11. (2 pts) Sodium Hydroxide, is also known as lye and was a critical component in
homemade soap. Now it is a commonly used drain cleaner because it chemically reacts
with fats (the typical cause of a clog) to form a soap that can be swept down the drain.
What is the molarity of 5.00 g Sodium Hydroxide in 750.0 mL of solution?

Answer:

3 fluorine atoms will be present

Answer:

3

Explanation:

The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.

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Answer:

1) 3-Ethylanisole

2) 2,3-Dibromobenzoic acid

3) 3-Fluorotoluene

Explanation:

Let us try to look at the structures of each compound one after the other as described in the question.

1) A ring with six vertices and alternating double and single bonds must refer to a benzene ring. A benzene ring having -OCH3 attached to the first vertex is called anisole. If a -CH2CH3 group is now attached at position 3, we now name the compound 3-Ethylanisole.

2) A ring with six vertices and alternating single and double bonds is a benzene ring. If the ring has -COOH attached to the first vertex, we call it benzoic acid. If bromine atoms are attached to the second and third vertices respectively, the compound is now named 2,3-Dibromobenzoic acid.

3) A ring with alternating single and double bonds is a benzene ring. If a -CH3 group is attached to the first vertex, we call the compound toluene. If a fluorine atom is now attached to position 3, the compound can now be named 3-Fluorotoluene

Answer:

The answer to this question can be defined as follows:

In question 1, the answer is "Option A".

In question 2, the answer is "[tex]\bold{CH_3COOH}[/tex]".

Explanation:

In the second question, there is mistype error in the choices so the correct answer to this question can be defined as follows:

Answer:

Explanation:

a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .  

C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O

O is provided by KMnO₄

b ) In this reaction isophthalic acid is formed .

C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂

c)

4-Propyl-3-t-butyltoluene

In this oxidation , three side chains of ring  are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .

The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )

Answer:

The new volume is 808L.

Explanation:

First, you need to set up your proportion, using Charles's law.

V1         V2

       =        

T1         T2

Then, substitute for the ones you already know.

625 L            V2

            =              

273 K          353 K

Then, you need to cross multiply.

625 x 353 = 273 x V2

Then, solve for V2.

220625 = 273 x V2

220625 ÷ 273 = V2

808 = V2

The final volume is 808L.

When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.

Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.

There are various methods of expressing the concentration of a solution.

Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.

Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

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Answer:

C₄H₁₀(g) + O₂(g) ⇒ CO₂(g) + H₂O(g)

2 C₄H₁₀(g) + 13 O₂(g) ⇒ 8 CO₂(g) + 10 H₂O(g)

Explanation:

Butane gas (C₄H₁₀) burns in oxygen gas to produce carbon dioxide gas and water vapor. The unbalanced equation is:

C₄H₁₀(g) + O₂(g) ⇒ CO₂(g) + H₂O(g)

First, we will balance carbon and hydrogen which are in just one compound on each side.

C₄H₁₀(g) + O₂(g) ⇒ 4 CO₂(g) + 5 H₂O(g)

Finally, we will balance the oxygen atoms.

C₄H₁₀(g) + 6.5 O₂(g) ⇒ 4 CO₂(g) + 5 H₂O(g)

In order to have integers, we will multiply everý compound by 2.

2 C₄H₁₀(g) + 13 O₂(g) ⇒ 8 CO₂(g) + 10 H₂O(g)

Answer:

Acid spills should be neutralized with sodium bicarbonate and then cleaned up with a paper towel or sponge.

Explanation:

Answer:

linear

Explanation:

If we look at the compound, we will understand that the underlined carbon is in sp hybridization. Recall that sp hybridization leads to a C-C bond angle of about 180°. When two chemical species are at a bond angle if 180°, then the both bonding groups are found to be on a straight line.

Hence, for any carbon in a triply bonded state or indeed any carbon atom that is in an sp will exhibit a linear geometry according to the Valence Shell Electron Pair repulsion Theory since there are two electron pairs present.

