A molecule from a new organism contains adenine, cytosine, guanine, and thymine. What molecules is being looked at?

Answer:

6.25×10¹⁹ e⁻

Explanation:

Let's apply a rule of three:

1 e⁻ has 1.60×10⁻¹⁹ C

There, we can think:

1.60×10⁻¹⁹ C of charge are made by 1 e⁻

If we want to produce 1 C, we would need ( 1 . 1) / 1.60×10⁻¹⁹

= 6.25×10¹⁹ e⁻

Answer: The statement is true

Explanation:

The half-life of a radioactive isotope is the time taken for half of the total number of atoms in a given sample of the isotope to decay.

For instance

The half-life of radium is 1622 years. This means that if we have 1000 radium atoms at the beginning, then at the end of 1622 years, 500 atoms would have disintegrated, leaving 500 undecayed radium atoms

Thus, the statement is true

Explanation:

a)

Moles of hydrogen gas = [tex]n_1=\frac{1.00 g}{2 g/mol}=0.5 mol[/tex]

Moles of oxygen gas = [tex]n_2=\frac{1.00 g}{32 g/mol}=0.03125 mol[/tex]

Total moles in container = [tex]n=n_1+n_2=0.5 mol+0.03125 mol=0.53125 mol[/tex]

Total pressure of mixture = P

Temperature of the mixture = [tex]T = 27^oC =27+273K= 300 K[/tex]

Volume of the container in which mixture is kept = [tex]2.00 dm^3 =2.00 L[/tex]

[tex]1 dm^3=1 L[/tex]

[tex]P=\frac{nRT}{V}[/tex] (from ideal gas equation )

[tex]P=\frac{0.53125 mol\times 0.0821 atm  L/mol K\times 300 K}{2.00 L}=6.54 atm[/tex]

Partial pressure of the hydrogen gas :

= [tex]p_1=P\times \chi_1=P\times \frac{n_1}{n}[/tex]

[tex]=6.54 atm\times \frac{0.5 mol}{0.53125 mol}=6.16 atm[/tex]

Partial pressure of the oxygen gas :

= [tex]p_2=P\times \chi_2=P\times \frac{n_2}{n}[/tex]

[tex]=6.54 atm\times \frac{0.03125 mol}{0.53125 mol}=0.38 atm[/tex]

hydrogen

= [tex]\frac{n_1}{n}\times 100=\frac{0.5 mol}{0.53125 }\times 100[/tex]

= 94.12%

oxygen :

= [tex]\frac{n_2}{n}\times 100=\frac{0.03125 mol}{0.53125  mol}\times 100[/tex]

= 5.88%

b)

Moles of nitrogen gas = [tex]n_1=\frac{1.00 g}{28 g/mol}=0.03571 mol[/tex]

Moles of oxygen gas = [tex]n_2=\frac{1.00 g}{32 g/mol}=0.03125 mol[/tex]

Total moles in container = [tex]n=n_1+n_2=0.03571 mol+0.03125 mol=0.06696 mol[/tex]

Total pressure of mixture = P

Temperature of the mixture = [tex]T = 27^oC =27+273K= 300 K[/tex]

Volume of the container in which mixture is kept = [tex]2.00 dm^3 =2.00 L[/tex]

[tex]1 dm^3=1 L[/tex]

[tex]P=\frac{nRT}{V}[/tex] (from ideal gas equation )

[tex]P=\frac{0.06696 mol\times 0.0821 atm  L/mol K\times 300 K}{2.00 L}=0.82 atm[/tex]

Partial pressure of the nitrogen gas :

= [tex]p_1=P\times \chi_1=P\times \frac{n_1}{n}[/tex]

[tex]=0.82 atm\times \frac{0.03571 mol}{0.06696 mol}=0.44 atm[/tex]

Partial pressure of the oxygen gas :

= [tex]p_2=P\times \chi_2=P\times \frac{n_2}{n}[/tex]

[tex]=0.82 atm\times \frac{0.03125 mol}{0.06696 mol}=0.38 atm[/tex]

Composition of each in mole percent :

nitrogen

= [tex]\frac{n_1}{n}\times 100=\frac{0.03571 mol}{0.06696 }\times 100[/tex]

= 53.33%

oxygen :

= [tex]\frac{n_2}{n}\times 100=\frac{0.03125 mol}{0.06696 mol}\times 100[/tex]

= 46.67%

c)

Moles of methane gas = [tex]n_1=\frac{1.00 g}{16 g/mol}=0.0625 mol[/tex]

Moles of ammonia gas = [tex]n_2=\frac{1.00 g}{17g/mol}=0.0588 mol[/tex]

Total moles in container = [tex]n=n_1+n_2=0.0625 mol+0.0588 mol=0.1213 mol[/tex]

Total pressure of mixture = P

Temperature of the mixture = [tex]T = 27^oC =27+273K= 300 K[/tex]

Volume of the container in which mixture is kept = [tex]2.00 dm^3 =2.00 L[/tex]

[tex]1 dm^3=1 L[/tex]

[tex]P=\frac{nRT}{V}[/tex] (from ideal gas equation )

