What is the standard gibbs free energy of formation, ∆gºf, of nh3(g) at 298 k?

The emf of the electrochemical cell has been calculated to be 0.0532 V.

The emf has been the potential of the cell in the reaction with the change in the electrons in the reaction. The emf of the cell has been given by the Nernst equation as;

[tex]emf=E^\circ _{cell}-\dfrac{0.059}{n}\;log\;\dfrac{1}{\rm concentration} [/tex]

The given cell has the number of electrons transfer, [tex]n=2[/tex]

The concentration of [tex]\rm Pb^2^+=0.83\;M[/tex]

The concentration of [tex]\rm H^+=0.064\;M[/tex]

The cell potential of the reaction has been, [tex]E^\circ =-0.126\;\text V[/tex]

Substituting the values for the emf of the cell:

[tex]emf=-0.126\;-\dfrac{0.059}{2}\;\times\;log\;\dfrac{1}{[0.83]\;\times\;[0.064]^2}\\ emf=0.0532\;\text V [/tex]

The emf of the electrochemical cell has been calculated to be 0.0532 V.

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Answer:

Magnesium hydroxide (Mg(OH)2): 58.33 g/mol

Iron(III) oxide (Fe2O3): 159.70 g/mol

Explanation:

To find the molar masses of both magnesium hydroxide and iron(III) oxide, add each  molar mass in each compound.

Solving:

[tex]\section*{Molar Mass of Magnesium Hydroxide (Mg(OH)_2):}\textbf{Identify Atomic Masses}\begin{itemize} \item Magnesium (Mg): 24.31 g/mol \item Oxygen (O): 16.00 g/mol (2 oxygen atoms) \item Hydrogen (H): 1.01 g/mol (2 hydrogen atoms)\end{itemize}[/tex]

[tex]\textbf{Calculate Molar Mass:}\[\text{Molar mass of Mg(OH)}_2 = \text{Mg} + 2 \times (\text{O} + \text{H})\]\[= 24.31 \, \text{g/mol} + 2 \times (16.00 \, \text{g/mol} + 1.01 \, \text{g/mol})\]\\\[= 24.31 \, \text{g/mol} + 2 \times 17.01 \, \text{g/mol}\]\[= 24.31 \, \text{g/mol} + 34.02 \, \text{g/mol} = \boxed{58.33 \, \text{g/mol}}\][/tex]

[tex]\hrulefill[/tex]

[tex]\section*{Molar Mass of Iron(III) Oxide (Fe_2\text{O}_3):}\textbf{Identify Atomic Masses}\begin{itemize} \item Iron (Fe): 55.85 g/mol (2 iron atoms) \item Oxygen (O): 16.00 g/mol (3 oxygen atoms)\end{itemize}[/tex]

[tex]\textbf{Calculate Molar Mass:}\[\text{Molar mass of Fe}_2\text{O}_3 = 2 \times \text{Fe} + 3 \times \text{O}\]\\\[= 2 \times 55.85 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol}\]\\\[= 111.70 \, \text{g/mol} + 48.00 \, \text{g/mol} = \boxed{159.70 \, \text{g/mol}}\][/tex]

[tex]\hrulefill[/tex]

The answer options about a nuclear reaction which are correct include the following:

A. Nuclear fission and fusion both affect the nucleus of an atom.

B. The final products of fission and fusion are elements that are different than the original.

D. Fission occurs mostly with elements heavier than lead on the periodic table.

A nuclear reaction refers to a reaction in which the nucleus of an atom is transformed or transmuted by either joining (fusion) or splitting (fission) the nucleus of another atom of a radioactive element. Also, a nuclear reaction is always accompanied by a release of energy.

Generally, the two (2) main types of nuclear reaction are:

From the above, we can deduce the following about a nuclear reaction:

I. The nucleus of an atom is affected by both nuclear fission and fusion.

II. The final products of both nuclear fission and fusion are chemical elements that are different than the original radioactive chemical elements.

III. Nuclear fission occurs mostly with chemical elements heavier than lead (Pb) on the periodic table because the nuclear force holding their atoms together is lesser than the electromagnetic force pushing the nucleus apart.

