How do the valence electron configurations of alkali metals compared with each other?

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

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: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 :

Latent heat : Latent heat is defined as the amount of heat required to convert the solid into liquid or vapor and a liquid into a vapor without changing the temperature.

As there are two types of latent heat which are :

(1) Latent heat of fusion

(2) Latent heat of vaporization

Latent heat of fusion : Latent heat of fusion is defined as the amount of heat energy released or absorbed when the solid converted to liquid at atmospheric pressure at its melting point.

Latent heat of vaporization : Latent heat of vaporization is defined as the amount of heat energy released or absorbed when the liquid converted to vapor at atmospheric pressure at its boiling point.

Answer: The given statement is true.

Explanation:

A pure substance is defined as the substance which is formed by same or different type of atoms which are chemically combined in a fixed ratio by mass.

Compounds and elements are considered as pure substances.

For Example:  Sodium and sodium chloride are both considered as pure substances because sodium is entirely made up of sodium atoms and sodium chloride is entirely made up of sodium chloride molecules.

Hence, the given statement is true.

B is the answer, hope this helps :D

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).

The percent by mass of water in iron (II) sulfate heptahydrate, FeSO4·7H2O, is 51.00%.

To calculate the percent by mass of water in iron (II) sulfate heptahydrate, FeSO4·7H2O, we need to determine the molar mass of the compound and the molar mass of water. The molar mass of iron (II) sulfate heptahydrate is calculated as follows:

Next, we determine the molar mass of the waters of crystallization:

Finally, we calculate the percent by mass of water in iron (II) sulfate heptahydrate:

(126.14 g/mol / 247.14 g/mol) x 100% = 51.00%

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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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

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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Without the exact heat of combustion per gram of octane (C8H18), it is impossible to calculate the specific number of kilocalories released by the combustion of 18.5 g of octane. The process would involve multiplying the heat of combustion per gram by the amount of substance burned and converting the energy from kilojoules to kilocalories if necessary.

The student is asking about the enthalpy change associated with the combustion of octane (C8H18). To answer this, we first need the heat of combustion per gram or per mole of octane, which can be found in a given table or chemistry reference. However, since this value is not provided in the information given, we can't calculate the specific number of kilocalories released without it.

To find the heat released by combustion, once the heat of combustion per gram is known, you would multiply this value by the mass of the substance burned (in this case, 18.5 g of octane). Remember to convert kilojoules to kilocalories if necessary, as there are 0.239006 kilocalories in one kilojoule.

For example, if we knew that the combustion of 1 gram of octane released X kJ, then the heat released by burning 18.5 grams would be 18.5 * X kJ, which could then be converted to kilocalories.

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Option (a) 3H⁺(aq) + Al(OH)₃(s) → Al³⁺(aq) + 3H₂O(l) is the net ionic equation for the reaction between nitric acid and aluminum hydroxide. This equation represents the primary species undergoing chemical change. It excludes spectator ions to show only the substances involved in the actual chemical reaction.

Among the given options, the correct net ionic equation is:

(a) 3H⁺(aq) + Al(OH)₃ (s) → Al³⁺(aq) + 3H₂O(l)

Here is the explanation:

Thus, option (a) is the net ionic equation for this reaction.

Complete Question:

Which one of the following is the net ionic equation for the reaction of nitric acid with aluminum hydroxide?

(a) 3H⁺(aq) + Al(OH)₃ (s) → Al³⁺(aq) + 3H₂O(l)

(b) 3HNO₃ (aq) + Al(OH)₃ (s) → 3Al(NO₃)₃ (aq) + 3H₂O(C)

(c) 2H⁺(aq) + Al(OH)₂ (s) → Al²⁺ (aq) + 2H₂O(l)

(d) H⁺(aq) + OH⁻(aq) → H₂O(l)

(e) 3NO₃⁻(aq) + Al³⁺(aq) → Al(NO₃)₃ (aq)

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.

The formula we will use is:

pH = 0.5 (pKa – log C)

where C is the concentration of sulfathiazole in molarity or mol / L

Since all values are given, we can compute directly for pH:

pH = 0.5 (7.12 – log 0.002)

pH = 4.91

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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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.

Answer: freezing point

Explanation:

Chemical property is defined as the property of a substance which is observed during a reaction where the chemical composition identity of the substance gets changed.

Physical property is defined as the property which can be measured and whose value describes the state of physical system. For Example: State, density etc.

Flammability is a chemical property as it alters the chemical composition of substance.

pH is a chemical property as it alters the chemical composition of substance.

Toxicity is a chemical property as it alters the chemical composition of substance.

Freezing point is a physical property as it deals only with phase changes.