Calculate the percent oxygen in sodium oxide.

Answer:

77 L of water can be made.

Explanation:

Molar mass of [tex]O_{2}[/tex] = 32 g/mol

So, 55 g of [tex]O_{2}[/tex] = [tex]\frac{55}{32}[/tex] mol of  [tex]O_{2}[/tex] = 1.72 mol of  [tex]O_{2}[/tex]

As hydrogen is present in excess amount therefore  [tex]O_{2}[/tex] is the limiting reagent.

According to balanced equation, 1 mol of  [tex]O_{2}[/tex] produces 2 mol of [tex]H_{2}O[/tex].

So, 1.72 mol of [tex]O_{2}[/tex] produce [tex](2\times 1.72)[/tex] mol of [tex]H_{2}O[/tex] or 3.44 mol of [tex]H_{2}O[/tex].

Let's assume [tex]H_{2}O[/tex] gas behaves ideally at STP.

Then, [tex]P_{H_{2}O}.V_{H_{2}O}=n_{H_{2}O}.R.T[/tex]   , where P, V, n, R and T represents pressure, volume, no. of moles, gas constant and temperature in kelvin scale respectively.

At STP, pressure is 1 atm and T is 273 K.

Here, [tex]n_{H_{2}O}[/tex] = 3.44 mol and R = 0.0821 L.atm/(mol.K)

So, [tex](1atm)\times V_{H_{2}O}=(3.44mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)[/tex]

  [tex]\Rightarrow[/tex]  [tex]V_{H_{2}O}=77L[/tex]

Option (b) is correct.

Final answer:

Using stoichiometry, 55 grams of oxygen gas can produce 77 liters of water at STP, considering the molar volume of a gas at STP is 22.4 liters per mole.

Explanation:

To determine how many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at STP, we must first convert the mass of oxygen to moles and then use stoichiometry to find the volume of water produced. The balanced chemical equation for the reaction is:

2 H2(g) + O2(g) → 2 H2O(g)

At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Therefore, the moles of oxygen (O2) can be calculated using its molar mass (32 g/mol):

Moles of O2 = 55 g / 32 g/mol = 1.71875 moles

The balanced equation shows us that 1 mole of O2 produces 2 moles of H2O.

Total moles of H2O produced = 1.71875 moles of O2 × 2 = 3.4375 moles

Volume of H2O (at STP) = 3.4375 moles × 22.4 L/mol = 76.92 L

Therefore, when rounded to the nearest whole number, 77 liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at STP, which corresponds to option b.

Answer:

[tex]m = 1\,g[/tex]

Explanation:

The molarity is the ratio of the amount of moles solvent to volume of the solute.

[tex]0.010\,M = \frac{0.010\,moles\,NaOH}{1\,L}[/tex]

The quantity of solute is determined by simple rule of three:

[tex]n = \left(\frac{2.5\,L}{1\,L} \right)\cdot (0.010\,mole)[/tex]

[tex]n = 0.025\,moles[/tex]

The molecular weight of NaOH is [tex]39.997\,\frac{g}{mole}[/tex], the mass of solute is:

[tex]m = 1\,g[/tex]

Answer:

2NaHCO3(aq)+H2SO4(aq)→Na2SO4(aq)+2CO2(g)+2H2O(l)

Explanation:

As we know that

acid + carbonate  →  salt + carbon dioxide + water

So, the general (un-balanced) equation would be-

NaHCO3(aq)+H2SO4(aq)→Na2SO4(aq)+CO2(g)+H2O(l)

Now we will write the net ionic reactions

[tex]HCO_3^-+ H3O^+[/tex] ----> CO2(g)↑+2H2O(l)

[tex]Na ---> Na^+ + e^-[/tex]

[tex]2H^+ + 2e^- ---> H2[/tex]

[tex]SO_4^{2-} ---- SO_4 + 2e^-[/tex]

Adding all the above equation, we get

2NaHCO3(aq)+H2SO4(aq)→Na2SO4(aq)+2CO2(g)+2H2O(l)

Answer: Concentration of N₂ is 4.8.[tex]10^{9}[/tex] M.

Explanation: [tex]K_{c}[/tex] is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For

N2(g) + 2 O2(g) ⇄ 2 NO2(g)

[tex]K_{c}[/tex] = [tex]\frac{[NO2]^{2} }{[N2][O2]^{2} }[/tex]

From the question concentration of NO2 is twice of O2:

[NO2] = 2[O2]

Substituting this into [tex]K_{c}[/tex]:

[tex]K_{c}[/tex] = [tex]\frac{[2O2]^{2} }{[N2][O2]^{2} }[/tex]

8.3.[tex]10^{-10}[/tex] = [tex]\frac{4O2^{2} }{[N2].O2^{2} }[/tex]

[N2] = [tex]\frac{4O2^{2} }{8.3.10^{-10}.O2^{2} }[/tex]

[N2] = [tex]\frac{4}{8.3.10^{-10} }[/tex]

[N2] = 4.8.[tex]10^{9}[/tex]

The concentration of N2 in the equilibrium is [N2] = 4.8.[tex]10^{9}[/tex]M.