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The oxidation state of the Ru metal in the complex [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex] is [tex]\boxed{{\text{zero}}}[/tex] .

Further explanation:

Oxidation number:

The oxidation number is used to represent the formal charge on an atom. It also shows that gain or loss of electrons by the atom. Oxidation number can be a positive or negative number but cannot be fractional.

The rules to identify the oxidation state:

(1) The oxidation state of an atom in the elemental form is zero.

(2) The total charge on the species is equal to the sum of the oxidation state of individual atoms.

(3) The oxidation state of halides is -1. For example, fluorine, chlorine, bromine, iodine have -1 oxidation state.

(4)Hydrogen has a +1 oxidation state.

(5) Oxygen has an oxidation state -2.

(6) In a coordination compound, neutral ligands have zero oxidation state and negative ligands such as CN have -1 oxidation.

The given compound is [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex]

Here, CN is a negative ligand thus oxidation state is -1 and CO is a neutral ligand thus it has 0 oxidation state. Also, the complex has -1 negative charge.

The expression to calculate the oxidation state in [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex] is,

[tex]\left[{\left({{\text{oxidation state of Ru}}}\right)+\left({{\text{oxidation state of CN}}}\right)+4\left({{\text{oxidation state of CO}}}\right)}\right]=-1[/tex]

…… (1)

Rearrange equation (1) for the oxidation state of Ru.

[tex]{\text{Oxidation state of Ru}}=\left[{-\left({{\text{oxidation state of CN}}}\right)-4\left({{\text{oxidation state of CO}}}\right)-1}\right][/tex]

…… (2)  

Substitute -1for the oxidation [tex]{\text{state}}[/tex]  of CN and 0 for the oxidation state of CO in equation (2).

[tex]\begin{aligned}{\text{Oxidation state of Ru}}&=\left[{-\left({-{\text{1}}}\right)-4\left({\text{0}}\right)-1}\right]\\&=0\\\end{aligned}[/tex]

The oxidation state of Ru is zero.

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Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Coordination complex

Keywords: Oxidation state, metal complex, ru(cn)(co)4-, formal charge, cynide, carbonyl, zero, hydrogen, oxygen, 0 and -1.

Answer : There is no change takes place in the oxidation number of Cl.

Explanation :

The given chemical reaction is,

[tex]AlCl_3+Na\rightarrow NaCl+Al[/tex]

This reaction is an unbalanced reaction because the chlorine atoms are not balanced.

In order to balance the chemical reaction, the coefficient 3 is put before the Na and NaCl.

The balanced chemical reaction will be,

[tex]AlCl_3+3Na\rightarrow 3NaCl+Al[/tex]

Now we have to calculate the oxidation number of all the elements.

In [tex]AlCl_3[/tex], the oxidation number of Al and Cl are, (+3) and (-1) respectively.

The oxidation number Na and Al are, zero (0)

In [tex]NaCl[/tex], the oxidation number of Na and Cl are, (+1) and (-1) respectively.

From this we conclude that the oxidation number of Cl changes from (-1) to (-1) that means remains same.

Therefore, there is no change takes place in the oxidation number of Cl.

Answer:

Magnetic fields.

Explanation:

Hello,

The presence of a magnetic fields best explains why a permanent magnet can act on another magnetic objects when are separated by a certain distance. There exist magnetic field lines which are emanated from the north pole to the south pole whose path is curved. Now, take into account that the longer the distance between the objects, the lower the magnetic field or force.

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Explanation:

Molecular compounds :  Molecular compounds are formed by the sharing of electrons with each elements.

Ionic compound : When compounds are formed by the transfer of electrons, they are called Ionic compounds.

Atomic species : Species comprised of only atom.

