The equilibrium constant (K p) for the interconversion of PCl 5 and PCl 3 is 0.0121:
PCl5 (g) → PCl3 (g) + Cl2 (g)
A vessel is charged with PCl 5 giving an initial pressure of 0.123 atm and yields PCl 3 and Cl 2. At equilibrium, the partial pressure of PCl 3 is ________ atm.
A) 0.0782.
B) 0.0455.
C) 0.0908.
D) 0.0330.
E) 0.123.

The question is incomplete, the complete question is shown in the image attached to this answer

Answer:

A

Explanation:

We can see from the conditions of the reaction that the incoming nucleophile is -SCH3 and there are two possible leaving groups in the substrate.

First of all, we have to look at the conditions of the reaction. We can see that the reaction is taking place in DMF, a polar aprotoc solvent. This condition favours the SN2 synchronous mechanism over the SN1 ionic mechanism.

Hence, the nucleophile at the 1-position is preferentially substituted owing to the conditions of the reaction.

Thus, option A is the major product of the reaction.

Answer:

The correct answer is Option c (0.163 g).

Explanation:

Given:

Heat energy,

Q = 0.21 J

Specific heat,

c = 4.184 J/g°c

Change in temperature,

ΔT = 0.308°C

As we know,

⇒ [tex]Q=mc \Delta T[/tex]

By substituting the values, we get

 [tex]0.21=m\times 4.184\times 0.308[/tex]

     [tex]m=\frac{0.21}{0.308\times 4.184}[/tex]

         [tex]=\frac{0.21}{1.28867}[/tex]

         [tex]=0.163 \ g[/tex]

Answer:

A fluid is a medium that has a defined mass and volume, but no fixed shape, at a constant temperature and pressure. This may include gases, liquids, plasmas, and to some extent plastic solids. A fluid can flow and deform, preventing it from carrying loads in a static equilibrium.  A fluid is always compressible and internal frictional forces always occur due to the viscosity of the fluid.

Answer:

41264 g of CO₂

Explanation:

Combustion reaction is:

C₃H₈ + 5O₂ →  3CO₂  +  4H₂O

1 mol of propane react to 5 moles of oxygen in order to proudce 3 moles of carbon dioxide and 4 moles of water.

In a combustion reaction, our reactant reacts to oxygen and the products are always CO₂ and water.

We have the volume of propane but we need moles of it, so we need to apply density.

Density = mass / volume so mass = density . volume.

Density of propane is: 493 g/L

Mass of propane is 493 g/L . 27.9L = 13754.7 g

We convert mass to moles: 13754.7 g . 1 mol/ 44g = 312.6 moles

According to reaction, 1 mol of propane can produce 3 moles of CO₂

Our 312.6 moles will produce 312.6 . 3 = 937.8 moles

We convert moles to mass: 937.8 mol . 44 g/mol = 41264 g

Answer:

(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms

Explanation:

Alcohols are poor leaving groups.

To make -OH group a better-leaving group, it should be treated with sulfonyl chlorides.

Then, methane sulfonyl group makes will be substituted on the -OH group and forms sulfonyl esters and makes it a better leaving group.

After that treating with KI proceeds through nucleophilic bimolecular substitution and the final product formed is shown below:.

Answer:

a) N2 (g) + H2 = 2 NH3

b) You have to state the mass of hydrogen

Answer:

the answer is 2.1 atm

Explanation:

the way people normally do it is by simply deciding the 325k with the 1.2

i think .117018

now to get to the 20charaeactrs minimum.

Answer:

The sample should be diluted

Explanation:

According to Beer Lambert's law, the absorbance of a sample depends on the concentration of the sample.

Hence, if the concentration of the sample is very high, the spectrophotometer will also report a very high value of absorbance.

When this is the case, the sample should simply be diluted and the readings are taken again using the spectrophotometer.

Answer:

a.draw 2,3 dicholoro octane

Explanation:

mag isip ka kung paano hehe

Answer:The green growing on the penny of copper and the rust forming on the nail of iron are chemical changes. Boiling away salt water, scraping iron filings from a mixture of sand with a magnet, and breaking a rock with a hammer, are physical changes.

Explanation:

Answer:

The material properties of nanostructures are different from the bulk due to the high surface area over volume ratio and possible appearance of quantum effects at the nanoscale. ... Yu; they found that the structural distortions on the quantum dots depend both on the kind of dopant and on the size of the dots.