[tex]P=\frac{0.1213 mol\times 0.0821 atm  L/mol K\times 300 K}{2.00 L}=1.49 atm[/tex]

Partial pressure of the methane gas :

= [tex]p_1=P\times \chi_1=P\times \frac{n_1}{n}[/tex]

[tex]=1.49 atm\times \frac{0.0625 mol}{0.1213mol}=0.77 atm[/tex]

Partial pressure of the ammonia gas :

= [tex]p_2=P\times \chi_2=P\times \frac{n_2}{n}[/tex]

[tex]=1.49 atm\times \frac{0.0588 mol}{0.1213mol}=0.72 atm[/tex]

Composition of each in mole percent :

Methane :

= [tex]\frac{n_1}{n}\times 100=\frac{0.0625 mol}{0.1213mol}\times 100[/tex]

= 51.52%

Ammonia

= [tex]\frac{n_2}{n}\times 100=\frac{0.0588 mol}{0.1213mol}\times 100[/tex]

= 48.47%

To calculate the partial pressure of each gas and the composition of the mixture, we can use the ideal gas law. By dividing the mass of each gas by its molar mass, we can calculate the number of moles. The total pressure can be obtained by summing up the partial pressures, and the mole percent composition can be calculated by dividing the number of moles of each gas by the total number of moles and multiplying by 100.

To calculate the partial pressure of each gas, we need to use the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. To calculate the total pressure, we can sum up the partial pressures of each gas. The mole percent composition can be calculated by dividing the number of moles of each gas by the total number of moles and multiplying by 100.

a) 1.00 g H2 and 1.00 g O2:

To calculate the number of moles, we divide the mass of each gas by its molar mass. The molar mass of H2 is 2 g/mol and the molar mass of O2 is 32 g/mol. So, the number of moles of H2 is 1 g / 2 g/mol = 0.5 mol, and the number of moles of O2 is 1 g / 32 g/mol = 0.03125 mol. The total number of moles is 0.5 mol + 0.03125 mol = 0.53125 mol.

The partial pressure of H2 can be calculated by multiplying the number of moles of H2 by the gas constant R and the temperature in Kelvin, and then dividing by the volume. The same process can be applied to calculate the partial pressure of O2. Finally, the total pressure can be calculated by summing up the partial pressures. The mole percent composition can be calculated by dividing the number of moles of each gas by the total number of moles and multiplying by 100.

b) 1.00 g N2 and 1.00 g O2:

Using the same process as before, we can calculate the number of moles of N2 and O2. The molar mass of N2 is 28 g/mol, so 1 g of N2 is equal to 1 g / 28 g/mol = 0.03571 mol of N2. The molar mass of O2 is 32 g/mol, so 1 g of O2 is equal to 1 g / 32 g/mol = 0.03125 mol of O2. The total number of moles is 0.03571 mol + 0.03125 mol = 0.06696 mol.

Using the ideal gas law, we can calculate the partial pressures of N2 and O2, the total pressure, and the mole percent composition.

c) 1.00 g CH4 and 1.00 g NH3:

Following the same calculations as above, the molar mass of CH4 is 16 g/mol, so 1 g of CH4 is equal to 1 g / 16 g/mol = 0.0625 mol of CH4. The molar mass of NH3 is 17 g/mol, so 1 g of NH3 is equal to 1 g / 17 g/mol = 0.05882 mol of NH3. The total number of moles is 0.0625 mol + 0.05882 mol = 0.12132 mol.

By using the ideal gas law, we can calculate the partial pressures of CH4 and NH3, the total pressure, and the mole percent composition.

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Answer:

a. the first pKa is lower because of the presence of an electron withdrawing 'OH' group attached to the carbon that is directly attached to it.

b. the third pKa is greater than that of acetic acid because of the presence of an electron donating methyl group which is directly attached to it.

Explanation:

Induction or Inductive effect is the electronic effects an atom or a group of atoms exert on a compound or a portion of a compound, which could either be electron donating or electron withdrawing, thereby affecting its acidity or basicity.

Electronegativity confers acidity on a compound. In a, the OH group withdraws electrons from the COOH group, conferring more electronegativity on the middle COOH group, thereby reducing the pKa and thus increasing acidity.

In b, the electron donating effect of the methyl group, decreases the electronegativity of the COOH group,p thereby increasing the pKa which also means decreased acidity.

The first pKa of citric acid is lower than acetic acid due to the presence of three COOH groups. The third pKa of citric acid is greater than acetic acid due to steric hindrance caused by the presence of two COOH groups at either end of the molecule.

a. The first pKa of citric acid is lower than the pKa of acetic acid due to the presence of three COOH groups. The proximity of the COOH groups in citric acid allows for easier liberation of the H+ ions, resulting in a lower pKa value. This is supported by the concept of resonance stabilization. In contrast, acetic acid has only one COOH group, resulting in a higher pKa value.

b. The third pKa of citric acid is greater than the pKa of acetic acid due to steric hindrance caused by the presence of two COOH groups at either end of the molecule. The close proximity of these two groups makes it difficult for the liberation of the H+ ion, resulting in a higher pKa value compared to acetic acid.

