A 10.00 mL sample of 8.50 M HNO3 solution is diluted to a new volume of 65.0 mL. What is the concentration of the dilute solution

Answer:

Explanation:

1) Glycolysis which occurs in the cytoplasm yield 2 ATP 2) Kreb cycle (critic acid cycle) yields 2 ATP per glucose molecule 3)electron transport chain yields the highest with 34 ATP both of which occurs in the mitochondria.

The question is incomplete, here is the complete question:

A chemist prepares a solution of aluminum chloride by measuring out of 11. g aluminum chloride into a 50. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's aluminum chloride solution.

Answer: The concentration of aluminium chloride solution is 1.65 mol/L

Explanation:

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

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

We are given:

Given mass of aluminium chloride = 11. g

Molar mass of aluminium chloride = 133.34 g/mol

Volume of solution = 50. mL

Putting values in above equation, we get:

[tex]\text{Molarity of solution}=\frac{11\times 1000}{133.34\times 50}\\\\\text{Molarity of solution}=1.65mol/L[/tex]

Hence, the concentration of aluminium chloride solution is 1.65 mol/L

Answer:

The question is not so clear, but there is a similar question that talks about reaction of alkanes with ammonia

Explanation:

The underlying factor dependent on whether alkane will react with ammonia or not is in their bond and reactivity. basically, alkanes are saturated hydrocarbons with a single bond existing between their chains. They are single bonded hydrocarbons and they majorly undergo SUBSTITUTION REACTION WITH HALOGENS

Alkanes are less reactive in this case, if other functional groups are attached or takes place alongside the reaction, the reaction may have a high chances of taking place.

The provided reaction notation is incorrect and does not correspond to known chemical species. Electron density alone does not determine a reaction's viability; conservation laws, such as those of mass-energy, charge, and nucleon number, are crucial in nuclear reactions.

The reaction provided in the question seems to be incorrectly or incompletely specified, as '327d1 + 327d2' does not correspond to any known chemical species. However, we can discuss the general rules that determine whether a nuclear reaction can occur, based on electron density and other factors.

In nuclear reactions, electron density alone is not sufficient to determine if a reaction will occur. Instead, factors such as conservation of mass-energy, conservation of charge, and conservation of nucleon number (protons and neutrons) are critical. Furthermore, the reaction must not violate any fundamental forces or principles, such as the conservation of lepton number in beta decay processes, where a neutron (n) can decay into a proton (p), an electron (e−), and an electron antineutrino (ve).

For example, in nuclear equations such as 238U → 234Ra + 4He, we can confirm the reaction is possible by checking these conservation laws. This is an alpha decay process, where the uranium nucleus emits an alpha particle (identical to a helium nucleus) and thus produces radium while observing all conservation laws.

Answer:

1) chemical change

2) filtration

Explanation:

A chemical change involves the formation of a new substance. In this case, the pure white solid formed is an entirely new substance, with a different chemical identity from those of the two solutions mixed to form it. The new solid is a precipitate. Precipitates are easily separated by filtration of the reaction mixture. Another name for chemical change is chemical reaction.

1) This is an example of chemical change.

2) The most convenient way to separate the newly formed white solid from the liquid mixture is filtration

A chemical change involves the formation of a new substance. In this case, the pure white solid formed is an entirely new substance, with a different chemical identity from those of the two solutions mixed to form it. The new solid is a precipitate. Precipitates are easily separated by filtration of the reaction mixture. Another name for chemical change is chemical reaction.

The most convenient way  to separate the newly formed white solid from the liquid mixture is filtration since precipitates are formed.

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Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)

Answer:

-3670.33 J/K

Explanation:

Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.

Mathematically,  change of Entropy can be expressed as,

ΔS = ΔH/T ....................................... Equation 1

Where ΔS = Change of entropy, ΔH = heat change, T = temperature.

ΔH = -(Lf×m).................................... Equation 2

Note: ΔH is negative because heat is lost.

Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg

Substitute into equation

ΔH = -(3.34×10⁵×3.0)

ΔH = - 1002000 J.

But T = 0 °C = (0+273) K = 273 K.

Substitute into equation 1

ΔS = -1002000/273

ΔS = -3670.33 J/K

Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C

The change in entropy for 3.0 kg of water changing into ice at 0 degrees Celsius is approximately 3658 J/K. This is calculated using the formula for entropy change and the known values for the heat of fusion for water and the temperature in Kelvin.

The change in entropy during a phase transition, such as the one from water to ice, can be calculated using the formula ΔS = Q/T, where ΔS is the change in entropy, Q is the heat involved during the phase transition, and T is the temperature in Kelvin. For the given scenario, the phase change is from liquid to solid (freezing), and the heat involved (Q) is equal to the mass multiplied by the heat of fusion for water. The heat of fusion for water is 333.55 J/g and the temperature (T) at which this change occurs is 0 degrees Celsius or 273.15 Kelvin. The mass is 3.0 kg or 3000g. Therefore, Q = 3000g x 333.55J/g = 999150 J. Substituting these values in our formula, ΔS = 999150 J / 273.15 K ≈ 3658 J/K.

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Answer:

The theoretical yield is 2.61 grams of aspirin

salicylic acid is the limiting reactant.

Explanation:

Step 1: Data given

Mass of salicylic acid = 2.0 grams

Volume of acetic acid = 5.0 mL

Density of acetic acid = 1.08 g/mL

Molar mass of acetic anhydride = 102.09 g/mol

Molar mass of salicylic acid = 138.12 g/mol

Step 2: The balanced equation

C7H6O3 + C4H6O3 → C9H8O4 + CH3COOH

Step 3: Calculate mass of acetic acid

Mass acid acid = density * volume

Mass acetic acid = 1.08 g/mL * 5.0 mL

Mass acetic acid = 5.5 grams

Step 4: Calculate moles salicylic acid

Moles salicylic acid = mass salicylic acid / molar mass salicylic acid

Moles salicylic acid = 2.00 grams / 138.12 g/mol

Moles salicylic acid = 0.0145 moles

Step 5: Calculate moles acetic anhydride

Moles acetic anhydride= 5.5 grams / 102.09 g/mol

Moles acetic anhydride = 0.0538 moles

Step 6: Calculate the limiting reactant

For 1 mol salicylic acid we need 1 mol acetic anhydride to produce 1 mol aspirine.

