The Fischer esterification mechanism is examined in the following two questions in the assignment. Part 1 involves MeOH addition to form the key tetrahedral intermediate. Part 2 will involve loss of H2O to form the ester. Part 1 of 2:

Answer:

Molarity → 0.410 M

Molality → 0.44 m

Percent by mass → 7.61 g

Mole fraction (Xm) = 7.96×10⁻³

Mole percent = 0.79 %

Explanation:

We analyse data:

28.8 g of glucose → To determine molarity, molality, mole fraction, mole percent we need to find out the moles:

28.2 g. 1mol / 180 g = 0.156 moles of glucose

350 g of water → Mass of solvent.

We convert from g to kg in order to determine molality = 350 g / 1000 = 0.350 kg

We also need the moles of solvent: 350 g / 18 g/mol = 19.44 moles

380 mL of solution → Volume of solution; to determine the molarity we need the volume in L → 380 mL / 1000 = 0.380L

Solution mass = Solute mass + Solvent mass

28.2 g + 350 g = 378.2 g

Total moles = Moles of solute + Moles of solvent

0.156 + 19.44 = 19.596 moles

Molarity → Moles of solute in 1L of solution → 0.156 mol / 0.380L = 0.410 M

Molality → Moles of solute in 1kg of solvent → 0.156 mol / 0.350 kg = 0.44 m

Percent by mass → Mass of solute in 100 g of solution

(28.8g /378.2g) . 100 = 7.61 g

Mole fraction (Xm)= Moles of solute/ Total moles → 0.156 mol / 19.596 moles = 7.96×10⁻³

Mole percent = Xm . 100 → 7.96×10⁻³ . 100 = 0.79 %

The molarity of the solution is 0.42 M. The molality of the solution is 0.46 m.

a)To obtain the molarity of the solution;

Number of moles = mass of glucose/molar mass of glucose = 28.8g/180 g/mol

= 0.16 moles

Volume of solution = 380mL or 0.38 L

Molarity = number of moles /volume =  0.16 moles/ 0.38 L  = 0.42 M

b) Molality of the solution;

Mass of solvent in Kg = 350g/1000 = 0.35 Kg

Molality = number of moles/mass of solution in kilogram = 0.16 moles/ 0.35 Kg = 0.46 m

c) percent by mass = mass of solute/mass of solution× 100

= 28.8g/(350g +28.8g)  × 100 = 7.6 %

d) Mole fraction

Number of moles of water = 350g/ 18 g/mol = 19.44 moles

Total number of moles = 19.44 moles + 0.16 moles = 19.6 moles

Mole fraction of glucose =  0.16 moles/19.6 moles = 0.0082

e) Mole percent

Mole fraction  × 100 =  0.0082  × 100 = 0.82%

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Final answer:

The concentration equilibrium constant (Kc) will not include CH₃OH because it is a pure liquid. Then, Kc = 68.

Explanation:

First, we will calculate the molar concentration (M) of each substance using the following expression.

M = mass of the substance / molar mass of the substance × volume of solution

CO₂

M = 2.25 g / 44.01 g/mol × 8.6 L = 0.0059 M

H₂O

M = 2.72 g / 18.02 g/mol × 8.6 L = 0.018 M

CH₃OH

M = 3.82 g / 32.04 g/mol × 8.6 L = 0.014 M

O₂

M = 1.98 g / 32.00 g/mol × 8.6 L = 0.0072 M

Let's consider the following reaction at equilibrium.

CO₂(g) + H₂O(g) ⇄ CH₃OH(l) + O₂(g)

The concentration equilibrium constant (Kc) will not include CH₃OH because it is a pure liquid. Then,

Kc = [O₂] / [CO₂] × [H₂O]

Kc = 0.0072 / 0.0059 × 0.018

Kc = 68

Answer:

51.1 g of  Ca₃(PO₄)₂(s) can be made in this reaction

Explanation:

The reactans are CaCl₂ and Na₃PO₄. Let's determine the reaction:

3CaCl₂(aq)  +  2Na₃PO₄(aq)  →  Ca₃(PO₄)₂(s) ↓ + 6NaCl(aq)

We convert the mass of chloride to moles:

55 g . 1 mol/ 110.98 g = 0.495 moles

Ratio is 3:1. Let's make a rule of three to find the answer in moles:

3 moles of chloride can produce 1 mol of phosphate

Therefore 0.495 moles will produce (0.495 . 1) / 3 = 0.165 moles

We convert the moles to mass:

0.165 mol .  310.18 g /1 mol = 51.1 g

The mass of calcium phosphate produced has been 51.1 g.

The balanced chemical equation of for the reaction has been:

[tex]\rm 3\;CaCl_2\;+\;2\;Na_3PO_4\;\rightarrow\;Ca_3(PO_4)_2\;+\;6\;NaCl[/tex]

The balanced chemical equation has been given that 3 moles of calcium chloride produces 1 moles of calcium phosphate.

The moles of calcium chloride in 55 g sample has been given as:

[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}[/tex]

Substituting the values:

[tex]\rm Moles=\dfrac{55}{110.98}\\Moles\;CaCl_2=0.495\;mol[/tex]

The moles of calcium chloride has been 0.495 mol.

The moles of calcium formed has been given by:

[tex]\rm 3\;mol\;CaCl_2=1\;mol\;Ca_3(PO_4)_2\\0.495\;mol\;CaCl_2=\dfrac{1}{3}\;\times\;0.495\;mol\; Ca_3(PO_4)_2\\0.495\;mol\;CaCl_2=0.165\;mol\;Ca_3(PO_4)_2\\[/tex]

The moles of calcium phosphate formed has been 0.165 mol.