Answer:

-36.6 kJ·mol⁻¹  

Explanation:

1. Calculate ΔG

[tex]\text{The relationship between $\Delta G^{\circ}$ and K is}\\\Delta G^{\circ} = -RT \ln K[/tex]

T = (250 + 273.15) K = 523.15 K

[tex]\begin{array}{rcl}\Delta G & = & -8.314 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1} \times \text{523.15 K} \ln (4.5 \times 10^{3}) \\& = & -4349\text{ J}\cdot \text{mol}^{-1} \times 8.412\\& = & \text{-36 590 J}\cdot \text{mol}^{-1}\\& = & \textbf{-36.6 kJ}\cdot \text{mol}^{-1}\\\end{array}[/tex]

2. Direction of spontaneity

ΔG is negative, so the reaction is spontaneous in the forward direction.

Answer:

heptan-2-one

Explanation:

In this case, the final product would be a ketone: heptan-2-one. To understand why this molecule is produced we have to check the reaction mechanism.

The first step is the protonation of the triple bond to produce the more stable carbocation (a secondary one) by the action of sulfuric acid [tex]H_2SO_4[/tex]. The next step is the attack of water to the carbocation to produce a new bond between C and the O, producing a positive charge in the oxygen. Then, a deprotonation step takes place to produce an enol. Finally, we will have a rearrangement (keto-enol tautomerism) to produce the final ketone.

See figure 1

I hope it helps!

Answer:

Explanation:

The substances are:

-) Ethane, [tex]C_2H_6[/tex]

-) 1-pentanol, [tex]C_5H_1_1OH[/tex]

-) Potassium chloride, [tex]KCl[/tex]

-) Propane,  [tex]C_3H_8[/tex]

For this question, we have to remember the structure of water. Due to the electronegativity difference between oxygen and hydrogen in this structure, we will have the formation of dipoles. The dipoles interact better with net charges, due to this, the Potassium chloride is the compound with highest solubility (due to the formation of a cation and an anion):

[tex]KC~l->~K^~+~Cl^-[/tex]

Then, in 1-pentanol we an "OH". This structure due to the presence of the hydroxyl group can form hydrogen bonds. Therefore,  this compound would be the second more soluble.

Finally,  the difference between propane and ethane is a carbon. In propane, we have an additional carbon. If we have more carbons we will have more area of ​​interaction. If we have more area we will have more solubility therefore propane is more soluble than ethanol.

In conclusion, the rank from most soluble to least soluble is:

1) Potassium chloride, [tex]KCl[/tex]

2) 1-pentanol, [tex]C_5H_1_1OH[/tex]

3) Propane,  [tex]C_3H_8[/tex]

4) Ethane, [tex]C_2H_6[/tex]

I hope it helps!

Order of solubility in water will be:

KCl > C₅H₁₁OH > C₃H₈ > C₂H₆

Any solvent soluble in water due to its polarity and ability to form hydrogen bonds. The presence of hydrogen bonding between molecules of a substance indicates that the molecules are polar. This means the molecules will be soluble in a polar solvent such as water.

Substances that are given:

Ethane(C₂H₆), 1-pentanol(C₅H₁₁OH), Potassium chloride(KCl) and propane(C₃H₈).

We will look at each compound one by one:

Order of solubility in water will be:

KCl > C₅H₁₁OH > C₃H₈ > C₂H₆

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Answer:

[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

[tex]2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}[/tex]

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

[tex]\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol[/tex]

[tex]\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol[/tex]

[tex]\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol[/tex]

[tex]\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})[/tex]

[tex]\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)[/tex]

[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]

The equilibrium constant determined from the partial pressure denoted as [tex]K_p[/tex] can be expressed as :

[tex]K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}[/tex]

[tex]K_p = \dfrac{1}{ (22.20)}[/tex]

[tex]K_p[/tex] = 0.045

[tex]\Delta G = \Delta G^0 _{rxn} + RT \ lnK[/tex]

where;

R = gas constant = 8.314 × 10⁻³ kJ

[tex]\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)[/tex]

[tex]\Delta G =207.6 + 2.4788191 \times \ ln(0.045)[/tex]

[tex]\Delta G =207.6+ (-7.687048037)[/tex]

[tex]\Delta G =[/tex] 199.912952  kJ

ΔG ≅ 199.91 kJ

Answer:

Explanation:

KHT is a salt which ionises in water as follows

KHT ⇄ K⁺ + HT⁻

Solubility product Kw= [ K⁺ ] [ HT⁻ ]

product of concentration of K⁺ and HT⁻ in water

In KCl solution , the solubility product of KHT will be decreased .

