What sublevels are filling across the transition elements

Answer:

B

Explanation:

Firstly, we will need to calculate the number of moles. To do this, we make use of the ideal gas equation

PV = nRT

n = PV/RT

The parameters have the following values according to the question:

P = 780mmHg, we convert this to pascal.

760mHG = 101325pa

780mmHg = xpa

x = (780 * 101325)/760 = 103,991 Pa

V= 400ml = 0.4L

T = 135C = 135 + 273.15 = 408.15K

n = ?

R = 8314.463LPa/K.mol

Substituting these values into the equation yields the following:

n = (103991 * 0.4)/(8314.463 * 408.15)

= 0.012 moles

Now we know 1 mole contains 6.02 * 10^23 molecules, hence, 0.012moles will contain = 0.012 * 6.02 * 10^23 = 7.38 * 10^21 molecules

Answer: The number of nitrogen molecules in the container are [tex]7.38\times 10^{21}[/tex]

Explanation:

To calculate the moles of gas, we use the equation given by ideal gas which follows:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 780 mmHg  

V = Volume of the gas = 400.0 mL = 0.4 L     (Conversion factor:  1 L = 1000 mL)

T = Temperature of the gas = [tex]135^oC=[135+273]K=408K[/tex]

R = Gas constant = [tex]62.364\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

[tex]780mmHg\times 0.4L=n\times 62.364\text{ L. mmHg }mol^{-1}K^{-1}\times 408K\\\\n=\frac{780\times 0.4}{62.364\times 408}=0.01226mol[/tex]

According to mole concept:

1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules

So, 0.01226 moles of nitrogen gas will contain = [tex](0.012\times 6.022\times 10^{23})=7.38\times 10^{21}[/tex] number of molecules

Hence, the number of nitrogen molecules in the container are [tex]7.38\times 10^{21}[/tex]

Answer:

d. n H2(g) = 0.034 mol

Explanation:

                              2 - Na - 2

                              4 - H - 4

                               2 - O - 2

∴ n Na(s) = 0.068 mol

⇒ n H2(g) = ( 0.068 mol Na(s) )( mol H2(g) / 2 mol Na(s) )

⇒ n H2(g) = 0.034 mol

Based on the stoichiometric relationship between sodium and hydrogen in the given balanced chemical equation, 0.068 mol of sodium would yield 0.034 mol of hydrogen gas.

The correct answer is (b) 0.034 mol H2. The balanced chemical equation given states that 2 moles of sodium metal (Na) react with 2 moles of water molecules (H2O) to produce 2 moles of sodium hydroxide (NaOH) and 1 mole of hydrogen gas (H2). This implies a stoichiometric ratio of 2:1 between sodium and hydrogen. Therefore, if we have 0.068 moles of sodium, we would obtain half of this number in moles of hydrogen because of this stoichiometric relationship. 0.068 mol ÷ 2 = 0.034 mol H2.

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Answer: a. Synthesis

Explanation:

a. Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

Example: [tex]Mg+O_2\rightarrow 2MgO[/tex]  

Thus magnesium in its elemental form is combining with oxygen to form magnesium oxide.

b. Double displacement reaction is one in which exchange of ions take place.  

Example: [tex]2NaOH(aq)+(NH_4)_2SO_4(aq)\rightarrow 2NH_4OH(aq)+Na_2SO_4(aq)[/tex]

c. Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Example: [tex]Li_2CO_3\rightarrow Li_2O+CO_2[/tex]

The density of a metal that crystallizes in a face-centered cubic unit cell can be calculated by determining the number of atoms per unit cell, calculating the volume of a unit cell, and finally calculating the mass of the atoms in the unit cell using the given molar mass. The density is then found by dividing the mass by the volume.

To answer this question, we need to know a couple of key pieces of information. First, a face-centered cubic unit cell consists of four atoms: one-eighth of an atom at each of the eight corners and half of an atom on each of the six faces (1/8*8+1/2*6=4 atoms).