Co2 -- Molecular

C --     Atomic

CaCl2  -- Ionic

C6H12O6 --- Molecular

KBr  ---  Ionic

PbS  ---- Ionic

The above compounds can be classified as;

CO₂- Molecular compound

C - Atomic compound

CaCl₂ - ionic compound

C₆H₁₂O₆ - Molecular compound

KBr - Ionic compound

PbS - Ionic compound

Molecular compounds

Atomic covalent compounds

Keywords: Compound, types of compounds, ionic compounds, molecular compounds, atomic compounds

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Level : High school

Subject: Chemistry

Topic: Structure and bonding

Sub-topic: Types of structures and compounds

Answer: The concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):  [tex]Zn(s)\rightarrow Zn^{2+}(0.100M,aq.)+2e^-[/tex]

Reduction half reaction (cathode):  [tex]Zn^{2+}(?M,aq.)+2e^-\rightarrow Zn(s)[/tex]

In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]_{anode}}{[Zn^{2+}]_{cathode}}[/tex]

where,

n = number of electrons in oxidation-reduction reaction = 2

[tex]E_{cell}[/tex] = 14.0 mV = 0.014 V    (Conversion factor:  1 V = 1000 mV)

[tex][Zn^{2+}]_{anode}[/tex] = 0.100 M

[tex][Zn^{2+}]_{cathode}[/tex] = ? M

Putting values in above equation, we get:

[tex]0.014=0-\frac{0.0592}{2}\log \frac{0.100M}{[Zn^{2+}]_{cathode}}[/tex]

[tex][Zn^{2+}]_{cathode}=0.295M[/tex]

Hence, the concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M

The concentration of Zn²⁺  ion at cathode is 0.295 M

The redox half equations are those of oxidation and reduction

Oxidation half reaction: Zn(s) ---> Zn²⁺ + 2 e⁻ [Zn⁺] = 0.100 M

reduction half reaction: Zn²⁺ (aq) + 2 e⁻ ----> Zn (s)  [Zn⁺] = y

The Ecell = 14.0mV = 0.014 V

Using the Nernst equation:

where n is number moles; n = 2

0.014 =  0 - 0.0592/2 * log (0.1/y)

y = 0.295 M

Therefore, the concentration of Zn²⁺  ion at cathode is 0.295 M

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In a 45.0 mg sample of vanillin, there are approximately 5.36 x 10^20 oxygen atoms. The calculation involves converting the mass of the sample to grams, calculating the number of moles, and using Avogadro's number to determine the number of oxygen atoms.

The question is asking how many oxygen atoms are present in a 45.0 mg sample of vanillin, which is a molecule responsible for the vanilla flavor in food. The molecular formula of vanillin is C8H8O3 and its molar mass is 152 g/mol.

First, we have to convert the mass of the sample from milligrams (mg) to grams (g) because the molar mass is in grams. Therefore, 45.0 mg equals 0.045 g.

Next, we calculate the number of moles of vanillin in the sample using the equation:

Number of moles = Mass / Molar mass

Substituting the given values:

Number of moles = 0.045 g / 152 g/mol = 2.96 x 10^-4 moles.

Each molecule of vanillin contains 3 oxygen atoms. Therefore, in one mole of vanillin, there are 3 moles of oxygen atoms. From the concept of Avogadro's number, we know that one mole of any substance contains 6.02 x 10^23 entities (atoms, molecules, ions etc.).

Therefore, in 2.96 x 10^-4 moles of vanillin, we have (3 moles O atoms / mole of vanillin) x (2.96 x 10^-4 moles of vanillin) x (6.02 x 10^23 O atoms / mole of O atoms) = 5.36 x 10^20 oxygen atoms.

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The molarity of a solution is calculated by the formula: Molarity = moles of solute divided by volume of solution in liters. For the mentioned solution, we have roughly 1 mole of sodium chloride. The molarity can be obtained by dividing this number by the volume of the solution in liters.

The molarity of a solution is given by the formula: Molarity (M) = moles of solute / volume of solution in liters.

Via the provided information, we know that one mole of sodium chloride (NaCl) weighs 58.44 g, and we have 58.4 g NaCl in our solution, roughly equivalent to 1 mole. The volume of the solution unfortunately is not given in the question. If this missing value can be retrieved, the molarity can be calculated by dividing the number of moles (which is 1 in this case) by the volume of the solution in liters.

