For dinner you make a salad with lettuce, tomatoes, cheese, carrots, and
croutons. Your salad would be classified as a(n)
O A. compound
OB. element
OC. homogeneous mixture
D. heterogeneous mixture​

Answer:

Your answer will either be 22.9897 or 22.990 !!

Explanation:

Answer:

Molarity of the sulfuric acid solution is 4.61M

Explanation:

The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:

2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻

That means, 2 moles of base react with 1 mole of sulfuric acid.

If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:

0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄

As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:

0.200 moles H₂SO₄ / 0.0434L = 4.61M

Answer:

Strontium is smaller

Strontium has the higher ionization energy

Strontium has more valence electrons

Explanation:

It must be understood that both elements belong to the same period i.e the same horizontal band of the periodic table

While Rubidium is an alkali metal(group 1) while Strontium is an alkali earth metal(group 2)

Since they are in the same period, periodic trends would be useful in evaluating their properties

In terms of atomic radius, rubidium is larger meaning it has a bigger atomic size

Generally, across the periodic table, atomic radius is expected to decrease and thus Rubidium which is leftmost is expected to have the higher atomic radius

Since strontium belongs to group 2 of the periodic table, it has 2 valence electrons which is more than the single valence electron that rubidium which is in group 1 has

In terms of ionization energy, the atom with the higher number of valence electrons will have the higher ionization energy which is strontium in this case

Answer:

Ionic Crystal

Explanation:

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T}[/tex]

We can solve for the temperature as follows:

[tex]T=\frac{\Delta H_{vap}}{\Delta S_{vap}}[/tex]

Thus, with the proper units, we obtain:

[tex]T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C[/tex]

Hence, answer is approximately 100 °C.

Best regards.

Answer:

[tex]Kp=0.0386[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]

For which the equilibrium expression is:

[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]

Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:

[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]

And the total pressure:

[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]

Yet it is 748 torr, for which the extent is:

[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]

Therefore, Kp turns out:

[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]

Best regards.

Answer:

Molar mass of the gas is 0.0961 g/mol

Explanation:

The effusion rate of an unknown gas = 11.1 min

rate of [tex]H_{2}[/tex] effusion = 2.42 min

molar mass of hydrogen = 1 x 2 = 2 g/m

molar mas of unknown gas = ?

From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.

from

[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]

where

[tex]R_{h}[/tex] = rate of effusion of hydrogen gas

[tex]R_{g}[/tex] = rate of effusion of unknown gas

[tex]M_{h}[/tex] = molar mass of H2 gas

[tex]M_{g}[/tex] = molar mass of unknown gas

substituting values, we have

[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587

[tex]\sqrt{M_{g} }[/tex] = 0.31

[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol

The molar mass of the unknown gas will be "0.0961 g/mol".

Given:

Effusion rate of unknown gas,

Effusion rate of [tex]H_2[/tex],

Molar mass of hydrogen,

              [tex]= 2 \ g/m[/tex]

According to the Graham's law, we get

→    [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]

By substituting the values, we get

→   [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]

→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]

→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]

   [tex]\sqrt{M_g} = 0.31[/tex]

       [tex]M_g = 0.0961 \ g/mol[/tex]

Thus the above solution is right.          

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Answer:

The process of slowly adding one solution to another until the reaction between the two is complete.

Explanation:

When you perform a titration, you are slowly adding one solution of a known concentration called a titrant to a known volume of another solution of an unknown concentration until the reaction reaches neutralization, in which the reaction is no longer taking place. This is often indicated by a color change.

Hope that helps.

Answer:

Approximately [tex]1.854\; \rm mol\cdot L^{-1}[/tex].

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

Look up the relative atomic mass of [tex]\rm Sr[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

Calculate the formula mass of [tex]\rm Sr(OH)_2[/tex]:

[tex]M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}[/tex].

[tex]M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}[/tex] means that each mole of [tex]\rm Sr(OH)_2[/tex] formula units have a mass of [tex]121.634\; \rm g[/tex].

The question states that there are [tex]10.60\; \rm g[/tex] of [tex]\rm Sr(OH)_2[/tex] in this solution.

