16. The concentration of a solution of potassium hydroxide is determined by titration with nitric
acid. A 30.0 mL sample of KOH is neutralized by 42.7 mL of 0.498 M HNO3. What is the
concentration of the potassium hydroxide solution?

Answer:

d. the conjugate base.

Explanation:

The general reaction of a weak acid, HA, with a strong base YOH, is:

HA + YOH → A⁻ + H₂O + Y⁻

Where A⁻ is the conjugate base of the weak acid and Y⁻ usually is a strong electrolyte.

That means after he complete reaction you don't have weak acid nor strong base, just conjugate base that will be in equilibrium with water, thus (Strong electrolyte doesn't change pH:

A⁻ + H₂O ⇄ HA + OH⁻

As the equilibrium is producing OH⁻, the pH of the solution is being affected for the conjugate base

Right option:

Answer:

-195.8º < -191.5º < 100º

Explanation:

Water, or H20, starts boiling at 100ºC.

Nitrogen, or N2, starts boiling at -195.8ºC.

Carbon monoxide, or C0, starts boiling at -191.5ºC.

When we place these in order from decreasing boiling point:

-195.8º goes first, then -191.5º, and 100º goes last.

Answer:

therefore, N2, CO, H20

Decreasing boiling point

Explanation:

the bond existing in H2O is hydrogen bond

bond existing in N2 is covalent bond, force existing is dipole-dipole-interaction

bond existing in CO is covalent bond , force existing between is induced -dipole- induced dipole-interaction

hydrogen bond is the strongest , followed by dipole-dipole-interaction and induced -dipole- induced dipole-interaction

the stronger the bond , the higher the boiling point

therefore, N2, CO, H20

-------------------------------------->

Decreasing boiling point

Answer:

Explanation:

For pH of a buffer solution , the formula is

pH = pKa + log [ Base ] / [ conjugate acid ]

=   pKa + log [ NH₃ ] / [ NH₄⁺ ]

Ka = Kw / Kb

Kb for NH₄OH = 1.8 x 10⁻⁵

Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵

= 5.6 x 10⁻¹⁰

pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2

= 10 - log 5.6

= 9.25

Effect of addition of HCl

H⁺ of HCl will react with NH₃ to produce NH₄⁺

25 mL of .1 HCl = 2.5 mM of HCl

25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺

65 mL of .2 M NH₃ = 13 mM of NH₃

65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺

NH₃ + H⁺ = NH₄⁺

NH₄⁺ formed = 2.5 + 13  mM

15.5 mM of NH₄⁺

NH₃ = 13 mM

Concentration of NH₃ = 13 / 90

Concentration of NH₄⁺ = 15.5 / 90

pH of final buffer mixture

= 9.6 + log 13 / 15.5

= 9.25 - .076

= 9.174

The pH value  is mathematically given as

pH= -6.332.

Question Parameters:

the pH of a 0.20 M NH3/0.20 M NH4Cl buffer

the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.

Generally, the equation for the Chemical Reaction  is mathematically given as

HCl + NH3 --> NH4^+ + Cl^-

Therefore

pH= pka + log(13/14).

pH= -6.3 + log 0.93.

pH= -6.3+ (-0.032).

pH= -6.332.

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Answer:

HBr, CH3C(OH)2 and CH3COOH, Br-

Explanation:

The conjugate acid-base pairs acid reacts with base to form a conjugate acid and conjugate base.

Conjugate acid is formed when a bases receives a proton  (H+) and a conjugate base is formed when an acid losses a proton (H+).

From the given equation:

HBr, CH3C(OH)2 and CH3COOH, Br- are conjugate acid-base pair, where HBr is an acid and CH3C(OH)2 is a conjugate acid while CH3COOH and Br- is the conjugate base.

Answer:

7.38 g/cm³ is the density of the metal

Explanation:

In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).

To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.

Volume = a³

a = √8×r

(r = 198x10⁻¹²m)

a = 5.6x10⁻¹⁰ m

Volume = 1.756x10⁻²⁸ m³

1m = 100cm → 1m³ = (100cm)³:

1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =

There are 4 atoms of gold:

4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold

As 1 mole weighs 195.08g:

6.64x10⁻²⁴ moles of gold × (195.08g / mol) =

1.296x10⁻²¹g / 1.756x10⁻²² cm³ =

The density of the metal is 7.40 g/cm³

In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.

It is expressed by using the formula:

[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]

where;

For face-centered cubic FFC;

The edge length  [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]

[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]

[tex]\mathbf{a =560.0285 \ pm }[/tex]

a = 5.60 × 10⁻⁸ cm

Replacing it into the previous equation, we have:

[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]

[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]

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Answer:

[tex]V_2=1.17L[/tex]

Explanation:

Hello,

In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:

[tex]V_1C_1=V_2C_2[/tex]

Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:

[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]

Best regards.

