Ethyl bromide reacts with sodium methoxide to form ?​

Answer:  The molar mass of butane will be [tex]11.2\times 10^{7}g[/tex]

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP, contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles. and weighs equal to the molar mass.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given molecules}}{\text {avogadro's number}}=\frac{4.49\times 10^{16}}{6.023\times 10^{23}}=0.74\times 10^{-7}moles[/tex]

Now [tex]0.74\times 10^{-7}moles[/tex]  weigh = 8.29 g

Thus 1 mole will weigh = [tex]\frac{8.29}{0.74\times 10^{-7}}\times 1=11.2\times 10^{7}g[/tex]

The molar mass of butane will be [tex]11.2\times 10^{7}g[/tex]

Answer:

All three titrations require the same volume of NaOH to reach the first equivalence point.

Explanation:

Statements are:

All three titrations have the same final pH.

All three titrations require the same volume of NaOH to reach the first equivalence point.

All three titrations have the same pH at the first equivalence point.

All three titrations have the same initial pH.

The pH of the titration depends of the nature of the acid: If the acid is a strong acid, pH at the equivalence of the titration is 7. For a weak acid equivalence point depends of the nature of the conjugate base and initial pH of the weak acid. For a diprotic acid also depends of the nature of the acid.

Thus:

All three titrations have the same initial pH, All three titrations have the same pH at the first equivalence point. and All three titrations have the same final pH.  are the three FALSE.

As the concentrations of the acids is 0.10M and are titrated with 0.100M NaOH, The volume to reach the first equivalence point is the same for all the three acids.

Thus:

Answer: DEAR THE ANSWER TO YOUR QUESTION IS,

Explanation: Consider the equation for the decay of radium-226 to radon-222, with the simultaneous loss of an alpha particle and energy in the form of a gamma ray. Radium-226 is the reactant; radon, an alpha particle, and a gamma ray are the products. The equation is:

shown in the attach figure

TYPE OF DECAY: as α-particle emmit in this reaction hence its the α-decay

It decays by emitting an alpha particle composed of two protons and two neutrons. The radium nucleus turns into radon-222 nucleus, itself radioactive, containing two protons and two neutrons less. The disintegration releases 4.6 million electronvolts of energy

PLS GIVE ME RATING IF YOU FIND IT HELPFULL SO THAT OTHER BE BENIFIT OF THAT ANSWEER

THANKS....

Answer:

16.0 g/mol (3 s.f.)

Explanation:

Molar mass is the mass of a mole of substance.

1 mole of CH₄ has 1 C atom and 4 H atoms.

Since the molar mass is numerically equal to the molecular mass of a compound, let's find the molecular mass of CH₄ first.

Molecular mass= sum of all the atomic mass in a molecule

Molecular mass of CH₄

= 12.011 amu +4(1.008 amu)

= 16.043 amu

Thus the molar mass of CH₄ is 16.043g/mol, or 16.0g/mol to 3 significant figures.

Answer:

0.02 moles of O₂ will be leftover.

Explanation:

The reaction is:

2NO(g) + O₂(g) → 2NO₂(g)    (1)

We have the mass of NO and O₂, so we need to find the number of moles:

[tex] n_{NO} = \frac{m}{M} = \frac{24.7 g}{30.01 g/mol} = 0.82 moles [/tex]

[tex] n_{O_{2}} = \frac{m}{M} = \frac{13.8 g}{31.99 g/mol} = 0.43 moles [/tex]

From equation (1) we have that 2 moles of NO reacts with 1 mol of O₂ to produce 2 moles of NO₂, so the excess reactant is:

[tex] n_{NO} = \frac{2}{1}*0.43 moles = 0.86 moles [/tex]          

[tex]n_{O_{2}} = \frac{1}{2}*0.82 moles = 0.41 moles[/tex]                                                                                

Hence, from above we can see that the excess reactant is O₂ since 0.41 moles react with 0.86 moles of NO and we have 0.43 moles in total for O₂.

The number of moles of excess reactant is:

[tex]n_{T} = 0.43 moles - 0.41 moles = 0.02 moles[/tex]  

Therefore, 0.02 moles of O₂ will be leftover.

I hope it helps you!                    

The number of moles of excess reactant that would be left over is 0.0197 mole

From the question,

We are to determine the number of moles of excess reactant that would be left over.