The concentration of [tex]N_2[/tex] is [tex]4.8.10^9M[/tex]

Since k_e represent the equilibrium constant and based on the concentrations of the reactants and the products of a balanced reaction.

Also, the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), And, the concentration of NO2 is twice the concentration of O2 gas

Kc = 8.3 × 10-10 at 25°C.

NO2 is twice of O2.

Now

[tex]8.310^{-10} = \frac{4O_2^2}{N_2O_2^2} \\\\N_2 = \frac{4O_2^2}{8.3.10^{-10}O_2^2}\\\\ = 4\div 8.3.10^{-10}\\\\= 4.8.10^9[/tex]

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Answer:

0.100 m AlCl3  will have the highest boiling point

Explanation:

Step 1: Data given

The molal boiling point elevation constant for water is 0.51°C/m

Since those are all  aqueous solutions, the have the same molal boiling point elevation constant

Step 2:

0.100 m C6H12O6

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr = 1

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT = 1 * 0.51 * 0.100

ΔT  = 0.051 °C

0.100 m NaCl

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr =  Na+ + Cl- = 2

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =2 * 0.51 * 0.100

ΔT = 0.102 °C

0.100 m AlCl3

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr =  Al^3+ + 3Cl- = 4

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =4 * 0.51 * 0.100

ΔT = 0.204 °C

0.100 m MgCl2

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr =  Mg^2+ +2Cl- = 3

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =3 * 0.51 * 0.100

ΔT = 0.153 °C

0.100 m AlCl3  will have the highest boiling point

Answer:

9.1 seconds

Explanation:

Given that for a second order reaction

1/[A]t = kt + 1/[A]o

Where [A]t= concentration at time = t= 0.340M

[A]o= initial concentration = 0.820M

k= rate constant for the reaction=0.190m-1s-1

t= time taken for the reaction (the unknown)

Hence;

(0.340)^-1 = 0.190×t + (0.820)^-1

t= (0.340)^-1 - (0.820)^-1/0.190

t= 9.1 seconds

Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.

Final answer:

To calculate the time it takes for the concentration of compound A to decrease in a second-order reaction, the integrated rate law 1/[A] - 1/[A]0 = kt is used with the provided rate constant and initial and final concentrations.

Explanation:

Calculating Time for a Second-Order Reaction:

The question pertains to the time it takes for the concentration of compound A to decrease from 0.820 M to 0.340 M in a second-order reaction with a rate constant of 0.190 M-1·s-1 at 300 °C. The integrated rate law for a second-order reaction is given by:

1/[A] - 1/[A]0 = kt

Where [A]0 is the initial concentration, [A] is the concentration at time t, k is the rate constant, and t is the time. Using the provided concentrations, we can solve for t:

1/0.340 M - 1/0.820 M = (0.190 M-1·s-1)t

After calculating the left side, we can isolate t:

t = (1/0.340 M - 1/0.820 M) / (0.190 M-1·s-1)

Hence:

(0.340)^-1 = 0.190×t + (0.820)^-1

t= (0.340)^-1 - (0.820)^-1/0.190

t= 9.1 seconds

Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.

Answer:

The mass of the solute and the volume of the solution.

Explanation:

Hello,

In this case, given the formula of molarity:

[tex]M=\frac{n_{solute}}{V_{solution}}[/tex]

In such a way, since the moles could not be directly measured, we must measure the mass of the solute and by using its molar mass, one could compute its moles. Moreover, since the solution is composed by the solvent (typically water) and the solute, we consequently must measure the volume of the solution needed for the preparation of such concentration-known solution. In such a way, we can actually prepare the required solution.

Best regards.

For the formation of 2 M solution of potassium nitrate, mass and volume of the solution has been measured.

Molarity can be defined as the mass of solute present in a liter of solution. The molarity has been used for the determination of the concentration of the compounds.

It can be expressed as mol/L. The molarity (M) has expression:

[tex]M=\rm \dfrac{solute\;mass}{solute\;molar\;mass}\;\times\;Solution\;Volume[/tex]

For the formation of 2 M potassium nitrate solution, the mass of the solute and the volume of the solution has to be measured.

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Answer:

1.22

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction. This is given below:

HCl + KOH —> KCl + H2O

From the balanced equation above,

The mole ratio of acid (nA) = 1

The mole ratio of base (nB) = 1

Step 2:

Data obtained from the question. This includes the following:

Volume of acid (Va) =

(62.35 + 62.40)/2 = 62.38 mL

Molarity of acid (Ma) =.?