Answer:

The boy must swim at an angle of 59°northwest to get to a position directly opposite his starting point.

Explanation:

To get to a point directly opposite his starting point, the boy must travel at an angle x, in a direction northwest of his starting point. The speed of the boy and the speed of the river current forms a right-angled triangle with an an opposite side of 1.8 m/a and a hypotenuse of 2.1 m/a having an angle x.

Sin x = opp/ hyp

Sin x = 1.8/2.1

x = sin⁻¹ (1.8/2.10

x = 58.99

x = 59°

Therefore, the boy must swim at an angle of 59° in the northwesterly direction to get to a position directly opposite his starting point.

Answer:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Explanation:

Hello,

In this case, since water and oil are immiscible due to the oil's nonpolarity and water's polarity, when mixed, the oil remains on the water since it is less dense than water. In such a way, for a plastic sunk in the oil and floating on the water (in middle of them) we can conclude that the plastic have a mid density, therefore, the required organization is:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Best regards.

Answer:

See explanation

Explanation:

First voltaic cell;

Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)

Anode;

Zinc

Cathode;

Copper

Oxidation half equation;

Zn(s)------> Zn^2+(aq) + 2e

Reduction half equation;

Cu^2+(aq) +2e -----> Cu(s)

Overall; Zn(s) + Cu^2+(aq) -----> Zn^2+(aq) + Cu(s)

E°cell = 0.34 -(-0.76) =1.1 V

Second voltaic cell;

Zn(s)|Zn^2+(aq)||Fe^2+(aq)|Fe(s)

Anode;

Zinc

Cathode;

Iron

Oxidation half equation;

Zn(s)------> Zn^2+(aq) + 2e

Reduction half equation;

Fe^2+(aq) +2e -----> Fe(s)

Overall; Zn(s) + Fe^2+(aq) -----> Zn^2+(aq) + Fe(s)

E°cell = (-0.44) -(-0.76) = 0.32 V

Third voltaic cell;

Fe(s)|Fe^2+(aq)||Cu^2+(aq)|Cu(s)

Anode;

Iron

Cathode;

Copper

Oxidation half equation;

Fe(s)------> Fe^2+(aq) + 2e

Reduction half equation;

Cu^2+(aq) +2e -----> Cu(s)

Overall; Fe(s) + Cu^2+(aq) -----> Fe^2+(aq) + Cu(s)

E°cell = 0.34 -(-0.44) = 0.78 V

Fourth voltaic cell

Cu(s)|Cu^2+(aq)||I2(aq)|C(s)|I^-(aq)

Anode;

Copper

Cathode;

Graphite rod

Oxidation half equation;

Cu(s)------> Cu^2+(aq) + 2e

Reduction half equation;

I2(aq) +2e -----> 2I^-(aq)

Overall; Cu(s) + I2(aq) -----> Cu^2+(aq) + 2I^-(aq)

E°cell = 0.54 -0.34 = 0.20 V

Answer:

Explanation:

In all calorimetric experiment , the calorimeter must be isolated from the surrounding . Otherwise the heat change in the experiment can not be determined with precision .

The reaction is endothermic . Hence, there is lowering of temperature due to absorption of heat in the reaction equal to ΔH°. The value of ΔH° can be calculated by measuring fall in the temperature of the content . The fall in the temperature will be less when heat is allowed to come from the surrounding . Less fall of temperature will result in less ΔH° to be calculated .

Hence in the given experiment , if the student neglects to put lid on the cup , the experiment will give less value of ΔH°.

Answer:

2HClO(aq) + Ba(OH)₂(aq) →  Ba(ClO)₂(aq) + 2H₂O(l)

Explanation:

The reaction corresponds to a neutralization reaction between an acid and a base, as follows:

2HClO(aq) + Ba(OH)₂(aq)  →  Ba(ClO)₂(aq) + 2H₂O(l)            

From the equation above we have that the acid HClO reacts with the base Ba(OH)₂ to obtain a salt Ba(ClO)₂ and water.

In the balanced reaction, we have that 2 moles of HClO react with 1 mol of Ba(OH)₂ to produce 1 mol of Ba(ClO)₂ and 2 moles of water.

I hope it helps you!    

Answer:

D. Reduction takes place at the cathode, which is negatively charged.

Explanation:

In an electrolytic cell there are two electrodes; the cathode and the anode. The anode is the positive electrode while the cathode is the negative electrode. Oxidation occurs at the anode while reduction occurs at the cathode.

At the anode, species give up electrons and become positively charged ions while at the cathode species accept electrons and become reduced.

Answer:

Check the Attachment.

