A solution of ammonia and water contains 2.60×1025 water molecules and 6.90×1024 ammonia molecules. how many total hydrogen atoms are in this solution

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]

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

Learn more: https://brainly.com/question/15178305

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

Learn more about:

Level : High school

Subject: Chemistry

Topic: Structure and bonding

Sub-topic: Types of structures and compounds

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.

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.

To know more about heat absorbed, visit the below link:
https://brainly.com/question/488376

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.

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

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.

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.

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 : 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:

[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]

Best regards.

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.

https://brainly.com/question/38723490

#SPJ2

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.

https://brainly.com/question/10070220

#SPJ3

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 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%.

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

Learn more about concentration cells at: https://brainly.com/question/15394851

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.

Learn more about emf of the cell, here:

https://brainly.com/question/10162133