Volume of base (Vb) = 25 mL

Molarity of base (Mb) = 0.150M

Step 3:

Determination of the molarity of the acid.

This is illustrated below:

MaVa/MbVb = nA/nB

Ma x 62.38/ 0.15 x 25 = 1

Cross multiply to express in linear form

Ma x 62.38 = 0.15 x 25

Divide both side by 62.38

Ma = (0.15 x 25) /62.38

Ma = 0.06M

The molarity of the acid is 0.06M

Step 4:

Determination of the concentration of Hydrogen ion, [H+] in the acid. This is illustrated below:

Hydrochloric acid (HCl) will dissociate to produce hydrogen ion as follow:

HCl —> H+ + Cl-

From the above equation,

1 mole of HCl produced 1mole of H+.

Therefore, 0.06M HCl will also produce 0.06M H+.

The concentration of Hydrogen ion, [H+] is 0.06M

Step 5:

Determination of the pH of HCl. This is illustrated below:

pH = – Log [H+]

[H+] = 0.06M

pH = – Log 0.06

pH = 1.22

Therefore, the pH of HCl is 1.22

Answer:

See explaination

Explanation:

H O / I MOH the OH da Anomer My nequatorial bovel SOH niematerial and CH2OH Yr axial boud - H o H o . H (Auro) 2 flip . note: 1)Fischer projection : L- Sugar ( OH group is in the left side ) at C-5 carbon (bottom most chiral center )

In the Haworth projection given above in the qsn it is Beta anomer( -CH2OH & C-OH are in same side i.e. Cis to each other ). It is a L -sugar .

I omitted C2 -H bond (axial ) in the chair conformation of alpha anomer.

See attachment for further solution

Transitioning from a Haworth to a Fischer projection, every carbon atom becomes a line end or intersection point. Correct position and orientation of the hydroxyl group determines the chair configuration. Carbohydrates, which contain carbon, hydrogen, and oxygen atoms in a 1:2:1 ratio, have monosaccharide, disaccharide, and polysaccharide subtypes.

The question is asking you to convert a Haworth projection of a carbohydrate into a Fischer projection and a chair conformation of its opposite anomer. This involves understanding several concepts in organic chemistry. The Haworth projection is a common way of representing cyclic sugars, which are typically found in ring forms in aqueous solutions. The Fischer projection is used to represent the stereochemistry of the molecule. In converting the Haworth projection to a Fischer projection, each carbon atom in the ring becomes the end of a line or a point where lines intersect in the Fischer projection. The most oxidized carbon (typically the one attached to an oxygen atom) is placed at the top.

Next, to draw the chair conformation, you must understand axial and equatorial bonds, which describe the orientation of chemical groups around the carbon atoms in the ring. Axial bonds are vertical and shunted to the side, while equatorial bonds are more horizontal and radiate outwards. For the opposite anomer, if the original has the hydroxyl group below the plane in an alpha position, the opposite anomer would have it above the plane in a beta position.

Carbohydrates bear the stoichiometric formula (CH₂O)n, indicating a carbon to hydrogen to oxygen ratio of 1:2:1. Classified into monosaccharides, disaccharides, and polysaccharides, carbohydrates are a crucial chemical group in biology.

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Answer:

decrease

Explanation:

Answer:

The correct answer is 0.00582 grams.

Explanation:

In order to solve the question, let us consider the vapor pressure of H2O, as hydrogen gas is collected over water, therefore, we have to consider the vapor pressure of water in the given case. Let us assume that the pressure is 760 torr or 1 atm.

It is known that the vapor pressure of water at 40 degree C is 53.365 torr (Based on the data).

Therefore, the pressure of H2 will be,

P = 760-55.365 = 704.635 torr or 704.635/760 = 0.9272 atm

The volume of the hydrogen gas collected in the tube is 80 ml or 0.08 L

Temperature in Kelvin will be 40+273 = 313 K

To calculate the moles of hydrogen (H2) gas, there is a need to use the ideal gas equation, that is, PV= nRT, in this R is the gas constant, whose value is 0.0821 L atm/molK, and n is the moles of the gas.

By inserting the values in the equation we get:

PV = nRT

n = PV/RT = 0.9272 *0.08 / 0.0821 * 313

n = 0.00289 moles

The mass of H2 will be moles * molar mass = 0.00289 * 2.016

= 0.00582 grams.