Half-reactions:

Anode: (OXIDATION) Ni --> Ni2+ + 2e-

Cathode: (REDUCTION) Cu2+ +2e- --> Cu

Overall reaction: Ni + Cu2+ --> Ni2+ + Cu

Explanation:

Overall, reaction is basically Anode + Cathode, where electrons on both sides cancel out  (if not, you need to multiply the equation in a way you can cancel them out).

Hope this helps.

Answer:

The temperature at which the vapor pressure would be 0.350 atm is 201.37°C

Explanation:

The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.

Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.

The equation for Clausius- Clapeyron Equation can be expressed as:

[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]

where ;

[tex]P_1[/tex] is the vapor pressure at temperature 1

[tex]P_ 2[/tex] is the vapor pressure  at temperature 2

∆Hvap = enthalpy of vaporization

R = universal gas constant

Given that:

[tex]P_1[/tex] = 1 atm

[tex]P_ 2[/tex]  = 0.350 atm

∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol

[tex]T_1[/tex] = 282 °C  = (282 + 273) K = 555 K

R = 8.314 J/mol/k

Substituting the above values  into the Clausius - Clapeyron equation, we have:

[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]

[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }[/tex]

[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }[/tex]

[tex]\dfrac{1}{T_2} = 0.002108048708[/tex]

[tex]T_2 = \dfrac{1}{0.002108048708}[/tex]

[tex]\mathbf{T_2 }[/tex] =  474.37 K

To °C ; we have [tex]\mathbf{T_2 }[/tex] =   (474.37 - 273)°C

[tex]\mathbf{T_2 }[/tex] =  201.37 °C

Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C

The temperature of the liquid at the given vapor pressure is 201.5 ⁰C.

The given parameters;

The temperature of the liquid will be determined by applying Clausius- Clapeyron Equation;

[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )[/tex]

where;

[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )\\\\ln(\frac{0.35}{1} ) = \frac{28.5 \times 10^3}{8.314} (\frac{T_2 - 555}{555T_2} )\\\\-1.049 = 6.176- \frac{3427.95}{T_2} \\\\\frac{3427.95}{T_2} = 6.176 + 1.049\\\\\frac{3427.95}{T_2} = 7.225\\\\T_2 = \frac{3427.95}{7.225} \\\\T_2 = 474.5 \ K\\\\T_2 = 474.5 - 273 = 201.5 \ ^0C[/tex]

Thus, the temperature of the liquid at the given vapor pressure is 201.5 ⁰C.

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Answer:

The answer is o2

Explanation:

I took the test

Answer:

(a )people separate mixtures in order to ger a specific substance that they need.

Answer:

(CH3)3N

Explanation:

Hydrogen bonding can be called a type of intracellular force of the attraction. It is the force that occur between molecules. It is the bonding between the molecules and of hydrogen and electronegative items in the covalent bond. This is called the hydrogen donor. An electro-negative hydrogen atoms may be a hydrogen bonded. It is also called a hydrogen acceptor.

Thus in (CH3)3N, the hydrogen atoms becomes bonded with carbon. Carbon is not electronegative atoms. Thus it does not play as donor. Nitrogen is electronegative and play as hydrogen acceptor. But there is no presence of hydrogen acceptor. Thus there is no molecules that exhibit hydrogen molecules bonding.

[tex]\bold {(CH_3)_3N}[/tex] does not have hydrogen bonding because of the absence of electronegativity difference.

Hydrogen bond:

It is an inter-molecular bond. It is due to the difference in electronegativities of constituent atoms. This creates dipole in the atoms so, atoms start to attract each other.

In [tex]\bold {(CH_3)_3N}[/tex], the hydrogen atoms are bonded with carbon. The difference between the electronegativities Carbon and hydrogen is very less.

Therefore, [tex]\bold {(CH_3)_3N}[/tex] does not have hydrogen bonding because of the absence of electronegativity difference.

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Answer:

D) 2.01 x 10⁻⁴ .

Explanation:

pH = 2.11

[ H⁺ ] = [tex]10^{-2.11}[/tex]  

Let the acid be HA

It will ionise as follows .

                                        HA       ⇄       H⁺       +        A⁻

in equilibrium                 .30               [tex]10^{-2.11}[/tex]          [tex]10^{-2.11}[/tex]         

Acid ionisation constant Ka  =   [tex]\frac{(10^{-2.11})^2}{0.3}[/tex]

= 2 x 10⁻⁴                

Answer:

D) 2.01 x 10⁻⁴ is correct!

Explanation:

I got it in class!

Hope this Helps!! :))

Answer:

Kp=1.68×10⁴∆1.7×10⁴

Kc=1.06∆1.1

Explanation:

Value of Kp at 227°C is 2.86×10² and value of Kc at 1000 K is 1.56.