Explanation:

hope it helps

Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction:    NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:  

[NH3] = 0.030M    [NH4+] = 0M    [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 =  x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:

a) pH = 10.86

b) pH = 9.66

c) pH = 9.15

d) pH = 7.70

e) pH = 5.56

f) pH = 3.43          

Initially, the pH of the solution is given by the dissociation of NH₃ in water.  

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻     (1)

The constant of the above reaction is:

[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex]   (2)

At the equilibrium, we have:  

   NH₃    +    H₂O   ⇄   NH₄⁺    +    OH⁻     (3)  

0.030 M - x                      x               x

[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]  

Now, we can calculate the pH of the solution as follows:

[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]

Hence, the initial pH is 10.86.

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻  (4)  

We can calculate the pH of the solution from the equilibrium reaction (3).            

[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)  

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

[tex] n_{b} = n_{i} - n_{HCl} [/tex]     (6)

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]          

[tex] n_{a} = n_{HCl} [/tex]   (7)

[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]

The concentrations are given by:

[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex]   (8)

[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex]      (9)

After entering the values of Ca and Cb into equation (5) and solving for x, we have:  

[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]

Hence, the pH is 9.66.

We can find the pH of the solution from the reaction of equilibrium (3).

The concentrations are (eq 8 and 9):

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]    

[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]    

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:  

[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]

So, the pH is 9.15.

We can find the pH of the solution from reaction (3).

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]      

[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]      

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]

x = 5.013x10⁻⁷ = [OH⁻]      

Then, the pH is:  

[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]  

So, the pH is 7.70.

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]    

[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:                

NH₄⁺    +    H₂O   ⇄   NH₃    +    H₃O⁺

Ca - x                             x               x

[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]  

[tex] Ka(Ca - x) - x^{2} = 0 [/tex]   (10)          

We can find the acid constant as follows:

[tex] Kw = Ka*Kb [/tex]

Where Kw is the constant of water = 10⁻¹⁴

[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]  

The concentration of NH₄⁺ is:

[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]      

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:  

[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]

Hence, the pH is 5.56.

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]      

[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]

Therefore, the pH is 3.43.

Find more about pH here:

brainly.com/question/491373

I hope it helps you!  

⇒b. Ksp = [Zn2+]3 [PO43]2

Hope It Helps You ✌️

Answer:

M=0.549M

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to perform this calculation by firstly assuming we have 1 kg of water as the solvent so that we have 0.550 moles of NaCl as well. Moreover, we realize we have 1000 grams of water and the correct mass of the solution can be calculated by converting 0.550 moles of NaCl to grams by using its molar mass:

[tex]m_{solute}=0.550mol*\frac{58.44 g}{1mol}= 32.14g\\\\m_{solution}=1000g+32.14g=1032.14g[/tex]

And subsequently, the volume in liters by using the density and the correct conversion factor:

[tex]V_{solution}=1032.14g*\frac{1mL}{1.03g} *\frac{1L}{1000mL} =1.002L[/tex]

Finally, the molarity will be:

[tex]M=\frac{0.550mol}{1.002L} =0.549M[/tex]

Regards!

The equation for the energy required to melt 2 kg of gold is 3440 kJ.

Energy is the ability to do work or cause change. It is an essential part of everyday life and is present in many forms, such as thermal energy, electrical energy, chemical energy, and mechanical energy. Energy can be converted from one form to another in order to do work.

The equation for calculating the energy required to melt a certain mass of material is Q = m x Lf, where Q is the energy required (in joules), m is the mass of the material (in kilograms), and Lf is the latent heat of fusion (in joules per kilogram).
Using the table below, we can see that the latent heat of fusion for gold is 1760 kJ/kg. Therefore, the equation for the energy required to melt 2 kg of gold is: Q = 2 kg x 1760 kJ/kg = 3440 kJ.

To learn more about energy
https://brainly.com/question/19666326
#SPJ1

Answer:

unsaturated fats, which are liquid at room temperature,are different from saturated fat because they contain one or more double bonds and fewer hydrogen atoms on their carbon chain.

Answer:

a) NH₄NO₃ ⇒ N₂O + 2 H₂O

b) 1.69 × 10²³ molecules

Explanation:

Step 1: Write the balanced equation

NH₄NO₃ ⇒ N₂O + 2 H₂O

Step 2: Convert 11.2 g of NH₄NO₃ to moles

The molar mass of NH₄NO₃ is 80.04 g/mol.