The mass of calcium phosphate has been:

[tex]\rm Mass=Moles\;\times\;molar\;mass\\Mass=0.165\;\times\;310.18\;g\\Mass=51.1\;g[/tex]

The mass of calcium phosphate produced has been 51.1 g.

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Answer:

Explanation:

The correct answer is 19, 20 DHDP is more polar than DHA. This is as a result of the presence of two hydroxyl groups.

Docosahexaenoate (DHA) is a long-chain polyunsaturated omega-3 fatty acid important for brain and heart health, whereas 19,20-dihydroxydocosapentaenoate (19,20-DHDP) is a hydroxylated derivative of DHA that is more polar due to additional oxygen-containing groups.

The difference between docosahexaenoate (DHA) and 19,20-dihydroxydocosapentaenoate (19,20-DHDP) primarily lies in their chemical structure and resulting properties. DHA is a long-chain polyunsaturated fatty acid (PUFA) with 22 carbon atoms and six double bonds, specifically an omega-3 fatty acid. It is known for its important role in brain development, cognitive function, and cardiovascular health. DHA can be synthesized in the body from alpha-linoleic acid (ALA), though the rate of conversion is limited, suggesting a dietary need for DHA-rich foods or supplements.

Meanwhile, 19,20-DHDP is a derivative of DHA that has been hydroxylated, containing additional oxygen groups which make it more polar than DHA. This increase in polarity could impact the compound's function and location within biological membranes.

The equilibrium constant (Kc) and reaction quotient (Qc) are calculated using the initial molar concentrations of the substances involved in the reaction. The Qc for the formation of nitrosyl chloride from nitric oxide and molecular chlorine is 0, given the initial conditions provided.

The student is asking about the equilibrium constant (Kc) and reaction quotient (Qc) for the formation of nitrosyl chloride from nitric oxide and molecular chlorine. To calculate the reaction quotient (Qc) of the reaction 2NO(g) + Cl2(g) ⇌ 2NOCl(g), you need to know the molar concentrations of the reactants and products.

First, find the initial concentrations of NO, Cl2, and NOCl by dividing their quantities by the volume of the flask, which is 4.40×10^-2 mol/2.60L for NO, 1.80×10^-3 mol/2.60L for Cl2, and 0 for NOCl because it's not initially present.

Then, we plug these concentrations into the Qc expression, which for this balanced equation is [NOCl]^2 / ([NO]^2 [Cl2]) = (0)^2 / ((4.40×10^-2 mol/2.60L)^2 * (1.80×10^-3 mol/2.60L)) = 0.

So, the reaction quotient Qc for the experiment is 0.

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Using the balanced equation and these concentrations, we find Qc to be 6.73 × 10⁷.

To determine the reaction quotient, Qc, for the given reaction:

2NO(g) + Cl₂(g) ⇌ 2NOCl(g)

We follow these steps:

Thus, the reaction quotient Qc for the experiment is 6.73 × 10⁷.

Answer:

The statement that is FALSE.

Entropy increases with dissolution.

Entropy generally increases with increasing molecular complexity.

For noble gasses, entropy increases with size.

The entropy of a gas is greater than the entropy of a liquid.

Free atoms have greater entropy than molecules.

Explanation:

Free atoms have greater entropy than molecules.

The energy is increased in faster moving particles, the opposite happens in slower ones. The entropy increases if there is randomized distribution of the energy,  the system becomes then more stable as energy is released in a cluster.

The false statement is: 'Free atoms have greater entropy than molecules'. Because complexity brings more possibilities for microstates, actual molecules usually have greater entropy than single atoms. All of the other statements provided in the original query are correct.

In the given context, the statement that is false is: "Free atoms have greater entropy than molecules". Entropy, which is a measure of randomness or disorder in a system, is influenced by the structure of the particles (atoms or molecules) that comprise a substance.

More complex molecules, with greater number of atoms, typically have a higher entropy than solitary atoms because they have more ways to vibrate and thus more potential microstates, increasing the entropy of the system. This goes contrary to the false statement made.All the other statements are true. For instance, entropy does increase with dissolution, reflecting the increase in molecular disorder and the number of potential microstates.

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Answer:

20.44% is the percentage of dimethylphthalate in the sample.

Explanation:

Molarity of NaOH = [tex]M_1=0.1215 M[/tex]

Volume of NaOH consumed in back titration = [tex]V_1=?[/tex]

Molarity of HCl =[tex]M_2=0.1251 M[/tex]

Volume of HCl = [tex]V_2=32.25 ml[/tex]

[tex]M_1V_1=M_2V_2[/tex]

[tex]V_1=\frac{M_2V_2}{M_1}=\frac{0.1251 M \times 50.0 mL}{0.1215 M}[/tex]

[tex]V_1=33.21 mL[/tex]

Volume of NaOH used in hydrolysis of ester = 50.00 mL - 33.21 mL = 16.79 mL[/tex]

Moles of NaOH in 16.79 ml of 0.1215 M solution : n

Volume of solution = 16.79 mL = 0.01679 L ( 1mL=0.001 L)

[tex]n=0.1215 M\times 0.01679 L =0.002040 mol[/tex]

1 mole of NaOH has 1 mole of hydroxide ions than 0.002040 moles of NaOH will have :

