What is the emf of a cell consisting of a pb2+ / pb half-cell and a pt / h+ / h2 half-cell if [pb2+] = 0.83 m, [h+] = 0.064 m and ph2 = 1.0 atm ?

The energy required to ionize a hydrogen atom in the n = 4 state is 0.85 eV.

The energy required to ionize a hydrogen atom when it is in the n = 4 state can be calculated using the energy levels of the hydrogen atom. The energy for any state n is given by En = -13.6 eV / n2, where 13.6 eV is the ionization energy of the hydrogen atom from its ground state. To find the ionization energy from n = 4, we substitute 4 for n.

E4 = -13.6 eV / 42

E4 = -13.6 eV / 16 = -0.85 eV

The negative sign indicates that this is the energy the electron has relative to the energy of a free electron at rest (which is defined as 0 eV). To ionize the atom, we need to provide enough energy to bring the electron's energy up to 0 eV. Therefore, the ionization energy required is the absolute value of E4.

Ionization Energy = |E4| = 0.85 eV

Answer : The pressure of the gas is, 8.42 atm

Explanation :

Using ideal gas equation,

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = ?

V = volume of the gas = 748 ml = 0.748 L

conversion used : (1 L = 1000 ml)

T = temperature of the gas = [tex]28^oC+273+28=301K[/tex]

n = number of moles of the gas = 0.255 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get  the pressure of the gas.

[tex]P\times (0.748L)=0.255mole\times (0.0821L.atm/mole.K)\times (301K)[/tex]

[tex]P=8.42atm[/tex]

Therefore, the pressure of the gas is, 8.42 atm

Answer:

[tex]n_{HCl}=1.906x10^{-3}molHCl[/tex]

Explanation:

Hello,

Titration is widely used to determine the neutralized moles of either an acid or base. In this case, the idea is to titrate (neutralize) hydrochloric acid with sodium hydroxide based on the following reaction:

[tex]NaOH+HCl-->NaCl+H_2O[/tex]

Thus, one computes the neutralized moles of hydrochloric acid (equivalence of moles) as long as the mole ratio between the acid and the base is 1 to 1 and the moles of both of them must be equal for the neutralization to be successfully carried out as shown below:

[tex]n_{HCl}=n_{NaOH}\\n_{HCl}=0.1250mol/L*0.01525L\\n_{HCl}=1.906x10^{-3}molHCl[/tex]

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Explanation:

Molarity is the number of moles present in a liter of solution.

Mathematically,     Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]

And, no. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

Molar mass of [tex]C_{6}H_{12}O_{6}[/tex] is 180.15 g/mol. Therefore, number of moles present will be as follows.

                No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]                                                                                             = [tex]\frac{28.33 g}{180.15 g/mol}[/tex]

                                      = 0.157 mol

Hence, calculate the molarity as follows.

                Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]

                       = [tex]\frac{0.157 mol}{1.28 L}[/tex]    

                       = 0.122 M

Thus, we can conclude that molarity of the solution is 0.122 M.

Answer : The balanced oxidation half reaction in basic medium will be :

[tex]Mn^{2+}(aq)+8OH^-(aq)\rightarrow MnO_4^-(aq)+4H_2O(l)+5e^-[/tex]

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Rules for the balanced chemical equation in basic solution are :

The balanced oxidation half reaction in basic medium will be :

[tex]Mn^{2+}(aq)+8OH^-(aq)\rightarrow MnO_4^-(aq)+4H_2O(l)+5e^-[/tex]

Answer : The boiling point of a solution is, [tex]100.42^oC[/tex]

Explanation :

Formula used for Elevation in boiling point :

[tex]\Delta T_b=k_b\times m[/tex]

or,

[tex]T_b-T^o_b=\frac{1000\times k_b\times w_2}{w_1\times M_2}[/tex]

where,

[tex]T_b[/tex] = boiling point of solution = ?

