * 2/7 = 2.06 moles

There will remain 5.055 - 2.06 = 2.995 moles ethane

2.995 moles ethane = 2.995 * 30.07 g/mol = 90.06 grams ethane

8 0

4 years ago

Read 2 more answers

At a wastewater treatment plant, FeCl3(s) is added to remove excess phosphate from the effluent. Assume the following reactions

RoseWind [281]

Answer : The concentration of $Fe^{3+}$ needed is, $2.37\times 10^4M$

Explanation :

First we have to calculate the mole of phosphate.

As we are given that, 1 mg P/L that means, 1 mg of phosphate present in 1 L of solution.

$\text{Moles of phosphate}=\frac{\text{Mass of phosphate}}{\text{Molar mass of phosphate}}$

Molar mass of phosphate = 94.97 g/mole

$\text{Moles of phosphate}=\frac{1mg}{94.97g/mol}=\frac{0.001g}{94.97g/mol}=1.053\times 10^{-5}mol$

Now we have to calculate the concentration of phosphate.

$\text{Concentration of phosphate}=\frac{\text{Moles of phosphate}}{\text{Volume of solution}}$

$\text{Concentration of phosphate}=\frac{1.053\times 10^{-5}mol}{1L}=1.053\times 10^{-5}mol/L$

Now we have to calculate the concentration of $Fe^{3+}$ .

The second equilibrium reaction is,

$FePO_4\rightleftharpoons Fe^{3+}+PO_4^{3-}$

The solubility constant expression for this reaction is:

$K_{sp}=[Fe^{3+}][PO_4^{3-}]$

Given: $K_{sp}=\frac{1}{4}$

$\frac{1}{4}=[Fe^{3+}]\times 1.053\times 10^{-5}mol/L$

$[Fe^{3+}]=2.37\times 10^4M$

Thus, the concentration of $Fe^{3+}$ needed is, $2.37\times 10^4M$

8 0

4 years ago

Calculate the total amount of energy required in calories to convert 50.0 g of ice at 0.00 degrees Celsius to steam at 100. degr

r-ruslan [8.4K]

Answer:

HFusion*mass + Spec.Heat*mass*ΔT + HVap*mass

80cal/g*50.0g + 1.00cal/g°C*50.0g*(100°C-0°C) + 540cal/g*50g

3.60x10⁵cal

Explanation:

Using the HFusion we can find the heat needed to convert the ice to liquid water.

With specific heat capacity we can find the heat needed to increase the temperature of water from 0 to 100°C.

With HVap we can find the heat to convert the liquid water into steam.

The equations are:

<h3>HFusion*mass + Spec.Heat*mass*ΔT + HVap*mass</h3><h3 />

Computing the values:

36000cal =

8 0

3 years ago