2HgCl2(aq) + C2O42–(aq) → 2Cl–(aq) +2CO2(g) + Hg2Cl2(s)

The above reaction was studied by the method of initial rates and the following results were obtained:

[HgCl2] (M)

[C2O42–] (M)

Initial Rate (M/min)

0.105

0.15

1.8x10–5

0.105

0.30

7.2x10–5

0.0525

0.30

3.6 x10–5

0.0525

0.15

9.0x10–6

How would you describe the kinetics of this reaction? What is the rate constant?

first-order wrt HgCl2, second-order wrt C2O42–, third-order overall

k = 7.6x10–3 M–2min–1

first-order with respect to (wrt) HgCl2, first-order wrt C2O42–, second-order overall

k = 1.1x10–3 M–1min–1

zero-order wrt HgCl2, second-order wrt C2O42–, second-order overall

k = 5.4x10–5 Mmin–1

first-order wrt HgCl2, second-order wrt C2O42–, second-order overall

k = 4.3x10–8 M–2min–1

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:

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.

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.

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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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: The number of electrons in the given amount of silver are [tex]2.36\times 10^{24}[/tex]

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of silver = 9.0 g

Molar mass of silver = 107.87 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of silver}=\frac{9.0g}{107.87g/mol}=0.0834mol[/tex]

We are given:

Number of electrons in one atom of silver = 47

According to mole concept:

1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of particles

So, 0.0834 moles of silver contains [tex](47\times 0.0834\times 6.022\times 10^{23})=2.36\times 10^{24}[/tex] number of electrons

Hence, the number of electrons in the given amount of silver are [tex]2.36\times 10^{24}[/tex]

Answer:

Mole fraction of [tex]CH_4O[/tex] = 0.58

Mole fraction of [tex]C_2H_6O[/tex] = 0.42

Explanation:

Let the mass of [tex]CH_4O[/tex] and [tex]C_2H_6O[/tex] = x g

Molar mass of [tex]CH_4O[/tex] = 33.035 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles_{CH_4O}= \frac{x\ g}{33.035\ g/mol}[/tex]

[tex]Moles_{CH_4O}=\frac{x}{33.035}\ mol[/tex]

Molar mass of [tex]C_2H_6O[/tex] = 46.07 g/mol

Thus,

[tex]Moles= \frac{x\ g}{46.07\ g/mol}[/tex]

[tex]Moles_{C_2H_6O}=\frac{x}{46.07}\ mol[/tex]

So, according to definition of mole fraction:

[tex]Mole\ fraction\ of\ CH_4O=\frac {n_{CH_4O}}{n_{CH_4O}+n_{C_2H_6O}}[/tex]

[tex]Mole\ fraction\ of\ CH_4O=\frac{\frac{x}{33.035}}{\frac{x}{33.035}+\frac{x}{46.07}}=0.58[/tex]

Mole fraction of [tex]C_2H_6O[/tex] = 1 - 0.58 = 0.42

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.

The alkyl bromide used is Tertiary-Butyl Bromide and the alcohol used is Methanol. The reaction involves reacting Tertiary-Butyl Bromide with Sodium Methoxide to form methyl t-butyl ether and Sodium Bromide as a byproduct.

The alkyl bromide and alcohol used to make methyl t-butyl ether using the Williamson ether synthesis are:

Alkyl Bromide: Tertiary-Butyl Bromide (C4H9Br)

Alcohol: Methanol (CH3OH)

Step 1: React Tertiary-Butyl Bromide (C4H9Br) with Sodium Methoxide (CH3ONa).

Step 2: Heating the reaction mixture will result in the formation of methyl t-butyl ether (MTBE) and Sodium Bromide (NaBr) as byproduct.

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LED light bulbs exhibit the highest energy efficiency among incandescent, fluorescent, and LED bulbs due to their ability to produce a significant amount of light energy while minimizing heat energy dissipation.

Looking at the graph, it is evident that LED light bulbs exhibit the highest efficiency among the three types. This conclusion can be drawn from the fact that LED bulbs produce a significant amount of light energy while minimizing heat energy dissipation. LEDs achieve this by converting a higher proportion of electrical energy into visible light, making them more energy-efficient than incandescent and fluorescent counterparts.

On the other hand, incandescent bulbs, characterized by the highest heat output on the graph, are the least efficient. This is because a substantial portion of the electrical energy is transformed into heat, contributing to energy loss rather than useful light output.

The claim that fluorescent bulbs are the most energy-efficient due to a balanced distribution of light and heat is not supported by the graph. While fluorescents fare better than incandescents, they fall short of the high efficiency demonstrated by LED bulbs. Therefore, the correct conclusion is that LED light bulbs provide superior energy efficiency by generating more light energy and less heat energy compared to incandescent and fluorescent alternatives.

