Use the following information on Cr to determine the amount of heat required to convert 159.6 g of solid Cr at 1760°C into liquid Cr at 2060°C. melting point = 1860°C; boiling point = 2672°C ΔHfus = 20.5 kJ/mol; ΔHvap = 339 kJ/mol; c(solid) = 44.8 J/g°C; c(liquid) = 0.94 J/g°C Enter your answer in units of kJ to three significant figures.

Answer:

Explanation:

The terms 'anode' and 'cathode' are applied to electrodes based on the chemistry occurring at a specified electrode. That is, the anode is ALWAYS the site of oxidation and the cathode is ALWAYS the site of the reduction reaction in electrochemical process. The charge (+ or -) results from the chemistry. In Voltaic Cells the anode is always the site of oxidation and is negative (-) and the cathode is the site of reduction and is always positive (+). However, in 'Electrolytic Cells' the charges are opposite at the electrodes. That is, at the electrode where oxidation is taking place in the electrolytic cell the charge on the electrode will be positive (+) and the charge on the electrode where reduction is taking place the charge will be negative (-).

As far as acronyms (or memory devices) one might consider each type cell separately...

Voltaic Cells

Anode => A Negative -ode (always site of oxidation) ... the other electrode is positive and site of reduction.                  

Electrolytic Cells

For electrolytic cells, just remember the electrode charges are opposite those of the Voltaic Cell but the chemistry is always the same. Charge is due to chemistry and type of cell, but anode is always oxidation and cathode is always reduction.      

Answer:

The solvent is Nitrogen and the solute is oxygen

Explanation:

Normally the solvent always contains higher amount of substance than the solute in a particular solution

Answer:

Sea water, a solution composed of a variety of salts dissolved in water. The salt is the SOLUTE

Soda water, a solution composed of carbon dioxide gas dissolved in water. The water is the SOLVENT

Air, a solution composed of 78% nitrogen and 21% oxygen. The nitrogen is the  SOLVENT

Explanation:

its right on EDGE

Answer:

The answer to your question is 0.04 M

Explanation:

Data

mass of Na₂SO₄ = 14.2 g

volume = 2.50 l

Molarity = ?

Process

1.- Calculate the molar mass of Na₂SO₄

Na₂SO₄ = (23 x 2) + (32 x 1) + (16 x 4) = 46 + 32 + 64 = 142 g

2.- Calculate the number of moles of Na₂SO₄

                    142 g of Na₂SO₄ ------------------- 1 mol

                     14.2 g of Na₂SO₄ ------------------  x

                                   x = (14.2 x 1) / 142

                                   x = 0.1 moles

3.- Calculate the molarity

Molarity = moles /volume

-Substitution

Molarity = 0.1 / 2.5

-Result

Molarity = 0.04

Answer:

200 mph

Explanation:

600/3=200

Answer:

The answer to your question is 9.034 x 10²³ molecules

Explanation:

Data

number of molecules = ?

number of moles = 1.50

Process

To solve this problem, use Avogadro's number. This number relates 1 mol of a substance to 6.023 x 10²³ molecules.

                          1 mol of CO₂ ---------------  6.023 x 10 ²³ molecules

                          1.5 moles of CO₂ ----------   x

                                x = (1.5 x 6.023 x 10²³) / 1

                                x = 9.034 x 10²³ molecules

Answer:

204.78 mL

Explanation:

- use Charle's law and rearrange formula

- change C to K

- Hope this helped! Let me know if you need further explanation.

Answer:

an electron is part of the 3 things that are in an atom, it has a negative charge

Answer:

Ag+(aq) + Cl-(aq) <--->AgCl(s)

Ksp = [Ag+] [Cl-]

aii) Pb2+(aq) + 2Cl-(aq) <--->PbCl2(s)

Ksp = [Pb+] [Cl-]^2

Explanation:

In this question, we are to write the net ionic equation and corresponding Ksp expression for the salts in the question.

We proceed as follows;

Ag+(aq) + Cl-(aq) <--->AgCl(s)

Ksp = [Ag+] [Cl-]

aii) Pb+2(aq) + 2Cl-(aq) <--->PbCl2(s)

Ksp = [Pb+] [Cl-]^2

Final answer:

To write net ionic equations for AgCl and PbCl2, the equations are AgCl(s)
ightarrow Ag+(aq) + Cl-(aq) with Ksp = [Ag+][Cl-], and PbCl2(s)
ightarrow Pb2+(aq) + 2Cl-(aq) with Ksp = [Pb2+][Cl-]^2. These represent the dissolution of silver chloride and lead (II) chloride, respectively.

Explanation:

To write the net ionic equations and corresponding Ksp expressions for the dissolution of solid silver chloride (AgCl) and lead (II) chloride (PbCl2), you should consider both substances dissolving separately in water:

AgCl

Net ionic equation: AgCl(s)
ightarrow Ag+(aq) + Cl-(aq)

Ksp expression: Ksp = [Ag+][Cl-]

This represents the solubility product for silver chloride, which is given as 1.8  imes 10^-10.