Answer:

The correct answer is 89.6 L

Explanation:

We have the following chemical equation and the molar masses for the reaction:

3H₂(g)   +   N₂ -->   2 NH₃

 6 g            28 g      34 g

That means that 3 moles of H₂ (6 g) reacts with 1 mol of N₂ (28 g) and gives 2 moles of NH₃ (34 g). In order to calculate how many liters of NH₃ result from the reaction of 12 grams of H₂ and 28 grams of N₂, we have to first figure out which reactant is the limiting reactant. According to the equation, if 6 grams of H₂ reacts with 28 g of N₂, and we have 12 grams:

6 g H₂------- 28 g N₂

12 g H₂-------- X = 12 g H₂ x 28 g N₂/6 g H₂ = 56 g N₂

We need 56 g of N₂ but we have 28 g of N₂, so N₂ is the limiting reactant. With the limiting reactant we can calculate the moles of product (NH₃) we will obtain:

We have 28 g N₂ -----> 28 g/14 g/mol = 2 moles N₂

1 mol N₂ ----------- 2 moles NH₃

2 mol N₂ --------- X = 2 mol N₂ x 2 moles NH₃/1 mol N₂ = 4 mol NH₃

Finally, we convert the moles of NH₃ to liters:

1 mol gas at STP = 22.4 L

Liters NH₃ obtained = 4 moles NH₃ x 22.4 L/1 mol = 89.6 L

Final answer:

When 12 grams of H2 and 28 grams of N2 react completely according to the equation 3H2(g) + N2(g) → 2NH3(g), 34.08 grams of NH3 are produced, with some H2 remaining unreacted due to it being in excess.

Explanation:

The question asks about the result when 12 grams of H2 and 28 grams of N2 react to completion according to the balanced chemical equation 3H2(g) + N2(g) -> 2NH3(g). This equation signifies that one mole of nitrogen gas (N2) reacts with three moles of hydrogen gas (H2) to produce two moles of ammonia (NH3). Given the molar masses (N2 = 28.02 g/mol, H2 = 2.02 g/mol), we can determine that the initial amounts provided are excess H2. Precisely, 28 grams of N2 is one mole, and 12 grams of H2 is six moles, which is enough to fully react according to the stoichiometry of the equation. The reaction of 28.02 g of N2 and 6.06 g of H2 produces 34.08 g of NH3, according to mass conservation principles. Since more H2 is provided than required, only 6.06 g will be consumed, leaving excess H2 unreacted. Therefore, the reaction yields 34.08 grams of NH3.

Final answer:

The reaction in question demonstrates the application of Le Chatelier's Principle, with shifts in equilibrium occurring in response to decreased pressure, increased temperature, and the removal of a reactant.

Explanation:

The reaction presented is dealing with changes in reaction conditions in a chemical equilibrium situation. This is directly related to Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In this scenario:

Answer:

See explanation below

Explanation:

This reaction is known as Ketone hydrolisis in acid medium. This involves the formation of an hemi cetal, and then, the acetal. This is often used to convert ketones or aldehydes in ethers.

The first step involves the reaction with the acid. The carbonile reacts with the acid and forms an alcohol there. The next step is the reaction of the alcohol, in this case, the methanol to form the hemi cetal. Then in the third step, we repeat the first step, using acid to turn the OH group into a great leaving group such water. Then the water leaves the molecule, leaving the space wide open in the next step for methanol, and the acetal is formed.

See picture for the curved arrow mechanism

Protonation of 4-methylpentan-2-one creates a positive carbon center. First, methanol attacks this center, then a proton transfer occurs. After water is lost forming an oxonium ion, a second methanol attacks the intermediate and deprotonation results in an acetal.

In order to draw the curved arrow mechanism for the formation of an acetal from acidic methanol and 4-methylpentan-2-one, we proceed as follows:

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Answer:

5 kJ/g  

Explanation:

There are two energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the solution

 q₁   +    q₂     = 0

m₁ΔH + m₂CΔT = 0

Data:

m₁ = 0.258 g

V₂ = 100 mL

  C = 4.184  J°C⁻¹g⁻¹

T_i = 22 °C

T_f = 25.1 °C

Calculations

(a) Mass of solution

[tex]\text{Mass} = \text{100 mL} \times \dfrac{\text{1.00 g}}{\text{1 mL}} = \text{100 g}[/tex]

(b) ΔT

ΔT = T_f - T_i = 25.1 °C - 22 °C = 3.1°C

(c) ΔH

[tex]\begin{array}{ccccl}m_{1}\Delta H & +& m_{2}C \Delta T& = &0\\\text{0.258 g}\times \Delta H& + & \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 3.1 \, ^{\circ}\text{C} & = & 0\\0.258 \Delta H \text{ g} & + & \text{1300 J} & = & 0\\&&0.258 \Delta H \text{ g} & = & \text{-1300 J} & & \\& &\Delta H & = & \dfrac{\text{-1300 J}}{\text{0.258 g}}\\\\& & & = & \text{-5000 J/g}\\& & & = & \textbf{-5 kJ/g}\\\end{array}[/tex]

[tex]\text{The reaction produces $\large \boxed{\textbf{5 kJ}}$ per gram of potassium.}[/tex]

Note: The answer can have only one significant figure because you measured the initial temperature of the water only to the nearest degree.

