Which of the following is a homogeneous mixture of two or more pure substances? Water alcohol coffee mercury

Answer: The periodic elements are arranged by their atomic number.

There are 3.7 moles in 100.0 g of aluminum

HOW TO CALCULATE NUMBER OF MOLES:

no. of moles = mass ÷ molar mass

Molar mass of Aluminum = 27g/mol

no. of moles = 100g ÷ 27g/mol

no of moles = 3.703mol

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Answer:

D

Explanation:

Bromine

Answer:

1.3 mol/kg is the molality of a solution.

Explanation:

[tex]Molality=\frac{\text{moles of solute}}{\text{Mass of the solvent (kg)}}[/tex]

Moles of solute = 4.0 moles

Mass of the solvent = 3.2 kg

[tex]Molality=\frac{4.0 mol}{3.2 kg}=1.25 mol/kg\approx 1.3 mol/kg[/tex]

1.3 mol/kg is the molality of a solution containing 4.0 moles of solute in 3.2 kg of solvent.

Final answer:

Weighing CaSO₄ precipitate before it is completely dried will result in an overestimation of the moles and concentration of the analyte, known as a positive determinate error in gravimetric analysis.

Explanation:

If a precipitate of CaSO₄ was weighed before it was completely dried, the likely impact of this error on the calculated moles of CaSO₄ would be that the measured mass is higher than the true mass of the dry CaSO₄. This is because the precipitate would contain additional water, which contributes to the overall mass. Since the stoichiometric calculation of moles is based on the mass of the precipitate, having extra mass from the water will result in an overestimation of the moles of CaSO₄. Therefore, the calculated concentration of the analyte in the sample will also be higher than it should be. This type of error is termed a positive determinate error in gravimetric analysis and it affects not just the stoichiometry, but also the reported percentage purity or concentration of the analyte within the sample.

The inferred pressure and temperature at the boundary of Earth's stiffer mantle and outer core are very high, reaching up to several million atmospheres and exceeding 5000 degrees Celsius, respectively.

At the boundary between Earth's stiffer mantle and outer core, the pressure and temperature are very high due to the immense weight and heat of the overlying material. The pressure can reach up to several million atmospheres, while the temperature can exceed 5000 degrees Celsius. These extreme conditions are necessary to maintain the liquid state of the outer core and the solid state of the inner core.These extreme conditions are necessary to maintain the liquid state of the outer core and the solid state of the inner core.These extreme conditions are necessary to maintain the liquid state of the outer core and the solid state of the inner core.

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2Cu(NO3)2 ---------->    2CuO (s) +4NO2 (g) + O2(g)

9.378g=0.05moles   

no of moles = weight / MW = 9.378/187.56 = 0.05moles

as per the above reaaction 2moles of Cu(NO3)2 can produce 4moles of N2

                                                    0.05moles Cu(NO3)2 can produce (0.05*4)/2 = 0.1moles of N2

and 2moles of Cu(NO3)2 can produce 1moles of O2

       0.05moles Cu(NO3)2 can produce (0.05*1)/2 = 0.025moles of O2

Total moles of gas i.e., N2 and O2 =0.1+0.025 = 0.125moles

From PV = nRT

       V = nRT/ P = 0.125*0.0821*273 = 2.80166Lit option is correct

To determine the volume of gas produced, we need to calculate the moles of gas using the molar mass of Cu(NO3)2 and then use the ideal gas law equation to determine the volume. The answer is 2.24 liters of gas produced.

To determine the volume of gas produced, we need to first calculate the moles of gas using the molar mass of Cu(NO3)2. Then, we can use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Solving for V, we can determine the volume of gas produced.

First, calculate the moles of Cu(NO3)2 using the given mass and molar mass:

moles of Cu(NO3)2 = mass / molar mass
moles of Cu(NO3)2 = 9.378 g / 187.56 g/mol
moles of Cu(NO3)2 = 0.050 moles

Since the stoichiometry of the reaction tells us that 2 moles of Cu(NO3)2 produces 4 moles of NO2 gas, we can determine the moles of NO2 gas produced:

moles of NO2 gas = (0.050 moles of Cu(NO3)2)(4 moles of NO2 / 2 moles of Cu(NO3)2)
moles of NO2 gas = 0.100 moles

Now we can use the ideal gas law equation to determine the volume:

V = (nRT) / P
V = (0.100 moles)(0.0821 L·atm/mol·K)(273 K) / 1 atm
V ≈ 2.24 L

Therefore, the answer is 2.24 liters of gas produced.

