STP is:

What is the pH of 9.01 x 10^-4 M Mg(OH)2 What is the pH of 2.33 x 10^-2M of NH4OH

const2013 [10]

Answer:

A. 11.26

B. 12.37

Explanation:

A. Step 1:

Dissociation of Mg(OH)2. This is illustrated below below:

Mg(OH)2 <==> Mg2+ + 2OH-

A. Step 2:

Determination of the concentration of the OH-

From the above equation,

1 mole of Mg(OH)2 produce 2 moles of OH-

Therefore, 9.01x10^-4 M Mg(OH)2 will produce = 9.01x10^-4 x 2 = 1.802x10^-3 M of OH-

A. Step 3:

Determination of the pOH. This is illustrated below:

pOH = - Log [OH-]

[OH-] = 1.802x10^-3 M

pOH = - Log [OH-]

pOH = - Log 1.802x10^-3

pOH = 2.74

A. Step 4:

Determination of the pH.

pH + pOH = 14

pOH = 2.74

pH + 2.74 = 14

Collect like terms

pH = 14 - 2.74

pH = 11.26

B. Step 1:

Dissociation of NH4OH. This is illustrated below below:

NH4OH <==> NH4+ + OH-

B. Step 2:

Determination of the concentration of the OH-

From the above equation,

1 mole of NH4OH produce 1 moles of OH-

Therefore, 2.33x10^-2M of NH4OH will also produce 2.33x10^-2M of OH-

B. Step 3:

Determination of the pOH. This is illustrated below:

pOH = - Log [OH-]

[OH-] = 2.33x10^-2M

pOH = - Log [OH-]

pOH = - Log 2.33x10^-2M

pOH = 1.63

B. Step 4:

Determination of the pH.

pH + pOH = 14

pOH = 1.63

pH + 1.63 = 14

Collect like terms

pH = 14 - 1.63

pH = 12.37

6 0

3 years ago

Which statement is true about real gases with the correct reason:

LenKa [72]

There are 2 possible answers here : b and d.

The Ideal Gas Equation is : PV = nRT

Here, when pressure is increased and temperature is lowered, the volume of the molecules will substantially decrease, which means it has deviated from ideal behavior.

4 0

2 years ago

Read 2 more answers

1) You have an aqueous solution where [OH-] = 1 x 10-4 mol/L.

Irina18 [472]

1. A. 1 x $10^{-10}$ x is the hydrogen ion concentration.

B. pH of the solution is 10

C. The solution is basic.

2. The molarity of NaCl in 2.8 litres of water is 1.18 M

3. 1.64 M is the molarity of new solution.

4. 0.64 M is the molarity of the acid or HCl used.

Explanation:

1. Data given [OH-] =1 x 10-4 mol/L.

A) $K_{w}$ = [ $H_{3}$ O+] [OH]- $K_{w}$ = 1× $10^{-14}$

[ $H_{3}$ O+]= 1× $10^{-14}$ ÷ 1 x 10-4

= 1 x $10^{-10}$ x is the hydrogen ion concentration.

B) pH =-log [ $H_{3}$ O+]

pH = -log[-1x $10^{-10}$ ]

pH = 10

c) pH value of 10 indicates that it is a basic solution because it is greater than 7 and on pH scale more than 7 value indicates basicity.

2. Molarity is calculated by the formula

Molarity = $\frac{number of moles}{volume}$

Number of moles can be calculated as:

n = $\frac{mass}{atomic mass of one mole of substance}$

n = $\frac{195}{58.44}$ ( atomic weight of NaCl is 58.44 gm/mole)

n= 3.33 moles

Now the molarity of NaCl in 2.8 litres of water is calculated as:

Molarity = $\frac{number of moles}{volume}$

M = $\frac{3.33}{2.8}$

M = 1.18

the molarity of NaCl in 2.8 litres of water is 1.18 M

3. Data given:

Initial volume of the HCl solution V1 = 152 ml, initial molarity M1 = 3

final volume of the diluted HCl V2= 750 ml, final molarity M2= Unknown

The initial and final volumes are converted into litres in calculation.

So, M1V1 = M2V2

3× $\frac{152}{1000}$ = M2 × $\frac{750}{1000}$

M2 = $\frac{0.75}{0.456}$

M2 = 1.64 M is the molarity of new solution.

4. In titration the formula used is: M acid x V acid = M base x V base

Volume of base V1= 20 ml Molarity of base M1 = 0.24 M

volume of the acid V2 = 31 ml Molarity of the acid = unknown

the volume will be converted to litres.

So applying the formula,

M acid x V acid = M base x V base

0.24 x $\frac{20}{1000}$ = Macid x $\frac{31}{1000}$

0.0048 = Macid x 0.031

Macid= $\frac{0.031}{0.048}$

M base= 0.64 M is the molarity of the acid or HCl used.

7 0

3 years ago

Calculate the temperature change in the water upon complete melting of the ice. Hint: Determine how much heat is absorbed by the

Anvisha [2.4K]

Answer:

The temperature change is 80°C

Explanation:

Heat absorbed by the melting ice (Q) = mL = 312×336 = 104,832 J

∆T = Q/mC = 104,832/312×4.2 = 80°C

4 0

3 years ago

PLEASE HELPP!!!!

Natasha_Volkova [10]

Answer:

The combined law is:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

See the derivation below.

Explanation:

1. Boyle's law:

$PV=constant\\\\P_1V_1=P_2V_2$

2. Charles' law:

$\dfrac{V}{T}=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

3. Gay-Lussac’s law

$\dfrac{P}{T}=constant\\\\\\\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

4. Summary:

PV = K

V/T = K'

P/T = K''

Mulitply PV by V/T and P/T

$PV\times \dfrac{V}{T}\times \dfrac{P}{T}=K\times K'\times K''\\\\\\\dfrac{P^2V^2}{T^2}=constant\\\\\\\sqrt{\dfrac{P^2V^2}{T^2}}=\sqrt{constant}\\\\\\\dfrac{PV}{T}=constant$

Thus:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

Which is the combined law.

4 0

4 years ago