From the calculation, the heat generated from the solution is -194.4 kJ/mol

A calorimeter is an instrument that is used to measure heat.

Now we know that number of moles of the potassium = 0.258 g /39 g/mol = 0.0066 moles

Total mass present = 0.258 g + 100 g = 100.258 g

Temperature change = 25.1°C  - 22°C = 3.1°C

Now;

H = -(100.258  * 4.128 *  3.1)/ 0.0066

= -194.4 kJ/mol

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Answer:

0.1 M KOH

Explanation:

The thymol blue indicator will appear blue in a basic solution, particularly in a solution with a pH greater than 8.0. A diluted sodium hydroxide solution is an example of such a basic environment where thymol blue would turn blue.

The thymol blue indicator will appear blue in a basic solution. Specifically, thymol blue changes color from yellow to blue over a pH range of approximately 8.0 to 9.6. This means that for thymol blue to exhibit a blue color, it needs to be in a solution with a pH greater than 8.0, indicating a basic environment.

An example of a solution in which thymol blue would appear blue is a diluted sodium hydroxide (NaOH) solution, as NaOH is a strong base that would increase the pH of the solution.

Answer:

The molecular weight of this compound is 386.4 g/mol

Explanation:

Step 1: Data given

MAss of the compound = 11.09 grams

Volume of diethyl ether = 180.9 mL

Osmotic pressure = 3.88 atm

Temperature = 298 K

The compound =  nonvolatile and non-electrolyte

Step 2: Calculate molar concentration

π = i*M*R*T

⇒with π = the osmotic pressure = 3.88 atm

⇒with i = the van't Hoff factor = 1

⇒with C = the molar concentration = = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = 298 K

C = 3.88 / (0.08206*298)

C = 0.1587 M

Step 3: Calculate moles compound

C = moles / volume

moles = 0.1587 M * 0.1809 L

Moles compound = 0.0287 moles

Step 4: Calculate molecular weight of the compound

Molar mass = mass / moles

Molar mass compound = 11.09 grams / 0.0287 moles

Molar mass compound = 386.4 g/mol

The molecular weight of this compound is 386.4 g/mol

Answer:

= 1.5 moles of carbon

Explanation:

Molar mass of Carbon = 12g

12g of carbon are weighed by 1 mole of carbon

18g of carbon will be weighed by [(18÷12)×1]

= 1.5 moles

The number of moles of carbon atoms in 18 grams is approximately 1.498 moles.

To calculate the number of moles of carbon atoms in 18 grams, we need to use the molar mass of carbon. The molar mass of carbon is 12.01 g/mol. We can use the formula:

Number of moles = Mass / Molar mass

So, for 18 grams of carbon:

Number of moles = 18 g / 12.01 g/mol = 1.498 moles

Therefore, there are approximately 1.498 moles of carbon atoms in 18 grams.

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Answer: The amount of heat released is 42739 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat released =?

c = heat capacity of granite = [tex]0.790J/g^0C[/tex]

Initial temperature  = [tex]T_i[/tex] = [tex]41.2^0C[/tex]

Final temperature of the calorimeter  = [tex]T_f[/tex]  = [tex]-12.9^0C[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i=(-12.9-41.2)^0C=-54.1^0C[/tex]

Putting in the values, we get:

[tex]Q=1000g\times 0.790J/g^0C\times -54.1^0C=-42739J[/tex]

As heat comes out to be negative, that means the heat has been released and the amount of heat released is 42739 Joules

Answer:

C) 3.40 x 10-3

Explanation:

Hello,

In this case, for the given pOH, we are able to compute the concentration of hydrixide ions by applying the following formula:

[tex]pOH=-log([OH^-])[/tex]

[tex][OH^-]=10^{-pOH}=10^{-2.468}\\[/tex]

[tex][OH^-]=0.0034M=3.4x10^{-3}M[/tex]

Therefore, answer is C).

Best regards.

Answer:

3.40 × 10⁻³ M

Explanation:

The pOH scale is used to express the acidity or basicity of a solution.

The pOH of this solution is 2.468, so it is basic. We can calculate the concentration of hydroxide ions in the solution using the following expression.

[tex]pOH = -log [OH^{-} ]\\\[[OH^{-}] = antilog-pOH = antilog-2.468 = 3.40 \times 10^{-3} M[/tex]

Answer:

The fraction of molecules with kinetic energy greater than Ea increases.

Explanation:

In the scenario above, it is primarily by the fraction of molecules with kinetic energy higher or bigger that Ea increases.

