A certain metal forms a soluble nitrate salt M(NO3)3. Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0mM solution of M(NO3)3 and the right half cell with a 3.0M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 C.

Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode.

Answer:

Hot material near the core is less dense and rises, when it cools, it becomes more dense and sinks.

Explanation:

This best explains how heat plays a role in the movement of materials within Earth's interior because it's how convection works. Convection is the circular motion that happens when warmer air or liquid which has faster moving molecules, making it less dense rises, while the cooler air or liquid drops down. Convection currents within the earth move layers of magma, and convection in the ocean creates currents.

Earth’s interior is like a closed chamber, hot material near Earth's surface is more denser and sinks, and when it cools, it becomes less dense and Rises.

Convection is the process of heat transfer by the bulk movement of molecules within fluids such as gases and liquids. The heat transfer take place from solid to no solid material. like air and water.

As we take example of home. In home heat take nearer the roof and cold air will be nearer to the flour. It is because of hot is less denser then the cold air. Hot air will become less denser and move towords upside. Cold air become denser move towards down.

Similarly, on earth. Nearer to earth hot materials are more dense and sinks. When it cools, it become less dense and rises.

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Answer:

p = 1.714 g/cm3

Explanation:

Density Equation

p=mV

Where:

p = density

m = mass  = 60g

V = volume = 35cm3

p = 60g x 35cm3

p = 1.714 g/cm3

p=1.714g/cm³

Explanation:

v=35cm³

m=60g

P=mass/volume (density formula)

=60/35

=1.714g/cm³

Answer:

Diamagnetic have paired electrons while paramagnetic have at least one unpaired electron.

Explanation:

F2, C2 and N2 are diamagnetic while O2 and B2 are paramagnetic. Diamagnetic are those atoms which have paired electrons while paramagnetic are those atoms which contain at least one unpaired electron so we can say that F2, C2 and N2 have paired electrons while O2 and B2 have unpaired electrons. When diamagnetic materials are allowed to contact with external magnetic field so they will be repelled while paramagnetic materials are attracted by magnetic field due to the presence of unpaired electrons.

Iron oxide (FeO) is the strongest paramagnetic material having the value of 720.

Answer:

Option C

The specific heat is the heat capacity for one mole of a substance.

Explanation:

The incorrect statement is  The specific heat is the heat capacity for one mole of a substance.

This is because the specific heat capacity is the amount of heat required to raise the temperature of 1 gramme of a substance by 1 degree Celcius.

Note that the unit in question here for the specific heat capacity of the substance is in grammes.

The definition given in the options is actually for the molar heat capacity of the substance, not the specific heat capacity.

Answer:

3.011x10^24 molecules

Explanation:

1 mole=6.022x10^23 molecules

5 moles*6.022x10^23 molecules/mole=3.011x10^24 molecules

Answer:

A. Solubility of calcium sulfite increases

B. Solubility of calcium fluoride increases

C. Solubility of Silver bromide decreases

Explanation:

The solubility factor is proportional to ions' concentration. The solubility of a solution can be predicted from Le Chatelier's principle which states that if an external constraint is imposed on a system in equilibrium, the equilibrium position will shift in order to annul the effect of the external constraint. So, If the reactant's concentration increases, the equilibrium shifts to the right indicating a higher solubility of the solution and if the product's concentration increases, the equilibrium shifts to the left indicating a lesser solubility of the solution.

Case 1. Calcium sulfite

The dissociation reaction of CaSO3 is given below:

CaSO3 ----> Ca²+ + SO3²-

SO3²- is the conjugate base of the weak acid, H2SO3. Therefore, on the addition of hydrobromic acid, some of the sulfite ion is removed from the solution by the following reaction;

H+ + SO3²- ---> HSO3-

This shifts the equilibrium to the right, more dissociation, thereby resulting in more solubility of the solute.

Case 2. Calcium fluoride

The dissociation reaction of calcium fluoride (CaF2) is shown below.