The given balanced chemical equation for the reaction is

2NO(g) + O₂(g) → 2NO₂ (g)

This means,

2 moles of NO is needed to completely react with 1 mole of O₂

Now, we will determine the number of moles of each reactant present

Mass = 24.7 g

Molar mass = 30.01 g/mol

From the formula

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

∴ Number of moles of NO present = [tex]\frac{24.7}{30.01}[/tex]

Number of moles of NO present = 0.823059 mole

Mass = 13.8 g

Molar mass = 32.0 g/mol

∴ Number of moles of O₂ present = [tex]\frac{13.8}{32.0}[/tex]

Number of moles of O₂ present = 0.43125 mole

Since,

2 moles of NO is needed to completely react with 1 mole of O₂

Then,

0.823059 mole of NO is will react with completely react with [tex]\frac{0.823059 }{2}[/tex] mole of O₂

[tex]\frac{0.823059 }{2} = 0.4115295[/tex]

∴ Number of moles of O₂ that reacted is 0.4115295 mole

This means O₂ is the excess reactant and NO is the limiting reactant

Now, for the number of moles of excess reactant left over

Number of moles of excess reactant left over = Number of moles of O₂ present - Number of moles of O₂ that reacted

∴ Number of moles of excess reactant left over = 0.43125 mole - 0.4115295 mole

Number of moles of excess reactant left over = 0.0197205 mole

Number of moles of excess reactant left over ≅ 0.0197 mole

Hence, the number of moles of excess reactant that would be left over is 0.0197 mole

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Answer:

identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas

identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas

Answer:

Pure substance:

Mixtures:

Explanation:

Pure substances exists as either elements or compounds.

A homogeneous mixture has even distribution and a single phase composition of its constituents; whereas heterogeneous mixtures are not uniform and constituents exist in different phases.

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.  

[tex]\text{calculating the initial HI}= \frac{mol}{V}[/tex]

                                       [tex]=\frac{9.3 \times 10 ^ -3}{2}[/tex]

                                      [tex]=0.00465 \ Mol[/tex]

[tex]\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }[/tex]

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

[tex]HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\[/tex]

It is defined that:

[tex]I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\[/tex]

[tex]HI \ eq= 0.00465 - 2x \\[/tex]

          [tex]=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M[/tex]

Now, we calculate the position:  

For the reaction [tex]H 2(g) + I 2(g)\rightleftharpoons 2HI(g)[/tex], you can calculate the value of Kc at 1000 K.  

data expression for Kc

[tex]2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}[/tex]

         [tex]= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386[/tex]

calculating the reverse reaction

[tex]H_2(g) + I_2(g)\rightleftharpoons 2HI(g)[/tex]

[tex]Kc = \frac{1}{Kc} \\\\[/tex]

     [tex]= \frac{1}{0.034386}\\ \\= 29.081\\[/tex]

The Kc of the reaction is 40.

Molarity of the HI = 9.30×10^−3mol/ 2.00 L = 4.65 × 10^-3 M

Let the concentrations of I2 and H2 be x, but we are told in the question that 6.29×10^−4M was present at equilibrium.

The molarity of HI at equilibrium now becomes; 4.65 × 10^-3 M - 6.29×10^−4M

= 4 × 10^-3 M

But;

Kc = [HI]^2/[H2] [I2]

Kc = ( 4 × 10^-3)^2/(6.29×10^−4)^2

Kc = 40

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Answer:

H₂SO₄  

Explanation:

A Brønsted acid is a proton donor.  It loses protons.

This equation may be easier to understand if we write it ionically.

[tex]\underbrace{\hbox{H$_{2}$SO$_{4}$}}_{\hbox{Br$\o{}$nsted acid}} + 2 \text{NH}_{3} \longrightarrow \, \underbrace{\hbox{SO$_{4}^{2-}$}}_{\hbox{Br$\o{}$nsted base}} + \text{2 NH}_{4}^{+}[/tex]

We see that the H₂SO₄ has lost two protons to become SO₄²⁻, so it is a Brønsted acid.

Answer:

THE MOLARITY OF THE SOLUTION IS 3.36 MOLE/L

Explanation:

First we must understand what molarity is.

Molarity is the number of mole per unit volume of solution. In this question, 9.25 mole of H2SO4 was given in 2.75 L of solution.

Molarity is written in mole per dm3 or L.