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Answer:

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Explanation:

The equation used to calculate the constant for first order kinetics:

[tex]t_{1/2}=\frac{0.693}{k}}[/tex] .....(1)

Rate law expression for first order kinetics is given by the equation:

[tex]t=\frac{2.303}{k}\log\frac{[A_o]}{[A]}[/tex] ......(2)

where,  

k = rate constant

[tex]t_{1/2}[/tex] =Half life of the reaction = [tex]2.42\times 10^3 s[/tex]

t = time taken for decay process = ?

[tex][A_o][/tex] = initial amount of the reactant = 0.163 M

[A] = amount left after time t =  66.8% of [tex][A_o][/tex]

[A]=[tex]\frac{66.8}{100}\times 0.163 M=0.108884 M[/tex]

[tex]k=\frac{0.693}{2.42\times 10^3 s}[/tex]

[tex]t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}[/tex]

t = 1,409.19 s

1 minute = 60 sec

[tex]t=\frac{1,409.19 }{60} min=23.49 min[/tex]

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Answer:

HCOOH + H2O <===> HCOO- + H3O+

Explanation:

The apparent "atomic mass" of the contaminated iodine is approximately 1111.9285 amu.

To determine the apparent "atomic mass" of contaminant iodine, we need to take into account the contribution of both the naturally occurring iodine ([tex]\rm ^1^2^7I[/tex]) and the man-made radioisotope ([tex]\rm ^1^2^9I[/tex]).

The formula for calculating the atomic mass is:

Atomic mass = (mass of isotope 1 * abundance of isotope 1) + (mass of isotope 2 * abundance of isotope 2) + ...

Lets calculating the abundances of [tex]\rm ^1^2^7I[/tex] and [tex]\rm ^1^2^9I[/tex]:

Abundance of [tex]\rm ^1^2^7I[/tex] = 1 - Abundance of [tex]\rm ^1^2^9I[/tex]

Given:

Mass of [tex]\rm ^1^2^7I[/tex] = 126.9045 amu

Mass of [tex]\rm ^1^2^9I[/tex] = 128.9050 amu

We have two isotopes: [tex]\rm ^1^2^7I[/tex] and [tex]\rm ^1^2^9I[/tex]. The masses of these isotopes and their abundances need to be considered:

Atomic mass = (mass of [tex]\rm ^1^2^7I[/tex] * abundance of [tex]\rm ^1^2^7I[/tex]) + (mass of [tex]\rm ^1^2^9I[/tex]* abundance of [tex]\rm ^1^2^9I[/tex])

We need to calculate the abundances of [tex]\rm ^1^2^7I[/tex] and [tex]\rm ^1^2^9I[/tex] before we can calculate the atomic mass.

Mass of contaminated iodine sample = 12.3849 g + 1.00070 g = 13.3856 g

Now we can calculate the abundances:

Abundance of [tex]\rm ^1^2^7I[/tex] = (mass of [tex]I-127[/tex] in sample) / (total mass of sample)

Abundance of [tex]\rm ^1^2^7I[/tex] = (12.3849 g * 126.9045 amu) / (13.3856 g) = 117.0997 amu / 13.3856 g ≈ 8.7411

Abundance of [tex]\rm ^1^2^9I[/tex] = (mass of I-129 in sample) / (total mass of sample)

Abundance of [tex]\rm ^1^2^9I[/tex]= (1.00070 g * 128.9050 amu) / (13.3856 g) = 128.9050 amu / 13.3856 g ≈ 9.6175

We can calculate the atomic mass:

Atomic mass = (126.9045 amu * 8.7411) + (128.9050 amu * 9.6175) ≈ 1111.9285

Therefore, the apparent "atomic mass" of the contaminated iodine is approximately 1111.9285 amu.

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The apparent atomic mass of the contaminated iodine is 126.9045 amu.

To calculate the apparent atomic mass of the contaminated iodine, we can use the formula:

Apparent atomic mass = (mass of naturally occurring iodine * atomic mass of naturally occurring iodine + mass of synthetic radioisotope * atomic mass of synthetic radioisotope) / total mass of contaminated iodine

Given:

Substituting the given values into the formula:

Apparent atomic mass = (12.3849 g * 126.9045 amu + 1.00070 g * 128.9050 amu) / (12.3849 g + 1.00070 g)

Simplifying the expression:

Apparent atomic mass = (1570.6078055 + 128.905070) / 13.3856

Apparent atomic mass = 1699.5128755 / 13.3856

Apparent atomic mass = 126.9045 amu

Therefore, the apparent atomic mass of the contaminated iodine is 126.9045 amu.

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Answer:

              1,1,1-trichloroethane being a saturated halogenated alkane will undergo substitution reaction via free radical mechanism. The mechanism is divided in three steps,

Step 1: Initiation:

                          In this step the reaction is started by treating the chlorine gas either with UV light or by sunlight. This results in the formation of free radical.

Step 2: Propagation:

                          In this step the radical formed will react with the hydrogen atom resulting in formation of HCl and generating free radical of corresponding alkane. Hence, the radical will agian react with Cl2 molecule generating another Chlorine radical and corresponding halogenated compound i.e. 1,1,1,2-tetrachloroethane.