CaF2 ----> Ca²+ + 2F-

Fluoride ion (F-) is a strong conjugate base of the weak acid. Therefore, some of fluoride ions is removed by the addition of hydrobromic acid as shown below:

H+ + F- ---->. HF

Hence, the concentration of fluoride ions reduces, shifting equilibrium in the forward direction. Therefore, the solubility will be more than in pure water solution.

Case 3: Silver bromide

The dissociation reaction of AgBr is as follows:

AgBr ----> Ag+ + Br-

The addition of HBr will increase the concentration of bromide ions. Hence, equilibrium will shift in backward direction resulting in a lesser solubility than in water.

The solubility of calcium sulfite and calcium fluoride is greater in 0.10 M hydrobromic acid solution than in pure water while the solubility of silver bromide is lesser in 0.10 M hydrobromic acid solution than in pure water.

Common ion effect refers to the decrease in solubility of a substance in a solution that contains another solute with which it has a common ion. If a substance is dissolved in a solution that contains a solute with which it has a common ion, the solubility of the substance in that solution is less than its solubility in pure water.

Considering the substances given, the solubility of calcium sulfite and calcium fluoride in 0.10 M hydrobromic acid solution is more than their solubility in pure water the equilibrium position is shifted in the forward direction.

However, solubility of silver bromide in 0.10 M hydrobromic acid solution is less than its solubility in pure water due to common ion effect.

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Answer:

The class of this compound is aldehyde or ketone (i).

Explanation:

Absorption peak at 1720 cm-1 shows the presence of a carbonyl group, possibly an aldehyde or ketone with C=O bond.

Further information on molecular formula would be required for structural elucidation.

Answer:

327.89g/mol

Explanation:

Step 1:

The following data were obtained from the question:

Van 't Hoff factor (i) = 1 (since the molecule is non-electrolyte)

Temperature (T) = 25°C = 25°C + 273 = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Mass of molecule = 241mg

Volume of water = 250mL

Molarity (M) =?

Osmotic pressure (Π) = 0.072 atm

Step 2:

Determination of the molarity of the molecule.

This can be obtained as follow:

Π = iMRT

0.072 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 0.072 / (0.0821 x 298)

M = 2.94×10¯³ mol/L

Step 3:

Determination of the number of mole of the molecule. This can be obtained as follow:

Molarity = 2.94×10¯³ mol/L

Volume of water = 250mL = 250/1000 = 0.25L

Mole of molecule =..?

Molarity = mole /Volume

2.94×10¯³ = mole / 0.25

Cross multiply

Mole of molecule = 2.94×10¯³ x 0.25

Mole of molecule = 7.35×10¯⁴ mole.

Step 4:

Determination of the molar mass of the molecule.

Mole of molecule = 7.35×10¯⁴ mole.

Mass of molecule = 241mg = 241×10¯³g

Molar mass of molecule =..?

Mole = Mass /Molar Mass

7.35×10¯⁴ = 241×10¯³/ Molar Mass

Cross multiply

7.35×10¯⁴ x molar mass = 241×10¯³

Divide both side by 7.35×10¯⁴

Molar Mass = 241×10¯³/7.35×10¯⁴

Molar Mass = 327.89g/mol

Therefore, the molar mass of the molecule is 327.89g/mol

The given question is not complete, the complete question is:

Suppose a reaction mixture, when diluted with water, afforded 300 mL of an aqueous solution of 30 g of the reaction product malononitrile [CH2(CN)2], which is to be isolated by extraction with ether. The solubility of malononitrile in ether at room temperature is 20.0 g/100 mL, and in water is 13.3 g/100mL. The ratio of these quantities is equal to the partition coefficient, k, which equals What weight of malononitrile would be recovered by extraction of (a) three 100-mL portions of ether and (b) one 300-mL portion of ether? SHOW WORK (Can be written in pen and attached to report). Suggestion: For each extraction, let x equal the weight extracted into the ether layer. In part (a), for the first of the three extractions, the concentration of malononitrile in the ether layer is x/100 and in the water layer is (30-x)/100.

Answer:

The correct answer is 10 grams and 18 grams.