So we can calculate the molarity:

9.25  ole of H2SO4 = 2.75 L of solution

The number of mole in 1 L of solution will be:

= 9.25 mole / 2.75 L

= 3.3636 mole/ L

In conclusion, the molarity of the solution is approximately 3.36 mole/L

Answer:

Protecting group

Explanation:

TMS groups can be used as protecting and leaving groups for the synthesis of siloxane-based molecules.   It is also used as protecting group for alcohols.

A protecting group is a temporary group added during organic synthesis to prevent a portion of molecule from reacting.

Answer:

1mg=0.001

210

210×0.001÷1

=0.21

Answer:

Acids do not react with bases in neutralization reactions to produce water.

Explanation:

Answer:

88.1% (to 3 s.f.)

Explanation:

Please see attached picture for full solution.

Explanation:

The percentage abundance of the following gases in the atmosphere is given as;

Carbon dioxide = 0.04 percent

Other trace gases =  about a tenth of one percent of the atmosphere.

Argon = 0.93 percent

Oxygen = 21 percent

Nitrogen = 78 percent

Te order from highest to lowest is given as;

Nitrogen

Oxygen

Argon

Carbon dioxide

Other trace gases

Answer:

Activation energy for the reaction is 39029J/mol

Explanation:

Arrhenius equation is an useful equation that relates rate of reaction at two different temperatures as follows:

[tex]ln\frac{K_2}{K_1} = \frac{-Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1} )[/tex]

Where K₁ and K₂ are rate of reaction, Ea is activation energy and R is gas constant (8.314J/molK

If the reaction at 400K is 50 times more faster than at 300K:

K₂/K₁ = 50 where T₂ = 400K and T₁ = 300K:

[tex]ln50 = \frac{-Ea}{8.314J/molK} (\frac{1}{400K} -\frac{1}{300K} )[/tex]

[tex]ln 50 = 1x10^{-4}Ea[/tex]

Ea = 39029 J/mol

The activation energy for this chemical reaction is equal to 39,029.24 J/mol.

Given the following data:

Ideal gas constant, R = 8.314 J/molK

To determine the activation energy for this chemical reaction, we would use the Arrhenius' equation:

Mathematically, Arrhenius' equation is given by the formula:

[tex]ln\frac{K_2}{K_1} = \frac{-E_a}{R} (\frac{1}{T_2} - \frac{1}{T_1})[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]ln50 = \frac{-E_a}{8.314} (\frac{1}{400} - \frac{1}{300})\\\\3.9120 = \frac{-E_a}{8.314} (\frac{-1}{1200})\\\\3.9120 = \frac{E_a}{9976.8} \\\\E_a = 9976.8 \times 3.9120\\\\E_a = 39,029.24 \;J/mol[/tex]

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Answer:

Equilibrium constant expression for [tex]\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq)[/tex]:

[tex]\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}[/tex].

Where

Explanation:

The equilibrium constant expression of a (reversible) reaction takes the form a fraction.

Multiply the activity of each product of this reaction to get the numerator.[tex]\rm H_2CO_3\; (aq)[/tex] is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be [tex]\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)[/tex].

Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient "[tex]2[/tex]" on the product side of this reaction. [tex]\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq)[/tex] is equivalent to [tex]\rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq)[/tex]. The species [tex]\rm H^{+}\, (aq)[/tex] appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:

[tex]\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right[/tex].

That's where the exponent "[tex]2[/tex]" in this equilibrium constant expression came from.

Combine these two parts to obtain the equilibrium constant expression:

[tex]\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}[/tex].

In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous "[tex](\rm aq)[/tex]" species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:

[tex]\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}[/tex].

Answer:

With the changes in atmospheric pressure, the gas laws also change accordingly, that is, the Charle's law, the Boyle's law, the Ideal Gas Law, and the Gay-Lussac's law changes with the change in atmospheric pressure. Therefore, the planets, which exhibits the constant value of atmospheric pressure will be more reliable, and those possessing different atmosphere or no atmosphere will be the least reliable.

Answer:

A. The neutral-charged neutrons and positive-charged protons are found within the nucleus, and the negative-charged electrons orbit outside the nucleus of the atom.

Explanation:

Before examining the options, it's important to know that the atom is made up of three sub atomic particles which are; Protons, Neutrons and Electrons.

The protons are positively charged and are found in the nucleus of an atom. The Neutron A=are neutral in terms of charge and are also found in the nucleus of an atom. The electrons on the other hand are negatively charged and are found outside the nucleus if an atom, more specifically on the orbitals.

The option that best describes this is; option A.