Naturally chemical reactions are everywhere around you and inside you. In addition to the biochemical reactions that allow you to transform the molecules you eat and breathe into usable energy, there are industrial laboratories in cities around the world producing chemicals as well as products that rely on chemicals for their manufacture.

In addition to the product or products produced and having a proper supply of reactants, is how quickly a reaction can be expected to proceed. This can have an impact on safety, product quality and other outcomes.

Final answer:

The reaction rate increases with temperature because more molecules possess the necessary kinetic energy to overcome the activation energy, leading to a higher number of effective collisions and thus an increased rate of reaction.

Explanation:

The rate of a reaction typically increases as the temperature increases because the fraction of molecules with kinetic energy greater than the activation energy (Ea) increases. As the temperature rises, molecules move more rapidly, leading to a higher frequency of collisions. More importantly, these collisions are more energetic which means more of them have enough energy to surmount the activation energy barrier. The Arrhenius equation helps in understanding this phenomenon, predicting that the rate constant (k) increases with temperature since a larger fraction of reactant molecules has enough kinetic energy to overcome Ea, resulting in an increased reaction rate.

Answer:

K-between the first and second ionization energies

Ca-between the second and third ionization energies

Explanation:

Ionization energy is the energy required to remove an electron from a gaseous atom or ion. The first ionization energy of an atom is the energy required to remove one mole of electrons from one mole of isolated gaseous atoms.

The second ionization energy is usually greater in magnitude than the first ionization energy. However, when the electrons on the valence shell of the atom are exhausted, further ionization will involve the core electrons of the atom. This requires the removal of electrons from a completely filled Shell, this process requires a very large amount of energy. The energy required is far larger than the energy required to remove valence electrons from the outermost shell of the atom. This sudden tremendous increase in the magnitude of ionization energy as electrons are removed from a completely filled shell is known as a jump.

For calcium, a jump occurs after the second electron has been removed, the third electron must be removed from a completely filled inner shell. For potassium, a jump occurs after the first electron has been removed. The second electron must be removed from a completely filled inner shell.

For potassium (K), the largest jump between successive ionization energies is between the first and second. For calcium (Ca), the largest jump between successive ionization energies occurs between the second and third.

For potassium (K), the largest jump between successive ionization energies occurs between the first and second ionization energies. The first ionization energy involves removing an electron from the higher energy level (4s), while the second involves removing an electron from a lower energy level (3p), so a much stronger attraction exists between the nucleus and the electron to be removed in the second ionization.

For calcium (Ca), the largest jump between successive ionization energies occurs between the second and third ionization energies. The third ionization requires removal of an electron from a lower energy level, which is closer to the nucleus and thus retains a stronger attraction. This removal requires significantly more energy than the first two ionizations, which involve removing the outermost 4s electrons.

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Answer:

pH = 8.93

Explanation:

In this case, this titration is the case of a weak acid and a strong base. Now, at the equivalence point, it's supposed that we have the same moles of each reactant in solution, and we will expect that the pH would have to be 7. However, as the acid is pretty weak, there's a little difference in the solution because of the grade of dissociation of the acid, and the pH will be higher than 7. To know this, we first need to calculate the volume of added base:

M₁V₁ = M₂V₂

With this expression, let's calculate the volume of the base:

V₂ = M₁V₁ / M₂

V₂ = 0.1917 * 220 / 0.1787 = 236 mL

So, at the equivalence point, 236 mL are needed to neutralize this reaction. As the moles are the same for each reactant, we just need to calculate the concentration of the acid in this part. This will be the sum between the initial volume of acid and the calculated volume of base:

V of solution = 236 + 220 = 456 mL or 0.456 L

Then, the new concentration of the acid is:

[C₂H₅COOH] = 0.1917 * 0.220 / 0.456 = 0.0924 M

Now, the reaction with the base is the following:

C₂H₅COOH + KOH --------> C₂H₅COOK + H₂O

This means that in the equivalence point we have the propionic potassium and water, so, if take this and dissociates into it's ions we can calculate the pH of the solution:

C₂H₅COO⁻ + H₂O <-------> C₂H₅COOH + OH⁻

With this reaction in solution in the equivalence point, we just need the Kb of propionate ion, and this can be calculated with the value of the pKa of the acid:

Ka = 10^(-pKa)

Ka = 1.29x10⁻⁵

Now the value of Kb can calculated using the following expression:;

Kb = Kw / Ka ---> replacing we have

Kb = 1x10⁻¹⁴ / 1.29x10⁻⁵

Kb = 7.75x10⁻¹⁰

Now, with this value and the above reaction we can write an ICE chart to calculate the [OH⁻] and then, the pH of solution:

     C₂H₅COO⁻ + H₂O --------> C₂H₅COOH + OH⁻   Kb = 7.75x10⁻¹⁰

i)       0.0924                                    0               0

e)     0.0924-x                                  x               x

The Kb expression:

Kb = [C₂H₅COOH] [OH⁻] / [C₂H₅COO⁻]

7.75x10⁻¹⁰ = x² / 0.0924-x ---> Kb is very small, so this substraction can be neglected to just 0.0924 assuming x will be very small too.