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Density is an intensive property as it does not depend on the quantity of the substances Whereas mass and volume are extensive property The  mass of the glucose solution that fills a 1.00 l intravenous bottle is  1500g.

Density tells about the compactness of the substances, how much dense is the substances in other words. Object that is more denser than water they just sink in the water.

Mathematically,

Density = Mass of the glucose solution ÷volume of  intravenous bottle

Volume of the intravenous bottle =  1.00l=  1.00 l*1000=1000ml

Density of glucose solution = 1.15 g/ml

Substituting all the values in above formula, we get

Mass of the glucose solution = Density*volume of  intravenous bottle

Mass of the glucose solution =1.15 g/ml*1000ml

Mass of the glucose solution = 1500g

Therefore, the mass of the glucose solution that fills a 1.00 l intravenous bottle is  1500g.

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Krypton in the ground state has a total of 4 completely filled sublevels, which are 1s, 2s, 2p, and 3s.

In the ground state, krypton has a total of 4 completely filled sublevels. The electron configuration for krypton is: 1s²2s²2p⁶3s²3p⁶3d¹4s²4p⁶. The completely filled sublevels are 1s, 2s, 2p, and 3s.

Answer:

b. unicellular prokaryotes

Explanation:

Most of the bacteria are primitive organisms that is why they are not very advanced in terms of cellular organelles and body functions. They do not have high level of organization in their body like presence of tissues, organs, organ systems etc. because they are only single celled organisms that is why they are known as unicellular organisms. A single cell performs all the functions required to sustain life. Their genetic material is very primitive type of structure which does not have any nuclear membrane or nuclear envelop. It is circular in shape and lies in the cytoplasm and it is known as nucleoid instead of nucleus.

For example: E. coli.

The majority of bacteria are classified as unicellular prokaryotes, meaning they are composed of a single cell and lack a nucleated cell structure.

Most bacteria are unicellular prokaryotes. Unicellular means that they are organisms composed of a single cell. Prokaryotes refer to organisms that do not have a nucleus and other specialized organelles in their cells, which is common in bacteria. One important distinction of prokaryotes, including bacteria, is that their DNA is not stored in a nucleus, unlike eukaryotes. In contrast, eukaryotes, whether unicellular or multicellular, have a nucleus to house their DNA in addition to other specialized organelles.

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Final answer:

The molarity of the HCl solution is 0.1174 M.

Explanation:

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

M1V1 = M2V2

Using the values given in the question:

M1 = Molarity of HCl solution = unknown

V1 = Volume of HCl solution = 25.00 mL = 0.025 L

M2 = Molarity of NaOH solution = 8.670x10^-2 M

V2 = Volume of NaOH solution = 34.30 mL = 0.0343 L

Plugging in these values into the equation, we get:

M1(0.025) = (8.670x10^-2)(0.0343)

M1 = (8.670x10^-2)(0.0343) / 0.025

M1 = 0.1174 M

B is the answer, hope this helps :D

Answer:Balanced half-reaction for the reduction of gaseous nitrogen to aqueous hydrazine is :

[tex]N_2(g)+4H^+(aq)+4e^-\rightarrow N_2H_4(aq)[/tex]

Explanation:

Half reaction will be written as:

[tex]N_2(g)\rightarrow N_2H_4(g)[/tex]

Balance the atoms in half reaction other than oxygen and hydrogen.

In acidic medium the hydrogen atom is balanced by adding [tex]H^+[/tex] ions.

[tex]N_2(g)+4H^+(aq)\rightarrow N_2H_4(aq)[/tex]

Now balance the charge on both the sides by adding electrons on that side where positive charge is greater.

[tex]N_2(g)+4H^+(aq)+4e^-\rightarrow N_2H_4(aq)[/tex]

Balanced half-reaction for the reduction of gaseous nitrogen to aqueous hydrazine is :

[tex]N_2(g)+4H^+(aq)+4e^-\rightarrow N_2H_4(aq)[/tex]

Final answer:

The balanced half-reaction for the reduction of gaseous nitrogen (N2) to aqueous hydrazine (N2H4) in an acidic solution is N2(g) + 4 H+(aq) + 4 e- → N2H4(aq). This reaction demonstrates the balance of both the number of atoms and the overall charge in a redox process.