Secondly, we need to find the volume of the atom that is engulfed in the cubic unit cell. Given that the radius of the metal is 234 picometers (or 234*10^-12 meters), the volume of the unit cell can be found by applying the formula for the volume of a cube (side^3) where the side equals 2*sqrt(2)*r.

We can then calculate the number of moles of atoms in the unit cell, convert them to grams using the molar mass, and finally calculate the density by dividing the calculated mass by the calculated volume.

We first convert the volume of the unit cell from cubic meters to cubic centimeters (1m^3 = 10^6 cm^3).

Then, we calculate the mass of the atoms in the unit cell using the molar mass (195.08 g/mol).

Finally, we calculate the density. Density = Mass/Volume.

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To calculate the density of a metal in a face-centered cubic structure, you need to determine the volume of the unit cell and the molar mass of the metal.

The density of a metal can be calculated using the formula:

Density = (Molar Mass ×Avogadro's Number) / (Volume of Unit Cell)

First, let's calculate the volume of the face-centered cubic (FCC) unit cell. The FCC unit cell consists of 4 atoms at the corners and 8 atoms at the face centers. The radius of the atom can be used to calculate the edge length of the unit cell using the formula:

Edge Length = 4×Radius / √2

Once we have the edge length, we can calculate the volume of the unit cell using the formula:

Volume of Unit Cell = Edge Length³

Finally, we can plug the values into the density formula and calculate the density of the metal.

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Answer:

The correct answer is: Oxidation number

Explanation:

According to the IUPAC nomenclature for the inorganic compounds, roman numerals are used to denote the oxidation number of a positively charged ion, called cation.

This rule is only applicable to elements that can form more than one cation with different oxidation state or charge.

The oxidation state or charge of each cation of such an element is denoted or represented by a Roman numeral in the parentheses followed by the name of the element.

Therefore, the oxidation number of a cation is denoted by a roman numeral which is written after the name of the element.

Answer:

The change in energy for the combustion of a mole of naphthalene is 79 kJ

Explanation:

An excersise where you have to apply the Heat Capacity formula

C =  Q . ΔT

where C is Heat Capacity and Q is Heat

ΔT means (T° final - T° initial)

5.11 kJ/°C = Q (32.33°C - 24.25°C)

5.11 kJ/°C = Q (32.33°C - 24.25°C)

5.11 kJ/°C = Q . 8.08°C

5.11 kJ / 8.08°C = Q

0.632 kJ = Q

This heat is released by 1.025 grams of naphtalene.

Molar mass Naphtalene : 128.17 g/m

1.025 g / 128.17 g/m = 0.008 moles

This are the moles, so we have to divide heat/moles to get the change in energy for one mole.

0.632 kJ/0.008 mol = 79 kJ/m

Answer:

A) Work (positive)

B) Heat (negative) and work (positive)

C) Work (negative) and heat (negative)

Explanation:

A) This energy change is work given that the first ball is using this energy to move the other one. The system (fist ball) is doing the work so it has + sign.

B) The book loses part of its energy by friction (heat) and part by applying a force to the floor (work). The heat is taken from the system so it has - sign and the work is done by the system and it has + sign.  

C) The daughter is receiving the work done by the father. This work done to the system has - sign. Also, you can say that the system (daughter and swing) loses energy by friction (heat) and because of that it slows down.

Note: the sign of work and heat is defined by convention.

Answer:

2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.

Explanation:

Molarity of sodium ions = [tex][Na^+][/tex] = 0.0500 M

Moles of sodium ions = n

Volume of the solution = V = 800.0 mL = 0.800 L

[tex]Molarity=\frac{n}{V(L)}[/tex]

[tex][Na^+]=\frac{n}{V}[/tex]

[tex]n=[Na^+]\times V=0.0500 M\times 0.800 L=0.04 mol[/tex]

[tex]Na_2SO_4(aq)\rightarrow 2Na^+(aq)+CO_3^{2-}(aq)[/tex]

1 mole sodium carbonates gives 2 moles of sodium ion and 1 mole of carbonate ions.