Explanation:

Based on the given question, 20 gram per 100 ml is the solubility of malononitrile in ether, and 13.3 gram per 100 ml is the solubility of malononitrile in water.  

Thus, the ration of the solubility is,  

Solubility in water/solubility in ether = 20/13.3 = 1.50

a) Let w be the weight of malononitrile extracted into water in every extraction. Then the concentration of the ether layer will be w/100. The concentration in the water layer will be 30-w/300. Now the ratio will be,  

Ratio = w/100 / (30-w)/300

1.50 = w/100 * 300 (30-w)

w = 10

Hence, the weight of malononitrile recovered by extraction is 10 grams.  

b) The concentration in the ether layer will be w/300. The concentration in the water layer will be (30-w) / 300. Now the ratio will be,  

Ratio = w/300 / (30-w) / 300

1.50 = w/300 * 300 (30-w)

w = 18

Hence, 18 grams is the weight of malononitrile recovered by extraction.  

Answer:

The rates of decay of radioactive elements

Explanation:

The age of a rock in years is called its absolute age. Geologists find absolute ages by measuring the amount of certain radioactive elements in the rock. When rocks are formed, small amounts of radioactive elements usually get included.

Answer:

1 particle of the gas would occupy a volume of 3.718*10⁻²³L

Explanation:

Hello,

1. The sample has a particle of 6.022×10²²particles

2. Volume of the sample = 22.4L

3. Temperature of the sample = 0°C = (0 +273.15)K = 273.15K

4. Pressure of the sample = 1.0atm

What volume would 1 particle of the gas occupy?

But we remember that 1 mole of any substance = 6.022×10²² molecules or particles or atoms

What would be the number of moles for 1 particule?

1 mole = 6.022×10²² particles

X moles = 1 particle

X = (1 × 1) / 6.022×10²² particles

X = 1.66×10⁻²⁴ moles

Therefore, 1 particle contains 1.66×10⁻²⁴ moles

Since we know our number of moles, we can proceed to use ideal gas equation,

Ideal gas equation holds for all ideal gas and is defined as

PV = nRT

P = pressure of the ideal gas

V = volume the gas occupies

n = number of moles of the gas

R = ideal gas constant = 0.082 L.atm / mol.K

T = temperature of the gas

PV = nRT

Solving for V,

V = nRT/ P

We can now plug in our values into the above

equation.

V = (1.66*10⁻²⁴ × 0.082 × 273.15) / 1

V = 3.718*10⁻²³L

Therefore, 1 particule of the gas would occupy a volume of 3.718*10⁻²³L.

Answer:

Approximately [tex]21\; \rm g[/tex].

Explanation:

[tex]\rm H_2SO_4[/tex] (a diprotic acid) reacts with [tex]\rm NaOH[/tex] (a monoprotic base) at a one-to-two ratio:

[tex]\rm 2\; NaOH\, (s) + H_2SO_4\, (aq) \to Na_2SO_4\; (aq) + 2\; H_2O\, (l)[/tex].

In other words, if [tex]n(\mathrm{NaOH})[/tex] and [tex]n(\mathrm{H_2SO_4})[/tex] represent the number of moles of the two compounds reacted, then:

[tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex].

Look up the relative atomic mass data on a modern periodic table:

Calculate the (molar) formula mass of [tex]\rm H_2SO_4[/tex] and [tex]\rm NaOH[/tex]:

[tex]M(\mathrm{H_2SO_4}) = 2 \times 1.008 + 32.06 + 4 \times 15.999 = 98.072\; \rm g \cdot mol^{-1}[/tex].

[tex]M(\mathrm{NaOH}) = 22.990 + 15.999 + 1.008 = 39.997\; \rm g \cdot mol^{-1}[/tex].

Calculate the number of moles of formula units in that [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:

[tex]\begin{aligned}n(\mathrm{NaOH}) &= \frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})} \\ &= \frac{33.8\; \rm g}{39.997\; \rm g \cdot mol^{-1}} \approx 0.845\; \rm mol\end{aligned}[/tex].