Explanation:

To balance the half-reaction for the reduction of gaseous nitrogen (N₂) to aqueous hydrazine (N₂H₄) in an acidic aqueous solution, we need to ensure that the number of nitrogen atoms and the overall charge is balanced on both sides of the reaction. The initial, unbalanced half-reaction is:

N₂(g) + H₂(g) → N₂H₄(aq)

First, we balance the nitrogen atoms by noting that there are still two nitrogen atoms present on both sides of the equation, so no change is needed here.

Then, to balance hydrogen atoms, we add 4 H⁺ ions to the right side:

N₂(g) + H⁺(aq) → N₂H₄(aq)

To balance the charge, we need to add electrons to the side with a positive charge. Since we have added four protons (4 H⁺ ions) on the right side, we must balance the charge by adding 4 electrons to the left side:

N₂(g) + 4 H⁺(aq) + 4 e⁻ → N₂H₄(aq)

All atoms and the charge are now balanced, making this the correct reduced half-reaction for the reduction of nitrogen to hydrazine in acidic solution.

Answer : The entropy change of the system is, 19.5 J/K

Solution :

Formula used :

[tex]\Delta S=\frac{n\times \Delta H_{vap}}{T_b}[/tex]

or,

[tex]\Delta S=\frac{\frac{w}{M}\times \Delta H_{vap}}{T_b}[/tex]

where,

[tex]\Delta S[/tex] = entropy change of the system = ?

[tex]\Delta H[/tex] = enthalpy of vaporization = 30.7 kJ/mole

n = number of moles of benzene

w = mass of benzene = 17.5 g

M = molar mass of benzene = 78 g/mole

[tex]T_b[/tex] = normal boiling point of benzene = [tex]80.1^oC=273+80.1=353.1K[/tex]

Now put all the given values in the above formula, we get the entropy change of the system.

[tex]\Delta S=\frac{\frac{17.5g}{78g/mole}\times (30.7KJ/mole)}{353.1K}=0.0195kJ/K=0.0195\times 1000=19.5J/K[/tex]

Therefore, the entropy change of the system is, 19.5 J/K

Final answer:

When aqueous sodium hydroxide is mixed with aqueous zinc(II) nitrate, crystallized zinc(II) hydroxide and aqueous sodium nitrate are formed. The balanced molecular equation for this precipitation reaction is Zn(NO₃)₂( aq) + 2NaOH(aq) → Zn(OH)₂(s) + 2NaNO₃(aq).

Explanation:

The question is about the chemical reaction where aqueous sodium hydroxide is mixed with aqueous zinc(II) nitrate, resulting in the formation of crystallized zinc(II) hydroxide as a precipitate. To find the other product of this precipitation reaction, we first write down the reactants and their formulas, which are NaOH(aq) for sodium hydroxide and (Zn(NO₃)₂(aq) for zinc nitrate. The reaction leads to the formation of zinc hydroxide, Zn(OH)₂, and sodium nitrate, NaNO₃, as products.

Writing the balanced molecular equation for this reaction:

\(Zn(NO₃)₂(aq) + 2NaOH(aq) \rightarrow Zn(OH)₂(s) + 2NaNO3(aq)\).

This equation shows the reactants on the left side and the products, including the solid zinc(II) hydroxide precipitate and aqueous sodium nitrate, on the right side. Note that it's crucial to balance the chemical equation to ensure the conservation of mass.

Final answer:

To calculate the grams of NiCO3 formed, the moles of Na2CO3 used are first calculated, then converted to moles of NiCO3 using the stoichiometry of the reaction, and finally, the mass of NiCO3 is found by multiplying its moles by its molar mass, resulting in 0.324 g of NiCO3.

Explanation:

To find out how many grams of NiCO3 will be formed, we first need to write the balanced chemical equation for the reaction between NiCl2 and Na2CO3.

However, the provided question does not specify the exact reaction formula; typically, such a reaction would produce NiCO3 and NaCl. Assuming a reaction of the form NiCl2 + Na2CO3 → NiCO3 + 2 NaCl, and that NiCl2 reacts completely with Na2CO3, the next step is to calculate moles of Na2CO3 since its volume and molarity are given (14.3 mL of 0.191 M).

Moles of Na2CO3 = Volume (L) * Molarity = 0.0143 L * 0.191 mol/L = 0.00273 mol. Since the reaction is in a 1:1 molar ratio, the moles of NiCO3 produced will also be 0.00273 mol. To find the mass of NiCO3 formed, multiply the moles by the molar mass of NiCO3 (approximately 118.7 g/mol).

Mass of NiCO3 = Moles * Molar mass = 0.00273 mol * 118.7 g/mol = 0.324 g (rounded to three significant figures).