7.75x10⁻¹⁰ = x² / 0.0924

7.75x10⁻¹⁰ * 0.0924 = x²

x = [OH⁻] = 8.46x10⁻⁶ M

With this value, we can calculate pOH and then the pH:

pOH = -log(8.46x10⁻¹⁰) = 5.07

Finally the pH:

pH = 14 - pOH

pH = 14 - 5.07

Answer:

The new volume is 1.62 L

Explanation:

Boyle's law says:

"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure." It is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

[tex]\frac{V}{T}=k[/tex]

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. So this law indicates that the quotient between pressure and temperature is constant.

Gay-Lussac's law can be expressed mathematically as follows:

[tex]\frac{P}{T}=k[/tex]

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law.

[tex]\frac{P*V}{T}=k[/tex]

Having an initial state 1 and a final state 2 it is possible to say that:

[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]

Standard temperature and pressure (STP) indicate pressure conditions P = 1 atm and temperature T = 0 ° C = 273 ° K. Then:

Replacing:

[tex]\frac{1 atm* 1.2 L}{273K} =\frac{0.8 atm*V2}{294K}[/tex]

Solving:

[tex]V2=\frac{1 atm*1.2 L}{273 K} *\frac{294 K}{0.8 atm}[/tex]

V2= 1.62 L

The new volume is 1.62 L

Answer:   11.6g of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.

Explanation:

To calculate the amount of solute, we use the equation for osmotic pressure, which is:

[tex]\pi=iMRT[/tex]

Or,

[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure of the solution = 4.45 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Let Mass of solute (estrogen)  = x g  

Volume of solution = 234 mL

R = Gas constant = [tex]0.0821Latmmol^{-1}K^{-1}[/tex]

T = temperature of the solution = [tex]298K[/tex]

Putting values in above equation, we get:

[tex]4.45=1\times \frac{x\times 1000}{272.4\times234}\times 0.0821Latmmol^{-1}K^{-1}\times 298K[/tex]

[tex]x=11.6g[/tex]

Hence, 11.6g of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.

The question is about calculating the quantity of estrogen needed to generate a specific osmotic pressure in a benzene solution using the Van 't Hoff equation. The moles of estrogen needed is calculated first, then converted into grams using the molar mass of estrogen.

This question is based on the concept of osmotic pressure and solution chemistry, for a nonvolatile, nonelectrolyte compound in solution. The Van 't Hoff equation (π= nRT/V) can be used to solve the problem, where π refers to the osmotic pressure, n is the amount of solute in moles, R is the gas constant (0.0821 L·atm/K·mol for this problem), T is the temperature in Kelvin, and V is the volume in liters. Given the osmotic pressure (4.45 atm), the temperature (298 K), and the volume (0.234 L), you can find the number of moles of estrogen needed. After calculating the amount in moles, use the molar mass of the estrogen (272.4 g/mol) to find the mass in grams. Hence, the quantity of estrogen needed can be calculated.

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Answer:

The new elution order expected will be the following:

Please see below for details and explanation.

Explanation:

Which compound will elute first depends on a number of factors. The compound with the lowest boiling point will elute before another compound with a higher boiling point and so on. By extension, the volatility of the compound will also be considered when predicting elution order. Thirdly, how the solutes interact with each other during the stationary phase. I've listed the boiling points below:

1-pentanol 138 °C

ethylene diamine 116 °C

diethylene glycol 245 °C

The advantage of using temperature programmed chromatogram is that it changes retention times (time needed for the solute to pass through the column). And it will be according to the respective boiling points.

Hope that answers the question, have a great day!

Given question is incomplete. The complete question is as follows.

When 72.8 g of benzamide ([tex]C_{7}H_{7}NO[/tex]) are dissolved in 600 g of a certain mystery liquid X, the freezing point of the solution is [tex]6.90^{o}C[/tex] less than the freezing point of pure X. Calculate the mass of ammonium chloride [tex](NH_{4}Cl)[/tex] that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor i = 70 for ammonium chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.

Explanation:

The given data is as follows.