1 × 0.002040 mol = 0.002040 mol of hydroxide ions

Moles of hydroxide ions = 0.002040 mol

[tex]C_6H_4(COOCH_3)_2 + 2OH^-\rightarrow C_6H_4(COO)^{2-} + 2H_2O[/tex]

According to reaction, 2 moles of hydroxide ion reacts with 1 mole of  dimethylphthalate , then 0.002040 moles of hydroxide ion swill react with ;

[tex]\frac{1}{2}\times 0.002040 mol=0.001020 mol[/tex] of dimethylphthalate

Mass of 0.001020 moles of dimethylphthalate :

194 g/mol × 0.001020 mol = 0.1979 g

Mass of sample = 0.9679 g

Mass of dimethylphthalate = 0.1979 g

Percentage of dimethylphthalate  in sample;

[tex]=\frac{0.1979 g}{0.9679 g}\times 100=20.44\%[/tex]

20.44% is the percentage of dimethylphthalate in the sample.

Final answer:

The percentage of dimethylphthalate in the sample is calculated by first determining the moles of NaOH that reacted with it, then converting those moles to grams of dimethylphthalate, and finally dividing this by the original sample mass. The calculated percentage is approximately 20.48%.

Explanation:

The question pertains to the calculation of the percentage of dimethylphthalate in a sample after a saponification reaction followed by a back titration. Firstly, we need to calculate the amount of NaOH that reacted with dimethylphthalate. This is done by determining the amount of NaOH remaining after the reaction, which is found through the back titration with HCl. The total amount of NaOH initially added to the reaction is 50.00 mL of 0.1215 M, which equals 6.075 mmol. The back titration used 32.25 mL of 0.1251 M HCl, which equals 4.033 mmol of NaOH remaining.

Thus, the amount of NaOH that reacted with dimethylphthalate is 6.075 mmol - 4.033 mmol = 2.042 mmol. Since the saponification of dimethylphthalate requires two moles of NaOH for each mole of ester, the moles of dimethylphthalate are half this amount, which is 1.021 mmol. Multiplying by the molar mass of dimethylphthalate (194.19 g/mol) gives 0.1983 g of dimethylphthalate in the sample.

Finally, the percentage of dimethylphthalate in the sample is calculated by dividing the mass of dimethylphthalate by the original sample mass and multiplying by 100. Therefore, it is (0.1983 g / 0.9679 g) × 100 = 20.48%.

Answer: The net chemical equation for the formation of manganese from manganese (II) carbonate, oxygen and aluminum is written above.

Explanation:

The given chemical equation follows:

Equation 1:  [tex]2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)[/tex]     ( × 3)

Equation 2:  [tex]3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)[/tex]     ( × 2)

As, the net chemical equation does not include manganese (IV) oxide. So, to cancel out from the net equation, we need to multiply equation 1 by (3) and equation 2 by (2)

Now, the net chemical equation becomes:

[tex]6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6Mn(s)+4Al_2O_3(s)+6CO_2(g)[/tex]

Hence, the net chemical equation for the formation of manganese from manganese (II) carbonate, oxygen and aluminum is written above.

Final answer:

To calculate the amount of heat energy that will be absorbed by the water, we need to use the specific heat capacity of water and convert the volume of water to liters. By using the formula Q = mcΔT, we can calculate that 143.32 kJ of heat energy will be absorbed by the water.

Explanation:

To calculate the amount of heat energy that will be absorbed by the water, we need to use the specific heat capacity of water, which is 4184 J/kg/°C. We also need to convert the volume of water from pint to liters. There are approximately 0.473 liters in a pintpint.

First, we convert the volume of water to liters: 0.900 pint × (0.473 liters/pint) = 0.426 liters.

Next, we calculate the temperature change: boiling point (100°C) - room temperature (20°C) = 80°C.

Finally, we use the formula Q = mcΔT, where Q is the heat energy absorbed, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change: Q = (0.426 kg) × (4184 J/kg/°C) × (80°C) = 143,315.52 J. To convert this to kilojoules (kJ), divide by 1000: 143,315.52 J / 1000 = 143.32 kJ.

The given question is incomplete. The complete question is as follows.

A chlorine atom is adsorbed on a small patch of surface (see sketch at right). This patch is known to contain 81 possible adsorption sites. The atom has enough energy to move from site to site, so it could be on any one of them. Suppose a Br atom also becomes adsorbed onto the surface. Calculate the change in entropy. Round your answer to significant digits, and be sure it has the correct unit symbol.

Explanation:

The change in entropy will be calculated by using the following formula.

           [tex]\Delta S = k_{B} ln (\frac{W}{W_{o}})[/tex]

Initially, it is given tha Cl atom could be adsorbed on any of the 81 sites. When Br is added then there will be 80 possible sites when the Br can be adsorbed. This means that total possible sites are as follows.

                   [tex]81 \times 80[/tex]

                   = 6480

This shows that there are 6480 microstates which are accessible to the system.

So, change in entropy will be calculated using the Boltzmann constant as follows.

      [tex]\Delta S = 1.381 \times 10^{-23} J/K \times ln (\frac{6480}{81})[/tex]

                 = [tex]1.381 \times 10^{-23} J/K \times 4.382[/tex]

                 = [tex]6.05 \times 10^{-23} J/K[/tex]

or,              = [tex]6.1 \times 10^{-23} J/K[/tex] (approx)

Thus, we can conclude that the change in entropy is [tex]6.1 \times 10^{-23} J/K[/tex].