Then 0.04 moles of sodium ions will be obtained from:

[tex]\frac{1}{2}\times 0.04 mol=0.02 mol[/tex] of sodium m carbonation.

Mass of 0.02 moles of sodium carbonate = 0.02 mol × 106 g/mol= 2.12 g

2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.

Answer:

mass water = 32.4 g

Explanation:

specific heat iron = 0.450 J/g°C

specific heat water = 4.18 J/g°C

32.8 x 0.450 ( 59.1 - 22.4) + mass water x 4.18 ( 59.1- 63.1)=0

541.7 - mass water x 16.7 = 0

mass water = 32.4 g

Answer:

[tex]K_{sp}[/tex] of [tex]Zn(CH_{3}COO)_{2}[/tex] is 0.0513

Explanation:

Solubility equilibrium of [tex]Zn(CH_{3}COO)_{2}[/tex]:

[tex]Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}[/tex]

Solubility product of [tex]Zn(CH_{3}COO)_{2}[/tex] ([tex]K_{sp}[/tex]) is written as-            [tex]K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}[/tex]

Where [tex][Zn^{2+}][/tex] and [tex][CH_{3}COO^{-}][/tex] represents equilibrium concentration (in molarity) of [tex]Zn^{2+}[/tex] and [tex]CH_{3}COO^{-}[/tex] respectively.

Molar mass of [tex]Zn(CH_{3}COO)_{2}[/tex] = 183.48 g/mol

So, solubility of [tex]Zn(CH_{3}COO)_{2}[/tex] = [tex]\frac{43.0}{183.48}M[/tex] = 0.234M

1 mol of [tex]Zn(CH_{3}COO)_{2}[/tex] gives 1 mol of [tex]Zn^{2+}[/tex] and 2 moles of [tex]CH_{3}COO^{-}[/tex] upon dissociation.

so,   [tex][Zn^{2+}][/tex] = 0.234 M and [tex][CH_{3}COO^{-}][/tex] = [tex](2\times 0.234)M=0.468M[/tex]

so, [tex]K_{sp}=(0.234)\times (0.468)^{2}=0.0513[/tex]          

Based on the data provided, the Ksp of zinc acetate is 0.051 M^{2}.

The solubility product, Ksp, of zinc acetate is derubed from the equation for the dissolution of Zinc acetate given below:

The solubility product, is given below:

Molar concentration of the zinc acetate = mass concentration/molar mass

Molar concentration of Zinc acetate = 43.0/183.48

Molar concentration of Zinc acetate = 0.234 M

From the equation of the reaction:

1 mole of Zn(CH_{3}COO)_{2} produces 1 mole Zn^{2+} and 2 CH_{3}COO^{-}

Hence;

Ksp = 0.234 × 0.468^{2}

Ksp = 0.051 M^{2}

Therefore, the Ksp of zinc acetate is 0.051 M^{2}.

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Answer:

-1.71 J/K

Explanation:

To solve this problem we use the formula

ΔS = n*ΔH/T

Where n is mol, ΔH is enthalpy and T is temperature.

ΔH and T are already given by the problem, so now we calculate n:

Molar Mass C₂H₅OH = 46 g/mol

2.71 g C₂H₅OH ÷ 46g/mol = 0.0589 mol

Now we calculate ΔS:

ΔS = 0.0589 mol * −4600 J/mol / 158.7 K

ΔS = -1.71 J/K

Answer:

Personal Referral.

Explanation:

This technique of finding employees is called Personal Referral. In personal referral the existing employee of the company uses hi/her own connections to find employees for the company. He/she may ask his/her asks current customers for names of friends, customers and other businesses that might be interested in job provided by the company.

