CHEMISTRY COLLEGE

Mg(OH)2 is a sparingly soluble salt with a solubility product, Ksp, of 5.61×10−11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams. What is the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.180 M NaOH solution? Express your answer numerically as the ratio of molar solubility in H2O to the molar solubility in NaOH.

Answers

Answer 1

Answer:

Answer:

The ration of the molar solubility is 165068.49.

Explanation:

The solubility reaction of the magnesium hydroxide in the pure water is as follows.

$Mg(OH)_(2)\Leftrightarrow Mg^(2+)(aq)+2(OH)^(-)(aq)$

$[Mg^(2+)][OH^(-)]$

Initial 0 0

Equili +S +2S

Final S 2S

$K_(sp)=[Mg^(2+)][OH^(-)]$

$5.61* 10^(-11)=(S)(2S)^(2)$

$S=((5.61* 10^(-11))/(4))^(1/3)=2.41* 10^(-4)M$

Solubility of $Mg(OH)_(2)$ in 0.180 M NaOH is a follows.

$Mg(OH)_(2)\Leftrightarrow Mg^(2+)(aq)+2(OH)^(-)(aq)$

$[Mg^(2+)][OH^(-)]$

Initial 0 0

Equili +S +2S

Final S 2S+0.180M

$K_(sp)=[Mg^(2+)][OH^(-)]$

$5.61* 10^(-11)=(S)(2S+0.180)^(2)$

$S=1.46* 10^(-9)M$

$Ratio\,of\,solubility=(2.41* 10^(-4))/(1.46* 10^(-9))=165068.49$

Therefore, The ration of the molar solubility is 165068.49.

Related Questions

What type of energy conversion takes place in a voltaic cell? A) Chemical energy is spontaneously converted to electrical energy. B) Chemical energy is converted to electrical energy only when there is an external power source. C) Electrical energy is spontaneously converted to chemical energy. D)Electrical energy is converted to chemical energy only when there is an external power source.
A well-insulated tank contains concentrated HCl solution in water. If more water is added to this tank the temperature of the system will __________.Select one:a. increaseb. decreasec. no changed. cannot tell
Pharmacists sometimes measure medicines in the unit of "grains," where 1 gr = 65 mg. The label on a bottle of aspirin reads "Aspirin, 5 gr." How many mg of aspirin are in a tablet containing 5 gr?
Answer the question please.
The entropy of a substance above absolute zero will always be: a. Negativeb. Positivec. Neither Negative nor positive

A solution is made by dissolving37.5g of sodium sulfide (Na2S) in

217g of water.

What is the molality of the solution?

Answers

Answer:

THE MOLARITY IS 2.22 MOL/DM3

Explanation:

The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water

Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol

Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;

37.5 g of Na2S = 217 g of water

( 37.5 * 1000 / 217 ) g = 1000 g of water

So, 172.81 g/dm3 of the solution

So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass

Molarity = 172.81 g/dm3 / 78 g/mol

Molarity = 2.22 mol/dm3

The molarity of the solution is 2.22 mol/dm3

Answer:

The answer is 2.22mol

Explanation:

Identify the sample and analyte in each of the scenarios.(a) Determination of the lead content in paint. lead paint
(b) Analysis of the nitrate content of soil near a local water source. soil nitrate
(c) Measurement of the citric acid found in a lime.

Identify the following as either sample or analyte.
(1) lead
(2) paint chips
(3) soil
(4) nitrate
(5) lime wedge
(6) citric acid

Answers

Answer:

a) Analyte: lead. Sample: paint.

b) Analyte: nitrate. Sample: soil.

c) Analyte: citric acid. Sample: Lime

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

Explanation:

A sample is a portion of material selected from a larger quantity of material while an analyte is the chemical of the system that will be analysed.

Thus:

a) Analyte is lead while you must take a sample of paint to analyze this lead.

b) Analyte is the nitrate while sample must be soil.

c) Analyte is citric acid and lime is the sample

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

A container was found in the home of the victim that contained 120 g of ethylene glycol in 550 g of liquid. How many drinks, each containing 100 g of this liquid, would a 85 kg victim need to consume to reach a toxic level of ethylene glycol

Answers

Answer:

0.432 drinks are toxic

Explanation:

The toxic dose of ethylene glycol is 0.1 mL per kg body weight (mL/kg). In grams (Density ethylene glycol = 1.11g/mL):

1.11g/mL * (0.1mL / kg) = 0.111g/kg

If the victim weighs 85kg, its letal dose is:

85kg * (0.111g/kg) = 9.435g of ethylene glycol

Using the concentration of ethylene glycol in the liquid:

9.435g of ethylene glycol * (550g liquid / 120g ethylene glycol) = 43.2g of liquid are toxic.

The drinks are:

43.2g of liquid * (1 drink / 100 g) =

0.432 drinks are toxic

What volume will 12 g of oxygen gas (O2) occupy at 25 °C and a pressure of 53 kPa?