Answer:

2.144 g of calcium hydride are needed to produce 2.5 l of hydrogen gas, collected over water at 26 degrees celcius and 760 torr.

Explanation:

The reaction of Calcium hydride and water is given by

CaH₂ + 2H₂O -----> Ca(OH)₂ + 2H₂

2 moles of Hydrogen gas are produced from 1 mole of Calcium hydride.

But we need to find out how much moles of Hydrogen are produced from this reaction first.

Using the ideal gas, equation,

PV = nRT

P = pressure = 760 torr = 101325 Pa

V = volume of hydrogen gas produced = 2.5 L = 0.0025 m³

n = number of moles of Hydrogen gas produced = ?

R = molar gas constant = 8.314 J/mol.K

T = absolute temperature in Kelvin = 26°C = 299.15 K

n = PV/RT = (101325×0.0025)/(8.314×299.15) = 0.102 moles

Back to the stoichiometric balance of the reaction

CaH₂ + 2H₂O -----> Ca(OH)₂ + 2H₂

2 moles of Hydrogen are produced from 1 mole of Calcium hydride

0.102 moles of Hydrogen will be produced from (0.102 × 1/2) moles of Calcium hydride.

Moles of Calcium hydride = 0.051 moles.

Mass of Calcium hydride that reacted = number of moles of Calcium hydride that reacted × Molar mass

Moles mass of Calcium hydride = 42.094 g/mol

Mass of Calcium hydride that reacted = 0.051 × 42.094 = 2.144g

Answer:

The answer is 181.437 ml of water to increase the soil's moisture content to 10%.

Explanation:

We have been given:

Soil weight = 5 lb = 2.26796 kg

2% of moisture ⇒ [tex]\frac{2}{100}[/tex] × 2.26796    =    0.045359 kg of water in the soil

10% of moisture ⇒ [tex]\frac{10}{100}[/tex] × 2.26796    = 0.226796 kg of water in the soil

The difference in moisture content = (0.226796 - 0.045359) kg

=0.181432 kg

But 1kg = 1000 ml

∴ 0.181437 kg = 181.437 ml of water required to increase the moisture content to 10%

Answer:

Unchanged

Explanation:

Epsom salt is also referred to as magnesium sulphate heptahydrate with the chemical formula [tex]MgSO_4.7H_2O.[/tex]

There are 7 moles of water and 1 mole of magnesium sulphate in every 1 mole of epsom salt. The water ratio remains the same irrespective of the amount or weight of epsom salt taken.

Hence, if 2.500 g of epsom salt is taken and heated to a constant mass instead of 1.500 g, the ratio of water molecules present will be the same.

Answer: The concentration of KOH solution is 1.215 M

Explanation:

For the given chemical equation:

[tex]2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)[/tex]

To calculate the concentration of base, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.

We are given:

[tex]n_1=2\\M_1=1.50M\\V_1=16.2mL\\n_2=1\\M_2=?M\\V_2=40.0mL[/tex]

Putting values in above equation, we get:

[tex]2\times 1.50\times 16.2=1\times M_2\times 40.0\\\\M_2=\frac{2\times 1.50\times 16.2}{1\times 40.00}=1.215M[/tex]

Hence, the concentration of KOH solution is 1.215 M

Answer: The molarity of bromine ions in the chemist's solution is [tex]4.06\times 10^{-3}M[/tex]

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]

Given mass of vanadium(III) bromide = 0.12 g

Molar mass of vanadium(III) bromide = 295.65 g/mol

Volume of solution = 300 mL

Putting values in above equation, we get:

[tex]\text{Molarity of solution}=\frac{0.12\times 1000}{295.65g/mol\times 300}\\\\\text{Molarity of solution}=1.353\times 10^{-3}M[/tex]

The chemical formula of vanadium(III) bromide is [tex]VBr_3[/tex]

1 mole of vanadium(III) bromide produces 1 mole of [tex]V^{3+}[/tex] ions and 3 moles of [tex]Br^-[/tex] ions

Molarity of bromine ions = [tex](3\times 1.353\times 10^{-3})=4.06\times 10^{-3}M[/tex]

Hence, the molarity of bromine ions in the chemist's solution is [tex]4.06\times 10^{-3}M[/tex]

Final answer:

The molarity of bromide ions (Br-) in the solution is 0.00179 M.

Explanation:

To calculate the molarity of bromide ions (Br-) in the solution, we need to determine the number of moles of VBr. Given that the mass of VBr is 0.12 g and its molar mass is 223.6 g/mol (51.9 g/mol for V and 79.9 g/mol for Br), we can calculate the number of moles:

moles of VBr = mass of VBr / molar mass of VBr

= 0.12 g / 223.6 g/mol

= 0.000536 mol

Since the solution is 300 ml, we need to convert the volume to liters:

volume in liters = volume in ml / 1000

= 300 ml / 1000

= 0.3 L

molarity of Br- = moles of VBr / volume in liters

= 0.000536 mol / 0.3 L

= 0.00179 M (rounded to significant digits)

Answer: Highest to lowest pH: Ba(OH)2, NaOH, N2H2, HOCL HCl

Explanation:

Stronger the acid, lower the pH, stronger the base, higher the pH.

N2H2 weak base, Ba(OH)2 strong base, HOCL weak acid, NaOH strong base, HCl strong acid.

Ba(OH)2 has produces more H ions, so it has a higher pH than NaOH.

Explanation:

The given data is as follows.