PbCl2

Net ionic equation: PbCl2(s)
ightarrow Pb2+(aq) + 2Cl-(aq)

Ksp expression: Ksp = [Pb2+][Cl-]^2

This represents the solubility product for lead (II) chloride, which is given as 1.6  imes 10^-5. Note that because there are two chloride ions for every lead ion that dissolves, the concentration of chloride appears squared in the expression.

Answer:

ΔSv = 0.1075 KJ/mol.K

Explanation:

Binary solution:

∴ a: solvent

∴ b: solute

in equilibrium:

⇒ Ln (1 - Xb) = ΔG/RT

∴ ΔG = ΔHv - TΔSv

⇒ Ln(1 -Xb) = ΔHv/RT - ΔSv/R

∴ Xb → 0:

⇒ Ln(1) = ΔHv/RT - ΔSv/R

∴ T = T*b....normal boiling point

⇒ 0 = ΔHv/RT*b - ΔSv/R

⇒ ΔSv = (R)(ΔHv/RT*b)

⇒ ΔSv = ΔHv/T*b

∴ T*b = 80°C ≅ 353 K

⇒ ΔSv = (38 KJ/mol)/(353 K)

⇒ ΔSv = 0.1075 KJ/mol.K

The equation which represents the chemical equilibrium is  2 NO₂ ⇄ N₂0₄

Explanation:

           2 NO₂ ⇄ N₂0₄.

Final answer:

The correct option is (a), which states that the precipitate formed when sodium sulfate and silver nitrate solutions are mixed is silver sulfate, and it provides the correct net ionic equation for this precipitation reaction.

Explanation:

When solutions of sodium sulfate (Na2SO4) and silver nitrate (AgNO3) are mixed together, a double-replacement reaction occurs. The balanced chemical equation for this reaction is:

Na2SO4(aq) + 2AgNO3(aq) → Ag2SO4(s) + 2NaNO3(aq)

The precipitate formed in this reaction is silver sulfate (Ag2SO4). The net ionic equation for the formation of the precipitate is:

2Ag+(aq) + SO42-(aq) → Ag2SO4(s)

Answer (a) is the correct option, as it accurately identifies the precipitate and provides the correct net ionic equation for its formation.

The following questions are answered below

Explanation:

1.a. 1.0 liter of a 1.0 M mercury (II) chloride (HgCl2) solution

  1.0 L 1 mol 271.49 g

    L        1 L    1 mol

= 271.49 g

= HgCl₂ = 271.49 g HgCl₂

1.b. 2.0 liters of a 1.5 M sodium nitrate (NaNO3) solution

[tex]\frac{2.0 L}{1}[/tex][tex]\frac{1.5 mol}{L}[/tex][tex]\frac{85.01 g}{1 mol}[/tex]

= 255.03 g

= 255 g NaNo₃

1.c. 5.0 liters of a 0.1 M Ca(OH)2 solution

= 37 g Ca(OH)2

1.d. 100 mL of a 0.5 M (NH4)3PO4 solution

= 7.5 g (NH₄)₃PO₄

2. To find the molarity of the following solutions

2.a. 12 g of lithium hydroxide (LiOH) in 1.0 L of solution

= 0.50 m (LiOH)

2.b. 198 g of barium bromide (BaBr2) in 2.0 L of solution

= 0.33 m (BaBr₂)

2.c. 54 g of calcium sulfide (CaS) in 3.0 L of solution

= 0.25 m (CaS)

3. To find the volume of each solution

3.a. 1.0 M solution containing 85 g of silver nitrate (AgNO3)

= 0.50 L (AgNO₃)

3.b. 0.5 M solution containing 250 g of manganese (II) chloride (MnCl2)

= 4.0 L MnCl2

3.c. 0.4 M solution containing 290 g of aluminum nitrate (Al(NO3)3)

= 3.4 L (Al(NO₃)₃)

Answer:

0.0174 mol

Explanation:

Molarity is (moles of solute)/(liters of solvent).  So, to find moles you need to multiply liters by molarity.

0.0437 L × 0.400 M = 0.0174 mol

The number of moles of CH₃OH present is 0.01748 mole

From the question,

We are to determine the number of moles of CH₃OH that are in 43.7 mL of 0.400 M CH₃OH

Using the formula

Number of moles = Concentration × Volume

From the question

Concentration of CH₃OH = 0.400 M

Volume of CH₃OH = 43.7 mL = 0.0437 L

∴ Number of moles of CH₃OH present = 0.400 × 0.0437

Number of moles of CH₃OH present = 0.01748 mole

Hence, the number of moles of CH₃OH present is 0.01748 mole

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Answer: An acid has more hydrogen ions in solution.

Explanation: Acids increase hydrogen/hydronium ion concentration in a solution, have pH's less than seven, have a low hydroxide concentration, and increase positive ions in a solution.

The correct definition of an acid is: An acid has more hydrogen ions in solution. Thus, the correct answer is option with 'An acid has more hydrogen ions in solution'.