Step 3: Termination:

                           This is the last step. In this step the reaction is stopped/terminated. The free radicals react with each other forming a single bonds and stopping the formation of further radicals.

The mechanism is shown below,

Chloroform-Chloroform and Acetone-Acetone interactions are weaker than chloroform-acetone interactions.

Explanation:

Raoult's law states, In a solution, vapor pressure is equal to the sum of the vapor pressures of individual component and this implies when it is multiplied by the mole fraction of that component in the solution. Raoult's Law is expressed by the formula:

                         P solution = Χ solvent [tex]\times[/tex] P0 solvent.

Raoult's law assumes that the components in the mixture are ideal, it assumes that the Chloroform-Chloroform interactions, Chloroform-Acetone, and Acetone-Acetone are the same. A negative deviation is shown by the measurement of observed vapor pressure which is less than that of calculated. Hence the interaction of Chloroform-Chloroform is weaker than the Chloroform-Acetone interactions.  

A solution of chloroform and acetone exhibits a negative deviation from Raoult's law, which means that the interactions between chloroform and acetone are stronger than those of the individual components

A solution of chloroform (CHCl3) and acetone((CH3)2CO) exhibits a negative deviation from Raoult's law. This means the solution has a lower vapor pressure than predicted by Raoult's Law due to the strength of the new intermolecular interactions being stronger than what was present in the pure components. The correct answer is X. chloroform-chloroform and acetone-acetone interactions are weaker than chloroform-acetone interactions. This is because in a solution exhibiting negative deviation, the interactions between the different molecules in the mixture are stronger than those of the pure components. This stronger interaction between chloroform and acetone molecules reduces the overall vapor pressure, leading to a negative deviation from Raoult's Law.

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Acetic acid, CH3CO2H, is composed of carbon, hydrogen, and oxygen atoms. The electron-dot structure is represented by placing valence electrons as dots around the symbol of each atom, showing their arrangement and the bonding between them.

Acetic acid, CH3CO2H, is composed of carbon, hydrogen, and oxygen atoms. The two carbon atoms are connected by a single bond, and both oxygen atoms are connected to the same carbon atom. To draw the electron-dot structure for acetic acid, we represent each atom using its symbol and show the valence electrons as dots surrounding the symbol.

In the case of acetic acid, the carbons are each connected to three hydrogens and one oxygen, and the oxygen is also connected to another carbon. Therefore, for each carbon atom, one dot is placed next to each hydrogen atom and three dots are placed next to the oxygen atom. Two additional dots are placed next to the oxygen atom connected to the other carbon. The arrangement of the dots represents the bonding between the atoms.

The resulting electron-dot structure for acetic acid would be represented as CH3CO2H.

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Answer:

Primary alcohol.

Explanation:

Jones reagent is mixture of chromium trioxide (CrO3) and sulfuric acid (H2SO4) dissolved in a mixture of acetone and water. Alternatively, potassium dichromate (K2CrO7) can be used in place of chromium trioxide because of its carcinogenic nature.

This oxidation reaction is an organic reaction for the oxidation of primary alcohols to aldehydes then carboxylic acid and secondary alcohols to ketones.

Lucas reagent is a solution of anhydrous zinc chloride (ZnCl) in concentrated hydrochloric acid. The reaction involves substitution reaction in which the chloride replaces a hydroxyl group.

A positive test is indicated by a change in appearance of the solution, from clear and colourless to fog-like, which shows the formation of a chloroalkane. Accurate results for this test are observed in tertiary alcohols, as they form alkyl halides the fastest due to the stability of their intermediate tertiary carbocation.

Therefore, an alcohol in which there was a positive Jones test and a negative Lucas test indicates the presence of primary alcohol.

This is because:

A primary alcohol would test positive to Jones test but in Lucas test, the substitution reaction is the slowest as compared to the secondary and tertiary alcohols.

1° alcohols < 2° alcohols < 3° alcohols

So a primary alcohol will give a negative result to Lucas reagent.

Answer:

a) Element = Chlorine

b) Element = Arsenic

Explanation:

The knowledge of Orbitals and Quantum number and the electronic configuration is applied as shown in the analysis in the attached file.

Answer:

The age of the ore is 4.796*10^9 years.

Explanation:

To solve this question, we use the formula;

A(t) =A(o)(1/2)^t/t1/2

where;

A(t) =3.22mg

A(o) = 6.731mg

t1/2 = 4.51*70^9 years

t = age of the ore

So,

A(t) =A(o)(1/2)^t/t1/2

3.22 = 6.73 (1/2)^t/4.51*10^9

Divide both sides by 6.73

3.22/6.73= (1/2)^t/4.51*10^9

0.47825= (0.5)^t/4.51*10^9

Log 0.4785 = t/4.51*10^9 • log 0.5

Log 0.4785/log 0.5 • 4.51*10^9 = t

t = 1.0634 * 4.51*10^9

t = 4.796*10^9

So therefore, the age of the ore is approximately 4.796*10^9 years.

Answer: The work required to compress helium gas is 540 kJ

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

[tex]W=-P\Delta V=-P(V_2-V_1)[/tex]

W = amount of work done = ?