         [tex]K_{k}[/tex] = 89 MPa,     [tex]\sigma[/tex] = 336 MPa

           Y = 0.92

Now, we will calculate the length of critical interior flaw as follows.

     [tex]a_{c} = \frac{1}{\pi}(\frac{K_{k}}{\sigma Y})^{2}[/tex]

                 = [tex]\frac{1}{\pi}(\frac{89}{336 \times 0.92})^{2}[/tex]

                 = [tex]\frac{656.38}{3.14}[/tex]

                = 209.04 mm

Thus, we can conclude that minimum length of a surface crack that will lead to fracture is 209.04 mm.

Answer:

It will lead to false positive (or negative results) for this classification tests.

Explanation:

In this tests, the functional group that is really being tested for is the Carbonyl group. In the iodoform, the presence of a Carbonyl group gives a reaction and in the dinitrophenylhydrazine test, aldehydes give a reaction and ketones do not.

So, rinsing the glassware with acetone is introducing ketone before the qualitative test has even began, thereby leading to false results for each of the two tests mentioned in the question.

Final answer:

Rinsing glassware with acetone before conducting qualitative tests such as the iodoform test or dinitrophenylhydrazine test should be avoided to prevent false positive results due to acetone's interference or reaction with the test compounds.

Explanation:

When performing qualitative tests in glass test tubes, such as the iodoform test or dinitrophenylhydrazine test, rinsing the glassware with acetone prior to the experiment should be avoided because it can lead to false positive results. These qualitative tests rely on specific chemical reactions to occur, and if residual acetone remained in the tube, it might react or interfere with the test reagents or the compound being tested, giving an inaccurate result. For instance, acetone itself can produce a yellow precipitate with dinitrophenylhydrazine, mimicking a positive result for the presence of certain carbonyl groups. Therefore, cleaning the glassware properly without using acetone, or ensuring that any acetone used is completely evaporated and the glassware is dried in a water-free environment, is crucial to the accuracy of these tests.

To quickly dry glassware without using acetone, rinsing with distilled water and allowing the glassware to dry overnight or using warm air or nitrogen gas for drying is recommended. Additionally, glassware cleanliness is paramount in chemical testing and experiments to avoid contamination that could affect the results.

Answer:

C

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4  +  Cl-

Firstly, we calculate the number of moles of  the ammonia  as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

                 NH4+      H2O     ⇄  NH3        H3O+

I                0.242                           0             0

C                 -X                              +x              +X

E             0.242-X                          X              X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b    = K w

, where  K w- the self-ionization constant of water, equal to  

10 ^-14  at room temperature

This means that you have

K a = K w.K b   = 10 ^− 14 /1.8 * 10^-5 =  5.56 * 10^-10

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

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈  0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

The pH at the equivalence point of the titration is:

C) 4.94

NH₃(aq)+HCl(aq)→NH₄Cl(aq)

Ionic chemical equation:

HCl + NH₃ ---> [tex]NH_4^+ + Cl^-[/tex]

c = n/v

n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Calculation for concentration of the ammonium ions = 0.36/1.486 = 0.242M

                [tex]NH_4^+[/tex]       H₂O     ⇄     NH₃        [tex]H_3O^+[/tex]

I                0.242                           0             0

C                 -X                              +x              +X

E             0.242-X                          X              X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value :Ka

To find this, we use the mathematical equation below:

K a ⋅ K b    = K w

Kw = [tex]10^{-14}[/tex]

[tex]K_a = K_w*K_b\\\\ 10^{− 14} /1.8 * 10^{-5} = 5.56 * 10^{-10}\\\\K_a = [NH_3][H_3O^+]/[NH_4^+]\\\\K_a = x * x/(0.242-x)[/tex]

Since, the value of Ka is small, we can say that 0.242-x ≈  0.242

[tex]K_a = x^2/0.242 = 5.56 * 10^{-10}\\\\x^2 = 0.242 * 5.56 * 10^{-10} = 1.35 * 10^{-10}\\\\x = 0.00001161895\\\\H_3O^+= 0.00001161895\\\\pH = -log[H_3O^+]\\\\pH = -log[0.00001161895 ] = 4.94[/tex]

Thus, correct option is C. The pH at the equivalence point of the titration is: 4.94

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Explanation:

A property that leads to changes in chemical composition of a substance is known as chemical property. So, when we move down a group in periodic table then chemical properties of the elements remain the same.

This is because elements of the same group tend to have same number of valence electrons. Therefore, they have similar reactivity which leads to change in their chemical composition upon reaction with another substance.

For example, lithium (Li) and potassium (K) are both group 1 elements. And, phosphorous (P) and arsenic (As) are both group 15 elements.

Thus, we can conclude that pair of Li and K will show similar chemical properties. Also, pair of P and As will show similar chemical properties.

The question is incomplete, complete question is ;

While idly tossing some keys from hand to hand one day, your friend Reuben (an expert chemist) says this: "Soluble metal oxides form hydroxides when dissolved in water." Using Reuben's statement, and what you already know about chemistry, predict the products of the following reaction. Be sure your chemical equation is balanced.

[tex]K_2O(aq) + H_2O(l)\rightarrow ?[/tex]

Answer:

The product will be potassium hydroxide.

Explanation:

When aqueous potassium oxide reacts with water it gives aqueous solution of potassium hydroxide as a product. And potassium hydroxide is a hydroxide of potassium metal with formula KOH.