Final answer:

Mary is employing a 'Word of Mouth' marketing strategy by asking current customers for referrals, tapping into the informal network to reach potential customers with the trust built through personal recommendations.

Explanation:

Mary, a sales representative, is utilizing a marketing strategy known as Word of Mouth. By asking current customers for referrals, she is tapping into her business's informal network to identify potential customers who might be interested in the cleaning services her company provides.

This approach takes advantage of the trust and reliability found within personal networks, as individuals are more likely to believe and try services recommended by people they know and trust. Additionally, Mary's strategy could involve engaging with information hubs or influencers within her network who are more likely to share information about the cleaning services with a broader audience.

In the context of increasing sales, connecting with customers on a more personal level and leveraging their networks can be an effective method to gain new leads. This is particularly effective as it has been shown that a vast majority of jobs, similar to customer opportunities, are found through social networks, highlighting the power of personal contacts and referrals. Mary's approach can be seen as a grassroots method to increase customer outreach and foster organic growth for the business.

Answer:

(a)The molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol

(b)The freezing point for the aqueous solution is [tex]-3.413\times10^{-5}[/tex] C

Explanation:

The osmotic pressure (π) is given by the following equation:

[tex]\pi =cRT[/tex]

[tex]c[/tex]= Concentration of solution

R = universal gas constant = 62.364 [tex]\frac{L\times torr}{mol\times K}[/tex]

T = temperature

Weight of solute = w = 10.0 mg

Let the molecular weight of the solute be m g/mol.

Concentration = [tex]c=\frac{n}{V}\\ n=\frac{w}{m}\\ n=\frac{10\times10^{-3}}{m}\\c=\frac{10\times10^{-3}}{m\times30\times10^{-3}}M[/tex]

[tex]m=\frac{RT}{3\pi}[/tex]

m = 18220.071g/mol

Therefore, the molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol

[tex]\Delta T_{f}=K_{f}m[/tex]

m is the molality of the solution.

m = [tex]1.835\times10^{-5}[/tex] mol/kg

[tex]\Delta T_{f}=1.86\times m[/tex]

[tex]\Delta T_{f}[/tex] = [tex]3.413\times10^{-5}[/tex] C

The freezing point for the aqueous solution is [tex]-3.413\times10^{-5}[/tex] C

Answer:

Covalent network solids typically have high melting points and high boiling points

Explanation:

Covalent solids are the solids which have covalent bonds as intermolecular or interatomic interactions.

As covalent bonds are very strong, these solids are generally characterized by high melting points and high boiling points.

The examples of such solids are: Diamond and Graphite.

The other classes of solids are

i) ionic

ii) molecular

iii) metallic.

Answer:can change as new researches and experiments are done

Explanation:

Answer:

The grams of MgCl₂ that are needed are 63.1 g

Explanation:

First of all, try to think the equation and ballance it. This is it:

3MgCl₂  +  2Na₃PO₄  →  Mg₃(PO₄)₂  +  6NaCl

Let's convert our f.u in mol

1 mol =  6.02x10²³ formula units (Avogadro's number)

So f.u / Avogadro = mol

1.33x10²³ / 6.02x10²³ = 0.221 moles.

So 1 mol of phosphate sodium comes from 3 mol of magnesium chloride.

How many mol of magnesium chloride are necessary, for 0.221 mol of phosphate.?

0.221 moles .3 = 0.663 moles.

Molar mass of MgCl₂ is 95.2 g/m

0.663 moles . 95.2 g/m = 63.1 g

The balanced chemical equation between magnesium chloride and sodium phosphate is 3 MgCl2 + 2 Na3PO4 -> Mg3(PO4)2 + 6 NaCl. To determine the grams of magnesium chloride needed to produce a specific number of formula units of magnesium phosphate, stoichiometry calculations can be used.