Kp and Kc are related by the formula Kp=Kc(RT).For part A , Kp is calculated as,

Kp=6.9×10⁵×8.314×500=28.683×10² and for part B Kc is calculated as,

Kc=1.3×10[tex]^-2[/tex]/(8.314×1000)=1.56

Kc and Kp are equilibrium constants of a mixture of ideal gases. Kp is equilibrium constant when concentrations at equilibrium are in atmospheric pressure and Kc is equilibrium constant when concentrations are in molarity. The relation is only valid for gaseous mixtures. The relation between these two parameters is obtained through ideal gas equation.

Kc and Kp of reaction change with temperature of reaction but remain unaffected by change in concentration , pressure and presence of catalyst.

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Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

The amount of heat to absorb is 6,261 J.

Heat required to raise the temperature of ice from -20 °C to 0 °C.

The formula for specific heat is used to calculate the amount of heat

Q = c * m * ΔT

Where,

Q =heat exchanged by a body,

m= mass of the body

c= specific heat

ΔT= change in temperature

Given:

m= 10 g,

specific heat of the ice= 2.1

ΔT=0 C - (-20 C)= 20 C

On substituting the values:

Q= 10 g*2.1  *20 C

Q=420 J

Heat required to convert 0 °C ice to 0 °C water.

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

Heat required to raise the temperature of water from 0 °C to 60 °C

m= 10 g,

Specific heat of the water= 4.18  

ΔT=60 C - (0 C)= 60 C

On substituting:

Q= 10 g*4.18  *60 C

Q=2,508 J

Thus, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

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Answer:

The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.  

Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams

The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,  

Moles of water = weight of water/molecular weight

= 42.1 grams / 18 = 2.3

The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,  

= 214.2 * 0.2190 = 46.91 grams

The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,

= 214.2 * 0.4314 = 92.40 grams

The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,  

= 214.2 * 0.3497 = 74.91 grams

The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17

The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,  

92.40/78.96 = 1.17

The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,  

74.91/15.999 = 4.68.  

Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O

Answer:

A. NaHCO₃

Explanation:

NaHCO₃ ⇒ NaOH + H₂CO₃

NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.

Hope that helps.

Answer:

  30. 5 planes are shown

  31. 1 plane

  32. CEF

  33. on line AB

  34. E or F

   35. ABCD or BCEF or CDEF or ACEF

Explanation:

30. Each of the surfaces of the rectangular pyramid is a plane. There are 5 planes.

__

31. 3 points define one plane only.

__

32. The only points shown on the same line segment are points E, F, and C.

__

33. If G is to be collinear with A and B, it must lie on line AB.

__

34. The only points shown that are not on plane N are points E and F. Either of those will do.

__

35. There are three planes that have 4 points shown on them. The four points that are on the same plane are any of ...

Plane ACEF is not shown on the diagram, but we know that those 4 points are also coplanar. (Any point not on line CE, together with the three points on that line, will define a plane with 4 coplanar points.)

Answer:

The 3rd one

Explanation:

Answer:

A

Explanation:

Absorbing moisture to protect goods from damage. Hence, option A is correct.

Silica gel is a desiccant, or drying agent, that manufacturers often place in little packets to keep moisture from damaging certain food and commercial products.

Silica Gel is a good drying agent for preventing corrosion, contamination, spoilage, and mould growth in many commodities and products due to its physical properties.

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Answer:

[tex]V_{HCl}=0.208L=208mL[/tex]

Explanation:

Hello,

In this case, since the chemical reaction is:

[tex]2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O[/tex]

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

[tex]n_{HCl}=2*n_{Mg(OH)_2}[/tex]

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

[tex]n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g} =0.00858mol[/tex]

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:

[tex][H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M[/tex]

Then, since the concentration and the volume define the moles, we can write:

[tex][HCl]*V_{HCl}=2*n_{Mg(OH)_2}[/tex]

Therefore, the neutralized volume turns out:

[tex]V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL[/tex]

Best regards.

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = 4.74

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

from

1moles=iatom

Mole=mass÷avogardos

Where

Avogadro's= 6.02×10²³

So moles = 65.0÷6.02×10²³

Atoms of zinc = 391.6 ×10²³

The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].

Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.

Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately  [tex]6.022 \times10^{23}[/tex]atoms per mole.

First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.

Number of moles = Mass / Molar mass

Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol

Next, multiply the number of moles by Avogadro's number to find the number of atoms.

Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]

Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol

Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.

Learn more about atoms here:

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Answer:

0.44

Explanation:

We know that the pH of any acid solution is given by the negative logarithm of its hydrogen ion concentration. Hence, if I can obtain the hydrogen ion concentration of any acid, I can obtain its pH.

For the acid, HNO3, [H^+] = [NO3^-]= 0.36 M

pH= -log [H^+]

pH= - log[0.36]

pH= 0.44