[tex]K_2O(aq) + H_2O(l)\rightarrow 2KOH(aq)[/tex]

According to recation , 1 mole of potassium oxide when recats with 1 mole of water to give 2 moles of potassium hydroxide.

Answer: The amount of heat absorbed by the solution is 56.98 kJ

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]    .....(1)

Molarity of barium hydroxide solution = 0.310 M

Volume of solution = 70 mL = 0.070 L   (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

[tex]0.310M=\frac{\text{Moles of }Ba(OH)_2}{0.070L}\\\\\text{Moles of }Ba(OH)_2=(0.310mol/L\times 0.070L)=0.0217mol[/tex]

Molarity of HCl solution = 0.620 M

Volume of solution = 70 mL = 0.070 L

Putting values in equation 1, we get:

[tex]0.620M=\frac{\text{Moles of HCl}}{0.070L}\\\\\text{Moles of HCl}=(0.620mol/L\times 0.070L)=0.0434mol[/tex]

The chemical equation for the reaction of NaOH and sulfuric acid follows:

[tex]Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O[/tex]

By Stoichiometry of the reaction:

2 moles of HCl produces 2 moles of water

So, 0.0434 moles of HCl will produce = [tex]\frac{2}{2}\times 0.0434=0.0434moles[/tex] of water

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of solution = 1 g/mL

Volume of solution = [70 + 70] = 140 mL

Putting values in above equation, we get:

[tex]1g/mL=\frac{\text{Mass of solution}}{140mL}\\\\\text{Mass of solution}=(1g/mL\times 140mL)=140g[/tex]

[tex]q=mc\Delta T[/tex]

where,

q = heat absorbed

m = mass of solution = 140 g

c = heat capacity of solution= 4.186 J/g°C

[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=(25.34-21.12)^oC=4.22^oC[/tex]

Putting values in above equation, we get:

[tex]q=140g\times 4.186J/g^oC\times 4.22^oC=2473.08J=2.473kJ[/tex]

To calculate the enthalpy change of the reaction, we use the equation:

[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]

where,

q = amount of heat absorbed = 2.473 kJ

n = number of moles of water = 0.0434 moles

[tex]\Delta H_{rxn}[/tex] = enthalpy change of the reaction

Putting values in above equation, we get:

[tex]\Delta H_{rxn}=\frac{2.473kJ}{0.0434mol}=56.98kJ/mol[/tex]

Hence, the amount of heat absorbed by the solution is 56.98 kJ

The total heat absorbed is approximately 2.47 kJ.

To find the heat absorbed by the solution in the given calorimetry problem, we use the equation q = mcΔT, where m is the mass, c is the specific heat, and ΔT is the temperature change.

where:

Step-by-Step Solution:

Thus, the heat absorbed by the solution is approximately 2.47 kJ.

Correct question is: In a constant‑pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)₂ was added to 70.0 mL of 0.620 M HCl . The reaction caused the temperature of the solution to rise from 21.12°C to 25.34°C . If the solution has the same density and specific heat as water, what is heat absorbed by the solution? Assume that the total volume is the sum of the individual volumes. (And notice that the answer is in kJ).

Answer:

151.4863 years

Explanation:

Half life, t1/2 = 100 years

Initial concentration,[A]o = 100%

Final concentration, [A] = 35% (after 65% have been decayed)

Time = ?

Half life for a first Order reaction is given as;

t1/2 = ln (2) / k

k = ln(2) / 100

k = 0.00693y-1

The integral rate law for first order reactions is given as;

ln[A] = ln[A]o − kt

kt = ln[A]o - ln[A]

t = ( ln[A]o - ln[A]) / k

t = [ln(100) - ln(35)] /0.00693

t = 1.0498 / 0.00693

t = 151.4863 years

It will take approximately 173.04 years for 65% of the nickel to undergo decay.

The decay of a radioactive substance is governed by the first-order kinetics equation:

[tex]\[ N(t) = N_0 e^{-\lambda t} \][/tex]

where:

- [tex]\( N(t) \)[/tex] is the number of un decayed nuclei at time [tex]\( t \)[/tex],

- [tex]\( N_0 \)[/tex] is the initial number of nuclei,

- [tex]\( \lambda \)[/tex] is the decay constant, and

- [tex]\( t \)[/tex] is the time elapsed.

The decay constant [tex]\( \lambda \)[/tex] is related to the half-life [tex]\( t_{1/2} \)[/tex] by the equation:

[tex]\[ \lambda = \frac{\ln(2)}{t_{1/2}} \][/tex]

Given that the half-life [tex]\( t_{1/2} \) of \( ^{63}Ni \)[/tex] is 100. years, we can calculate [tex]\( \lambda \)[/tex]:

[tex]\[ \lambda = \frac{\ln(2)}{100} \approx \frac{0.693}{100} \][/tex]

Now, we want to find the time [tex]\( t \)[/tex] when 65% of the nickel has decayed, which means that 35% of the original nickel remains un decayed. Let's denote this time as [tex]\( t_{65\%} \)[/tex]. We can set up the equation:

[tex]\[ 0.35N_0 = N_0 e^{-\lambda t_{65\%}} \][/tex]

Dividing both sides by[tex]\( N_0 \)[/tex] and taking the natural logarithm gives us:

[tex]\[ \ln(0.35) = -\lambda t_{65\%} \][/tex]

[tex]\[ -\lambda t_{65\%} = \ln(0.35) \][/tex]

[tex]\[ t_{65\%} = -\frac{\ln(0.35)}{\lambda} \][/tex]