The balanced chemical equation between magnesium chloride (MgCl2) and sodium phosphate (Na3PO4) is:

3 MgCl2 + 2 Na3PO4 → Mg3(PO4)2 + 6 NaCl

To determine the grams of magnesium chloride needed to produce 1.33 x 10^23 formula units of magnesium phosphate (Mg3(PO4)2), we need to use stoichiometry.

The enthalpy of formation, ΔHf, for CsCl, can be determined using the Born-Haber cycle by summing up the energy changes in various steps including the sublimation of Cs, ionization of Cs, dissociation of Cl2, and the formation of CsCl. The lattice energy of CsCl is an exothermic process and is equal to the negative of the enthalpy of formation.

The enthalpy of formation, ΔHf, for CsCl(s) can be determined using the Born-Haber cycle. The cycle involves several steps including the sublimation of Cs(s), ionization of Cs(g), dissociation of Cl2(g), and the formation of CsCl(s). The lattice energy of CsCl is an exothermic process and is equal to the negative of the enthalpy of formation. By summing up the energy changes in all the steps, we can calculate the enthalpy of formation for CsCl.

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Final answer:

To find the enthalpy of formation for CsCl, the enthalpy changes from sublimation, ionization, bond dissociation, electron affinity, and lattice energy are combined resulting in an enthalpy of formation of -390.5 kJ/mol.

Explanation:

To determine the enthalpy of formation (ΔHf) of cesium chloride (CsCl), we must use the Born-Haber cycle which involves several energy changes related to the formation of ionic compounds. These are the steps to calculate ΔHf for CsCl(s):

The enthalpy of sublimation of Cs (ΔHsub)

The first ionization energy of Cs (IE1)

The bond energy of Cl2 (BE)

The electron affinity of Cl (ΔHEA)

The lattice energy of CsCl (ΔHlattice)

To calculate the enthalpy of formation, first, we need to break the Cl2 bond energy into two Cl atoms which takes 1/2 of the bond energy (1/2 x BE) for one mole of Cl atoms. The total enthalpy change of formation is:

ΔHf = ΔHsub + IE1 + (1/2 x BE) - ΔHEA + ΔHlattice

Substituting the given values:

ΔHf = 76.5 + 376 + (1/2 x 243) - (-349) - 657
=76.5 + 376 +121.5 + 349 - 657

= -657 + 266.5 kJ/mol

= -390.5 kJ/mol

Answer:

a. If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g).

Explanation:

The answer is a) because the value for Kp is really close to zero (having x10⁻¹⁵), this means that at equilibrium O₂ and F₂ will be significantly more present than OF₂.

Final answer:

For the equilibrium O2(g) + 2F2(g) ↔ 2OF2(g) with Kp = 2.3×10^-15, the reaction favors the reactants, making the correct answer that a mixture initially containing only OF2(g) will consist of essentially only O2(g) and F2(g) at equilibrium.

Explanation:

The question considers the equilibrium O2(g) + 2F2(g) ↔ 2OF2(g); Kp = 2.3×10-15 and asks which statement is true. Given the extremely low value of Kp, this indicates a strong preference for reactants at equilibrium. Therefore, the correct answer is: If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g). This is because a very small Kp value means the equilibrium lies heavily on the side of the reactants, making statement (a) true.

Answer:

55.7%

Explanation:

The reaction that takes place is:

With the volume and concentration of NO₃⁻ solution, we can calculate the moles of Sn²⁺ that reacted:

Now we convert moles of Sn to mass, using its atomic weight:

Finally we calculate the percent by mass of Sn:

Answer:

coefficient 3 is needed t with water molecule to balance the equation.

Explanation:

Chemical equation:

Fe₂O₃ + H₂SO₄ →  Fe₂(SO₄)₃ + H₂O

Balanced chemical equation:

Fe₂O₃ + 3H₂SO₄ →  Fe₂(SO₄)₃ + 3H₂O

The equation is balanced. There are two iron three sulfate six hydrogen and three oxygen atoms are present on both side of equation.

