Acid chlorides can be converted into primary amines with the loss of a single carbon atom via the Curtius rearrangement. The reaction involves treating an acid chloride with sodium azide to form an acyl azide. The acyl azide then loses N2 to give an isocyanate, which is hydrolyzed to release CO2 and the primary amine. Draw curved arrows to show the movement of electrons in this step of the mechanism.

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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The bonding in BeF2 involves the promotion of a 2s electron to a 2p orbital on Be, followed by sp hybridization of the Be orbitals, and overlap with the F atom 2p orbitals to form two sigma bonds. So the correct statements are iv, iii, and i.

To explain the bonding in BeF2, we must select the appropriate statements and arrange them in a logical order. The correct sequence of statements is:

Firstly, a 2s electron in Be is promoted to a 2p orbital (Statement iv), creating two singly occupied orbitals. These orbitals then undergo hybridization to form two equivalent sp hybrid orbitals (Statement iii), which are oriented at a 180° angle to each other. Each sp hybrid orbital overlaps with a singly occupied 2p orbital on a fluorine atom to form a sigma bond (Statement i), involving the pairing up of the Be valence electrons with the unpaired electron on each F atom. This results in the formation of two identical Be-F sigma bonds, giving BeF2 a linear molecular geometry.

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 Answer:

Delta H for endothermic reaction is positive-True. This is because an endothermic reaction absorbs heat energy, therefore more energy is retained inside the product of the reaction system than the reactants, as the value of deltaH is greater than Zero.

Delta H for an exothermic reaction is positive. This is false. Because in an exothermic reaction  heat is liberated to the  surrounding environment. therefore the value of thus the outer environment contains more energy than the internal environments, thus the enthalpy of the reactants is greater than that of the products.

when the energy is transferred as heat from system to surroundings, deltaH is negative. True . This is true because the surrounding environment gain heat energy, (positive)while  the system loses it,(negative) therefore delta H is negative.

when the energy is transferred as heat from surroundings to system, deltaH is negative.  False. This is  positive, because now the environments loses heat, (negative) while the systems gains heat,( positive) therefore delta H of the system is positive. endothermic

the evaporation of water is an exothermic process-False, This is an endothermic reaction in which water molecules need to gain heat energy from the surrounding environments to increase the average kinetic energy  of collusion to escape the intermolecular forces to escape as steam.

Combustion reaction is exothermic. True., because heat energy is transferred to the surrounding from the internal system. The energy needed for the formation  of  new bonds in the products is higher than the energy for breaking of original bonds in the reactants. Thus more heat is liberated.

Explanation:

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.

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:

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:

0.3M

Explanation:

Step 1:

Data obtained from the question. This include the followingb:

Volume of acid (Va) = 90mL

Concentration of acid (Ca) = 0.2M

Volume of base (Vb) = 60mL

Concentration of base (Cb) =....?

Step 2:

The balanced equation for the reaction. This is given below:

HBr + NaOH —> NaBr + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 3:

Determination of the concentration of the base, NaOH.

The concentration of the base can be obtained as follow:

CaVa /CbVb = nA/nB

0.2 x 90 / Cb x 60 = 1

Cross multiply

Cb x 60 = 0.2 x 90

Divide both side by 60

Cb = 0.2 x 90 /60

Cb = 0.3M

Therefore, the concentration of the base, NaOH is 0.3M

Answer:

[tex]M_{NaOH}=0.3M[/tex]

Explanation:

Hello,

In this case, since we are talking about a neutralization reaction, the moles of acid must equal the moles of base as shown below:

[tex]n_{HBr}=n_{NaOH}[/tex]

Thus, in terms of molarities we've got:

[tex]M_{HBr}V_{HBr}=M_{NaOH}V_{NaOH}[/tex]

This is possible since HBr reacts with NaOH in a 1:1 molar ratio:

[tex]HBr+NaOH\rightarrow NaBr+H_2O[/tex]

Hence, for the given concentration and volume of hydrobromic acid and the volume of sodium hydroxide, we compute its concentration as shown below:

[tex]M_{NaOH}=\frac{M_{HBr}V_{HBr}}{V_{NaOH}} =\frac{90mL*0.2M}{60mL} \\\\M_{NaOH}=0.3M[/tex]

Best regards.

Answer:

The ovaries and the testes

Explanation:

The organs responsible for the development of secondary sex characteristics would be the ovaries and the testes.

Both ovary and testes are associated with the sex organs in the body and are responsible for the synthesis of hormones such as androgen and estrogen. While the latter (estrogen) is produced by the ovary, the former (androgen) is produced by the testes.

More specifically, both androgen and estrogen are hormones responsible for the growth of hairs in areas such as the armpit and the pubic regions of the body.

The correct options are the ovaries and the testes.

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:

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.

For more information about molarity, refer to the link:

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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:

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:

120g

Explanation:

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

Sn + 2HF —> SnF2 + H2

Next, we shall determine the number of mole of HF needed to react with 3 moles of Sn.

From the balanced equation above, 1 mole of Sn reacted with 2 moles of HF.