The limiting reactant is salicylic acid. It will completely be consumed. (0.0145 moles). Acetic anhydride will be in excess. There will react 0.0145 moles . There will remain 0.0538 - 0.0145 = 0.0393 moles

Step 7: Calculate moles aspirine

For 1 mol salicylic acid we need 1 mol acetic anhydride to produce 1 mol aspirine.

For 0.0145 moles salicylic acid, we'll have 0.0145 moles aspirine

Step 8: Calculate mass of aspirin

Mass aspirin = moles aspirin * molar mass aspirin

Mass aspirin = 0.0145 moles * 180.158 g/mol

Mass aspirin = 2.61 grams = Theoretical yield

The theoretical yield is 2.61 grams of aspirin

salicylic acid is the limiting reactant.

Answer:

2H2S + 3O2 → 2SO2 + 2H2O

Explanation:

Step 1: Data given

Hydrogen sulfide = H2S

Oxygen = O2

sulfur dioxide = SO2

water = H2O

Step 2: The unbalanced equation

H2S + O2 → SO2 + H2O

Step 3: Balancing the equation

H2S + O2 → SO2 + H2O

On the left side we have 2x O (in O2) and on the right side we have 3x O (2x in SO2 and 1x in H2O). To balance the amount of O, we have to multiply O2 (on the left side) by 3 and SO2 and H2O on the right side by 3.

H2S + 3O2 → 2SO2 + 2H2O

On the right side we have 4x H and on the left side we have 2x H. To balance the amount of H, we have to multiply H2S by 2.

Now the equation is balanced.

2H2S + 3O2 → 2SO2 + 2H2O

Final answer:

The concentration of the nickel(II) chloride contaminant in the groundwater sample is 5.0 x 10^-5 M, calculated by determining the moles of AgCl precipitate and its relation to NiCl2 through the stoichiometry of the reaction.

Explanation:

To calculate the concentration of nickel(II) chloride in the original groundwater sample, we need to use the information about the precipitate of silver chloride formed during the reaction. First, we must determine the moles of AgCl produced using its molar mass. The molar mass of AgCl is 143.32 g/mol, and we have 3.6 mg, or 0.0036 g of AgCl:

Moles of AgCl = mass (g) / molar mass (g/mol) = 0.0036 g / 143.32 g/mol = 2.51 x 10-5 moles

From the balanced chemical equation, we know that 1 mole of NiCl2 reacts with 2 moles of AgNO3 to produce 2 moles of AgCl. Therefore, the moles of NiCl2 present in the reaction will be half the moles of AgCl:

Moles of NiCl2 = 1/2 * Moles of AgCl = 1/2 * 2.51 x 10-5 moles = 1.255 x 10-5 moles

To find the concentration of NiCl2, we divide the moles of NiCl2 by the volume in liters:

Concentration of NiCl2 = moles / volume (L) = 1.255 x 10-5 moles / 0.250 L = 5.02 x 10-5 M

Therefore, the concentration of the nickel(II) chloride contaminant in the groundwater sample is 5.0 x 10-5 M, rounded to two significant digits.

Final answer:

The concentration of nickel(II) chloride in the original groundwater sample is calculated to be approximately 1.63 mg/L, based on the mass of the AgCl precipitate produced in the reaction with silver nitrate.

Explanation:

Calculating the Concentration of Nickel(II) Chloride

To calculate the concentration of nickel(II) chloride in the original groundwater sample, we first need to understand the reaction that takes place between nickel(II) chloride and silver nitrate to form silver chloride (AgCl) precipitate:

NiCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Ni(NO3)2(aq).

From the balanced equation, we see that one mole of NiCl2 reacts with two moles of AgNO3 to yield two moles of AgCl. The mass of the collected AgCl precipitate is given as 3.6 mg.

The molar mass of AgCl is 143.32 g/mol. Since we have 3.6 mg (or 0.0036 g) of AgCl, we can calculate the number of moles:

Moles of AgCl = mass (g) / molar mass (g/mol) = 0.0036 g / 143.32 g/mol ≈ 2.51 x 10-5 mol AgCl.

Since the ratio of AgCl to NiCl2 in the reaction is 2:1, there were half as many moles of NiCl2 as there were of AgCl in the original sample:

Moles of NiCl2 = 0.5 x Moles of AgCl = 0.5 x 2.51 x 10-5 mol ≈ 1.255 x 10-5 mol NiCl2.

The volme of the groundwater sample was 250.0 mL, or 0.250 L. The concentration (C) of NiCl2 in moles per liter (mol/L) is:

Concentration of NiCl2 = Moles of NiCl2 / Volume (L) = 1.255 x 10-5 mol / 0.250 L = 5.02 x 10-5 mol/L.

To convert this into mg/L (knowing that the molar mass of NiCl2 is 129.6 g/mol), we have

Mass (mg) = Moles x molar mass (g/mol) x 1000 mg/g = 1.255 x 10-5 mol x 129.6 g/mol x 1000 mg/g ≈ 1.63 mg.

Therefore, the concentration of NiCl2 in the groundwater sample is roughly 1.63 mg/L.