11.2 g × 1 mol/80.04 g = 0.140 mol

Step 3: Calculate the moles of H₂O produced

0.140 mol NH₄NO₃ × 2 mol H₂O/1 mol NH₄NO₃ = 0.280 mol H₂O

Step 4: Calculate the number of molecules in 0.280 moles of water

We will use Avogadro's number.

0.280 mol × 6.02 × 10²³ molecules/1 mol = 1.69 × 10²³ molecules

Answer:

1) 16.0 moles Al

24.0 moles S

96.0 moles O

2)In 6.10 moles magnesium perchlorate, (Mg(CIO4)2 we have:

6.10 moles Mg

12.2 moles Cl

48.8 moles O

3)4.6 moles of propane (total) contains 13.8 moles of carbon and 36.8 moles of hydrogen atoms

4)The gold coin contains 7.8 *10^22 atoms

Explanation:

Step 1: Data given

Number of moles of aluminum sulfate, Al2(SO4)3 = 8.00 moles

Step 2: Calculate the number of moles

In 1 mol of aluminum sulfate, Al2(SO4)3 we have:

2 moles of Al

3 moles of S

12 moles of O

This means that in 8.00 moles of aluminum sulfate, Al2(SO4)3 we have:

2*8.00 = 16.0 moles Al

3*8.00 = 24.0 moles S

12*8 = 96.0 moles O

2. Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 6.10 moles of magnesium perchlorate, (Mg(CIO4)2

1 mol of magnesium perchlorate, (Mg(CIO4)2 has:

1 Mol of Mg

2 moles of Cl

8 moles of O

In 6.10 moles magnesium perchlorate, (Mg(CIO4)2 we have:

1 * 6.10 moles = 6.10 moles Mg

2*6.10 = 12.2 moles Cl

8*6.10 = 48.8 moles O

3. A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain?

In 1 mol of propane, C3H8 we have:

3 moles of C and 8 moles of H

This means if we have 13.8 moles of carbon, we have 13.8/3 = 4.6 moles of propane, C3H8 and 4.6 *8 = 36.8 moles of H

So 4.6 moles of propane contains 13.8 moles of carbon and 36.8 moles of hydrogen atoms

4. A rare gold coin (24 karat, or 100% gold) has a mass of 25.54 g. How many atoms of gold are present in this coin?

Calculate moles of gold:

Moles = mass of gold / molar mass gold

Moles = 25.54 grams / 196.97 g/mol

Moles = 0.1297 moles

Calculate atoms:

Number of atoms = moles * number of Avogadro

0.1297 * 6.022 *10^23 = 7.8 *10^22 atoms

The gold coin contains 7.8 *10^22 atoms

Answer:

1.433g of HgS are produced

Explanation:

A Solution Containing 2.0 Grams Of Hg(NO3)2 Was Added To A Solution Containing 2.0 Grams Of Na2S. Calculate the mass of the HgS that was formed (it is a precipitate) according to this reaction:

Based on the reaction:

Na2S + Hg(NO3)2 → HgS + 2NaNO3

To solve this question we need to find the moles of each reactant in order to find the limiting reactant. The moles of limiting reactant = moles of HgS:

Moles Na2S -Molar mass: 78.0452 g/mol-

2.0g * (1mol / 78.0452g) = 0.0256 moles Na2S

Moles Hg(NO3)2 -324.7g/mol-

2.0g * (1mol / 324.7g) = 0.006159 moles Hg(NO3)2

As the reaction is 1:1, and moles of Hg(NO3)2 < moles Na2S

The moles of Hg(NO3)2 = Moles HgS = 0.006159 moles

The mass is:

Mass HgS -Molar mass: 232.66g/mol-:

0.006159 moles * (232.66g/mol) =

Answer:

14 moles of oxygen needed to produce 12 moles of H2O.

Explanation:

We are given that balance eqaution

[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]

We have to find number of moles of O2 needed  to produce 12 moles of H2O.

From given equation

We can see that

6 moles of   H2O produced by Oxygen =7 moles

1 mole of   H2O produced by Oxygen=[tex]\frac{7}{6}[/tex]moles

12 moles of H2O produced by Oxygen=[tex]\frac{7}{6}\times 12[/tex]moles

12 moles of H2O produced by Oxygen=[tex]7\times 2[/tex]moles

12 moles of H2O produced by Oxygen=14 moles

Hence, 14 moles of oxygen needed to produce 12 moles of H2O.