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.

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:

It will take 950 million years for this amount of urea to be hydrolyzed under the same conditions in the absence of urease.

Explanation:

Given that:

Urease enhances the rate of hydrolysis by = 10^14

Time taken in hydrolysis is 5 min

Time taken in hydrolysis will be = 5 min x 10^14

Now, converting minutes into years

 Time   =  (5 min x 10^14) / (60 min/hr x 24 hr/day x 365 days/year)

Time = 9.50 x 10^8 years

Time = 950 X 10^6 years

Time = 950 million years

Answer:

Explanation:

The enzyme increases the rate of urea hydrolysis by a factor of [tex]10^{14}[/tex]

So in absence of enzyme, reaction will take [tex]10^{14}[/tex] times more time for completion.

Therefore time taken,

[tex]\frac{rate catalysed}{rate uncatalysed} = 10^{14}[/tex]

[tex]\frac{t uncatalysed}{t catalysed} = 10^{14}\\\\t uncatalysed = 10^{14} * 5min\\\\t uncatalysed = 5*10^{14}min[/tex]

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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.

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:

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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Answer:

344 nm is the longest wavelength of radiation with enough energy to break carbon-carbon bonds.

Explanation:

[tex]C-C(g)\rightarrow 2C(g)[/tex] ,ΔH = 348 kJ/mol

Energy required to break 1 mole of C-C bond = 348 kJ

Energy required to break 1 C-C bond = E

[tex]E = \frac{348,000J}{6.022\times 10^{23}}=5.779\times 10^{-19} J[/tex]

Energy related with the wavelength of light is given by Planck's equation:

[tex]E=\frac{hc}{\lambda }[/tex]

[tex]\lambda =\frac{hc}{E}[/tex]

[tex]=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{5.779\times 10^{-19} J}[/tex]

[tex]\lambda =3.44\times 10^{-7} m = 344 nm[/tex]

[tex]1 m =10^9 nm[/tex]

344 nm is the longest wavelength of radiation with enough energy to break carbon-carbon bonds.

Ultraviolet radiation and radiation of shorter wavelengths can break carbon-carbon bonds within biological molecules. The longest wavelength that can break carbon-carbon bonds is approximately 1.72 x 10^-6 meters or 1720 nm.

Ultraviolet radiation and radiation of shorter wavelengths can damage biological molecules because they carry enough energy to break bonds within the molecules. A carbon-carbon bond requires 348 kJ/mol to break. To find the longest wavelength of radiation with enough energy to break carbon-carbon bonds, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength. Rearranging the equation to solve for λ, we get λ = hc/E. Substituting the given energy of 348 kJ/mol (which is equivalent to 348,000 J/mol) and solving for λ, we find that the longest wavelength is approximately 1.72 x 10^-6 meters or 1720 nm.

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Answer:

Mass of heptane = 102g

Vapor pressure of heptane = 454mmHg

Molar mass of heptane = 100.21

No of mole of heptane = mass/molar mass = 102/100.21

No of mole of heptane = 1.0179

Therefore the partial pressure of heptane = no of mole heptane *Vapor pressure of heptane

Partial pressure of heptane = 1.0179*454mmHg

Partial pressure of heptane = 462.1096 = 462mmHg

the partial pressure of heptane vapor above this solution = 462mmHg

The partial pressure of the heptane vapor above the solution is calculated using Raoult's law as 214.5 mmHg.

In order to calculate the partial pressure, we will use Raoult's law, which states that the partial pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. First, we need to calculate the mole fraction, which is calculated as:

moles of component/total moles in the mixture.

We first need to determine the number of moles of heptane and chloroform, using their respective molecular weights (C₇H₁₆ is approximately 100.21 g/mol, CHCl₃ is about 119.38 g/mol).

Moles of C₇H₁₆ = 102 g / 100.21 g/mol = 1.018 mol

Moles of CHCl₃= 135 g / 119.38 g/mol = 1.131 mol

To determine the mole fraction of the heptane, we take the moles of heptane and divide by the total moles.

X_C₇H₁₆ = 1.018 mol / (1.018 mol + 1.131 mol) = 0.473

Applying Raoult's law, the partial pressure of heptane P_C₇H₁₆ = X_C₇H₁₆ ₓ P(C₇H₁₆, pure) = 0.473 ₓ 454 mmHg = 214.5 mmHg.

So, the partial pressure of heptane vapor above the solution is 214.5 mmHg.

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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:

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)

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:

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.

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: [tex]2NaOH(aq)+MgCl_2(aq)\rightarrow NaCl(aq)+Mg(OH)_2(s)[/tex]

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

[tex]2NaOH(aq)+MgCl_2(aq)\rightarrow NaCl(aq)+Mg(OH)_2(s)[/tex]

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.  