The given equation completely follow the law of conservation of mass.

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

Coefficient with reactant and product:

Fe₂O₃             1

H₂SO₄             3

Fe₂(SO₄)₃        1

H₂O                 3

So coefficient 3 is needed t with water molecule to balance the equation.

Answer:

pH = 7.45

Explanation:

This is a buffer solution and we can solve it by using the Henderson-Hasselbalch equation:

pH = pKa + log ((A⁻)/(HA))

Here we will first have to calculate the  A⁻ formed  in the 1. 0 L solution which is formed by the reaction of  HClO with the strong base NaOH and add  it to the original mol of NaClO

mol NaClO = mole NaCLO originally present in the 1L of M solution + 0.030 mol produced in the reaction of HCLO with NaOH

0.350 mol + .030 mol = 0.380 mol

New concentrations :

HClO = 0. 350 mol-0.030 mol  = 0.320 M (have to sustract the 0.030 mol reacted with NaOH)

NaClO = 0.380 mol/ 1 L = 0.380 M

Now we have all the values required and we can plug them into the equation

pH = -log (2.9 x 10^-8) + log (0.380/.320) = 7.45

The pH of 1.00 L of the solution is 7.45.

This is defined as the power of hydrogen and it measures how acidic or basic a substance is.

Using Henderson-Hasselbalch equation:

pH = pKa + log ((A⁻)/(HA))

mol of NaClO = mole NaCLO initially present in the 1L of M solution + 0.030 mol produced in the reaction of HClO with NaOH

0.350 mol + .030 mol = 0.380 mol

We can then calculate the new concentrations below:

HClO = 0. 350 mol-0.030 mol  = 0.320 M

NaClO = 0.380 mol/ 1 L = 0.380 M

Substitute the values into the equation

pH = -log (2.9 x 10⁻⁸) + log (0.380/.320)

     = 7.45

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The thermosphere is a layer of Earth's atmosphere located 80-700km above sea level. It supports long-distance communication by reflecting radio waves back to Earth and is also where auroras form due to the collision of charged particles and gas molecules.

The thermosphere is the fourth layer of Earth's atmosphere, located above the mesosphere and extend from about 80 to 700 kilometers above sea level. This layer holds a unique property due to the presence of electrically charged particles, or ions which allows it to support long-distance communication by reflecting radio waves back to Earth, bypassing the need for satellites. This is also the layer in which auroras form. Auroras occur when these charged particles from the sun collide with gas molecules in the Earth's atmosphere, releasing energy that manifests as visible, colorful light.

The Earth's atmosphere is divided into five main layers: The troposphere, the stratosphere, the mesosphere, the thermosphere and the exosphere. The Troposphere being the one closest to the Earth's surface and is where weather is generally observed. This layer expands to a height of roughly 12 km above the sea level and makes up around 80% of the atmosphere's mass. The Thermosphere is not the layer of atmosphere closest to the Earth’s surface where weather typically occurs, this property represents the troposphere.

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The third option is correct.

3. The thermosphere is the layer where auroras form when electrically charged particles from the sun collide with gas molecules releasing energy visible as light of different colors.

Answer: The increasing wavelength of colors:

Red > Green > Blue

Explanation:

Wavelength: This is the property of wave which includes the distance between two consecutive crests or trough. This is denoted by the Greek letter Lambda and it is found by dividing the velocity of the wave with its frequency.

Wavelength of colours are

Violet: 400 - 420 nm

Indigo: 420 - 440 nm

Blue: 440 - 490 nm

Green: 490 - 570 nm

Yellow: 570 - 585 nm

Orange: 585 - 620 nm

Red: 620 - 780 nm

In visible light, the wavelength increases from violet to red. For the colors mentioned, in order of increasing wavelength, it is blue < green < red and blue < yellow < red for the additional colors provided.