Assuming a 1 L solution for simplicity, the molarity will be: 1 mol NaCl / 1 L solution = 1 M NaCl solution.

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The electron dot structure for Cl is B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1).

The electron dot structure for Cl is B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1). The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons.

To calculate the percent yield of silver, divide the experimental yield (2.55 g) by the theoretical yield (2.78 g) and multiply by 100, resulting in a percent yield of 91.73%.

The concept of percent yield is a key aspect of laboratory work in chemistry that compares what was actually obtained from a reaction to what could be obtained based on stoichiometry. The percent yield is calculated using a simple formula: (actual yield/theoretical yield)  imes 100. In the student's experiment where the theoretical yield was 2.78 grams of silver and the experimental yield was 2.55 grams, we calculate the percent yield by dividing 2.55 by 2.78 and then multiplying by 100.

Therefore, the calculation is:
(2.55 g / 2.78 g) imes 100 = 91.73%

So, the percent yield for the reaction is 91.73%.

Explanation:

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

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

It is given that no. of moles present into the solution are 0.175 mol and volume is 150 ml.

As 1 ml = 0.001 L. So, 150 ml will be equal to 0.15 L.

Hence, calculate the molarity as follows.

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

                  = [tex]\frac{0.175 mol}{0.15 L}[/tex]                

                  = 1.16 M

Thus, we can conclude that molarity of the given solution is 1.16 M.

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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The amount of heat needed to boil 5.25 grams of ice when we take into account three changes as melting, heating and vaporizing the water is  15797.93 J.

Total heat for the given condition will be calculated by the addition of the heat of fusion, heat of vaporization and specific heat of water.

In the question it is given that,

mass of ice = 5.25 grams

Change in temperature = 100 - 0 = 100 degree celsius

We know that heat of fusion of water = 334 J/g

Required heat for the melting of ice = 334 × 5.25 = 1753.5 J

Amount of heat will be used by using the formula as,

Q = mcΔT, where

c = specific heat of water = 4.18 J/gK

Required heat for heating of water = 4.18 × 5.25 × 100 = 2195.18 J

We know that heat of vaporization of water = 2257 J/g

Required heat for vaporization of ice = 2257 × 5.25 = 11849.25 J

Total amount of heat involved = 1753.5 + 2195.18 + 11849.25 = 15797.93 J

Hence total amount of heat is 15797.93 J.

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Answer:

Right

Left

Left

Right

Explanation:

For the equilibrium system described by this equation, what will happen if SO3 is removed?

The equilibrium shifts to the

✔ right

What will happen if NO is added?

The equilibrium shifts to the  

✔ left

.For the equilibrium system described by this equation, what will happen if SO2 is removed?

The equilibrium shifts to the  

✔ left

.

What will happen if NO2 is added?

The equilibrium shifts to the  

✔ right

To calculate the volume of the gas sample at standard temperature and pressure (STP), we find the moles of each gas (CH4 and He) using their molar masses and then multiply by the molar volume (22.4 L/mol) to get their individual volumes at STP. Adding these volumes together gives us the total volume of the gas sample, which is 16.8 L.

To calculate the volume of a gas sample at standard temperature and pressure (STP), we can use the molar volume concept where one mole of any gas occupies 22.4 liters at STP. Methane (CH4) has a molar mass of 16.00 g/mol, and helium (He) has a molar mass of 4.00 g/mol.

To find the moles of CH4, we divide 4.00 g by its molar mass, 16.00 g/mol, which gives us 0.250 moles. Similarly, for He, we divide 2.00 g by its molar mass, 4.00 g/mol, which gives us 0.500 moles. To find the total volume, we sum the volumes of CH4 and He after multiplying their mole quantities by the molar volume (22.4 L/mol).

Total volume of CH4 = 0.250 moles × 22.4 L/mol = 5.60 L

Total volume of He = 0.500 moles × 22.4 L/mol = 11.2 L

Therefore, the total volume of the gas sample at STP is 5.60 L + 11.2 L = 16.8 L.