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katen-ka-za [31]

3 years ago

8

If the pH of a solution of magnesium chloride gets too high, magnesium hydroxide (insoluble metal hydroxide) could precipitate f

rom solution. If you prepare a buffer solution, it is possible to prevent the precipitation of Magnesium Hydroxide by controling the pH. Which of the following buffers solutions would not result in the precipitation of Magnesium Hydroxide from a solution that is 0.0173 M in Magnesium chloride . Check all that apply.

1 answer:

Tasya [4]3 years ago

7 0

Answer:

Buffers would not result in precipitation are:

*1.0M HNO2 with 1.0M NaNO2 (Ka = 4.5x10⁻⁴; pKa = 3.347)

*5.6M C6H5OH with 1.0M KC6H5O (Ka = 1.3x10⁻¹⁰; pKa = 9.89)

*2.0M HCN with 1.0M NaCN (Ka = 4.0x10⁻¹⁰; pKa = 9.40)

Explanation:

The Ksp of magnesium hydroxide, Mg(OH)₂, is:

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Ksp = [Mg²⁺] [OH⁻]² = 1.2x10⁻¹¹

The precipitation will occurs when the product [Mg²⁺] [OH⁻]² > Ksp. If [Mg²⁺] is 0.0173M:

[0.0173M] [OH⁻]² > 1.2x10⁻¹¹

[OH⁻]² > 6.936x10⁻¹⁰

<em>[OH⁻] > 2.63x10⁻⁵M</em>

pOH (-log [OH]) is 4.58, and pH (pH = 14-pOH) is:

pH = 9.42, <em>if pH > 9.42, the precipitation will occurs</em>

For the following options, using H-H equation, pH is:

*1.0M HNO2 with 1.0M NaNO2 (Ka = 4.5x10⁻⁴; pKa = 3.347)

pH = 3.347 + log (1.0M / 1.0M)

<em>pH = 3.347. As pH < 9.42; </em>the buffer solution will not result in precipitation.

*2.0M C6H5OH with 1.32M KC6H5O (Ka = 1.3x10⁻¹⁰; pKa = 9.89)

pH = 9.89 + log (1.32M / 2.0M)

<em>pH = 9.71. As pH > 9.42; </em>the buffer solution will result in precipitation.

*1.0M HCN with 2.0M NaCN (Ka = 4.0x10⁻¹⁰; pKa = 9.40)

pH = 9.40 + log (2.0M / 1.0M)

<em>pH = 9.70. As pH > 9.42; </em>the buffer solution will result in precipitation.

*5.6M C6H5OH with 1.0M KC6H5O (Ka = 1.3x10⁻¹⁰; pKa = 9.89)

pH = 9.89 + log (1.0M / 5.6M)

<em>pH = 9.14. As pH < 9.42; </em>the buffer solution will not result in precipitation.

*2.0M HCN with 1.0M NaCN (Ka = 4.0x10⁻¹⁰; pKa = 9.40)

pH = 9.40 + log (1.0M / 2.0M)

<em>pH = 9.10. As pH < 9.42; </em>the buffer solution will not result in precipitation.

*1.0M C6H5OH with 1.0M KC6H5O (Ka = 1.3x10⁻¹⁰; pKa = 9.89)

pH = 9.89 + log (1.0M / 1.0M)

<em>pH = 9.89. As pH > 9.42; </em>the buffer solution will result in precipitation.

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Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2

mina [271]

<u>Answer:</u> The amount of energy released per gram of $B_5H_9$ is -71.92 kJ

<u>Explanation:</u>

For the given chemical reaction:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]$

Taking the standard enthalpy of formation:

$\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ$

We know that:

Molar mass of pentaborane -9 = 63.12 g/mol

By Stoichiometry of the reaction:

If 2 moles of $B_5H_9$ produces -9078.57 kJ of energy.

Or,

If $(2\times 63.12)g$ of $B_5H_9$ produces -9078.57 kJ of energy

Then, 1 gram of $B_5H_9$ will produce = $\frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ$ of energy.

Hence, the amount of energy released per gram of $B_5H_9$ is -71.92 kJ

8 0

3 years ago

SOMEONE PLEASE HELP ME WITH THIS QUESTION!!!!!!!!!!!!

Advocard [28]

Answer:

KOH

Explanation:

Chemical reaction:

2K + 2H₂O → 2KOH + H₂

Element ratio of K.

K = 1

Because only potassium is present.

Element ratio of H₂O.

2 : 1

in water ratio of element is 2 : 1 because two hydrogen and one oxygen atom present.

Element ratio of KOH

1 : 1 : 1

in KOH elemental ratio is 1 : 1 : 1 because one potassium one hydrogen and one oxygen atom are present.

Element ratio of H₂.

2

Just two atoms of hydrogen are present.

8 0

3 years ago

How is the periodic table constructed ?

jasenka [17]

Before the periodic table, there were a bunch of symbols, number, letters etc (In all kinds of languages) that represented the elements. Scientists around the world saw that a chart of the elements needed to be universally accepted and finalized. A guy named Mendeleev presented this idea to the scientific community. Mendeleev was also the first to order elements according to atomic number rather than atomic weight. The modern day periodic table was not published by him, it was developed with the help of the entire scientific community. Honestly, there isn't a specific way to tell you how the periodic table was constructed, scientists developed thousands of tables that represented the elements. And just to let you know, the modern day periodic table is constantly going through changes as we discover more and more about elements, atoms, molecules etc. so in the near future it wouldn't be surprising if we saw something completely different than what we see today.

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3 years ago

Read 2 more answers

one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40

Ronch [10]

<u>Answer:</u> The volume of chlorine gas produced in the reaction is 2.06 L.

<u>Explanation:</u>

<u>For potassium permanganate:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

$\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol$

<u>For hydrochloric acid:</u>

To calculate the moles of hydrochloric acid, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of HCl = 6.00 M

Volume of HCl = 45.0 mL = 0.045 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol$

For the reaction of potassium permanganate and hydrochloric acid, the equation follows:

$2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O$

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with = $\frac{2}{16}\times 0.27=0.033moles$ of potassium permanganate.

As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with = $\frac{5}{16}\times 0.27=0.0843moles$ of chlorine gas.

To calculate the volume of gas, we use the equation given by ideal gas equation:

$PV=nRT$

where,

P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = $40^oC=[40+273]K=313K$

Putting values in above equation, we get:

$1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L$

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.

3 0

4 years ago

What is the ph of a 0.001 62 m naoh solution?

marin [14]

The answer
first of all, we should know that NaOH is a strong base. For such a product, the conentration of the OH - is equivalent to the concentration of the NaOH itself.
that means:
[ OH -] = [ NaOH] =<span>0.001 62
and for a strong basis, pH can be calculated as pH = 14 + log </span>[ OH -]
first we compute log [ OH -] :
log [ OH -] = log (0.001 62)= -2.79

finally pH = 14 -2.79 = 11.20

4 0

3 years ago