Apply the ratio [tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex] to find the (maximum) number of moles of [tex]\rm H_2SO_4[/tex] that would react with the [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:

[tex]\begin{aligned}n(\mathrm{H_2SO_4}) &= \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} \cdot n(\mathrm{NaOH})\\ &= \frac{1}{2} \times 0.845 \approx 0.4225\; \rm mol\end{aligned}[/tex].

Calculate the mass of that [tex]0.4225\; \rm mol[/tex] of [tex]\rm H_2SO_4[/tex]:

[tex]\begin{aligned}m(\mathrm{H_2SO_4}) &= n(\mathrm{H_2SO_4}) \cdot M(\mathrm{H_2SO_4})\\ &= 0.4225 \; \rm mol \times 98.072\; \rm g \cdot mol^{-1} \approx 41.435\; \rm g \end{aligned}[/tex].

When the maximum amount of [tex]\rm H_2SO_4[/tex] is reacted, the minimum would be in excess. Hence, the minimum mass of

[tex]62\; \rm g - 41.435\; \rm g \approx 21\; \rm g[/tex] (rounded to two significant figures.)

Answer:

5-methylhex-1-yne//5-methylhex-2-yne//2-methylhex-3-yne

Explanation:

We have to start with the information on the IR spectrum. The signal at 3300 is due to a C-H bend sp carbon and the signal in 2200 is due to the stretch carbon-carbon. Therefore we will have an alkyne. Now if we have 2-methylhexane as the product of hydrogenation we have several options to put the triple bond. Between carbons 1 and 2 (5-methylhex-1-yne), between carbons 2 and 3 (5-methylhex-2-yne) and between carbons 3 and 4 (2-methylhex-3-yne). On carbon 5 we have a tertiary carbon therefore we dont have any other options.

See figure 1

I hope it helps!

Answer:

38.493 KJ/mol

Explanation:

Equation of reaction; HBr + KOH ---> KBr + H2O

Heat evolved = mass * specific heat capacity * temperature rise

Mass of solution = density * volume

Mass = 1.00 g/ml*50 ml = 50g

Temperature rise = 31.9 - 22.7 = 9.2 °C

Heat evolved = 50 * 4.184 * 9.2 = 1924.64 J

From the equation of reaction, 1 mole of HBr reacts with 1mole of KOH to produce 1 mole of H20

Number of moles of HBr involved in the reaction = molar concentration * volume (L)

Molar concentration = 2.0 M, volume = 25 ml = 0.025 L

Number of moles = 2.0 M * 0.025 L= 0.05 moles

Therefore, 0.05 moles of HBr reacts with 0.05 moles of KOH to produce 0.05 moles of H20

Enthalpy change per mole of HBr = 1924.64 J/0.05 moles = 38492.8 J/mol = 38.493 KJ/mol

Answer:

Explanation:

The objective here is to prepare  1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.

Given that :

the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml since 100 mg/ml = 100000  μg/ml

However, using formula C₁V₁=C₂V₂ (Ampicilin),

where:

C₁ = 100000 μg/ml,

V₁=?,

C₂= 50  μg/ml,

V₂=1000 ml

100000 μg/ml × V₁ = 50  μg/ml × 1000 ml

V₁ =  50  μg/ml × 1000 ml/100000 μg/ml

V₁ =   0.5 ml

Given that:

the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml , 25 mg/ml = 25000 μg/ml

Now by using formula C₁V₁=C₂V₂ (Kanamycin),

C₁ = 25000 μg/ml,

V₁=?,

C₂= 100  μg/ml,

V₂=1000 ml

25000 μg/ml × V₁ = 100  μg/ml × 1000 ml

V₁ =  100  μg/ml × 1000 ml/25000 μg/ml

V₁ =   4 ml

Thus; in 1 lite of Lb+ Kan+Amp preparation;

0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.

Finally;

Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.