To know more about Free energy,

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Complete Question:

To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?

Answer:

2.23x10⁶ g

Explanation:

The concentration of the fluoride (F⁻) must be 0.800 ppm, which is 0.800 parts per million, so the water must have 0.800 g of F⁻/ 1000000 g of the solution. The density of the water at room temperature is 997 kg/m³ = 997x10³ g/m³. So, the concentration of the fluoride will be:

0.800 g of F⁻/ 1000000 g of the solution * 997x10³ g/m³

0.7976 g/m³

The volume of the reservoir is the volume of the cylinder: area of the base * depth. The base is a circumference, which has an area:

A = πR², where R is the radius = 1.01x10² m (half of the diameter)

A = π*(1.01x10²)²

A = 32047 m²

The volume is then:

V = 32047 * 87.32

V = 2.7983x10⁶ m³

The mass of the F⁻ is the concentration multiplied by the volume:

m = 0.7976 * 2.7983x10⁶

m = 2.23x10⁶ g

The Haber process increased agricultural productivity and human support through the large-scale production of fertilizers, but it also enabled the creation of harmful chemical weapons and resulted in environmental issues such as algal blooms and dead zones.

The Haber process, developed by Fritz Haber in the early 20th century, had both positive and negative effects. The positive effects of this process include the mass production of nitrogen fertilizers, which increased agricultural productivity and the sustainable support of human populations. An example of this is the jump in the number of humans per hectare that arable land could support, increasing from 1.9 in 1908 to 4.3 by 2008.

However, there are also negative consequences associated with the Haber process. It enabled the creation of harmful chemical weapons during the World War, marking a dark turn in its utilization. Also, the excessive use of nitrogen-based fertilizers, majorly produced through this process, has led to detrimental environmental consequences such as algal blooms, red tides, and dead zones in water bodies. These effects end up destroying marine life and disrupting ecosystems.

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Answer:

79.0 mL

Explanation:

Given data

In order to find the volume required of the concentrated solution, we will use the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂/C₁

V₁ = 1.00 M × 790.0 mL/10.0 M

V₁ = 79.0 mL

Explanation:

In the previous question, you calculated the amount of CO2 that was required to heat the air in your room. Which of the following are true statements?

You didn't complete the above question, please complete the question and reupload, thanks for your anticipated cooperation.

Percent ionic character of bonds is estimated from the electronegativity difference between the bonded atoms - TiO2 has moderate ionic character, ZnTe has low, CsCl has an ionic character, InSb less ionic character and MgCl2 has significant ionic character.

The percent ionic character of the bonds in a compound can be estimated based on the electronegativity difference between the bonded atoms. Ionic character increases with the increasing difference in electronegativity, i.e., higher the electronegativity difference, higher the ionic character.

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The percent ionic character of a bond is estimated using the difference in electronegativity between the bonded atoms. For the given compounds, the calculated percent ionic characters are: TiO₂ (75%), ZnTe (9.41%), CsCl (75.15%), InSb (4%), and MgCl₂ (59.4%). This demonstrates the varying degrees of ionic versus covalent character in each compound.

The percent ionic character of a bond is determined by the difference in electronegativity between the two atoms involved. Linus Pauling proposed an empirical relation to calculate it:

Percent Ionic Character =  10.16(Δχ)²  - 2(57.9 - Δχ)²

Now, let's calculate the percent ionic character for each compound:

1. TiO2

Electronegativity of Ti (Titanium) = 1.54

Electronegativity of O (Oxygen) = 3.44

Δχ = 3.44 - 1.54 = 1.90

Using the formula, Percent Ionic Character = 1-(6-2 × (1.9)^2) x 100 = 75%

2. ZnTe

Electronegativity of Zn (Zinc) = 1.65

Electronegativity of Te (Tellurium) = 2.1

Δχ = 2.1 - 1.65 = 0.45

Using the formula, Percent Ionic Character = 9.41%

3. CsCl

Electronegativity of Cs (Caesium) = 0.79

Electronegativity of Cl (Chlorine) = 3.16

Δχ = 3.16 - 0.79 = 2.37

Using the formula, Percent Ionic Character = 75.15%

4. InSb

Electronegativity of In (Indium) = 1.78

Electronegativity of Sb (Antimony) = 2.05

Δχ = 2.05 - 1.78 = 0.27

Using the formula, Percent Ionic Character = 4%

5. MgCl2

Electronegativity of Mg (Magnesium) = 1.31

Electronegativity of Cl (Chlorine) = 3.16

Δχ = 3.16 - 1.31 = 1.85

Using the formula, Percent Ionic Character = 59.4%