In KCl solution , there is already presence of K⁺  ion in the solution . So

in the equation  

[ K⁺ ] [ HT⁻ ]  = constant

when K⁺ increases [ HT⁻ ] decreases . Hence less of KHT dissociates due to which its  solubility decreases . It is called common ion effect . It is so because here the presence of common ion that is K⁺ in both salt to be dissolved and in solvent , results in decrease of solubility of the salt .

Answer:

pH = 4.52

Explanation:

A monoprotic acid, HA, reacts with NaOH as follows:

HA+ NaOH → A⁻ + H₂O

When the weak acid HA, is in solution with its conjugate base, A⁻, a buffer is produced. That means in the titration of the weak acid with NaOH you are producing a buffer.

The pH of a buffer can be found using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is -logKa = 4.52 and [] can be understood as the moles of A⁻ and HA.

pH = 4.52 + log [A⁻] / [HA]

In the half-equivalence point, the half of HA was converted in A⁻ and the other half still remains as HA.

That means moles of A⁻ and HA are: 0.480/2 = 0.24 moles of both A⁻ and HA

Replacing in H-H equation:

pH = 4.52 + log [A⁻] / [HA]

pH = 4.52 + log [0.24] / [0.24]

-In the half-equivalence point of a titration of a weak acid, pH = pKa-

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

[tex]w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln} *100\%\\\\w/w\%=3.15\%[/tex]

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

For the w/w% of the solution, information about the molecular mass of the solute, and density of the solution has been required.

Molarity can be defined as the moles of the solute per liter of the solution. The molarity can be used for the determination of the weight of the solute, by the information about the molecular weight of the compound.

Thus, for the w/w% of the solution, the weight of the solute has been determined with information about the molecular mass of the solute.

The weight of the solvent has been determined with the density of the solution. The density has been defined as the mass per unit volume.

Thus, for the w/w% of the solution, the weight of the solvent has been determined by the density of the solution.

For more information about the w/w% of the solution, refer to the link:

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Answer:

24.6g of CO₂ is theoretical yield

Explanation:

The reaction of 8.00g of octane with 38.9g of oxygen.

The reaction of octane with oxygen is:

C₈H₁₈(l) + 25/2O₂ → 9H₂O + 8CO₂

1 mole of octane reacts with 25/2 moles of oxygen to produce 8 moles of CO₂

Theoretical yield is the amount of carbon dioxide formed assuming a yield of 100%. To calculate theoretical yield, first, we need to find limiting reactant and, with the chemical reaction, we can obtain the theoretical moles of CO₂ produced and its mass to obtain theoretical yield.

Limiting reactant:

Moles octane (Molar mass: 114.23g/mol) in 8.00g:

8.00g × (1mol / 114.23g) = 0.0700 moles octane.

Moles oxygen (Molar mass: 32g/mol) in 38.9g:

38.9g × (1mol / 32g) = 1.2156 moles oxygen.

For a complete reaction of 1.2156 moles of O₂ there are necessaries:

1.2156 moles O₂ ₓ (1mol C₈H₁₈ / 25/2 moles O₂) = 0.0973 moles octane

As we have just 0.0700 moles,

Moles and mass of carbon dioxide:

As limiting reactant is octane, 0.0700 moles of C₈H₁₈ will produce:

0.0700mol C₈H₁₈ × (8 moles CO₂ / 1 mol C₈H₁₈) = 0.56 moles of CO₂ are theoretically produced. In mass (Molar mass CO₂ = 44.01g/mol):

0.56moles CO₂ × (44.01g / mol) =

-Theoretical yield because we are assuming all octane is reacting. In real life, never happens like that-

Answer:

The equilibrium position will shift to the left, thus, favouring the backward reaction.

Explanation:

The equation for the reaction is given below:

3C(s) + 3H2(g) <==> CH4(g) + C2H2(g)

According to Le Chatelier's principle, if an external constrain such as change in concentration, temperature or pressure is imposed on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.

From Le Chatelier's principle, removing some C(s) implies removing some of the concentration of the reactants.

This will shift the equilibrium position to the left, thus, favouring the backward reaction (i.e forming more reactants) because removing C(s) implies that the products are now more than the reactants and as such, they will react to form more reactants.