Complete Question

The complete question is shown on the first uploaded image

Answer:

For reaction 1

    [tex]K_{eq} = 10^{29}[/tex]

     The equilibrium position is to the right

For reaction 2

        The equilibrium position is to the left

Explanation:

Generally  [tex]pKa[/tex] is mathematically evaluated as  

[tex]pKa = pKa _ \ {left }} - pKa _ \ {right }}[/tex]

And equilibrium position [tex]K_a[/tex] is mathematically evaluated as [tex]K_{eq} = 10^\ {-pK_a}[/tex]

From the question we are told that

For reaction 1

         [tex]pKa_\ {left}} \ = 9[/tex]

        [tex]pKa_\ {right }} \ = 38[/tex]

So

       [tex]pKa = 9-38[/tex]

       [tex]pKa =-29[/tex]

So  [tex]K_{eq} = 10^{-(-29)}[/tex]

      [tex]K_a = 10^{29}[/tex]

This implies that the equilibrium position is to the right

   For reaction 2

       [tex]pKa_\ {left}} \ = 15.9[/tex]

       [tex]pKa_\ {right }} \ = 9.24[/tex]

So

       [tex]pKa = 15.9-9.24[/tex]

       [tex]pKa = 6.66[/tex]

So  [tex]K_{eq} = 10^{-(6.66)}[/tex]

      [tex]K_{eq} = 10^{-6.66}[/tex]

This implies that the equilibrium position is to the left

Answer:

CONVERT IT:

50°F is equal to 10°C

Answer:

10 degrees Celsius

Explanation:

(50°F − 32) × 5/9 = 10°C

Answer:

Anode half reaction;

Co(s) ----> Co^2+(aq) + 2e

Cathode half reaction;

2Ag^+(aq) + 2e-------> 2Ag(s)

Explanation:

A voltaic cell is an electrochemical cell that spontaneously produces electrical energy from chemical reactions. A voltaic cell comprises of an anode (where oxidation occurs) and a cathode (where reduction occurs). The both electrodes are connected with a wire . A salt bridge ensures charge neutrality in the anode and cathode compartments. Electrons flow from anode to cathode.

For the cell referred to in the question;

Anode half reaction;

Co(s) ----> Co^2+(aq) + 2e

Cathode half reaction;

2Ag^+(aq) + 2e-------> 2Ag(s)

Answer:

A. The cell will undergo crenation

B. The cell will undergo hemolysis

C. The cell will undergo hemolysis

D. The cell will undergo crenation.

E. The cell will undergo neither crenation nor hemolysis

Explanation:

A hypotonic solution is a solution in which the concentration of solutes is greater inside the cell than outside the cell.

A hypertonic solution is a solution in which the concentration of solutes is greater outside the cell than inside the cell.

An isotonic solution is a solution in which the concentration of solution is the same outside and inside of the cell. A solution with 5% (m/v) glucose and 0.9% (m/v) NaCl is an isotonic solution.

When a red blood cell is placed in a hypotonic solution, it will swell and burst. This is known as hemolysis.

When placed in a hypertonic solution, a red blood cell will lose water and shrivel. This is is known as cremation.

When a red blood cell is placed in an isotonic solution, neither hemolysis or crenation occurs as there is no net movement of water across the cell's membrane.

A: 3.75 % (m/v) NaCl Solution is a hypertonic solution. The cell will undergo crenation

B: 1.92 % (m/v) glucose Solution is hypotonic. The cell will undergo hemolysis

C: Distilled H2O Solution is a hypotonic solution. The cell will undergo hemolysis

D: 9.03 % (m/v) glucose Solution is a hypertonic solution. The cell will undergo crenation

E: 5.0% (m/v) glucose and 0.9% (m/v) NaCl are both isotonic solutions. The cell will undergo neither hemolysis not crenation.

Solution A (3.75% NaCl): Crenation

To find out if a red blood cell will shrink, burst, or stay the same when placed in a solution, we have to think about how concentrated the solution is compared to the red blood cell.

A red blood cell has a normal concentration of about 0. 9% salt or 0. 3% sugar Solutions that have more concentrated substances are called hypertonic, while solutions that have less concentrated substances are called hypotonic.