Answer: the empirical formula is C3H4O3

Explanation:Please see attachment for explanation

Answer:

The answer to your question is  C₃H₄O₃

Explanation:

Data

CxHyOz

mass of sample = 0.150 g

mass of CO₂ = 0.225 g

mass of H₂O = 0.0614 g

Reaction

                     CxHyOz + O₂   ⇒   CO₂   + H₂O

Process

1.- Calculate the moles of C

                            44 g of CO₂ ----------------- 12 g of C

                            0.225 g    ---------------- x

                            x = (0.225 x 12) / 44

                            x = 0.0614 g of C

                            12 g of C -------------------- 1 mol

                           0.0614 g of C --------------- x

                                 x = 0.0051 moles of Carbon

2.- Calculate the moles of hydrogen

                            18 g of H₂O ------------------ 2 g of H

                             0.0614 g --------------- x

                             x = 0.0068 g of H

                             1 g of H ----------------------- 1 mol of H

                            0.0068 g --------------------- x

                             x = 0.0068 moles of H

3.- Calculate the mass of Oxygen

    Mass of oxygen = 0.150 - 0.0614 - 0.0068

                               = 0.0818 g

                       16 g of O -------------------  1 mol

                      0.0818 g -------------------- x

                         x = (0.0818 x 1) / 16

                         x = 0.0051 moles of O

4.- Divide by the lowest number of moles

Carbon       0.0051 / 0.0051  = 1

Hydrogen   0.0068 / 0.0051 = 1.33

Oxygen       0.0051 / 0.0051 = 1

Multiply these numbers by 3

Carbon 3

Hydrogen = 4

Oxygen = 3

5.- Write the empirical formula

                                      C₃H₄O₃

Answer:

Explanation:Isobaric is a thermodynamic process in which the pressure of the system is zero. It is a process where there is no work done in the system. Therefore the temperature of the system according to the ideal gas law will vary linearly with pressure and inversely with volume.

Answer:

Dipole-Dipole force

Explanation:

Dipole - Dipole force -

These are the force of attraction , that occurs between two dipole , i.e. ,a species with two poles , hence , the attraction between the delta positive charge of first species with the delta negative charge of the second species , arises to a dipole - dipole force of attraction.

Hence, from the question,

SO₂ , is a polar compound , where O is more electronegative in comparison to S , thus , O attains a delta negative charge and S attains a delta positive charge and therefore , generates a dipole , and interacts with the dipole of the second molecule of SO₂ , arising a  dipole - dipole force of attraction .

Answer:

46.9mg of oxygen

Explanation:

From Henry's law,

Concentration of oxygen (C) = Henry's constant (K) × partial pressure of oxygen in air (p)

K = 1.3×10^-3M/atm O2, p = mole fraction of oxygen in air × pressure of air = 0.21×1.08atm = 0.2268atm

C = K×p = 1.3×10^-3 × 0.2268 = 0.00029484M of O2

Concentration (C) = number of moles of oxygen (n)/volume of water (V)

Volume of water (V) = 4.97L

n = CV = 0.00029484 × 4.97 = 0.001465mole

number of moles (n) = mass of O2/MW of O2

mass of O2 = number of moles of O2 × MW of O2 = 0.001465mole × 32g/mole = 0.0469g = 0.0469×1000mg = 46.9mg (to three significant figures)

The volume of a potassium iodide solution with a concentration of 0.0380 M and containing 150 g of potassium iodide is 3.95 L, rounded to three significant digits.

To calculate the volume V of a solution, you can use the formula:

[tex]\[ V = \frac{\text{mass of solute}}{\text{molarity}} \][/tex]

Given that the molarity M is [tex]\(0.0380 \, \text{M}\)[/tex] and the mass of potassium iodide is [tex]\(150 \, \text{g}\)[/tex], substitute these values into the formula:

[tex]\[ V = \frac{150 \, \text{g}}{0.0380 \, \text{M}} \]\[ V = 3947 \, \text{mL} \][/tex]

Since the answer should have the correct number of significant digits, the volume is [tex]\(3.95 \, \text{L}\)[/tex] (rounded to three significant digits).

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The complete question is:

Calculate the volume in liters of a 0.0380M potassium iodide solution that contains 150 g of potassium iodide. Be sure your answer has the correct number of significant digits.

Answer: The vapor pressure of water at [tex]50^0C[/tex] is 93.8 torr

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

[tex]ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]

where,

[tex]P_1[/tex] = initial pressure at [tex]25^oC[/tex] = 23.8 torr

[tex]P_2[/tex] = final pressure at [tex]50^oC[/tex] = ?

[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 43.9 kJ/mol = 43900 J/mol

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]

[tex]T_2[/tex] = final temperature = [tex]50^oC=273+50=323K[/tex]

Now put all the given values in this formula, we get

[tex]\log (\frac{P_2}{23.8}=\frac{43900}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{323K}][/tex]

[tex]\frac{P_2}{23.8}=antilog(0.5955)[/tex]

[tex]P_2=93.8torr[/tex]

Therefore, the vapor pressure of water at [tex]50^0C[/tex] is 93.8 torr

Explanation:

It is given that the total units of polymixn B silfate are present as follows.

              (10 + 250) ml = 260 ml

Therefore, total volume necessary to inject the whole volume is as follows.

              [tex]260 \times 15[/tex] drops

               = 3900

It is given that total time required to inject the solution is 2 hours. Converting it into minutes as follows.

                     [tex]2 hrs \times \frac{60 min}{1 hr}[/tex]

                       = 1200 minutes

Hence, rate of injection is calculated as follows.

      Rate of injection (r) = [tex]\frac{\text{total drops}}{\text{total time}}[/tex]

                   r = [tex]\frac{3900}{120}[/tex]

                     = 32.5 drops/min

Thus, we can conclude that the rate of injection is 32.5 drops/min.

To infuse 260 mL of solution over 2 hours with a drop factor of 15 drops/mL, the flow rate should be adjusted to 32.5 drops per minute.