Answers

ANSWER

The volume of the oxygen gas is 17.5 L

EXPLANATION

Given that;

The mass of oxygen gas is 12 grams

The temperature of the gas is 25 degrees Celcius

The pressure of the gas is 53 kPa

To find the volume of the oxygen gas, follow the steps below

Step 1; Assume the gas behaves like an ideal gas

Therefore, apply the ideal gas equation to find the volume of the gas

$\text{ PV }=\text{ nRT}$

Where

P is the pressure of the gas

V is the volume of the gas

n is number of moles of the gas

R is the universal gas constant

T is the temperature of the gas

Step 2: Find the number of moles of the oxygen gas using the below formula

$\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}$

Recall, that the molar mass of the oxygen gas is 32 g/mol

$\begin{gathered} \text{ mole }=\text{ }\frac{12}{\text{ 32}} \n \text{ mole }=\text{ 0.375 mol} \end{gathered}$

Step 3; Convert the temperature to degree Kelvin

$\begin{gathered} \text{ T }=\text{ t }+\text{ 273.15} \n \text{ t }=\text{ 25}\degree C \n \text{ T }=25\text{ }+\text{ 273.15} \n \text{ T }=\text{ 298.15K} \end{gathered}$

Step 4; Substitute the given data into the formula in step 1

Recall, that R is 8.314 L kPa K^-1 mol^-1

$\begin{gathered} \text{ 53 }*\text{ V }=\text{ 0.375}*\text{ 8.314}*\text{ 298.15} \n \text{ 53V }=\text{ 929.557} \n \text{ Divide both sides by 53} \n \text{ }\frac{\cancel{53}V}{\cancel{53}}\text{ }=\text{ }(929.557)/(53) \n \text{ V }=\text{ }(929.557)/(93) \n \text{ V }=\text{ 17.5 L} \end{gathered}$

Hence, the volume of the oxygen gas is 17.5 L

Find the initial velocity for an enzymatic reaction when Vmax = 6.5 x 10–5 mol•sec–1 , [S] = 3.0 x 10–3 M, and KM = 4.5 x 10–3 M. A) not enough information is given to make this calculation B) 2.6 x 10–5 mol•sec–1 C) 1.4 x 10–2 mol•sec–1 D) 8.7 x 10–3 mol•sec–1 E) 3.9 x 10–5 mol•sec–1

Answers

Answer: correct option is the option B= 2.6 × 10^-5 mol.sec^-1

Explanation:

Check attached file/picture for the plot of graph of reaction velocity against substrate concentration.

Initial velocity can be defined as the velocity at the beginning of an enzyme-catalyzed reaction. It is usually denoted as V°.

In the equation of reaction below;

A + B----------> [AB] ------------>Product

<--------

Where the forward reaction is k1 that is the rection producing the intermediate [AB].

And, the backward reaction is k-1 that is the arrow pointing back to the Reactants.

K2 is the arrow from the intermediate [AB] to the products.

Note that k1 and k-1

represent rapid non-covalent dissociation of substrate from enzyme active site. Also, K2 is the rate constant for the formation of the products.

In order to calculate the initial velocity,V° we need to use the Michaelis- Menteen equation.

V° = Vmax ×(S)/ Km + (S)-------------------------------------------------------------(1).

From the question the parameters given are;

Vmax = 6.5 x 10–5 mol•sec–1 , [S] = 3.0 x 10–3 M, and KM = 4.5 x 10–3 M.

Therefore, slotting in the above parameters into the equation (1) above, we have;

V° = 6.5 x 10–5 mol•sec–1 × 3.0 x 10–3 M / 4.5 x 10–3 M + 3.0 x 10–3 M.

Initial velocity,V° = 2.6 x 10-5 moles per sec..

The higher the pH, the less acidic the solution

Answers

If the pH is higher the concentration of hydrogen ions becomes less and the solution becomes less acidic.

As the pH becomes lower, the concentration of hydrogen ions becomes greater, and the solution becomes more acidic.

HOPEIT'SHELPFUL:)

Answer:

yh thats true lol, ty for that very interesting fact

Other Questions

If 1.20 g of aluminum hydroxide reacts with 3.00 g of sulfuric acid, what is the mass of water produced? Al(OH)3(s)+H2SO4(l)→Al2(SO4)3(aq)+H2O(l)

Consider four small molecules, A–D, which have the following binding affinities for a specific enzyme (these numbers are the equilibrium constants Kd for the dissociation of the enzyme/molecule complex). Which binds most tightly to the enzyme? Which binds least tightly?A) 4.5 μMB) 13 nMC) 8.2 pMD) 6.9 mM

In general, atomic radii: A. decrease down a group and remain constant across a period. B. decrease down a group and increase across a period. C. increase down a group and increase across a period. D. increase down a group and remain constant across a period. E. increase down a group and decrease across a period.

Consider the following reaction at 298K.I2 (s) + Pb (s) = 2 I- (aq) + Pb2+ (aq)Which of the following statements are correct?Choose all that apply.ΔGo > 0The reaction is product-favored.K < 1Eocell > 0n = 2 mol electronsB-