How many moles of [tex]\rm Sr(OH)_2[/tex] formula units would that be?

[tex]\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}[/tex].

There are [tex]8.71467\times 10^{-2}\; \rm mol[/tex] of [tex]\rm Sr(OH)_2[/tex] formula units in this [tex]47\; \rm mL[/tex] solution. Convert the unit of volume to liter:

[tex]V = 47\; \rm mL = 0.047\; \rm L[/tex].

The molarity of a solution measures its molar concentration. For this solution:

[tex]\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

(Rounded to four significant figures.)

Answer:

C

Explanation:

Turning liquid to a solid is like freezing water to ice and requires the water to LOSE (release) heat causing an exothermic reaction.

Answer:

1.3 mL

Explanation:

First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:

1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.

It should have 2 significant figures, because 1.2kg has 2 and we are dividing.

The volume of olive oil will be nearly 1300mL or 1.30 L as per the given data.

Volume is a measurement of three-dimensional space that is occupied. It is frequently numerically quantified using SI derived units or various imperial units. The definition of length is linked to the definition of volume.

Volume is, at its most basic, a measure of space. The units liters (L) and milliliters (mL) are used to measure the volume of a liquid, also known as capacity.

This measurement is done with graduated cylinders, beakers, and Erlenmeyer flasks.

Here, it is given that mass of olive oil is 1.2kg.

We know that,

Density of olive oil = 0.917kg/l.

Volume = mass/density

Volume = 1.2/0.917.

Volume = 1.30 lit.

Volume = 1300mL.

Thus, the volume of olive oil will be 1300 mL.

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Answer:

3.00 L

Explanation:

Convert the pressure to Pascals.

P = 82 atm × (101325 Pa/atm)

P = 8,308,650 Pa

Convert temperature to Kelvins.

T = 27°C + 273

T = 300 K

Use ideal gas law:

PV = nRT

(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)

V = 0.00300 m³

If desired, convert to liters.

V = (0.00300 m³) (1000 L/m³)

V = 3.00 L

Answer:

[tex]\large \boxed{\text{3.0 L}}[/tex]

Explanation:

[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]

Answer:

[tex]\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl[/tex].

One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.

Explanation:

Start by finding the formula for each of the compound.

Therefore, the ratio between [tex]\rm K[/tex] atoms and [tex]\rm Br[/tex] atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula [tex]\rm KBr[/tex]. Similarly, the ratio between

The formula for chlorine gas is [tex]\rm Cl_2[/tex], while the formula for bromine gas is [tex]\rm Br_2[/tex].

Write down the equation using these chemical formulas.

[tex]\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl[/tex].

Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both [tex]\rm KBr[/tex] and [tex]\rm KCl[/tex] features two elements each.

Assume that the coefficient of [tex]\rm KCl[/tex] is one. Hence:

[tex]\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl[/tex].

Note that [tex]\rm KBr[/tex] is the only source of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms among the reactants of this reaction.

There would thus be one [tex]\rm K[/tex] atom and one [tex]\rm Br[/tex] atom on the reactant side of the equation.

Because atoms are conserved in a chemical equation, there should be the same number of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms on the product side of the equation.

In this reaction, [tex]\rm Br_2[/tex] is the only product with [tex]\rm Br[/tex] atoms.

One [tex]\rm Br[/tex] atom would correspond to [tex]0.5[/tex] units of [tex]\rm Br_2[/tex].

Similarly, in this reaction, [tex]\rm KCl[/tex] is the only product with [tex]\rm K[/tex] atoms.

One [tex]\rm K[/tex] atom would correspond to one formula unit of [tex]\rm KCl[/tex].

Hence:

[tex]\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].

Similarly, there should be exactly one [tex]\rm Cl[/tex] atom on either side of this equation. The coefficient of [tex]\rm Cl_2[/tex] should thus be [tex]0.5[/tex]. Hence:

[tex]\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].

That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:

[tex]\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl[/tex].