We need to calculate the rate at which the solution should be administered in order to complete the infusion over 2 hours. First, we look at the total volume of the solution to be infused which is 250 mL of 5% dextrose injection plus the initial 10 mL, making it 260 mL in total. To calculate the rate, we'll use the formula:

Rate (drops/min) = Total volume (mL) × Administration set drop factor (drops/mL) ÷ Time (min)

Since we are told that the administration set delivers 15 drops/mL and that the infusion needs to be administered over 2 hours (which is 120 minutes), we can substitute the values into the formula:

Rate (drops/min) = (260 mL) × (15 drops/mL) ÷ (120 min)

After calculating, we find:

Rate (drops/min) = 32.5 drops/min

Note: This calculation is based on the student's given values and intended to demonstrate the process of determining the flow rate for an IV infusion.

To create a 15% solution of sodium chloride in 750 milliliters of water, you would need to add 112.5 grams of sodium chloride.

To create a 15% solution of sodium chloride (NaCl) in 750 milliliters (ml) of water, you need to know the concept of weight/volume percentage. A 15% w/v solution means 15g of solute (in this case, NaCl) is present in 100 ml of solution. To find how much NaCl is needed for 750ml, we simply multiply the percentage by the volume of solution (in this case, 750ml) and divide by 100.

So, to answer your question, the calculation would be: (15g/100ml)*750ml = 112.5g

Therefore, you would need to add 112.5 grams of sodium chloride to 750 ml of water to create a 15% solution.

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Answer:

179Kj/mol

Explanation:

The equation for the reaction is:

CaO + 3C     ⇒   CaC₂ + CO ₂

The formula for calculating the heat absorbed is:

      ∆Hf (Products) - ∆Hf (Reactants)

Using the heat of formation (∆Hf) for each component in the reaction

∆Hf Cao = -635kj/mol

∆Hf C = 0kj/mol

∆Hf CaC2 = -63kj/mol

∆Hf CO2 = -393.5

The heat absorbed = ( -63 + -393.5) – (-635)

                                  = -  456 + 635

                                  = 179kj/mol

Answer:

1. For 0.11 m [tex]FeBr_3[/tex]  : Lowest boiling point

2. For 0.15 [tex]CuBr_2[/tex]

: Second highest boiling point

3. For 0.24 [tex]AgNO_3[/tex]

: Third highest boiling point

4.  0.51 m glucose : Highest boiling point

Explanation:

Elevation in boiling point:

[tex]\Delta T_b=ik_b\times m[/tex]

where,

[tex]T_b[/tex] = change in boiling point

i= vant hoff factor

[tex]k_b[/tex] = boiling point constant

m = molality

1. For 0.11 m [tex]FeBr_3[/tex]

[tex]FeBr_3\rightarrow Fe^{3+}+3Br^{-}[/tex]  

, i= 4 as it is a electrolyte and dissociate to give 4 ions and concentration of ions will be [tex]1\times 0.11+3\times 0.11=0.44[/tex]

2. For 0.15 [tex]CuBr_2[/tex]

[tex]CuBr_2\rightarrow Cu^{2+}+2Br^{-}[/tex]  

, i= 3 as it is a electrolyte and dissociate to give 3 ions, concentration of ions will be [tex]1\times 0.15+2\times 0.15=0.45[/tex]

3. For 0.24 [tex]AgNO_3[/tex]

[tex]AgNO_3\rightarrow Ag^{+}+NO_3^{-}[/tex]  

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]1\times 0.24+1\times 0.24=0.48[/tex]

4. 0.51 m glucose

i= 1 as it is a non electrolyte and does not dissociate to give ions, concentration will be [tex]1\times 0.51=0.51[/tex]

Thus as boiling point depends on the concentration of solutes, the solution having highest concentration will have highest boiling point.

Answer:

mm protein = 15365.8183 g/mol

Explanation:

∴ molar mass (mm) ≡ g/mol

∴ π = 0.118 atm

∴ T = 25°C ≅ 298 K

∴ concentration (C ) [=] mol/L

∴ mass protein = 371 mg = 0.371 g

∴ volume sln = 5.00 mL = 5 E-3 L

⇒ C = π / RT = (0.118 atm)/((0.082 atm.L/K.mol)(298 K))

⇒ C = 4.8289 E-3 mol/L

⇒ mol protein = (4.8289 E-3 mol/L)×(5 E-3 L) = 2.4145 E-5 mol

⇒ mm protein = (0.371 g)/(2.4145 E-5 mol) = 15365.8183 g/mol

Final answer:

To find the molar mass of the protein, we begin by converting mass to grams and volume to liters, and then apply the van 't Hoff equation for osmotic pressure. After rearranging the equation, we calculate the molar mass using the given osmotic pressure, volume, temperature, and ideal gas constant.

Explanation:

To calculate the molar mass of the protein using the osmotic pressure, we can use the van 't Hoff equation for osmotic pressure, Π = MRT, where Π is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin. In this scenario, the student provided the mass of the protein (371 mg), the volume of the solution (5.00 mL), the osmotic pressure (0.118 atm), and the temperature at which the osmotic pressure was measured (25 °C which is 298.15 K).

First, convert the mass of the protein to grams:

371 mg = 0.371 g

Next, convert the volume from mL to liters:

5.00 mL = 0.005 L

Now, calculate the molarity (M) of the protein:

Number of moles = mass (g) / molar mass (g/mol)

As we do not know the molar mass yet, let's call it 'Mm'. The molarity (concentration) of the solution is:

Molarity (M) = Number of moles / Volume (L) = (mass (g) / Mm (g/mol)) / Volume (L)

Convert the temperature to Kelvin:

25 °C = 298.15 K

Using the van 't Hoff equation, we solve for the molar mass (Mm):

Π = (mass (g) / Mm (g/mol)) / Volume (L) × R × T

0.118 atm = (0.371 g / Mm (g/mol)) / 0.005 L × 0.0821 (L·atm/K·mol) × 298.15 K

Rearranging the equation and solving for Mm gives us:

Mm = (0.371 g / 0.005 L) / (0.118 atm / (0.0821 L·atm/K·mol × 298.15 K))

After calculating, we find the molar mass of the protein.

