The mineral enargite is 48.41% cu, 19.02% as, and 32.57% s by mass. what is the empirical formula of enargite?

Cows can derive energy from cellulose-rich grass due to the presence of symbiotic bacteria in their digestive system, which secrete an enzyme called cellulase, breaking down cellulose into usable glucose monomers. Their multiple-chambered stomach and appendix further facilitate cellulose digestion. This unique adaptation allows cows and other ruminants to utilize cellulose as an energy source.

Cows and other herbivores are able to derive energy from cellulose-rich grass due to symbiotic bacteria that reside in their rumen, a part of their digestive system. These bacteria secrete an enzyme called cellulase which breaks down cellulose into glucose monomers, providing a viable energy source for the animal. Herbivores, such as cows, buffalos, and horses, possess a multiple-chambered stomach where cellulose digestion mostly occurs, and also an appendix where bacteria help in further breaking down cellulose.

Cellulose is composed of glucose monomers packed tightly as extended long chains, giving it its high tensile strength and rigidity, an important attribute for plant cells. It's interesting to note that every other glucose monomer in cellulose is flipped over, which adds to its complexity and indigestibility in many animals including humans. However, the unique digestive system of ruminants, rooms for cellulase-secreting bacteria such as those in the rumen and appendix, enable them to utilize cellulose as an energy source.

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The question concerns Boyle's Law in physics, which states that pressure and volume of a gas are inversely related at a constant temperature. Using the provided initial conditions and the final pressure, the final volume of the gas can be calculated using Boyle's Law to be approximately 2.33 L.

The subject matter of this question falls under the concept of gases in Physics, especially where Boyle's Law is applicable. Boyle's Law states that the pressure (P) and volume (V) of a gas have an inverse relationship when the temperature is kept constant. In other words, an increase in pressure causes a decrease in volume and vice versa.

To solve the student's question, we'll use the equation of Boyle's Law: P1*V1 = P2*V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

In this case, the standard pressure (P1), which is generally 760 mm Hg, and the initial volume (V1) is 3.5 L. The final pressure, P2, is given as 1140 mm Hg, and we need to find the final volume (V2).

Substituting the values we know into the Boyle's Law equation: (760mm Hg * 3.5 L) = (1140 mm Hg * V2) Solving for 'V2' gives us a final volume of approximately 2.33 L. Therefore, the volume of the gas when the pressure is 1140 mm Hg will be about 2.33 L.

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The Lewis structure for MgI2 shows 3 lone pairs of electrons on each I atom and no lone pairs on the Mg atom.

The Lewis structure for MgI2 can be drawn by following a few steps:

The final Lewis structure for MgI2 will have 3 lone pairs of electrons on each I atom and no lone pairs on the Mg atom.

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Final answer:

The conversion of pyruvate to acetyl CoA is represented by the equation: C3H3O3- + CoA-SH + NAD+ → CH3-C-S-CoA + CO2 + NADH + H+. This step catalyzed by the pyruvate dehydrogenase complex involves oxidative decarboxylation and connects glycolysis to the Krebs cycle.

Explanation:

The conversion of pyruvate to acetyl CoA is a critical step that links glycolysis, the initial phase of glucose catabolism, to the Krebs cycle, where further energy extraction occurs. The overall chemical equation representing the conversion of pyruvate (a three-carbon molecule) to acetyl CoA (a two-carbon molecule) is:

C3H3O3- (pyruvate) + CoA-SH + NAD+ → CH3-C-S-CoA (acetyl CoA) + CO2 + NADH + H+

This process, known as pyruvate oxidation, is catalyzed by the multi-enzyme complex pyruvate dehydrogenase. It involves the oxidative decarboxylation of pyruvate, resulting in one molecule of carbon dioxide and the reduction of NAD+ to NADH. The transferred acetyl group is then bound to Coenzyme A, forming acetyl CoA, which subsequently enters the Krebs cycle within the mitochondria.

O2 experiences London dispersion forces, NBr3 experiences dipole-dipole interaction, CH4 experiences London dispersion forces, and HF experiences hydrogen bonding due to the presence of a hydrogen atom bonded to a highly electronegative atom.

The given molecules show different types of intermolecular forces. Oxygen gas (O2) is a non-polar molecule and it interacts through London dispersion forces, the weakest form of forces. Nitrogen tribromide (NBr3) has polar bonds, so this compound's main intermolecular force would be dipole-dipole interaction. Methane (CH4) is non-polar, so the intermolecular forces are due to London dispersion forces. Hydrogen Fluoride (HF) experiences hydrogen bonding because it contains a hydrogen atom bonded to a more electronegative atom, fluorine.