[tex]T^o_b[/tex] = boiling point of pure water = [tex]100^oC[/tex]

[tex]k_b[/tex] = boiling point constant  for water = [tex]0.512^oC/m[/tex]

m = molality

[tex]w_2[/tex] = mass of solute (sucrose) = 4.27 g

[tex]w_1[/tex] = mass of solvent (water) = 15.2 g

[tex]M_2[/tex] = molar mass of solute (sucrose) = 342.3 g/mole

Now put all the given values in the above formula, we get the boiling point of a solution.

[tex]T_b-100^oC=\frac{1000\times 0.512^oC/m\times 4.27g}{15.2g\times 342.3g/mole}[/tex]

[tex]T_b=100.42^oC[/tex]

Therefore, the boiling point of a solution is, [tex]100.42^oC[/tex]

The temperature of the solution is 100.42°C.

Given that;

ΔT = K m i

Where;

ΔT = boiling point elevation

K = boiling point constant for water

m = molality of the solution

i = Van't Hoff factor

Number of moles of solute = 4.27 g /342 g/mol = 0.0125 moles

Molality of the solution = 0.0125 moles/15.2 × 10^-3 Kg = 0.822 m

Since the boiling point of pure water = 100°C

Let the boiling point of pure water be Ta

Let the boiling point of the solution be Tb

ΔT = Tb - Ta = Tb - 100

Substituting values;

Tb - 100 = 0.512c/m × 0.822 m × 1

Note that the Van't Hoff factor (i) = 1 because the solute is molecular

Tb = [0.512°C/m × 0.822 m × 1] + 100°C

Tb = 100.42°C

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To create 400 grams of a 2% mass/mass glucose solution, you need 8 grams of glucose.

You're trying to find out how many grams of glucose are needed to prepare a 400 gram 2% mass/mass glucose solution. A 2% w/w glucose solution means that for every 100 grams of solution, 2 grams are glucose. Therefore, if you have 400 grams of solution, the amount of glucose required will be 2% of 400 grams.

To calculate this, you will multiply 400 grams by 0.02 (which is the decimal equivalent of 2%). So, 400 grams * 0.02 = 8 grams. Therefore, you need 8 grams of glucose to prepare 400 grams of a 2% w/w glucose solution.

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Final answer:

The mass of water produced when 10.00 g of copper is formed by the reduction of Cu2O with hydrogen gas is 1.4131 g. This is calculated using stoichiometry and the molar mass of water.

Explanation:

The mass of water produced when 10.00 g of copper is formed from the reduction of copper(I) oxide (Cu2O) by hydrogen gas (H2) can be determined by using stoichiometry. The reaction is as follows:

Cu2O(s) + H2(g) → 2 Cu(s) + H2O(g)

First, calculate the moles of copper produced using its molar mass. Since the copper produced is 10.00 g, and the molar mass of copper is approximately 63.55 g/mol, the moles of copper formed are:

10.00 g Cu × (1 mol Cu / 63.55 g Cu) = 0.157 mol Cu

According to the reaction, 1 mole of Cu2O produces 2 moles of Cu, so we have:

0.157 mol Cu × (1 mol Cu2O / 2 mol Cu) = 0.0785 mol Cu2O

For each mole of Cu2O reduced, 1 mole of water (H2O) is produced:

0.0785 mol Cu2O × (1 mol H2O / 1 mol Cu2O) = 0.0785 mol H2O

Finally, to find the mass of water produced:

0.0785 mol H2O × (18.015 g H2O / 1 mol H2O) = 1.4131 g H2O

Therefore, the mass of water produced is 1.4131 g.

The property from the provided options that increases down the periodic table is the atomic radius, due to the addition of electron shells as we move down groups.

In the periodic table, as we move down a group, there are certain properties that increase, and others that decrease. Among the options provided: the number of valence electrons, electronegativity, ionization energy, and atomic radius, the property that increases down the periodic table is the atomic radius.

This is because as we go down each group in the periodic table, there's an additional electron shell added to the atoms. So, even though the positive charge in the nucleus also increases, it's largely screened or shielded by the inner-shell electrons from interacting with the outer-shell or valence electrons. The result is that the atomic size or radius increases.

On the other hand, electronegativity and ionization energy generally decrease down a group because the increasing atomic radius means the outer electrons are less tightly bound to the nucleus.

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