Answer:

The answer is Inductive effect

Explanation:

To determine the acidity or alkalinity of an organic compound. We have to keep in mind that the whole analysis is based on the comparison between the compounds, and we must work with the conjugated base of the molecule. Keeping in mind, the more unstable the base, the less acidic the molecule is. Thus, to determine instability, the Inductive Effect of the molecule can be used.

This type of effect occurs when atoms of different electronegativities are linked or very close in the compound. The most electronegative atom has a tendency to bring electrons close to it, thus creating a dipole. This dipole can have a stabilizing effect on the molecule, as it “relieves” the excessive charge on some occasions, better accommodating the charges.

However, in some cases, instead of chains with chlorine radicals, we may have chains with methyl radicals. This has a major impact on the inductive effect, keeping in mind that alkyl groups are electron donors.

Answer:

29.3 g HClO₄

Explanation:

We have 39.1 grams of 74.9 wt% aqueous perchloric acid solution, that is, there are 74.9 grams of perchloric acid in 100 grams of perchloric acid solution. The mass, in grams, of perchloric acid contained in 39.1 grams of perchloric acid solution is:

39.1 g Solution × (74.9 g HClO₄/100 g Solution) = 29.3 g HClO₄

Answer:

a) 1.44 kg

b) 1.47 kg

Explanation:

a) By the reaction given, the stoichiometry, the molar ratio, is:

2 moles of C3H6 : 2 moles of NH3 : 3 moles of O2 : 2 moles of C3H3N : 6 moles of H2O, or 2:2:3:2:6.

By the mixture given, one or two of the reactants may be in excess, so, we must found which of them is the limiting reactant, the reactant that will be totally consumed.

The molar masses of the compounds are:

C3H6 = 42.0 g/mol

NH3 = 17.0 g/mol

O2 = 32.0 g/mol

C3H3N = 53.1 g/mol

H2O = 18.0 g/mol

Thus, the mass ratio between the reactants will be the molar ratio multiplied by the molar mass:

2 moles of C3H6*42.0 g/mol --- 2 moles of NH3*17.0 g/mol -- 3 moles of O2* 32.0 g/mol

84.0 g of C3H6 -- 34.0 g of NH3 -- 96.0 g of O2

Thus, assuming C3H6 as limiting, let's use the rule of three to find out which would be the masses of the other reactants needed:

84.0 g of C3H6 -- 34.0 g of NH3

1.14 kg -- x

84.0x = 38.76 kg

x = 0.46 kg of NH3

Because there's more NH3 than it's required, NH3 is in excess, and C3H6 is limiting. Let's test using O2 as reference:

84.0 g of C3H6 -- 96.0 f of O2

1.14 kg -- x

84.0x = 109.44

x = 1.30 kg of O2

So, O2 is also in excess, and the limiting reactant is C3H6.

Thus, the molar ratio between C3H6 and C3H3N is 2:2, which is simplified by 1:1, and so the mass ratio is 42 g of C3H6 -- 53.1 g of C3H3N, so the mass of acrylonitrile can be found by a rule of three:

42 g of C3H6 -- 53.1 f of C3H3N

1.14 kg -- x

42x = 60.534

x = 1.44 kg

b) The molar ratio between C3H6 and water is 2:6, which can be simplified by 1:3, thus the mass ratio is 1*42g/mol of C3H6 -- 3*18 g/mol of H2O

42 g of C3H6 -- 54 g of H2O

By a rule of three the mass of water is:

42 g of C3H6 -- 54 g of H2O

1.14 kg -- x

42x = 61.56

x = 1.47 kg

Final answer:

The mass of acrylonitrile that can be produced from the given mixture of reactants is approximately 1437.73 g, and the mass of water produced is 1464.68 g, assuming a 100% yield and propylene as the limiting reactant.

Explanation:

Calculating Mass of Acrylonitrile and Water Produced

To answer your questions about the production of acrylonitrile and water, we first need to use the provided stoichiometry and the masses of the reactants. The balanced chemical reaction is:

2C₃H₆(g) + 2NH₃(g) + 3O₂(g) → 2C₃H₃N(g) + 6H₂O(g)

This informs us that two moles of propylene (C₃H₆) react with two moles of ammonia (NH₃) and three moles of oxygen (O₂) to produce two moles of acrylonitrile (C₃H₃N) and six moles of water (H₂O).

Firstly, we determine the number of moles of each reactant:

Propylene (C₃H₆): Molar mass = 42.08 g/mol, Moles = 1140 g / 42.08 g/mol = 27.11 moles

Ammonia (NH₃): Molar mass = 17.031 g/mol, Moles = 1650 g / 17.031 g/mol = 96.91 moles

Oxygen (O₂): Molar mass = 32.00 g/mol, Moles = 1850 g / 32.00 g/mol = 57.81 moles

Since the limiting reactant determines the amount of product formed, we can find it by comparing the mole ratios. In this case, propylene is the limiting reactant. Using the stoichiometry, we calculate the mass of acrylonitrile produced:

Moles of acrylonitrile produced = Moles of propylene used = 27.11 moles

Mass of acrylonitrile produced = Moles of acrylonitrile produced x Molar mass of acrylonitrile = 27.11 moles x 53.06 g/mol = 1437.73 g

For the mass of water produced:

Moles of water produced = 3 x Moles of propylene used = 3 x 27.11 mol = 81.33 moles

Mass of water produced = Moles of water produced x Molar mass of water = 81.33 moles x 18.015 g/mol = 1464.68 g

Assuming 100% yield, we can produce approximately 1437.73 g of acrylonitrile and 1464.68 g of water from the given reactants.