Answer:

Explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

.02M

CH₃COOH  = CH₃COO⁻ + H⁺

C                       xC             xC

Ka = xC . xC / C = x² C

1.8 x 10⁻⁵ = x² . .02

x² = 9 x 10⁻⁴

x = 3 x 10⁻²

= .03

concentration of H⁺ = xC = .03 . .02

= 6 x 10⁻⁴ M , volume =  45 x 10⁻³ L

moles of H⁺  = 6 X 10⁻⁴  x 45 x 10⁻³

= 270 x 10⁻⁷ moles

= 2.7 x 10⁻⁵ moles

concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L

moles of Na OH = 2 X 10⁻²  x 35 x 10⁻³

= 70 x 10⁻⁵ moles

=  

NaOH is a strong base so it will dissociate fully .

there will be neutralisation reaction between the two .

Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles

= 67.3 x 10⁻⁵ moles of NaOH

Total volume = 45 + 35 = 80 x 10⁻³

concentration of NaOH after neutralisation.= 67.3  x 10⁻⁵ / 80 x 10⁻³ moles / L

= 8.4125  x 10⁻³ moles / L

OH⁻ = 8.4125  x 10⁻³

H⁺ = 10⁻¹⁴ / 8.4125  x 10⁻³

= 1.1887 x 10⁻¹²

pH = - log (  1.1887 x 10⁻¹² )

= 12 - log 1.1887

= 12 - .075

= 11.925 .

The given question is incomplete. The complete question is :

Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?​

a) [tex]O^{2-}[/tex]

b)  [tex]F^{-}[/tex]

c)  [tex]N^{3-}[/tex]

d)  [tex]S^{2-}[/tex]

Answer: b)  [tex]F^{-}[/tex]

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

Here potassium is having an oxidation state of +1 called as  cation and thus is an anion must have an oxidation state of -1 if they have to combine in 1: 1 ratio to  give neutral ionic compound.

Thus the anion has to be [tex]F^-[/tex] which combines with [tex]K^+[/tex] in 1: 1 ratio to give [tex]KF[/tex]

Answer:

Option (b) Hydrocyanic acid, 4.9×10^-10

Explanation:

Data obtained from the question include:

Ka of Hydrofluoric acid = 3.5×10^-4

Ka of Hydrocyanic acid = 4.9×10^-10

Ka of Nitrous acid = 4.6×10^-4

To know which acid is least acidic, we shall determine the the pKa value for each acid.

This is illustrated below:

For Hydrofluoric acid

Ka = 3.5×10^-4

pKa =..?

pKa = –Log Ka

pKa = –Log 3.5×10^-4

pKa = 3.5

For Hydrocyanic acid

Ka = 4.9×10^-10

pKa =..?

pKa = –Log Ka

pKa = –Log 4.9×10^-10

pKa = 9.3

For Nitrous acid

Ka = 4.6×10^-4

pKa =..?

pKa = –Log Ka

pKa = –Log 4.6×10^-4

pKa = 3.3

Summary:

Acid >>>>>>>>>>>>> Ka >>>>>>>> pKa

Hydrofluoric acid >> 3.5×10^-4 >> 3.5

Hydrocyanic acid >> 4.9×10^-10 > 9.3

Nitrous acid >>>>>>> 4.6×10^-4 >> 3.3

NB: The smaller the pKa value, the more acidic the compound is and the larger the pKa value, the less acidic the compound will be.

From the above calculations, Hydrocyanic acid has the highest pKa value.

Therefore, Hydrocyanic acid is the least acidic compound

Answer:

0.373 M

Explanation:

The balanced equation for the reaction is given below:

HC2H3O2 + NaOH —> NaC2H3O2 + H2O

From the balanced equation above, the following were obtained:

Mole ratio of the acid, HC2H3O2 (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, we shall write out the data obtained from the question. This include:

Volume of base, NaOH (Vb) = 32.17 mL

Molarity of base, NaOH (Mb) = 0.116 M

Volume of acid, HC2H3O2 (Va) = 10 mL

Molarity of acid, HC2H3O2 (Ma) =..?

The molarity of the acid solution can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 0.116 x 32.17 = 1

Cross multiply

Ma x 10 = 0.116 x 32.17

Divide both side by 10

Ma = (0.116 x 32.17) /10

Ma = 0.373 M

Therefore, the concentration of the acetic acid is 0.373 M.