P = pressure = 180 kPa

[tex]V_1[/tex] = initial volume = [tex]5m^3[/tex]

[tex]V_2[/tex] = final volume = [tex]2m^3[/tex]

Putting values in above equation, we get:

[tex]W=-180kPa\times (2-5)m^3=540kPa.m^3[/tex]

To convert this into joules, we use the conversion factor:

[tex]1kPa.m^3=1kJ[/tex]

So, [tex]540kPa.m^3=540kJ[/tex]

The positive sign indicates that work is done by the system.

Hence, the work required to compress helium gas is 540 kJ

The work required to compress 1 kg of helium from 5 m³ to 2 m³ at a constant pressure of 180 [tex]K_p_a[/tex] is 540 kJ.

This problem involves the calculation of work done during the compression of helium gas in a piston-cylinder device at a constant pressure. The work required to compress the gas can be determined using the formula for work done by a gas at constant pressure:

Work (W) = P x ΔV

Where:

Thus, ΔV =  [tex]V_i_n_i_t_i_a_l[/tex] - [tex]V_f_i_n_a_l[/tex]

Given:

Therefore:

ΔV = 5 m³ - 2 m³ = 3 m³

Substituting the values into the work formula:

Work (W) = 180 [tex]K_p_a[/tex] x 3 m³ = 540 kJ

Therefore, the work required to compress the helium is 540 kJ.

Answer: The pressure of carbon dioxide needed is 2.94 atm

Explanation:

To calculate the partial pressure of carbon dioxide, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]3.4\times 10^{-2}mol/L.atm[/tex]

[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas = [tex]0.10mol/L[/tex]

[tex]p_{CO_2}[/tex] = pressure of carbon dioxide = ?

Putting values in above equation, we get:

[tex]0.10mol/L=3.4\times 10^{-2}mol/L.atm\times p_{CO_2}\\\\p_{CO_2}=\frac{0.10mol/L}{3.4\times 10^{-2}mol/L.atm}=2.94atm[/tex]

Hence, the pressure of carbon dioxide needed is 2.94 atm

Answer:

0.681 atm

Explanation:

To solve this problem, we make use of the General gas equation.

Given:

P1 = 785 torr

V1 = 2L

T1 = 37= 37 + 273.15 = 310.15K

P2 = ?

V2 = 3.24L

T2 = 58 = 58+273.15 = 331.15K

P1V1/T1 = P2V2/T2

Now, making P2 the subject of the formula,

P2 = P1V1T2/T1V2

P2 = [785 * 2 * 331.15]/[310.15 * 3.24]

P2 = 515.715 Torr

We convert this to atm: 1 torr = 0.00132 atm

515.715 Torr = 515.715 * 0.00132 = 0.681 atm

Answer:

Al 72.61%

Mg 27.39%

Explanation:

To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.

If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.

To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.

Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g

The mass of magnesium is 0.0381 * 24 = 0.9144g

We can now calculate the mass percentage.

The total mass is 0.9144 + 2.4246 = 3.339g

% mass of Al = 2.4246/3.339 * 100 = 72.61%

% mass of Mg = 0.9144/3.39 * 100 = 27.39%

Answer:

Mass % Al = 72.3 %

Mass % Mg = 27.7 %

Explanation:

Step 1: Data given

Number of moles Al = 0.0898 moles

Number of moles Mg = 0.0381 moles

Molar mass Al = 26.98 g/mol

Molar mass Mg = 24.3 g/mol

Step 2: Calculate mass Al

Mass Al = moles Al * molar mass Al

Mass Al = 0.0898 moles * 26.98 g/mol

Mass Al = 2.42 grams

Step 3: Calculate mass Mg

Mass Mg = 0.0381 moles * 24.3 g/mol

Mass Mg = 0.926 grams

Step 4: Calculate total mass

Total mass = mass Al + mass Mg

Total mass = 2.42 grams + 0.926 grams

Total mass = 3.346 grams

Step 5: Calculate mass %

Mass % Al = (mass Al/ total mass) * 100%

Mass % Al = (2.42 grams / 3.346 grams ) *100%

Mass % Al = 72.3 %

Mass % Mg = (mass Mg/ total mass)*100%

Mass % Mg = (0.926 / 3.346) *100 %

Mass % Mg = 27.7 %

Answer:

The wavelength in nm = 5120

The wavelength in A°= 51,200

The frequency of the absorbed radiation is [tex]5.859\times 10^{13} Hertz[/tex]

Explanation:

1) The wave number of the CO bond = [tex]\bar v=1953 cm^{-1}[/tex]

The wavelength corresponding to this wave number =[tex]\lambda [/tex]

[tex]\lambda =\frac{1}{\bar v}=\frac{1}{1953 cm^{-1}}=5.12\times 10^{-4} cm[/tex]

[tex]1 cm = 10^7 nm[/tex]

[tex]5.12\times 10^{-4}\times 10^7 nm=5120 nm[/tex]

[tex]1 cm = 10^8 \AA[/tex]

[tex]5.12\times 10^{-4}\times 10^8 \AA=51,200 \AA[/tex]

The wavelength in nm = 5120

The wavelength in A°= 51,200

2)