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Chlorine (Cl2) exhibits dispersion forces as its strongest intermolecular force. HBr exhibits dipole-dipole forces, and Co and NaCl show stronger forms of bonding (metallic and ionic, respectively). Larger, heavier atoms or molecules generally display stronger dispersion forces due to their greater number of electrons.

The molecule or compound that exhibits dispersion forces as its strongest intermolecular force among Co, HBr, Cl2, and NaCl is Cl2. Dispersion forces are the only type of intermolecular force that occurs between noble gas atoms and nonpolar molecules such as Cl2. They arise as a result of fluctuations in electron density in an atom or molecule, that create temporary regions of positive and negative charge. These forces can attract nearby oppositely charged regions in neighboring molecules.

In the provided list, Hydrogen Bromide (HBr) exhibits dipole-dipole forces which are stronger than dispersion forces because it's a polar molecule. Cobalt (Co) is a metal and exhibits metallic bonding which is stronger than the dispersion forces. Sodium chloride (NaCl) forms an ionic compound and hence shows ionic bonding, which is much stronger than the dispersion forces.

Molecules with larger and heavier atoms, like chlorine (Cl2) , generally exhibit stronger dispersion forces compared to the smaller and lighter atoms, because of the presence of more electrons that can create temporary poles. At room temperature, Cl2 is a gas, but its dispersion forces allow it to condense into a liquid or solid under the right conditions.

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the molecular egometry around each carbon atom in a staurated hydro carbon is tetrahydral

When 352 g of Fe₂O₃ reacts with excess CO, it produces 4.41 mol of Fe.

To determine how much Fe (iron) is produced when 352 g of Fe₂O₃ reacts with excess CO, follow these steps:

Calculate moles of Fe₂O₃:

The molar mass of Fe₂O₃ is 159.7 g/mol.

Moles of Fe₂O₃ = 352 g / 159.7 g/mol = 2.20 mol Fe₂O₃.

Determine the mole ratio of Fe₂O₃ to Fe using the balanced equation:

The balanced equation is Fe₂O₃ + 3CO → 3CO₂ + 2Fe.

1 mol of Fe₂O₃ produces 2 mol of Fe.

Calculate moles of Fe produced:

Moles of Fe = 2.20 mol Fe₂O₃ * 2 mol Fe / 1 mol Fe₂O₃ = 4.40 mol Fe.

Convert moles of Fe to grams:

The molar mass of Fe is 55.85 g/mol.

Mass of Fe = 4.40 mol * 55.85 g/mol = 245.74 g of Fe.

The correct answer is option 1: 4.41 mol Fe.

Answer:

True

Explanation:

The pH of a 0.45 M solution of aniline (C6H5NH2) is 9.4.

The pH of a 0.45 M solution of aniline (C6H5NH2) can be calculated using the given pKb value of 9.40.

To find the pH, we need to find the pOH first using the pKb value:

pOH = 14 - pKb = 14 - 9.40 = 4.6

Then, we can use the equation pH + pOH = 14 to find the pH:

pH = 14 - pOH = 14 - 4.6 = 9.4

Therefore, the pH of the solution is 9.4.

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The pH of a 0.45 M solution of aniline, given that its pKb is 9.40, is calculated to be approximately 10.99 after taking into account the equilibrium constant Kb for aniline and the concentration of OH- ions introduced on the ionization of aniline.

To calculate the pH of a 0.45 M solution of aniline (C6H5NH2), we need to take into account its pKb value, 9.40. Aniline is a weak base, and we can use the formula pOH = pKb - log[(base)/(acid)]. First, we have to calculate the equilibrium constant Kb for aniline which is Kw/Ka = 10^-14 / Ka. We know Ka equals 10^-5.6, hence Kb for aniline = 10^-14 / 10^-5.6 = 4.3 × 10^-10.

Next, using an approximation approach and the ICE table, we calculate the concentration of OH- introduced by aniline when it ionizes, and find it to be roughly 9.7 × 10-4 M. The pOH of the solution is then -log(9.7 × 10-4) which equals 3.01. To get the pH, we subtract this from 14 (since pH + pOH = 14), yielding a pH for the solution of approximately 10.99.

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Explanation:

It is known that density of water at [tex]20^{o}C[/tex] is 998,29 [tex]kg/m^{3}[/tex] or 0.998 g/ml (as 1 [tex]kg/m^{3}[/tex] = 0.001 g/ml).

Therefore, calculate mass of water as follows.