The question asks to arrange the colors of visible light (green, red, and blue) in order of increasing wavelength and then provides a similar task with the colors yellow, blue, and red. For visible light, wavelengths increase from violet through to red. Hence, using the mnemonic ROY G BIV (Red, Orange, Yellow, Green, Blue, Indigo, Violet), we can deduce that blue has a shorter wavelength than green, which in turn has a shorter wavelength than red.

Part A: For the colors yellow, blue, and red, in order of increasing wavelength, it would be: blue < yellow < red.

Part B: Since the frequency of light waves is inversely proportional to their wavelength, the order according to frequency, from lowest to highest, would inversely mirror the wavelengths: red < yellow < blue.

Answer: [tex]156.4^0C[/tex]

Explanation:

[tex]2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542kJ[/tex] [tex]\Delta H=-6542kJ[/tex]

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.700g}{78.11g/mol}=0.9858moles[/tex]

According to stoichiometry :

2 moles of [tex]C_6H_6[/tex] releases = 6542 kJ of heat

0.9858 moles of [tex]C_6H_6[/tex]  release =[tex]\frac{6542}{2}\times 0.9858=3224kJ[/tex] of heat

Thus heat given off by burning 7.700 g of [tex]C_6H_6[/tex]  will be absorbed by water.

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

Q = Heat absorbed= 3224kJ = 3224000J   (1kJ=1000J)

m= mass of water = 5691 g

c = specific heat capacity = [tex]4.184J/g^0C[/tex]

Initial temperature of the water = [tex]T_i[/tex] = 21.0°C

Final temperature of the water = [tex]T_f[/tex]  = ?

Putting in the values, we get:

[tex]3224000J =5691g\times 4.184J/g^0C\times (T_f-21)[/tex]

[tex]T_f=156.4^0C[/tex]

The final temperature of the water is [tex]156.4^0C[/tex]

The jar contains A Mixture

Explanation:

A mixture consists of different atoms that are not chemically bonded. It is of two types of the heterogeneous mixture and homogeneous mixture. The chemical substances can be differentiated visually in the heterogeneous mixture.

In a mixture, the proportion of atoms can be slightly changed without changing  or modifying any substance. This is because the individual substances in mixture keep their properties when mixed together. Further, mixtures have variable compositions and substances. these mixtures can be are separated using physical methods.  

The jar contains A Mixture

Explanation:

A mixture consists of different atoms that are not chemically bonded. It is of two types of the heterogeneous mixture and homogeneous mixture. The chemical substances can be differentiated visually in the heterogeneous mixture.

In a mixture, the proportion of atoms can be slightly changed without changing  or modifying any substance. This is because the individual substances in mixture keep their properties when mixed together. Further, mixtures have variable compositions and substances. these mixtures can be are separated using physical methods.

Answer:

2Fe + 3Br2 = 2FeBr3

Explanation:

The vapor pressure of the solution can be calculated using Raoult's law. First, we need to calculate the mole fraction of glycerol and water in the solution. Then, we can use the mole fraction to calculate the vapor pressure of the solution.

In order to find the vapor pressure of the solution, we can use Raoult's law. Raoult's law states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.

First, we need to calculate the mole fraction of glycerol in the solution. To do this, we need to convert the mass percent of glycerol to moles. Since the molar mass of glycerol is 92.1 g/mol and the solution contains 34.4% glycerol by mass, we can calculate the moles of glycerol:

Moles of glycerol = (34.4 g / 92.1 g/mol)

Next, we need to calculate the mole fraction of water in the solution. Since the solution is 34.4% glycerol, the mass percent of water is 100% - 34.4% = 65.6%. Using the molar mass of water (18.0 g/mol), we can calculate the moles of water:

Moles of water = (65.6 g / 18.0 g/mol)

Now we can calculate the mole fraction of glycerol and water:

Mole fraction of glycerol = (moles of glycerol) / ((moles of glycerol) + (moles of water))

Mole fraction of water = (moles of water) / ((moles of glycerol) + (moles of water))

Finally, we can calculate the vapor pressure of the solution using Raoult's law:

Vapor pressure of solution = (mole fraction of water) * (vapor pressure of water)

Answer:

The answer is NO₂-NO₂, N₂O₄

Explanation:

If 23.0 g of this gas were found to occupy 5.6 L at STP we can apply the Ideal gas Law to determinate the quantity of mols of the oxide.