Answer:

a. 14.9g/L

b. 1.49x10⁻⁶

c. 1.49x10⁷

Explanation:

You can write the buffer Ksp of Co(OH)₂ as follows:

Co(OH)₂(s) ⇄ Co²⁺ + 2OH⁻

Ksp = 1.6x10⁻¹⁵ = [Co²⁺] [OH⁻]²

To have buffered the solutions means  [OH⁻] is fixed. From the equilibrium of water we can relate  [OH⁻] with pH as follows:

[OH⁻] = 10^[14-pH]

With [OH⁻] and Ksp we can solve for [Co²⁺]. Its concentration is equal to solubility (That is the amount of Co(OH)₂ that can be dissolved).

[Co²⁺] is in mol/L. With molar mass of Co(OH)₂ -92.948g/mol-, We can obtain, in the end, its solubility in g/L.

-Molar concentration of [Co²⁺] and solubility:

a. [OH⁻] = 10^[14-7.00] = 1x10⁻⁷

[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻⁷]²

[Co²⁺] = 0.16mol / L = Solubility.

In g/L = 0.16mol / L ₓ(92.948g/mol) =

14.9g/L

b. [OH⁻] = 10^[14-10.00] = 1x10⁻⁴

[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻⁴]²

[Co²⁺] = 1.6x10⁻⁸mol / L = Solubility.

In g/L = 1.6x10⁻⁸mol / L ₓ(92.948g/mol) =

1.49x10⁻⁶g/L

c.[OH⁻] = 10^[14-4.00] = 1x10⁻¹⁰

[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻¹⁰]²

[Co²⁺] = 1.6x10⁵mol / L = Solubility.

In g/L = 1.6x10⁵mol / L ₓ(92.948g/mol) =

1.49x10⁷g/L

As you can see, and as general rule, all hydroxides are solubles in acids.

Answer:

the expected absorbance of the solution = 0.17

Explanation:

From the information given:

Using Beer's Lambert Law, we have

A = ∈CL

where;

A = Absorbance

∈ = extinction coefficient

C = concentration

L = cell length

Since Absorbance is associated with concentration.

Assuming the measurement were carried out in the same solution; Then  ∈ and L will be constant and A  ∝ C ----- (1)

Let consider the concentration to be C  (mol/L)

5.0 mL of a Sports Drink = 5.0 mL × C (mol)/1000 mL

= 5C/1000 mL

was diluted with water to 10.0 mL

So, when diluted with water to 10.0 mL; we have:

The new concentration to be : [tex]\dfrac{(5 C \times 1000) \ mol }{(1000 \times 10 \times 1000)\ mL}[/tex]

Since :1000mL = 1 L

The new concentration = [tex]\dfrac{C \ mol }{2 \ L}[/tex]

As stated that the initial absorbance reading [tex]A_1[/tex] = 0.34

The expected absorbance reading will be [tex]A_2[/tex] = ???

From (1)

A ∝ C

∴

[tex]\dfrac{A_2}{A_1}=\dfrac{C_2}{C}[/tex]

[tex]A_2 = \dfrac{A_1}{C}[/tex]

[tex]A_2 = \dfrac{0.34}{2}[/tex]

[tex]A_2 = 0.17[/tex]

Thus ; the expected absorbance of the solution = 0.17

Answer:

Negligible

Explanation:

According to the kinetic theory of gases, the degree of intermolecular interaction between gases is minimal and gas molecules tend to spread out and fill up the volume of the container.

If the attraction between gas molecules increases, then the volume of the gas decreases accordingly. This is because, gas molecules become highly attracted to each other.

This intermolecular attractive force may be so strong, such that the actual volume of the gas become negligible compared to the volume of the container.

Answer:[tex]2CH_3COONH_4(aq)+K_2S(aq)\rightarrow 2CH_3COOK(aq)+(NH_4)_2S(aq)[/tex]

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The balanced molecular reaction between aqueous solutions of ammonium acetate and potassium sulfide will be

[tex]2CH_3COONH_4(aq)+K_2S(aq)\rightarrow 2CH_3COOK(aq)+(NH_4)_2S(aq)[/tex]

The balanced molecular equation for the reaction between aqueous solutions of ammonium acetate and potassium sulfide is

[tex]2 CH_3COONH_4(aq) + K_2S(aq) \rightarrow 2 CH_3COOK(aq) + 2 NH_3(g) + H_2S(g)[/tex]

Let's consider the unbalanced molecular equation for the reaction between aqueous solutions of ammonium acetate and potassium sulfide. This is originally a double displacement reaction that would produce potassium acetate and ammonium sulfide. However, ammonium sulfide is unstable and will rapidly decompose into hydrogen sulfide and ammonia.