Wavelength of the wave = [tex]\lambda =5120 nm = 5120\times 10^{-9} nm[/tex]

[tex]1 nm = 10^{-9} m[/tex]

Frequecy of the wave = [tex]\nu [/tex]

[tex]\nu=\frac{c}{\lambda }[/tex]

c = Speed of light = [tex]3\times 10^8 m/s[/tex]

[tex]\nu=\frac{3\times 10^8 m/s}{5120\times 10^{-9} m}=5.859\times 10^{13} s^{-1}[/tex]

The frequency of the absorbed radiation is [tex]5.859\times 10^{13} Hertz[/tex]

The bond between hemoglobin and CO absorbs radiation of 1953 cm⁻¹. The corresponding wavelength is 5120 nm and 51200 Å. The corresponding frequency is 5.855 × 10¹³ Hz.

The bond between hemoglobin and CO absorbs radiation of 1953 cm⁻¹, that is, the wavenumber (ν) is 1953 cm⁻¹.

We can calculate the wavelength (λ) using the following expression.

[tex]\lambda = \frac{1}{\nu } = \frac{1}{1953cm^{-1} } = 5.120 \times 10^{-4} cm[/tex]

We will convert 5.120 × 10⁻⁴ cm to nm using the following conversion factors.

[tex]5.120 \times 10^{-4} cm \times \frac{1m}{100cm} \times \frac{10^{9}nm }{1m} = 5120 nm[/tex]

We will convert 5.120 × 10⁻⁴ cm to Å using the following conversion factors.

[tex]5.120 \times 10^{-4} cm \times \frac{1m}{100cm} \times \frac{10^{10}A }{1m} = 51200 nm[/tex]

We can calculate the frequency (f) of the absorbed radiation using the following expression.

[tex]f = \frac{c}{\lambda } = \frac{2.998 \times 10^{8}m/s }{5.120\times 10^{-6}m } = 5.855 \times 10^{13} Hz[/tex]

where,

The bond between hemoglobin and CO absorbs radiation of 1953 cm⁻¹. The corresponding wavelength is 5120 nm and 51200 Å. The corresponding frequency is 5.855 × 10¹³ Hz.

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Answer:

+1725

Explanation:

The specific rotation can be calculated using the formula:

Where l is the path length in decimeters (20 cm / 10), 2 dm; and C is the concentration of the compound in mg/mL (2.0g / 50mL), 0.04 g/mL.

Putting the data we're left with:

The specific rotation of the compound is [tex]\( +1725\° \text{ dm}^{-1} \text{ (g/mL)}^{-1} \).[/tex]

The specific rotation of the compound is given by the formula:

[tex]\[ [\alpha] = \frac{\alpha}{l \cdot c} \][/tex]

where:

- [tex]\( [\alpha] \)[/tex] is the specific rotation,

- [tex]\( \alpha \)[/tex] is the observed rotation in degrees,

- l is the length of the polarimeter tube in decimeters,

- c  is the concentration of the solution in grams per milliliter.

Given:

- The observed rotation [tex]\( \alpha = +138\° \)[/tex],

- The length of the polarimeter tube l = 20cm= 2 dm (since 1 dm = 10 cm),

- The mass of the compound m = 2.0 g,

- The volume of the solution V = 50mL.

First, we need to calculate the concentration c of the solution:

[tex]\[ c = \frac{\text{mass of compound}}{\text{volume of solution}} = \frac{2.0 \text{ g}}{50 \text{ mL}} = 0.04 \text{ g/mL} \][/tex]

Now we can calculate the specific rotation:

[tex]\[ [\alpha] = \frac{\alpha}{l \cdot c} = \frac{+138\°}{2 \text{ dm} \cdot 0.04 \text{ g/mL}} = \frac{+138\°}{0.08 \text{ dm} \cdot \text{g/mL}} \]\\[/tex]

[tex]\[ [\alpha] = \frac{+138\°}{0.08} \] \[ [\alpha] = +1725\° \text{ dm}^{-1} \text{ (g/mL)}^{-1} \][/tex]

Therefore,

The answer is: [tex]+1725\° \text{ dm}^{-1} \text{ (g/mL)}^{-1}.[/tex]

Answer:

(a)    ⁴¹₁₉K

(b)  ¹⁰⁶₄₆Pd

(c)  ¹²⁵₅₂Te

(d)   ⁸⁸₃₈Sr

Explanation:

The identity of an element  is its atomic number, by convention we write the atomic mass as superscript and the and the atomic number as subscript to the left of the element .

(a) Z = 19   A = 41         symbol:      ⁴¹₁₉K

(b) Z = 46  A = 106       symbol:   ¹⁰⁶₄₆Pd

(c) Z = 52   A = 125       symbol:  ¹²⁵₅₂Te

(d) Z = 38   A  = 88       symbol:    ⁸⁸₃₈Sr

The isotopes correspond to Potassium-41 (41K), Palladium-106 (106Pd), Tellurium-125 (125Te), and Strontium-88 (88Sr). Z and A denote atomic and mass numbers, which help in identifying particular isotopes.

The appropriate symbol for each of these isotopes would be:

Z and A represent atomic and mass numbers respectively. These isotopes are written with a mass number (A) preceding the symbol and atomic number (Z) is typically understood from the symbol itself.