As I have the mass, I can find out the molar weight.

STP are 1 atm and 273, 1 K so:

P . V = n . R . T

1 atm .  5.6L = n . 0.082 L.atm/mol.K . 273.1K

(1 atm .  5.6L)/ 0.082 mol.K/L.atm . 273.1K = n

0.250 mol = n

Molar weight = mass/mols

23 g /0.250 mol = 92 g/m

The percentages are:  30.4% N and 69.6%O

100% ____ 92 g

30.4% _____ (30.4 .92)/100 = 28 g → 2 mols of N

100% _____ 92 g

69.6% ____ (69.6 . 92)/ 100 = 64 g → 4 mols of O

Molecular formula N₂O₄

Empirical formula: NO₂-NO₂

To determine the spontaneity of the reaction between N2 and O2 to form 2NO and the equilibrium constant K, one would use the Gibbs Free Energy (ΔG) formula and the equilibrium constant (K) formula. While exact values can't be computed without data for the standard enthalpy and entropy changes, it can be inferred that this reaction is typically not spontaneous at room temperature (298 K), but becomes more spontaneous as temperature increases toward 2000 K.

The given chemical reaction is between nitrogen and oxygen to form nitrogen monoxide: N2(g) + O2(g) = 2 NO(g). To answer your question, we'll need to use the Gibbs Free Energy (ΔG) formula and the equilibrium constant (K) formula which are linked by the equation ΔG = -RTlnK. These formulas are what we use to determine the spontaneity of a reaction and the equilibrium state at a given temperature.

Without the specific values of the standard enthalpy (ΔH) and entropy (ΔS) changes for this reaction, it's impossible to calculate ΔG and K exactly. However, at room temperature (298 K), the reaction is typically not spontaneous because the formation of NO requires high-energy conditions - usually above 2000 K. This is inferred by the presence of NO only in extreme conditions such as lightning strikes or in high-temperature combustion processes.

At 2000 K, although I can't give an exact numerical estimate without the ΔH and ΔS values, we can say that the reaction becomes more spontaneous as temperature increases since high energy and high temperature favor the formation of NO. It's always important to remember that whether a chemical reaction is spontaneous or not depends not only on the change in enthalpy (ΔH), but also on the change in entropy (ΔS) and the absolute temperature (T).

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The reaction N2(g) + O2(g) = 2NO(g) is not spontaneous at 298 K, with a very small equilibrium constant (K). However, at 2000 K, the reaction becomes more spontaneous as the free energy change (ΔG) is negative.

The reaction N2(g) + O2(g) = 2NO(g) has a standard free energy change (ΔG°) of 173.4 kJ/mol at 298 K. To find the equilibrium constant (K) at this temperature, we use the equation ΔG° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin. Rearranging the equation, we get K = e-ΔG°/(RT).

Substituting the values into the equation and solving, we find K to be 6.95 x 10-31, which is a very small value. Since K is less than 1, the reaction is not spontaneous at 298 K.

To estimate ΔG° at 2000 K, we can use the equation ΔG = ΔH - TΔS, where ΔH is the enthalpy change and ΔS is the entropy change. Assuming ΔH and ΔS do not change significantly with temperature, we can use the same values as at 298 K.

Substituting the values into the equation, we find ΔG to be -545.4 kJ/mol at 2000 K. Since ΔG is negative, the reaction becomes more spontaneous as temperature increases.

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