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Answer:

142 grams

Explanation:

To find the molar mass of a molecule or compound, you simply need to add together the molar masses of all of the atoms that comprise it. Phosphorus has a molar mass of about 31, while oxygen has one of about 16, meaning that the molar mass of this molecule is:

2(31)+5(16)=62+80=142

Hope this helps!

Answer:

Li>Na>K>Rb

Explanation:

Answer:

c. Li > Na > K > Rb

Explanation:

edge 2021

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Answer:

The type of salt to be added to the water is not known from the question but no worries, I will try to give you the step by step procedure to answer any type of question similar to this.

To answer this question, we should know some facts.

1. the molar freezing point depression constant of water (Kf) = 1.86 K kg/mol

2. the molar mass of the salt if NaCl = 58.5 g/mol ; KCl = 74.5 g/mol

3. since the salt can dissociate if NaCl or KCl into two ions, the Van't Hoff factor ( i )= 2

Note that: the change in freezing point, molarity, deepression constant and van't Hoff factor are related by this formula;

                ΔTf = i Kf m

So lets take NaCl as the salt:

Molar mass = 58.5 g/mol

Van't Hoff factor = 2

1. calculate the number of moles

So we can calculate the molarity of the salt NaCl from the formula;

m = ΔTf / i Kf

m = 5 / 2 * 1.86

m = 5 / 3.72

m = 1.344 mol/kg

2. calculate the number of moles of the salt required

Next is to multiply the molarity by the mass of water. Density of water = 1kg/L

number of moles = 1.344 mol/kg * 1 Kg/L * 1 kg water

number of moles = 1.344 moles.

3. calculate the mass of the salt.

numner of moles = mass / molar mass

mass = number of moles * molar mass

mass = 1.344 * 58.5

mass = 78.624 g of NaCl salt.

You can follow these steps to solve for the type of salt you are given in the question.

Answer:

In this problem the correct thing would be to use the ideal gas equation.

Explanation:

Well in this exercise we will use the following equation:

(P x V) / T = (p x v) / t

On the right side of the equation we will find the initial values, that is, the values ​​with which the reaction begins and in general they are always the first to write in the problems.

Instead on the left side of the equation, the letters that are in lowercase are the final values, that is to say at the end of the reaction that the values ​​of pressure, temperature and volume are reached.

P is pressing, just like p, T and t are temperature, and V and v are volume.

We use this equation so we consider the behavior of said gas to be an IDEAL gas, a constant volume.

That is why the given pressures require an atmosphere to pass, which is another unit used to press the pressure ... Much needed in this equation! An atmosphere is equivalent to 760 millimeters of mercury ...

Then the final and initial pressures would be:

initial pressure: 1.15 atm

final pressure: 1.38 atm

In this way you already have the values ​​to be able to solve in the equation your unknown that would be the final temperature:

Considering that the volume is constant, we cancel it from the equation, 1.15 atm would be in the value of P and 1.38 in the value of p ... In this way it considers that 20 degrees Celsius is the initial temperature or ses T, we would only have to clear the t.

Answer:

not 100% but i think its 1.57x10^20

Explanation:

5.25x10^-4g / 2.016g

2.60x10^-4 x 6.022x10^23= 1.56x10^20 molecules

Answer:

Explanation:

You will investigate the breakdown of starch by amylase at different pHs.

The different pHs under investigation will be produced using buffer solutions. Buffer solutions produce a particular pH, and will maintain it if other substances are added.

The amylase will break down the starch.

A series of test tubes containing a mixture of starch and amylase is set up at different pHs.

A sample is removed from the test tubes every 10 seconds to test for the presence of starch. Iodine solution will turn a blue/black colour when starch is present, so when all the starch is broken down, a blue-black colour is no longer produced. The iodine solution will remain orange-brown.

A control experiment must be set up - without the amylase - to make sure that the starch would not break down anyway, in the absence of an enzyme. The result of the control experiment must be negative - the colour must remain blue-black - for results with the enzyme to be valid.