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:

(a) The diode voltage,  [tex]V_D =[/tex]  0.776 V

(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A

Explanation:

Given;

saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A

nonideality factor, n = 1.05

(a) the diode voltage

Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A

Diode voltage is calculated as;

[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]

Where;

[tex]V_T[/tex] is thermal voltage at 25°C = 0.025

[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]

b) the diode current for VD = 0.1 mV

[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]

Answer:

An alkyl halide can undergo SN2 reaction with an amine

Explanation:

The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.

The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).

This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.

The detailed mechanism of this reaction has been attached to this answer.

Answer:

6

Explanation:

Alkanes undergo substitution reaction so the number of replacement reaction hydrogen is 6

Answer:

Water

Explanation:

Solid potassium hydrogen tartrates (KHT) is soluble in water. This is especially at room temperature.

The solvent for KHT is water.

Answer:

Lewis acid

Explanation:

In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.

If we look at the formation of PF6^-, the process is as follows;

PF5 + F^- -----> PF6^-

We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.

Hence in the formation of PF6^-, PF5 acts a Lewis acid.

Answer:

Partial pressure (Ar) = 316.1mmHg

Explanation:

In the mixture of Ar and H₂ you can find the total moles of both gases using general gas law and with the mass of the sample and molar weight of each gas find the mole fraction of Argon and thus, its partial pressure.

Moles of gases:

PV = nRT

P = 910.0mmHg ₓ (1atm / 760mmHg) = 1.1974atm

V = 1.66L

n = Moles gases

R = 0.082atmL/molK

T = 54.9°C + 273.15K = 328.05K

PV = nRT

1.1974atm*1.66L = n*0.082atmL/molK*328.05K

Knowing H₂ = 2.016g/mol and Ar = 39.948g/mol you can write:

1.13g = 2.016X + 39.948Y (1)

Where X = moles of hydrogen and Y = moles of Argon.

Also we can write:

0.0739moles = X + Y (2)

Total moles of the sample are moles of hydrogen + moles Argon

Replacing 2 in 1:

1.13g = 2.016(0.0739-Y) + 39.948Y

1.13 = 0.1564 - 2.016Y + 39.948Y

0.9736 = 37.932Y

0.02567 = Y = moles of Argon

As total moles are 0.0739moles, mole fraction of Ar in the sample are:

XAr = 0.02567mol / 0.0739mol

X Ar = 0.347

Last, partial pressure of Ar = X Ar * total pressure.

Partial pressure (Ar) = 0.347*910.0mmHg

Answer:

Bleaching Powder's chemical formula is CaOCl2 and is called Calcium Oxychloride. It is prepared on dry slaked lime by chlorine gas. 2. ... It gives calcium chloride, chlorine and water when bleaching powder reacts with hydrochloric acid.

Explanation:

Complete question:

Write the condensed formula from left to right, starting with (CH3)x where x is a number.

See attached image for the structure formula of the compound

Answer:

(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane

Explanation:

If we number the longest chain of the carbon starting from the left, we will observe that there are four carbons in the straight chain as shown in the image.

Starting from first carbon from the left of the carbon chain, at carbon number number 2, there two alkyl group, that is two methyl (CH3 is two). Also at carbon number 3, there are three alkyl group, that is three methyl (CH3 is three).

The condensed formula will be written as;

(CH₃)₂CHC(CH₃)₃

This compound is named as 2,2,3-Trimethylbutane, an isomer of Heptane

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:

1,4-hexanediamine contains two [tex]-NH_{2}[/tex] functional groups.

Explanation:

1,4-hexanediamine is an organic molecule which contains two [tex]-NH_{2}[/tex] functional groups at C-1 and C-4 position.

The longest carbon chain in 1,4-hexanediamine contains six carbon atoms.

Molecular formula of 1,4-hexanediamine is [tex]C_{6}H_{16}N_{2}[/tex].

1,4-hexanediamine used as a bidentate ligand in organometallic chemistry.

The structure of 1,4-hexanediamine is shown below.