The equilibrium position will shift to the left, thus, favouring the backward reaction.

The following information should be considered:

[tex]3C(s) + 3H2(g) <==> CH4(g) + C2H2(g)[/tex]

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Answer:

Wave A has a higher frequency because it has a shorter wavelength.

Explanation:

The frequency of a wave and the wave length are related by the following equation:

Velocity (v) = wave length (λ) x frequency (f)

v = λf

If we make frequency (f) the subject of the above equation, we will have:

f = v/λ

Let the velocity (v) be constant.

f = v/λ

f & 1/λ

From the equation above,

We can see that the frequency (f) is inversely proportional to the wavelength (λ).

This implies that a wave with a high frequency, will have a short wavelength and a wave with a short frequency will have a longer wavelength.

Now considering wave A and B in the diagram above,

Wave A will have a higher frequency because it has a shorter wavelength as explained above.

Answer:

it is the second option

Explanation:

Answer:

[tex]\boxed{Heptene}[/tex]

Explanation:

Double Bond => An Alkene molecule

So, the suffix will be "-ene"

7 Carbons => So, we'll use the prefix "Hept-"

Combining the suffix and prefix, we get:

=> Heptene

Answer:

[tex]\boxed{\mathrm{Heptene}}[/tex]

Explanation:

Alkenes have double bonds. The molecule has one double bond.

Suffix ⇒ ene

The molecule has 7 carbon atoms and 14 hydrogen atoms.

Prefix ⇒ Hept (7 carbons)

The molecule is Heptene.

[tex]\mathrm{C_7H_{14}}[/tex]

Answer:

1.01 V

Explanation:

From Nernst equation;

Ecell= E°cell- 0.0592/n log Q

Where;

Ecell= observed emf of the cell

E°cell= standard emf of the cell

n= number of moles of electrons transferred

Q= reaction quotient

Q= [Ag^+]^3/[MnO4^-] [H^+]^4

Q= [0.01]^3/[1.20] [1.50]^4

Q= 1.65×10^-7

Ecell= 0.88 - 0.0592/3 log 1.65×10^-7

Ecell= 0.88 - [0.0197×(-6.78)]

Ecell= 0.88 + 0.134

Ecell= 1.01 V

Answer:

(a) [tex]Ksp=4.50x10^{-7}[/tex]

(b) [tex]Ksp=1.55x10^{-6}[/tex]

(c) [tex]Ksp=2.27x10^{-12}[/tex]

(d) [tex]Ksp=1.05x10^{-22}[/tex]

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) [tex]BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)[/tex]

[tex]Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}[/tex]

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

[tex]Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}[/tex]

(B) [tex]Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)[/tex]

[tex]Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}[/tex]

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

[tex]Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}[/tex]

(C) [tex]NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)[/tex]

[tex]Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}[/tex]

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

[tex]Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}[/tex]

(D) [tex]La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)[/tex]

[tex]Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}[/tex]

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

[tex]Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}[/tex]

Best regards.

Answer:

Pressure exerted by the gas is 574.85 torr

Explanation:

Atmospheric pressure = 751 torr

but 1 torr = 1 mmHg

therefore,

atmospheric pressure = 751 mmHg

1 mmHg = 133.3 Pa

therefore,

atmospheric pressure = 751 x 133.3 = 100108.3 Pa

distance labeled (tube section with mercury) = 176 mm

the pressure within the tube will be

[tex]P_{tube}[/tex] = ρgh

where ρ is the density of mercury = 13600 kg/m^3

h is the labeled distance = 176 mm = 0.176 m

g is acceleration due to gravity = 9.81 m/s^2

[tex]P_{tube}[/tex]  = 13600 x 9.81 x 0.176 = 23481.216 Pa

The general equation for the pressure in the manometer will be

[tex]P_{atm}[/tex] = [tex]P_{tube}[/tex] + [tex]P_{gas}[/tex]

where [tex]P_{atm}[/tex]  is the atmospheric pressure

[tex]P_{tube}[/tex]  is the pressure within the tube with mercury

[tex]P_{gas}[/tex] is the pressure of the gas

substituting, we have

100108.3 = 23481.216 + [tex]P_{gas}[/tex]

[tex]P_{gas}[/tex] = 100108.3 - 23481.216 = 76627.1 Pa

This pressure can be stated in mmHg as

76627.1 /133.3 = 574.85 mmHg

and also equal to 574.85 torr

Answer:

k = 100 mol⁻² L² s⁻¹, r= k[A][B]²

Explanation:

A + B + C --> D

[A] [B] [C] IRR

0.20 0.10 0.40 .20

0.40 0.20 0.20 1.60

0.20 0.10 0.20 .20

0.20 0.20 0.20 .80

Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.