When heated, 25 grams of sodium bicarbonate (NaHCO3) will decompose and produce approximately 13.1 grams of carbon dioxide (CO2) gas.

The decomposition of sodium bicarbonate (NaHCO3) can be represented by the equation: NaHCO3 → Na2CO3 + H2O + CO2. The ratio of NaHCO3 to CO2 is 1:1, so in theory, one mole of NaHCO3 will produce one mole of CO2. Based on molecular weights, 84 grams of NaHCO3 (the weight of one mole) will produce 44 grams of CO2 (the weight of one mole).

First, we need to find how many moles are in 25 grams of NaHCO3. Using the formula number of moles=(given mass/molar mass), we get approximately 0.297 moles.

Now, applying the molar relationship between NaHCO3 and CO2, we will also produce 0.297 moles of CO2. Converting this to grams: number of CO2 grams = number of moles * molecular weight, we get approximately 13.1 grams of CO2 gas.

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To determine the mass of CO2 produced from the decomposition of 25.0 g NaHCO3, we use stoichiometry based on the chemical equation, resulting in 6.55 grams of CO2.

To calculate the mass of CO2 produced from the decomposition of 25.0 g NaHCO3, we need to use the stoichiometry of the balanced chemical equation:

2 NaHCO3(s) ightarrow Na2CO3(s) + H2O(g) + CO2(g)

First, we calculate the molar mass of NaHCO3, which is (23+1+12+3(16)) = 84 g/mol. Next, we determine the moles of NaHCO3 in 25.0 g:

Moles of NaHCO3 = 25.0 g \/ 84 g/mol = 0.2976 moles

According to the balanced equation, 2 moles of NaHCO3 produce 1 mole of CO2. Thus, 0.2976 moles of NaHCO3 would produce:

Moles of CO2 = 0.2976 moles NaHCO3 \/ 2 = 0.1488 moles CO2

The molar mass of CO2 is (12 + 2(16)) = 44 g/mol, so the mass of CO2 produced is:

Mass of CO2 = 0.1488 moles \/ 44 g/mol = 6.5472 g

Therefore, 6.55 grams of CO2 are produced from the decomposition of 25.0 g NaHCO3.

Answer:

D is the true statement.

Explanation:

Recall that during a phase change the temperature remains constant until all the material has changed its phase. The energy put into the system is utilized to make the phase change, and afterward the temperature of the vliquid will start to increase.

In this case we have a mixture of liquid water and ice at  0ºC  ( assume standard pressure) which is the temperature in which liquid water and ice coexist. The ice will  melt until consumed at constant T = 0ºC, and then the temperature of the liquid water will start to increase at a uniform rate since we are heating at a constant rate.

Now we are in position to answer this question.

A- False the temperature does not increase as the ice melts.

B- False for the reasons given above.

C- False the temperature does not increase slowly as the ice melts, but remains constant.

D-True the temperature of the mixture does not change until all the ice has melted, and then the temperature of liquid water will start to increase uniformly.

Heat applied to a water-ice mixture initially goes into breaking the hydrogen bonds in ice (latent heat of fusion), without causing a temperature increase. Only after all the ice has melted, the added heat increases the water temperature.

The correct option is D: The temperature of the mixture does not change at all until all the ice has melted, at which point it increases at a constant rate. This is because the energy from the heat being applied is first used to break the hydrogen bonds in ice, a process known as latent heat of fusion, which does not involve an increase in temperature. Only after all the ice has melted will the added heat then lead to an increase in the temperature of the water.

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Answer:

the transition state will resemble the products more than the reactants

Explanation:

Since the free-radical bromination of methane involves the following reactions

Br₂ → 2 Br•  , ΔH⁰ (per mole) = +192 kJ , Ea (per mole) = + 192 kJ

CH₄ +Br•   → CH₃• + HBr  , ΔH⁰ (per mole) = +67 kJ , Ea (per mole) = + 75 kJ

CH₃• +Br₂  → CH₃Br  + Br•  , ΔH⁰ (per mole) = -101 kJ , Ea (per mole) = -4 kJ

without considering the Br dissociation (initiation reaction) the rate-determining step is the second equation ( highest Ea) .

Then since the reaction is endothermic (ΔH⁰ (per mole) = +67 kJ) , the transition state will resemble the products more than the reactants ( Hammond postulate)

The free-radical bromination of methane's transition state resembles both the products and reactants equally. This similarity arises due to the formation and subsequent decay of the high-energy transition state that exists between reactants and products. The reaction's activation energy is always positive and its exothermic nature results in a decrease in system enthalpy.

In the free-radical bromination of methane, the nature of the transition state of the rate-determining step is closest to option b. The transition state resembles both the products and reactants equally. This is because during the process, reactant molecules with enough energy collide to form a high-energy activated complex or transition state. This transition state then decays to yield stable products, maintaining similarities to both the initial reactants and final products.

As per chemical kinetics, the reaction diagram indicating this process highlights the activation energy, Eå, as the energy difference between the reactants and the transition state. The enthalpy change of the reaction, ΔH, is evaluated as the energy difference between the reactants and products. Here, the reaction is exothermic (ΔH < 0) as it results in a decrease in system enthalpy.

Understanding that the activation energy is always positive in these reactions, regardless of whether the reaction is exergonic (releases energy) or endergonic (absorbs energy), further supports the resemblance between the transition state and both reactants and products.

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Answer:

Check explanation

Explanation:

There are two ways to test for calcite in rocks. Calcite is popularly known as calcium carbonate and it is found almost everywhere. There are many minerals that looks like calcite/calcium carbonates, so we there is a need for a good test for proper identification.

The tests are:

(1). ACID TEST: the calcite/calcium carbonate reacts vigorously with Hydrogen Chloride acid,HCl with the evolution of carbondioxide,CO2. The equation of reaction is given below;

CaCO3 + 2 HCl ---------> CaCl2 + CO2 + H2O.