[tex]CH_3COONH_4(aq) + K_2S(aq) \rightarrow CH_3COOK(aq) + NH_3(g) + H_2S(g)[/tex]

We will balance it using the trial and error method.

We will:

The balanced molecular equation is:

[tex]2 CH_3COONH_4(aq) + K_2S(aq) \rightarrow 2 CH_3COOK(aq) + 2 NH_3(g) + H_2S(g)[/tex]

The balanced molecular equation for the reaction between aqueous solutions of ammonium acetate and potassium sulfide is

[tex]2 CH_3COONH_4(aq) + K_2S(aq) \rightarrow 2 CH_3COOK(aq) + 2 NH_3(g) + H_2S(g)[/tex]

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ProccessA

Explanation:

Ethanol Heavy Oils

Answer:

B. CH_3 CH_2 CH_2 F

Explanation:

The alkyl group contains all univalent group derived from alkanes by the loss of a hydrogen atom. When an alkyl group combines or is bonded with an halogen  family (such as Flourine, Chlorine , Bromine , Iodine) , we have an alkyl halide such as the options given in the question.

Alkyl  halide bond with an sp³ Carbon. Depending on where the halide is bonded to , we can have the primary halide, secondary halide and the tertiary halide.

When alkyl halide react with magnesium metal, we have a reaction that is known as Grignard Reaction. The order of reactivity of alkyl halide with grignard reagents is :

F < Cl < Br < I .

This is because of the nature of the electronegativity and the bond strength.

As we move than the group from the top to the bottom, the electronegativity decreases and the bond strength increase, so as the size. Thus, this is exactly what the reactivity of alkyl halide with Grignard reaction is established for.

Thus since the fluorine have the highest electronegativity and the smallest bond size in the halogen family , then it will be the least reactive alkyl halide in reacting with magnesium metal.

Answer:

1) evolution of gas

2) evolution of heat

Explanation:

In this reaction, glucose is broken down into its constituents; carbon dioxide and water. The question is to decipher indicators of a chemical reaction from the equation.

If we look at the equation carefully, we will notice that a gas was evolved (CO2). The evolution of a gas indicates that a chemical reaction must have taken place. Secondly, energy is given off as heat. This is another indication that a chemical reaction has taken place.

Answer:

Normality N = 0.2 N

Explanation:

Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.

Normality is represented by N.

Mathematically, we have :

[tex]\mathbf{Normality \ N = \dfrac{Number \ of \ gram \of \ equivalent\ of\ solute }{volume \ of \ solution}}[/tex]

Given that:

number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent

volume of solution (HCl) = 450 mL 450 × 10⁻³ L

[tex]\mathbf{Normality \ N = \dfrac{90 \times 10^{-3}}{450 \times 10^{-3}}}[/tex]

Normality N = 0.2 N

Answer:

16.7 mL

Explanation:

Convert 250 mL to L.

250 mL = 0.250 L

Calculate the amount of moles of NaOH in 250 mL of 0.300 M NaOH.

0.250 L × 0.300 M = 0.075 mol

Using this amount of moles, you need to find out what volume of 4.50 M will give you that many moles.  You can do this by dividing the amount of moles by the molarity.

(0.075 mol)/(4.50 M) = 0.0167 L

Convert from L to mL.

0.0167 L = 16.7 mL

Answer:

1.7 ppm

Explanation:

Original amount N' = 2.6 ppm

time to testing t = 24 hr

final amount N = 2.1 ppm

Using exponential inhibited decay, we have

N = N'e^(-kt)

Where

N is the new reading

N' is the original reading

t is the decay time

k is the decay constant

Substituting, we have

2.1 = 2.6 x e^(-k x 24)

2.1 = 2.6 x e^(-24k)

0.808 = e^(-24k)

We take the natural log of both sides of the equation

Ln 0.808 = Ln (e^(-24k))

-0.213 = - 24k

K = 0.213/24 = 0.00886

After 48 hrs, the reading of free chlorine will be

N = 2.6 x e^(-0.00886 x 48)

N = 2.6 x e^(-0.425)

N = 2.6 x 0.654

N = 1.7 ppm