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Answer:

a) Germanium = 5.76 x 〖10〗^11 〖cm〗^(-3) , Semiconductor is n-type.

b) Silicon = 2.25 x 〖10〗^5 〖cm〗^(-3) , Semiconductor is n-type.

For clear view of the answers: Please refer to calculation 5 in the attachments section.

Explanation:

So, in order to find out the concentration of holes and electrons in a sample of germanium and silicon which have the concentration of donor atoms equals to 〖10〗^15 〖cm〗^(-3). We first need to find out the intrinsic carrier concentration of silicon and germanium at room temperature (T= 300K).

Here is the formula to calculate intrinsic carrier concentration: For calculation please refer to calculation 1:

So, till now we have calculated the intrinsic carrier concentration for germanium and silicon. Now, in this question we have been given donor concentration (N_d) (N subscript d), but if donor concentration is much greater than the intrinsic concentration then we can write: Please refer to calculation 2.

So, now we have got the concentration of electrons in both germanium and silicon. Now, we have to find out the concentration of holes in germanium and silicon (p_o).  (p subscript o)

Equation to find out hole concentration: Please refer to calculation 3. and Calculation 4. in the attachment section.

Good Luck Everyone! Hope you will understand.  

(a) The germanium with 10^15 cm^-3 donor atoms: Concentration of electrons = 10^15 cm^-3, Concentration of holes ≈ 0, Semiconductor is n-type. (b) For silicon with 10^15 cm^-3 donor atoms: Concentration of electrons = 10^15 cm^-3, Concentration of holes ≈ 0, Semiconductor is n-type.

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The formula weight of the unknown copper compound is approximately 13.90145 amu.

To find the formula weight (F.W.) of the unknown copper compound, we need to consider the percent copper composition and the atomic weight of copper. The atomic weight of copper is approximately 63.55 amu.

We can use the percent composition to calculate the mass of copper in one mole of the compound:

(0.2190) * F.W. = (0.2190) * (63.55 amu) = 13.90145 amu

Since one mole of the compound contains one mole of copper, we can equate the mass of copper to the formula weight:

13.90145 amu = F.W.

Therefore, the formula weight of the unknown copper compound is approximately 13.90145 amu.

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Answer:

The reaction requires anti-Markovnikov product without sigmatropic rearrangement.

Explanation:

The reaction is known to begin with the concerted syn addition of B and H across the double bond, with the boron adding to the less substituted carbon atom.

The second step of the reaction involves hydrogen peroxide and a base such as NaOH are added, NaOH deprotonates the hydrogen peroxide.

The resulting NaOOH then attacks the boron and sets up the key migration step, where the carbon-boron bond migrates to the oxygen bound to boron, breaking the weak oxygen-oxygen bond . Then the -OH expelled then returns to form a bond on the boron resulting in a deprotonated alcohol (alkoxide). The alkoxide is then protonated by water or some other comparably acidic species.

Hydroboration is a syn addition that gives an anti-Markovnikov product without sigmatropic rearrangement.

The reasons why hydroboration–oxidation reagents were chosen to effect the transformation will be:

It should be noted that hydroboration–oxidation is vital for the production of alcohol. It's needed for the transformation rather than using other reagents because hydroboration-oxidation reagents yield the anti-Markovnikov product of addition.

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Answer: Inittiation: Br2 -> Br. + Br.

Propagation: CH3CH(CH3)CH3+Br. -> CH3CH(CH3)CH2. +HBr

CH3CH(CH3)CH2. + Br. -> CH3CH(CH3)CH2Br

CH3CH(CH3)CH2Br +Br. -> CH3C.(CH3)CH2Br +HBr

CH3C.(CH3)CH2Br + Br. -> CH3CBr(CH3)CH2Br

Termination: Br. + Br. ->Br2

Answer:

C flyash = 10677789.55 μg/m³

Explanation:

A coal-fired power plant:

∴ rate = 50 ft³/s = (1.416 m³/s)(3600 s/h) = 5097.6 m³/h

hot gas-flyash:

∴ rf = 120 Lb/h = 54431.1 g/h = 54431100000 μg/h

⇒ C flyash = ? [=] μg/m³

⇒ C flyash = (54431100000 μg/h)/(h/5097.6 m³)

⇒ C flyash = 10677789.55 μg/m³

Final answer:

The average atomic mass of titanium on the hypothetical planet is calculated by using the weighted average of the abundances and masses of its isotopes, resulting in 46.8989 amu.

Explanation:

The average atomic mass of titanium on that hypothetical planet can be calculated by multiplying the abundance of each isotope by its mass (in atomic mass units, amu), then summing these products. The calculation will look as follows:

To find the average atomic mass, we add these values together to get the sum which should give us the average atomic mass of titanium on the planet.

The correct calculation would be:

Hence, the average atomic mass of titanium for this planet is 46.8989 amu. It's important to note that the average atomic mass of an element is the weighted average of all the isotopes of that element.