(2). DOUBLE REFRACTION METHOD: There are other minerals in the rock that can react with Hydrochloric acid,HCl to evolve Carbondioxide,CO2(although their response to HCl differs). There is a need for a confirmatory test. This double refraction method is a confirmatory test. In this test, one will pass light through the calcite, if the light splits into two rays and reflected twice then, it is a calcite.

The acid test for calcite involves applying dilute hydrochloric acid to a rock sample. If the rock contains calcite, effervescence (the release of bubbles) occurs due to the reaction between the acid and calcite, forming calcium chloride, water, and carbon dioxide. This test is a useful tool in determining the presence of calcite, an essential mineral in the identification and classification of various types of rocks and minerals.

To determine whether a rock contains the mineral calcite, one common test that geologists use is the acid test, specifically the reaction of calcite with dilute hydrochloric acid (HCl). This test helps identify calcite due to its distinctive reaction with the acid. Here's a description of the test:

Acid Test for Calcite in Rocks:

Materials Needed:

Dilute hydrochloric acid (HCl)

The rock sample in question

Procedure:

Safety Precautions: Before conducting the test, ensure that you are wearing appropriate safety gear, including gloves and eye protection, as HCl is a corrosive acid.

Select a Representative Sample: Choose a representative sample of the rock you want to test. It's essential to select an area that appears to contain calcite or where you suspect its presence.

Apply the Acid: Place a small drop of dilute hydrochloric acid (typically 10% HCl solution) onto the surface of the rock. You can also use a dropper or a small pipette for precision.

Observe the Reaction: Watch the reaction between the acid and the rock. If the rock contains calcite, you will observe effervescence or the release of bubbles of carbon dioxide (CO₂). This effervescence is a result of the chemical reaction between the HCl and the calcite, producing soluble calcium chloride (CaCl₂), water (H₂O), and carbon dioxide:

CaCO₃ (calcite) + 2HCl → CaCl₂ + H₂O + CO₂↑

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Explanation:

According to the Handerson equation,  

          pH = [tex]pK_{a} + log \frac{\text{salt}}{\text{acid}}[/tex]

or,      pH = [tex]pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}[/tex]

Putting the given values into the above equation as follows.

     pH = [tex]pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}[/tex]

       5.0 = 6.0 + log \frac{\text{conjugate base}}{\text{acid}}[/tex]

      [tex]log \frac{\text{conjugate base}}{\text{acid}}[/tex] = -1.0

or,      [tex]\frac{\text{conjugate base}}{\text{acid}} = 10^{-1.0}[/tex]

                            = 0.1

Therefore, we can conclude that molar ratios of conjugate base to weak acid for given solution is 0.1.

The molar ratio of conjugate base to acid for a weak acid at pH 5.0 with a pKa of 6.0 is 0.1. Among formic acid, acetic acid, and ethylamine, the best buffer at pH 5.0 would be acetic acid due to its pKa being closest to 5.0.

The calculation of molar ratios of conjugate base to weak acid from pH and the choice of a weak acid for a buffer can be accomplished by using the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]).

Firstly, for a weak acid with a pKa of 6.0 at a pH of 5.0, the ratio of conjugate base ([A-]) to acid ([HA]) can be calculated by rearranging this equation to [A-]/[HA] = 10^(pH-pKa) = 10^(5.0-6.0) = 0.1. So the ratio of conjugate base to acid is 0.1.

Secondly, the choice of weak acid for a buffer at pH 5.0 is ideally the one with a pKa closest to the desired pH. In this case, acetic acid with a pKa of 4.76 would be the best buffer since its pKa is closest to the pH of 5.

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The desired compound using a stepwise approach from dicyclopentadiene and incorporating a Diels-Alder reaction .

Step 1: Diels-Alder Reaction

In the Diels-Alder reaction, dicyclopentadiene (DCPD) will react with a suitable dienophile to form the desired cycloadduct. The dienophile you choose should have a functional group that can be further manipulated to achieve the final compound. Let's select maleic anhydride (C4H2O3) as the dienophile, which can react with DCPD to yield a cycloadduct.

The reaction would look like this:

Dicyclopentadiene + Maleic Anhydride → Cycloadduct

Step 2: Functionalization of the Cycloadduct

The cycloadduct formed in the Diels-Alder reaction will contain the necessary functional groups for further manipulation. In this case, you want to introduce additional substituents or functional groups.

For example, you can introduce an alcohol group (-OH) through a nucleophilic addition reaction. You can do this by treating the cycloadduct with a strong base (e.g., sodium hydroxide, NaOH) and water (H2O) to open the anhydride ring and form a carboxylic acid intermediate. Then, you can reduce the carboxylic acid using a reducing agent (e.g., lithium aluminum hydride, LiAlH4) to obtain the alcohol group.

The reaction sequence would look like this:

Cycloadduct + NaOH + H2O → Carboxylic Acid Intermediate

Carboxylic Acid Intermediate + LiAlH4 → Alcohol Group

Step 3: Further Functionalization

Depending on your specific requirements, you can further modify the alcohol group by various organic reactions such as esterification, acylation, or oxidation, among others, to achieve the desired final compound.

This stepwise synthesis should allow you to obtain the desired compound from dicyclopentadiene, using a Diels-Alder reaction as one of the key steps. Keep in mind that the reaction conditions and reagents may need to be optimized based on the specific compound you are targeting. Always follow safety guidelines and consult relevant chemical literature for detailed reaction conditions.

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To synthesize the given compound from dicyclopentadiene, a stepwise approach involving a Diels-Alder reaction can be followed. First, dicyclopentadiene is converted to cyclopentadiene. Then, cyclopentadiene reacts with maleic anhydride to form the Diels-Alder adduct.