        Mass of solute (benzamide), [tex]w_{B}[/tex] = 72.8 g

        Mass of solvent (X), [tex]w_{A}[/tex] = 600 g

          [tex]\Delta T_{f} 6.90^{o}C[/tex]

  Molar mass of benzamide, [tex]M_{w_{B}}[/tex] = 121.14 g/mol

We know that,

              [tex]\Delta T_{f} = k_{f} \times X \times m[/tex]   (for non-dissociating)

   [tex]6.90 = k_{f} \times \frac{72.8 \times 1000}{121.14 \times 600}[/tex] ...... (1)

For other experiment, when [tex]NH_{4}Cl[/tex] is taken :

       Mass of [tex]NH_{4}Cl[/tex], ([tex]w_{NH_{4}Cl}[/tex]) = ?

  Molar mass of [tex]NH_{4}Cl[/tex] = 53.491 g/mol

  Mass of solvent (X) = 600 g

        [tex]\Delta T_{f} = 6.90^{o}C[/tex]

           i = Van't Hoff factor = 1.70

As,     [tex]\Delta T_{f} = i \times k_{f} \times m[/tex]

      [tex]6.90 = 1.70 \times k_{f} \times \frac{w_{NH_{4}Cl} \times 1000}{53.491 \times 600}[/tex] ........... (2)

Now, we will divide equation (1) by equation (2) as follows.

         [tex]w_{NH_{4}Cl} \times 1 = \frac{72.8 \times 53.491}{1.70 \times 121.14}[/tex]  

                = 18.90 g

Therefore, we can conclude that the mass of ammonium chloride [tex](NH_{4}Cl)[/tex] that must be dissolved in the same mass of X to produce the same depression in freezing point is 18.90 g.

Final answer:

The mass of ammonium chloride required to produce the same freezing point depression as benzamide can't be calculated without additional data. The van't Hoff factor is crucial for understanding the effects of ionic compounds in freezing point depression calculations.

Explanation:

To calculate the mass of ammonium chloride that must be dissolved to produce the same depression in freezing point as benzamide, we will need to use the freezing point depression concept, which states that the change in freezing point (ΔTf) is equal to the molal freezing point depression constant of the solvent (Kf) multiplied by the molality (m) of the solution. The given data is as follows.

       Mass of solute (benzamide), w_(B) = 72.8 g

       Mass of solvent (X), w_(A) = 600 g

         \Delta T_(f) 6.90^(o)C

 Molar mass of benzamide, M_{w_(B)} = 121.14 g/mol

We know that,

             \Delta T_(f) = k_(f) * X * m   (for non-dissociating)

  6.90 = k_(f) * (72.8 * 1000)/(121.14 * 600) ...... (1)

For other experiment, when NH_(4)Cl is taken :

      Mass of NH_(4)Cl, (w_{NH_(4)Cl}) = ?

 Molar mass of NH_(4)Cl = 53.491 g/mol

 Mass of solvent (X) = 600 g

       \Delta T_(f) = 6.90^(o)C

          i = Van't Hoff factor = 1.70

As,     \Delta T_(f) = i * k_(f) * m

     6.90 = 1.70 * k_(f) * \frac{w_{NH_(4)Cl} * 1000}{53.491 * 600} ........... (2)

Now, we will divide equation (1) by equation (2) as follows.

        w_{NH_(4)Cl} * 1 = (72.8 * 53.491)/(1.70 * 121.14)  

               = 18.90 g

Therefore, we can conclude that the mass of ammonium chloride (NH_(4)Cl) that must be dissolved in the same mass of X to produce the same depression in freezing point is 18.90 g.

Answer:

Saponification and fermentation .

Explanation:

Hello,

In this case, we define:

- Sublimation as such physical change (the substance does not change its composition) from solid to gases by adding heat.

- Deposition as such physical change (the substance does not change its composition) from gas to solid by heat withdrawal.

- Saponification as such chemical change (the substance does changes its composition) which consists on the conversion of a fat, oil or lipid  by adding  heat in the presence of aqueous strong base such as NaOH or KOH into soap and alcohol.

- Fermentation as a biochemical change that produces chemical changes in organic substrates by the action of biocatalysts called enzymes. It typically turn glucose into piruvate, ethanol or short organic acids.

For that reason, both fermentation and saponification are chemical reactions, more specifically organic reactions since the reactants are mainly carbon-based.

Best regards.

Saponification and fermentation represent two types of organic reaction

Organic reactions are chemical reactions involving organic compounds. The basic organic chemistry reaction types are addition reactions, elimination reactions, substitution reactions.

In the question it is mentioned to identify two types of organic bamong the options

In the options we see that

1. Sublimation : The process in which a solid directly turns into gases without becoming a liquid. It is only physical change

2.Deposition : Deposition is the laying down of sediment carried by wind, water, or ice.

3.Saponification : The process of conversion of fatty acids and oils into soap, it is a chemical process as chemical reaction is taking place.

4.Fermentation : The process of microbial action to break down complex molecules into simpler molecules , chemical change occur during this process

Therefore The option D is the correct answer .

To know more about Organic Reaction

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