Substituting the value of [tex]\( \lambda \)[/tex] we calculated earlier:

[tex]\[ t_{65\%} = -\frac{\ln(0.35)}{\frac{\ln(2)}{100}} \][/tex]

[tex]\[ t_{65\%} = -\frac{100 \cdot \ln(0.35)}{\ln(2)} \][/tex]

[tex]\[ t_{65\%} = -\frac{100 \cdot \ln(0.35)}{0.693} \][/tex]

[tex]\[ t_{65\%} \approx -\frac{100 \cdot (-1.0498)}{0.693} \][/tex]

[tex]\[ t_{65\%} \approx 150.48 \][/tex]

However, we must consider that the time calculated is the time for 35% of the nickel to remain, not for 65% to decay. Since the half-life is the time for 50% to decay, and we have calculated the time for less than 50% to remain, the actual time for 65% to decay must be longer than one half-life. Therefore, we need to find the additional time [tex]\( \Delta t \)[/tex] it takes for the remaining 35% to decay to 15% (since 85% decayed in the first half-life).

We can use the same formula, but this time for the remaining 35% to decay to 15%:

[tex]\[ 0.15N_0 = 0.35N_0 e^{-\lambda \Delta t} \][/tex]

Solving for [tex]\( \Delta t \)[/tex]:

[tex]\[ \Delta t = -\frac{\ln(0.15/0.35)}{\lambda} \][/tex]

[tex]\[ \Delta t = -\frac{\ln(0.4286)}{\lambda} \][/tex]

[tex]\[ \Delta t = -\frac{100 \cdot \ln(0.4286)}{\ln(2)} \][/tex]

[tex]\[ \Delta t \approx -\frac{100 \cdot (-0.8473)}{0.693} \][/tex]

[tex]\[ \Delta t \approx 122.56 \][/tex]

Adding this to the initial half-life:

[tex]\[ t_{total} = 100 + 122.56 \approx 222.56 \][/tex]

However, since we are looking for the time it takes for 65% of the original amount to decay, we need to subtract the time it took for the first 50% to decay from the total time calculated:

 [tex]\[ t_{65\%} = t_{total} - t_{1/2} \][/tex]

[tex]\[ t_{65\%} = 222.56 - 100 \][/tex]

[tex]\[ t_{65\%} \approx 122.56 \][/tex]

This calculation is incorrect because we did not consider that the decay process is continuous and the time for the first 50% to decay is already included in the total time. Therefore, the correct total time for 65% to decay is the initial calculation of 150.48 years plus the additional time of 122.56 years:

[tex]\[ t_{65\%} = 150.48 + 122.56 \][/tex]

[tex]\[ t_{65\%} \approx 273.04 \][/tex]

Again, we must correct for the fact that the initial half-life is already part of the decay process. The correct additional time needed for 65% of the nickel to decay is:

[tex]\[ t_{additional} = t_{65\%} - t_{1/2} \][/tex]

[tex]\[ t_{additional} = 273.04 - 100 \][/tex]

[tex]\[ t_{additional} \approx 173.04 \][/tex]

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

[tex]C=18.13gCO_2*\frac{12gC}{44gCO_2} =4.945gC\\H=4.639gH_2O*\frac{2.016gH}{18.0152gH_2O}=0.519gH\\N=2.885gN_2\\O=10.0g-4.945g-0.519g-2.885g=1.651gO[/tex]

Next, the corresponding moles:

[tex]C=4.945gC*\frac{1molC}{12gC}=0.412mol\\H=0.519gH*\frac{1molH}{1gH}=0.519mol\\N=2.885gN*\frac{1molN}{14gN}=0.206molN\\O=1.648gO*\frac{1molO}{16gO} =0.103molO[/tex]

Then, each element's subscripts is found to be:

[tex]C=\frac{0.412}{0.103}=4\\H=\frac{0.519}{0.103}=5\\N=\frac{0.206}{0.103} =2\\O=\frac{0.103}{0.103}=1[/tex]

Therefore, the empirical formula is:

[tex]C_4H_5N_2O[/tex]

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

[tex]C_8H_10N_4O_2[/tex]

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

The molar mass of the compound is found by finding the empirical and

molecular formula of the compound.

Reasons:

Molar mass of CO₂ = 44.01 g/mol

Number of moles of CO₂ produced = [tex]\dfrac{18.13}{44.01}[/tex] ≈ 0.412 moles

Number of moles of produced C = 0.412 moles

Mass of C = 12 × 0.412 = 4.944 g

Molar mass of H₂O = 18.015 g/mol

Moles of H₂O produced = [tex]\dfrac{4.639}{18.015 }[/tex] = 0.2575 moles

Molar mass of N₂ = 28.0134 g/mol

Mass of oxygen = 10 - 4.944 - 2.885 - 0.519 = 1.652

Therefore, we get;

Number of moles of produced C = 0.412 moles

Number of moles of produced H = 0.515 moles

Number of moles of oxygen, O ≈ 0.103 moles

Number of moles of N produced = 0.206 moles

Dividing by 0.103 gives;

Molar mass of the compound is between 150 g/mol and 210 g/mol (given)

The molar mass of C₄H₅N₂O = 4×12 + 5×1.00784 + 2×14 + 16 ≈ 97

The molar mass of C₄H₅N₂O ≈ 97 g/mol

Molar mass of the compound is between 150 and 210 g/mol, therefore, n in

(C₄H₅N₂O)ₙ = 2, which gives;

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Answer:

1.56 M

Explanation:

This is a dilution process and so a dilution formula is suitably used as follows C1V1 = C2V2 where

C1 = concentration of the stock solution

V1 = volume of the stock solution

C2 = concentration of the resulting (dilute) solution and

V2 = the volume of the resulting (dilute) solution

C1V1 = C2V2 (Making C2 subject of the formula)

C2 = C1V1/V2

Given: C1 = 5.736 M; V1 = 3 Ml; V2 = (3+8) 11 Ml

C2 = 5.736 x 3 / 11

C2 = 1.56 M

The concentration of the resulting solution will be "1.56 M".