A stepwise synthesis of the given compound from dicyclopentadiene using a Diels-Alder reaction as one step can be achieved as follows:

Note that the specific reagents and conditions for each step may vary depending on the desired compound and the available starting materials. Finally, the adduct can be further modified to obtain the desired compound.

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To estimate the surface-to-volume ratio of a C60 fullerene, treat the molecule as a hollow sphere and calculate its surface area and volume. The surface area is found using the formula A = 4πr² and the volume is found using the formula V = (4/3)πr³, where r is the radius of the sphere. Dividing the surface area by the volume gives the surface-to-volume ratio.

To estimate the surface-to-volume ratio of a C60 fullerene, we can treat the molecule as a hollow sphere and calculate its surface area and volume.

The surface area of a sphere is given by the formula A = 4πr^2, where r is the radius of the sphere.

The volume of a sphere is given by the formula V = (4/3)πr^3.

Since the C60 fullerene is composed of 60 carbon atoms, we can divide the atomic radius of carbon by 2 to get the radius of the molecule. Using the given atomic radius of 77pm, the radius of the C60 fullerene is 38.5pm (or 0.385nm).

Using these values, we can calculate the surface area and volume of the C60 fullerene:

Now we can calculate the surface-to-volume ratio by dividing the surface area by the volume:

Surface-to-Volume Ratio = A / V

Substituting the calculated values into the formula, we get:

Surface-to-Volume Ratio = (4π(0.385nm)^2) / ((4/3)π(0.385nm)^3)

Simplifying the equation, we find that the surface-to-volume ratio of the C60 fullerene is approximately 0.649 nm-1.

This question is incomplete and the full question can be seen below:  

b) Calculate ΔG for ATP hydrolysis in liver at 18° C. Use the liver concentration From part a.

The equation for the ATP hydrolysis is:  

           H₂0  

ATP ------------> ADP + P₁                      ΔG = -30.5 [tex]\frac{KJ}{mol}[/tex]  

a) calculate ΔG for ATP hydrolysis to rank the following conditions from most favorable to least favorable. Assume a temperature of 37.0° C, R = 8.315 [tex]\frac{J}{mol.k}[/tex]  

muscle: [ATP]= 8.1mM; [ADP]= 0.9mM; [P₁]= 8.1mM  

liver: [ATP]= 3.4mM; [ADP]= 1.3mM; [P₁]= 4.8mM  

brain: [ATP}= 2.6mM; [ADP}= 0.7mM; [P₁]= 2.7mM  

b) Calculate ΔG for ATP hydrolysis in liver at 18° C. Use the liver concentration From part a.

Answer:  

−45.8 KJ/mol

Explanation:  

Equilibrium constant (k) is defined as a measure of the ratio of the equilibrium concentration of the products of a reaction, to the Equilibrium concentration of the reactants with each concentration raised to the power corresponding to the coefficient in the balanced equation of the reaction.

In the reaction in the question given above;  

[tex]K=\frac{[ADP][P_1]}{ATP}[/tex]

For muscle;

ADP= 0.9 × 10⁻³

P₁= 8.1 × 10⁻³

ATP= 8.1 × 10⁻³

∴ K = [tex]\frac{(0.9*10^-3)(8.1*10^-3)}{(8.1*10^-3)}[/tex]

  K = 0.9 × 10⁻³

For liver;

ADP= 1.3 × 10⁻³

P₁= 4.8 × 10⁻³

ATP= 3.4 × 10⁻³

∴ K = [tex]\frac{(1.3*10^-3)(4.8*10^-3)}{(3.4*10^-3)}[/tex]

  K = 1.8 × 10⁻³

For brain;

ADP= 0.7 × 10⁻³

P₁= 2.7 × 10⁻³

ATP= 2.6 × 10⁻³

∴ K = [tex]\frac{(0.7*10^-3)(2.7*10^-3)}{(2.6*10^-3)}[/tex]

  K = 7.3 × 10⁻⁴

b) Since we are concerned about calculating ΔG for ATP hydrolysis in liver at 18° C and we've already obtained the liver concentration from part a; we can therefore calculate ΔG as:

ΔG = ΔG° + RTInK

ΔG° = -30.5

R= 8.315 [tex]\frac{J}{mol.k}[/tex]  

T= 18° C = 18 + 273.15k = 291.15k

K= 1.8 × 10⁻³

InK = In(1.8 × 10⁻³ )

      ≅ -6.32

∴ ΔG = -30.5 +  8.315 [tex]\frac{J}{mol.k}[/tex] × 291.15k × (-6.32)

   ΔG = -30.5 +  (−15.30016542)

   ΔG = −45.80016542

   ΔG ≅ −45.8 KJ/mol

Change in free energy for ATP hydrolysis at 18 degree Celsius   is -45.8 kJ.  

The equation for the ATP hydrolysis is,    

[tex]\bold {ATP \rightarrow ADP + Pi }[/tex]                     ΔG = -30.5 kJ/mol at STP  

Change in free energy for ATP hydrolysis in liver at 18° C can be calculated by the formula,  

[tex]\bold {\Delta G = \Delta G^o + RT \times lnK}[/tex]  

Where,

ΔG° - Free energy at STP = -30.5  

R - gas constant = 8.315    

T - temperature in Kelvin = 18° C = 18 + 273.15k = 291.15k  

K - Equilibrium constant = 1.8 × 10⁻³  

In K = In(1.8 × 10⁻³ ) =      -6.32  

Put the values in the formula,

[tex]\bold {\Delta G = -30.5 + 8.315 \times 291.15 \times (-6.32)}\\\\\bold {\Delta G = -30.5 + (- 15.3)}\\\\\bold {\Delta G = - 45.8 KJ}\\\\[/tex]

Therefore, change in free energy for ATP hydrolysis at 18 degree Celsius   is -45.8 kJ.  

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