        Mass of water = Final mass - initial mass

                                = 20.2144 g - 10.2634 g

                                = 9.951 g

As we known that density is the amount of mass present in a unit volume.                            

Mathematically,       Density = [tex]\frac{mass}{volume}[/tex]

Therefore, putting the given values into the above formula we will calculate the volume as follows.

                         Density = [tex]\frac{mass}{volume}[/tex]

                         0.998 g/ml = [tex]\frac{9.951 g}{Volume}[/tex]

                        Volume = 9.97 ml

Thus, we can conclude that volume of the flask at [tex]20^{o}C[/tex] is 9.97 ml.

The true volume of the flask at 20°C is 8.30145 mL.

The true volume of the flask can be calculated using the difference in mass before and after adding water.

Therefore, the true volume of the flask at 20°C is 8.30145 mL.

Correct question is: An empty 10 ml volumetric flask weighs 10.2634 g. when filled to the mark with distilled water and weighed again in the air at 20°C , the mass is 20.2144 g. What is the true volume of the flask at 20°C ?

Trans-2-butene's C=C double bond can be indicated by IR absorption signals of C-H stretching near 3100 cm⁻¹ for the =CH₂ group, 3020 cm⁻¹ for the –CH= group and C-H out-of-plane bending near 915 cm⁻¹ and 1000 cm⁻¹

To indicate the presence of a C=C double bond in trans-2-butene despite the absence of a distinct signal in the double bond region of the infrared spectrum (1600–1850 cm−1), one would look for other characteristic IR absorption signals. These include the C-H stretching vibrations associated with alkenes, found around 3100 cm⁻¹ for the =CH2 group and around 3020 cm⁻¹ for the –CH= group. Additionally, the out-of-plane bending vibrations of the C-H bonds at the terminal =CH2 group occur near 915 cm⁻¹ and those of the –CH= grouping near 1000 cm⁻¹. These peaks would be indicative of the alkenic C-H bonds as opposed to alkane C-H stretches, which occur at slightly different frequencies.

0.499 mg will remain after 13.0 days. So, the correct answer is (E).

The time which is required for a quantity to reduce to half of its initial value is defined as Half life. This is commonly used in nuclear physics to for the description of unstable atoms undergo radioactive decay or how much time stable atoms survive.

Formula for Half Life,

t1/2 = ln 2 / λ

where,

t1/2 is half-life

λ is radioactive decay constant

For given information,

here, t1/2 = 8.04 days

so, 8.04 days    = ln 2 / λ                        

λ  = ln 2 / 8.04 days

As per Radioactive decay law,

[tex]N_t = N_o[/tex]e∧(-λt)

where,

[tex]N_t[/tex] is amount of compound at t time,

[tex]N_o[/tex] is amount of compound at  t = 0 time

t is time taken to decay

λ is radioactive decay constant.

[tex]N_o[/tex] = 1.53 mg

λ  = ln 2 / 8.04 days = 0.693 / 8.04 days

t  = 13.0 days

By substituting,

[tex]N_t[/tex]= 1.53 mg e∧{(-0.693/8.04 days) x 13.0 days)}

[tex]N_t[/tex] = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

Thus, 0.499 mg will remain after 13.0 days. So, the correct answer is (E).

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Answer:

[tex]m_{Na_2CO_3}=2.0gNa_2CO_3[/tex]

Explanation:

Hello,

In this case, we consider the equation defining molarity, in order to compute the mass that is present into 125 mL of an aqueous 0.15M of sodium carbonate (solute), Na₂CO₃ as shown below:

[tex]M=\frac{n_{solute}}{V_{solution}}[/tex]

Now, since the unknown is the mass which comes from the moles, by solving for it and subsequently using its molar mass, one obtains:

[tex]m_{Na_2CO_3}=125mL*\frac{1L}{1000mL}*0.15\frac{mol{Na_2CO_3}}{L}*\frac{105.98gNa_2CO_3}{1mol{Na_2CO_3}}   \\m_{Na_2CO_3}=2.0gNa_2CO_3[/tex]

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The ability to attract an electron for bonding is called Electronegativity. Hence, option C is correct.

An atom consists of a central nucleus that is usually surrounded by one or more electrons.

Electronegativity is a measure of an atom's ability to attract shared electrons to itself.

The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity.

Hence, option C is correct.

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Answer:

Aluminium.

Explanation:

Hello,

In this case, the most appropriated metal turns out being the Aluminum since  it is higher in the activity series than nickel. Don't forget that activity series is an empirical tool used to predict products in displacement reactions and reactivity of metals with water and acids in replacement reactions and ore extraction.

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