Therefore, 3 moles of Sn will react with = 3 x 2 = 6 moles of HF.

Finally, we shall convert 6moles of HF to grams

This is illustrated below:

Number of mole of HF = 6moles

Molar Mass of HF = 1 + 19 = 20g/mol

Mass of HF =..?

Mass = number of mole x molar Mass

Mass of HF = 6 x 20

Mass of HF = 120g

Therefore, 120g of HF is needed to react with 3 moles of Sn

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:

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 : The volume of water added are, 15 mL

Explanation :

Formula used :

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume of HCl.

[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume of water.

We are given:

[tex]M_1=12M\\V_1=50mL\\M_2=40M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]12M\times 50mL=40M\times V_2\\\\V_2=15mL[/tex]

Hence, the volume of water added are, 15 mL

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:

5.74g of NO

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

4NH3 + 5O2 —> 4NO + 6H2O

Step 2:

Determination of the masses of NH3 and O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar Mass of NO = 14 + 16 = 30g/mol

Mass of NO from the balanced equation = 4 x 30 = 120g

From the balanced equation above,

68g of NH3 reacted with 160g of O2 to produce 120g of NO.

Step 3:

Determination of the limiting reactant.

We need to determine the limiting because it will be used to calculate the maximum yield of the reaction. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Therefore, 3.25g of NH3 will react with = (3.25 x 160)/68 = 7.65g of O2.

From the simple illustration above, we can see that lesser mass of O2 is needed to react with 3.25g of NH3. Therefore, NH3 is the limiting reactant while O2 is the excess reactant.

Step 4:

Determination of the mass of NO produced from the reaction.

In this case the limiting reactant will be used because all of it were used in the reaction.

The limiting reactant is NH3.

From the balanced equation above,

68g of NH3 reacted to produce 120g of NO.

Therefore, 3.25g of NH3 will react to produce = (3.25 x 120)/68 = 5.74g of NO.

From the calculations made above, 5.74g of NO is produced.

Answer:  Cu

Explanation: It is Cu because the origin of the word Copper comes from the latin word "Cuprum".

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:

3.25

Explanation:

pH = -log[H3O+]

-log(0.000558)=3.25

Answer:

C. Al2O3

Explanation:

The empirical formula is an expression that represents the simplest proportion in which the atoms that form a chemical compound are present. It is therefore the simplest representation of a compound.

For options A, B and D the empirical formulas are as follows:

A. CH2

B. BH3

D. CH

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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 .

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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:

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:

F2(g)/HF(aq)>Co3 (aq)/Co2 (aq)> H2O2(aq)/H2O(l)> PbO2(s)/Pb(s)>Br2(l)/Br-(aq)

Explanation:

The tendency of any specie to function as oxidizing agent is a highly dependent on the reduction potential of the couple. The more positive the value of the reduction potential of the couple, the better it does as an oxidizing agent.

This implies that we could know a good oxidizing agent by looking at their respective reduction potentials. The couple having the greatest (most positive) reduction potential is selected as the best oxidizing agent. If there are a number of couples at having different reduction potentials, the order of oxidizing ability can be obtained by arranging the species in order of decreasing positive reduction potentials just as we have done in the answer above.

Decreasing strength will be:

F₂(g)/HF(aq)>Co₃ (aq)/Co₂ (aq)> H₂O₂(aq)/H₂O(l)>PbO₂(s)/Pb(s)>Br₂(l)/Br-(aq)

The tendency of any specie to function as an oxidizing agent is highly dependent on the reduction potential of the couple. The more positive the value of the reduction potential of the couple, the better it does as an oxidizing agent.

The couple having the greatest (most positive) reduction potential is selected as the best oxidizing agent.

Thus, the decreasing strength for oxidizing agents will be:

F₂(g)/HF(aq)>Co₃ (aq)/Co₂ (aq)> H₂O₂(aq)/H₂O(l)>PbO₂(s)/Pb(s)>Br₂(l)/Br-(aq)

Find more information about Reduction potential here:

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Mirrors reflect light without separating it into colors, unlike prisms or water droplets which cause dispersion, leading to the formation of rainbows. Dispersion occurs when light passes through a medium that slows down colors differentially. Historical scientists like al-Farisi and Theodoric of Freiberg demonstrated this with water droplets.

The answer is no; a typical mirror reflects light without separating it into its constituent colors. This property allows a mirror to create an image with the same color composition as the original scene.

Now, the separation of white light into its constituent colors—an effect known as dispersion—occurs when light passes through a medium that causes different colors to travel at different speeds. This is how prisms or water droplets in the atmosphere create beautiful rainbows. It relies on the refraction of light, which bends the light as it moves from one medium to another, like air into water, and does so at varying degrees depending on the wavelength (color) of the light. However, while mirrors can exhibit total reflection due to refraction, they do not inherently disperse white light.

Historically, figures like Kamal al-Din al-Farisi and Theodoric of Freiberg conducted experiments which confirmed that water droplets are responsible for decomposing white light into the colors of the rainbow. Their work built on foundational optics principles from Ibn al-Haytham.

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.