The procedure of decreasing the concentration of such a particular solute through its solution has been known as dilution.

This same chemist could essentially add extra solvent to the mixture.

According to the question,

Stock solution's concentration, C₁ = 5.736 M

Stock solution's volume, V₁ = 3 mL

Resulting solution's volume, V₂ = 3 mL + 8 mL

                                                    = 11 mL

By using the dilution equation, we get

→ C₂ = [tex]\frac{C_1 V_1}{V_2}[/tex]

By substituting the above values,

       = [tex]\frac{5.736\times 3}{11}[/tex]

       = [tex]\frac{17.208}{11}[/tex]

       = [tex]1.56[/tex] M

Thus the above answer is right.

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Explanation:

Hence, compound A will be Acetic acid.

[tex]CH_3OH + CO_2 +H_2[/tex] → [tex]CH_3COOH + H_2O[/tex]

        The product formed is methane and carbon monoxide.

    2. [tex]CH_3COOH[/tex] → [tex]CH_2CHO + H_2O[/tex]

         The product formed is formaldehyde and water.

    3. [tex]CH_3COOH + NaHCO_3[/tex] → [tex]CH_3COONa +CO_2 +H_2O[/tex]

         The product formed is sodium acetate, carbon dioxide, and water.

Answer:

53.9 g

Explanation:

We must  use the Henderson-Hasselbach equation to answer this question:

pH = pKa + log [A⁻]/[HA]

we know the pH,  pKa  (pKa = -log Ka), thus we can compute the ratio  [A⁻]/[HA], and from this the mass of  KC6H5CO2 knowing that M = mol/L.

therefore,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA]

4.63 = - ( - 4.20 ) +  log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking inverse log to both sides of this equation

2.69 = [A⁻]/[HA]

Now [A⁻] =2.69 x [HA] =2.69 x 1.00 M =  2.69 M

We know the molarity is equal to mol per liter of solution, so

mol  KC6H5CO2  = 2.69 mol/L x 0.125 L = 0.36 mol

and using the molecular weight of  KC6H5CO2   we get that the mass is

0.336 mol x 160.21 g/mol = 53.9 g

The student should take 53.9 g of KC6H5CO2

Answer:

He should dissolve 53.9 grams of KC6H5CO2

Explanation:

Step 1: Data given

Volume of A 1.00 M benzoic acid solution = 125 mL = 0.125 L

Ka of benzoic acid = 6.3*10^-5

pH = 4.63

Step 2: Calculate concentration of conjugate base

pH = pKa + log ([A-]/[HA])

4.63 = 4.20 + log ([A-]/[HA])

0.43 = log ([A-]/[HA])

10^0.43 = [A-]/[HA])

2.69 = [A-]/[HA])

2.69 = [A-]/ 1.00

[A-] = 2.69 M

Step 3: Calculate moles KC6H5CO2

Moles molarity * volume

Moles = 2.69 M * 0.125 L

Moles = 0.33625 moles

Step 4: Calculate mass KC6H5CO2

Mass of KC6H5CO2 = moles * molar mass

Mass of KC6H5CO2 = 0.33625 moles * 160.21 g/mol

Mass of KC6H5CO2 = 53.9 grams

He should dissolve 53.9 grams of KC6H5CO2

Answer:

Mg+2 (aq) + CO3-2 (aq) ---> MgCO3 (s)

Explanation:

The net ionic equation for the reaction of magnesium iodide and sodium carbonate is Mg⁺² (aq) + CO₃⁻²(aq) ---> MgCO₃ (s)

The ionic equations are those in which the electrolytes are dissociated into the ion of a solution.

The equations in which the written in dissociated ions of an electrolyte solution.

Thus, the ionic equation is Mg⁺² (aq) + CO₃⁻²(aq) ---> MgCO₃ (s).

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Answer:

The number of moles of gas in the sample = 0.916 moles

Explanation:

Step 1: Data given

Volume of the container = 8.5 L

Temperature = 55 °C = 328 K

Pressure = 2.9 atm

Step 2: Calculate the number of moles of gas in the sample.

p*V = n*R*T

⇒with p = the pressure = 2.9 atm

⇒with V = the volume of the container = 8.5L

⇒with n = the number of moles of gas = ?

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 328 K

n = (p*V)/(R*T)

n = (2.9*8.5)/(0.08206*328)

n = 24.65 / 26.91568‬

n =  0.916 moles

The number of moles of gas in the sample = 0.916 moles

Answer:

1. Na3[Fe(CN)6]

Oxidation number of iron is +3

Sodium hexacyanoferrate (III)

2. [Cr (en)2Cl2]+

Oxidation number of chromium is +3

Dichlorobis (ethylenediamine)chromium (III) ion

3. [Co (en)3]Cl3

Oxidation number of cobalt is +3

Tris (ethylenediamine)cobalt (III) chloride