This means the rate of reaction is second order with respect to B.

Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.

This means the rate of reaction is first order with respect to A.

Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.

This means the rate of reaction is zero order with respect to C.

The rate expression for this reaction is given as;

r = k [A]¹[B]²[C]⁰

r= k[A][B]²

In order to obtain the value of the rate constant, let's work with the first reaction.

r = 0.20

[A] = 0.20 [B] = 0.10

k = r / [A][B]²

k = 0.20 / (0.20)(0.10)²

k = 100 mol⁻² L² s⁻¹

Explanation:

The electric force between two charged particles is given by the formula as follows :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Here,

k is electrostatic constant

[tex]q_1,q_2[/tex] are charges

r is the distance between charges

The above formula shows that the electric force is inversely proportional to the square of the distance between charges. It means that as the distance increases by a factor, the electric force decreases by the square of that factor. Hence, the correct option is (e).

Answer:

A

Explanation:

Answer:

Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.

Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10 genes.

Answer:

Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.

Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10

Explanation:

Answer:

171 mL of HCl

Explanation:

The first thing we want to do is consider the reaction between Al(OH)3 and water - as that is the expected reaction that is taking place,

Al(OH)3 + 3HCl → AlCl3 + 3H2O

Knowing this, let's identify the mass of Al(OH)3. Aluminum = 27 g / mol, Oxygen( 3 ) = 16 [tex]*[/tex] 3 = 48, Hydrogen ( 3 ) = 1 [tex]*[/tex] 3 = 3 - 27 + 48 + 3 = 78 g / mol. This value is approximated however ( 78 g / mol ), as the molar mass of each substance is rounded as well. Another key thing we need to do here is to convert 340 mg → grams, considering that that unit is a necessity with respect to moles, as you might know - 340 mg = 0.340 g.

Now we can calculate how much moles of HCl will be present in solution, provided we have sufficient information for that,

(0.340 g Al(OH)3) / (78.0036 g / mol Al(OH)3) [tex]*[/tex] (3 mol HCl / 1 mol Al(OH)3)

⇒ (.004358773185 g^2 / mol Al(OH)3) [tex]*[/tex] (3 HCl / Al(OH)3 )

⇒ .01307632 mol HCl

We can apply this same concept on the reaction of Mg(OH)2 and water, receiving the number of moles of HCl when that takes place. Then we can add the two ( moles of HCl ) and divide by the value " 0.18 mol / L " given to us.

" Mg(OH)2 + 2HCl → MgCl2 + 2H2O "

Molar mass of Mg(OH)2 = 58.3197 g / mol,

516 mg = 0.516 g

(0.516 g Mg(OH)2) / (58.3197 g / mol Mg(OH)2) [tex]*[/tex] (2 mol HCl / 1 mol Mg(OH)2)

= .017695564 mol HCL

___________

( .01307632 + .017695564 ) / ( 0.18 M HCl )

= 0.170954911 L

= 171 mL of HCl

Answer:

6.5 mL

Explanation:

Step 1: Write the balanced reaction

Ca(OH)₂ + 2 HNO₃ ⇒ Ca(NO₃)₂ + 2 H₂O

Step 2: Calculate the reacting moles of nitric acid

25.0 mL of 0.50 M  nitric acid react.

[tex]0.0250L \times \frac{0.50mol}{L} = 0.013 mol[/tex]

Step 3: Calculate the reacting moles of calcium hydroxide

The molar ratio of Ca(OH)₂ to HNO₃ is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 0.013 mol = 6.5 × 10⁻³ mol

Step 4: Calculate the volume of calcium hydroxide

To answer this, we need the concentration of calcium hydroxide. Since the data is missing, let's suppose it is 1.0 M.

[tex]6.5 \times 10^{-3} mol \times \frac{1,000mL}{1.0mol} = 6.5 mL[/tex]