The amount of oxygen required for the combustion of ethane to produce 12 moles of water is 14 moles.

The moles of oxygen produced in the reaction can be given from the stoichiometric law of the balanced chemical equation.

The balanced chemical equation for the combustion of ethane is:

[tex]\rm 2\;C_2H_6\;+\;7\;O_2\;\rightarrow\;4\;CO_2\;+\;6\;H_2O[/tex]

The 6 moles of water are produced from 7 moles of oxygen. The moles of oxygen required to produce 12 moles of water are:

[tex]\rm 6\;mol\;H_2O=7\;mol\;Oxygen\\12\;mol\;H_2O=\dfrac{7}{6}\;\times\;12\;mol\;O_2\\ 12\;mol\;H_2O=14\;mol\;O_2[/tex]

The moles of oxygen required to produce 12 moles of water are 14 moles.

Learn more about moles produces, here:

https://brainly.com/question/10606802

Answer:

True

Explanation:

The word Organic refers to the methods used to cultivate and process farm agricultural products. Organic foods are edible and nutritious substances consumed (both plants and animals) that are free from the use of synthetics and chemicals. In plants, the include the use of organic manure that serves as fertilizers and carrying out the weeding process by hand weeding. In animals, diseases can be prevented by maintaining a clean house or rotational grazing.

The benefit of organic foods are to produce food substances with no chemical substances.

hope it helps to everyone

Answer:

False

Explanation:

Answer:

Explanation:

From the given information:

We are to make use of the spinach absorbance extract which is the corrected absorbance  (y) = 0.306

And also the trendline equation:

y = 1609x + 0.0055

where,

x = absorbance of the spinach extract.

∴

0.306 = 1609x + 0.0055

collecting the like terms

0.306 - 0.0055 = 1609x

0.3005  = 1609x

x = 0.3005/1609

x = 1.8676 × 10⁻⁴

x ≅ 0.0002 M

No. of grams for the chlorophyll can be computed as follows:

recall that:

molar mass of chlorophyll = 893.5 g/mol

the volume = 25ml = (25/1000) L = 0.025 L

∴

In spinach solution, the no. of grams for the chlorophyll:

= (0.0002) mol/L × (893.5 g/mol) × (0.025) L

= 0.0044675 g

≅ 0.0045 g

In the spinach, the concentration of chlorophyll = no of grams of chlorophyll/ mass of the spinach

= 4.5 mg/0.1876 g

= 23.987 mg/g

≅ 24 mg/g

From the given information:

Chlorophyll is any member of the class of the green pigments involved in the photosynthesis process.

And also the trendline equation:

y = 1609x + 0.0055

where,

x = absorbance of the spinach extract.

so 0.306 = 1609x + 0.0055

collecting the like terms

0.306 - 0.0055 = 1609x

0.3005  = 1609x

x = 0.3005/1609

x = 1.8676 × 10⁻⁴

x ≅ 0.0002 M

2. No.of grams for the chlorophyll can be computed as follows:

recall that:

molar mass of chlorophyll = 893.5 g/mol

The volume = 25ml = (25/1000) L = 0.025 L

Therefore:

In spinach solution, the no. of grams for the chlorophyll:

= (0.0002) mol/L × (893.5 g/mol) × (0.025) L

= 0.0044675 g

≅ 0.0045 g

3. In the spinach, the concentration of chlorophyll = no of grams of chlorophyll/ mass of the spinach

= 4.5 mg/0.1876 g

= 23.987 mg/g

≅ 24 mg/g

Read more about chlorophyll here:

https://brainly.com/question/3529377

Answer: The molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Explanation:

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol[/tex]

Now, molarity of ethanol solution is as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M[/tex]

(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol[/tex]

Now, molarity of sucrose solution is as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L} (1 mL = 0.001 L)\\= 0.94 M[/tex]

(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol[/tex]

Now, molarity of sodium chloride solution is as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M[/tex]

Thus, we can conclude that the molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Answer:

Yes. The solution would be optically active.

Explanation:

Diastereomer are defined as the image that is non mirror and non -identical. It is made up of two stereoisomers. They are formed when the two stereoisomers or more than two stereoisomers of the compound have the same configuration at the equivalent stereocenters.

In the given context, as the product given is a diastereomeric mixture, the product would have an optical activity in total.

So the answer is Yes.

C. It can help us track how quickly element of the climate are changing