Answer:

4.98 × 10⁻³ mol

Explanation:

Given data for Cl₂

[tex]805torr.\frac{1atm}{760torr} =1.06atm[/tex]

First, we will calculate the moles of Cl₂ using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.06 atm × 0.115 L / (0.0821 atm.L/mol.K) × 298 K

n = 4.98 × 10⁻³ mol

Let's consider the balanced equation.

MnO₂(s) + 4 HCl(aq) ⟶ MnCl₂(aq) + 2 H₂O(l) + Cl₂(g)

The molar ratio of MnO₂ to Cl₂ is 1:1. The required moles of MnO₂ are 4.98 × 10⁻³ moles.

Answer:

Net work done overall = sum of work done for all the processes = 16,995.84 J

Explanation:

From the start, P₁ = 10bar = 1 × 10⁶ Pa, T₁ = 600K, V₁ = ?

We can obtain V from PV = nRT; n = 2, R = 8.314 J/mol.K

V = 2 × 8.314 × 600/(1000000) = 0.009977 m³

P₁ = 10bar = 1 × 10⁶ Pa, T₁ = 600K, V₁ = 0.009977 m³

For an adiabatic process for an ideal gas,

P(V^γ) = constant

γ = ratio of specific heats = Cp/CV = 1.4,

P₂ = 1 bar = 10⁵ Pa

P₁ (V₁^1.4) = P₂ (V₂^1.4) = k

10⁶ (0.009977^1.4) = 10⁵(V₂^1.4) = 1579.75 = k

V₂ = 0.0517 m³

Work done for an adiabatic process

W = k((V₂^(1-γ)) - (V₁^(1-γ))/(1-γ)

W = 1579.75 ((0.0517^0.4) - (0.009977^0.4))/0.4

W = 582.25 J

We still need T₂

PV = nRT

T₂ = P₂V₂/nR = 100000×0.0517/(2×8.314) = 310.92K

Step 2, constant volume heating,

Work done at constant volume is 0 J.

T₂ = 310.92K, T₃ = 600K

V₂ = 0.0517 m³, V₃ = V₂ = 0.0517 m³ (Constant volume)

P₂ = 1bar, P₃ = ?

PV = nRT

P₃ = nRT₃/V₃ = 2 × 8.314 × 600/0.0517 = 192974.85 Pa = 1.93bar

Step 3, isothermally returned to the initial state.

P₃ = 1.93bar, P₄ = P₁ = 10bar

T₃ = 600K, T₄ = T₁ = 600K (Isothermal process)

V₃ = 0.0517 m³, V₄ = V₁ = 0.009977 m³

Work done = nRT In (V₃/V₁) = 2 × 8.314 × 600 In (0.0517/0.009977) = 16413.59 J

Net work done = W₁₂ + W₂₃ + W₃₁ = 582.25 + 0 + 16413.59 = 16995.84 J

Hope this helps!!

Answer: The given statement is true.

Explanation:

When we increase the amount of solvent which is water in this case then it means there will occur an increase in the molecules. Hence, there will be more number of collisions to take place with increase in number of molecules.

Therefore, more is the amount of interaction taking place between the molecules of a solution more will be its rate of hydrolysis.

Thus, we can conclude that the statement increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis, is true.

The molarity of the 10.5% by mass glucose solution is 0.600 M.

To find the molarity of the glucose solution, we need to determine the number of moles of glucose present in the solution first. We can use the percent by mass to calculate this.

Given that the solution is 10.5% by mass, we know that 10.5 grams of glucose is present in a 100 gram solution. We can convert this to moles by dividing the mass of glucose by its molar mass, which is 180.16 g/mol.

So, the number of moles of glucose is 10.5 g / 180.16 g/mol = 0.0583 mol. To find the molarity, we divide the number of moles by the volume of the solution in liters. The volume of the solution can be determined by multiplying the density of the solution by its mass: 100 g / (1.03 g/mL) = 97.09 mL = 0.0971 L.

Therefore, the molarity of the glucose solution is 0.0583 mol / 0.0971 L = 0.600 M.

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The molarity of a 10.5% by mass glucose solution with a density of 1.03 g/mL can be calculated by assuming a 100 g sample of the solution. The mass of glucose per liter is found to be 108.1395 g, and with the molar mass of glucose, the molarity is determined to be 0.600 M.

The question pertains to determining the molarity of a 10.5% by mass glucose solution with a given density of 1.03 g/mL. To calculate the molarity, we need to use the given mass percentage and density to find out how many moles of glucose are present in a liter of solution.

First, assume you have 100 g of this solution. Because it's a 10.5% by mass solution, this means there are 10.5 g of glucose (C6H12O6) and 89.5 g of water in the mixture.

Using the density, we find the volume of 100 g of solution:
100g / 1.03g/mL = 97.09 mL
Because we want to know the molarity per liter, it's important to work with a liter of the solution:
(1000 mL/L) / (97.09 mL) = 10.299 L^-1 multiplication factor
Now, we will use the multiplication factor to scale up the mass of glucose to what would be in one liter:
10.5 g * 10.299 = 108.1395 g glucose per liter

The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol, so the number of moles in one liter would be:
108.1395 g / 180.16 g/mol = 0.600 mol/L

Therefore, the molarity of the glucose solution is 0.600 M.