When the starch solution is added:

For each pH investigated, record the time taken for the disappearance of starch, ie when the iodine solution in the spotting tile remains orange-brown.

The time taken for the disappearance of starch is not the rate of reaction.

It will give us an indication of the rate, but is the inverse of the rate - the shorter the time taken, the greater the rate of the reaction.

We can calculate the rate of the reaction by calculating  \frac{1}{t}, obtaining a measure of the rate of reaction by dividing one by the time taken for the reaction to occur.

A similar experiment can be carried out to investigate the effect of temperature on amylase activity.

Set up a series of test tubes in the same way and maintain these at different temperatures using a water bath - either electrical or a heated beaker of water.

Depending on the chemical reaction under investigation, you might monitor the reaction in a different way. If investigating the effect of temperature on the breakdown of lipid by lipase, you could monitor pH change - lipids are broken down into fatty acids and glycerol. As the reaction begins, the release of fatty acids will mean that the pH will decrease.

good luck :)

Answer:

The ranks is

Ge: 3d10 4s2 4p2 (6 electrons in the outer shell)

Br: 3d10 4s2 4p5 (7 electrons in the outer shell)

Kr: 3d10 4s2 4p6 (8 electrons in the outer shell)

Explanation:

Electronic configuration reffers to the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It gives us the understanding of the shape and energy of its electrons. The electronic configuration explain the The electron affinity or propensity to attract electrons

It Should be noted that the most stable configuration in an electronic configuration is attributed to when the last shell is full, i.e. when the last shell has 8 electrons.

When an atom is closer to reach the 8 electrons in the outer shell, then it's electron affinity big.

Considering the given three configuration of the elements above, we can see that "Br"needs requires only 1 electron to have 8 electrons in the outer shell, therefore, it is considered to have the biggest electron affinity among them which is reffers to as the LEAST NEGATIVE.

Ge: with the electronic configuration 3d10 4s2 4p2 has 6 electrons in the outer shell which means it still requires 2 electrons to complete 8 electrons in its outer shell, so it can be deducted that it posses an atom that is more negative than Br.

Kr: with the electronic configuration 3d10 4s2 4p6 which is a noble gas has 8 electrons in the outer shell cannot add more electrons to its outer shell because the 8 electrons is complete posses the least electron affinity among the three elements and it is the MOST NEGATIVE

Answer:

C = 98%

Explanation:

Hello,

To determine the percentage of salt recovered, we'll divide the mass of the salt recovered over by the original mass of the salt.

Mass of salt recovered = 1.47g

Initial mass of salt = 1.50g

Percentage of salt recovered = (mass recovered/ initial mass of salt) × 100

Percentage of salt recovered = (1.47 / 1.50) × 100

Percentage of salt recovered = 0.98 × 100

Percentage of salt recovered = 98%

The percentage of salt recovered is equal to 98%

Explanation:

We know that more is the [tex]pK_{a}[/tex] value, weaker will be the acid. Also, an acid completely dissociates into ions in an aqueous base solution when [tex]pK_{a}[/tex] of conjugate acid of base is greater than acid.

4-methylphenol [tex](CH_{3}C_{6}H_{4}OH)[/tex] ([tex]pK_{a} = 10.26[/tex]) is quite soluble in its sodium salt. In NaOH, the dissociation will be as [tex]Na^{+}[/tex] and [tex]OH^{-}[/tex] ions as NaOH is a strong base.

Therefore, 4-methylphenol will readily dissolve in NaOH solution.

As, [tex]NaHCO_{3}[/tex] is not a strong base but as 4-Methylphenol forms a sodium salt hence, it will have a low solubility as compared to NaOH.

Whereas [tex]Na_{2}CO_{3}[/tex] is not a base but when dissolved in water it shows basic character as it produces NaOH (strong base) and [tex]H_{2}CO_{3}[/tex] (weak acid). As a result, the solution gets basic. Hence, 4-methylphenol will readily dissolve in [tex]Na_{2}CO_{3}[/tex].