An acid is characterized by its ability to increase the concentration of hydrogen ions (H⁺) when dissolved in water. This results in a pH lower than 7, indicating increased acidity. The pH scale measures the acidity or basicity of a solution; acids have pH values less than 7.

Let's understand other options given:

An acid has a pH of 7:

An acid has hydroxide ions:

An acid has fewer positive ions in solution:

Therefore, the defining characteristic of an acid is the presence of more hydrogen ions (H⁺) in solution.

Answer:

[tex] = 1.92*10^-^3^0 C.m[/tex]

Explanation:

Given:

[tex] r_ca_^2_^+ = 0.100 nm[/tex]

[tex] r_o_^2_^- = 0.14 nm [/tex]

Let's find the distance of separation between cation and anion when there is no applied electric field with the formula:

[tex] d = r_ca_^2_^+ + r_o_^2_^- [/tex]

d = 0.100 nm + 0.140 nm

= 0.240 nm

Let's also calculate the distance of separation between anion and cation when there is an applied electric field.

We use the formula:

∆d = 5%d => 0.05d

= 0.05 * 0.024 nm

∆d = 0.0120 nm

[tex]0.120* 10^-^9m[/tex]

[tex]= 1.20*10^-^1^1m[/tex]

Given magnitude of each dipole= [tex]1.602*10^-^1^9 C [/tex]

Let's find the dipole moment, with the formula:

p = q∆d

Substituting figures in the formula, we have:

[tex]p = 1.602*10^-^1^9 * 12*10^-^2^1 [/tex]

[tex] = 1.92*10^-^3^0 C.m[/tex]

Answer:

Explanation:

Answer:

= 1.92*10^-^3^0 C.m

Explanation:

Given:

r_ca_^2_^+ = 0.100 nm

r_o_^2_^- = 0.14 nm

Let's find the distance of separation between cation and anion when there is no applied electric field with the formula:

d = 0.100 nm + 0.140 nm = 0.240 nm

Let's also calculate the distance of separation between anion and cation when there is an applied electric field.

We use the formula:

∆d = 5%d => 0.05d

= 0.05 * 0.024 nm

∆d = 0.0120 nm

0.120* 10^-^9m

= 1.20*10^-^1^1m

Given magnitude of each dipole= 1.602*10^-^1^9 C

Let's find the dipole moment, with the formula:

p = q∆d

Substituting figures in the formula, we have:

p = 1.602*10^-^1^9 * 12*10^-^2^1

= 1.92*10^-^3^0 C.m

1.95  or 2  is the molarity of a 45.3g sample of KNO3 (101g) dissolved in enough water to make a 0.225L solution.

The correct answer is option b

Explanation:

Data given:

mass of KN[tex]O_{3}[/tex] = 45.3 grams

volume = 0.225 litre

molarity =?

atomic mass of KNO3 = 101 grams/mole

molarity is calculated by using the formula:

molarity = [tex]\frac{number of moles}{volume of the solution}[/tex]

first the number of moles present in the given mass is calculated as:

number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]

number of moles = [tex]\frac{45.3}{101}[/tex]

0.44 moles of KNO3

Putting the values in the equation of molarity:

molarity = [tex]\frac{0.44}{0.225}[/tex]

molarity = 1.95

It can be taken as 2.

The molarity of the potassium nitrate solution is 2.

The answer for the following problem is mentioned below.

Explanation:

 Given:

Initial pressure ([tex]P_{1}[/tex]) = 600 mm of Hg

Final pressure ([tex]P_{2}[/tex]) = 1200 mm of Hg

Initial volume ([tex]V_{1}[/tex]) = 200 ml      

To find:

Final volume ([tex]V_{2}[/tex])

We know;

According to the ideal gas equation,

    P × V = n × R × T

Where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So,

 From the above mentioned equation,

        P × V = constant

[tex]\frac{P_{1} }{P_{2} }[/tex] = [tex]\frac{V_{1} }{V_{2} }[/tex]

Where,

([tex]P_{1}[/tex]) represents the initial pressure of the gas

([tex]P_{2}[/tex]) represents the final pressure of the gas

([tex]V_{1}[/tex])  represents the initial volume of the gas

([tex]V_{2}[/tex])  represents the final volume of the gas

So;

[tex]\frac{600}{1200}[/tex] = [tex]\frac{V_{2} }{200}[/tex]    

[tex]V_{2}[/tex] = 100 ml

Therefore the final volume of the gas is 100 ml.                                                                                                                                                                              

Final answer:

Using Boyle's Law, which states that the product of pressure and volume for a gas is constant at a fixed temperature, it is calculated that the new volume of the CO2 (g) sample when the pressure is doubled to 1200 mmHg at constant temperature is 100 mL.

Explanation:

The question asks about the change in volume of a gas sample when the pressure is increased while keeping the temperature constant. This scenario is described by Boyle's Law, which states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with the pressure. The initial condition for the CO2 (g) is 200 mL at 600 mmHg, and after the pressure is increased to 1200 mmHg, we want to find the new volume.

Boyle's Law can be expressed as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively. Using this relationship, when P1 = 600 mmHg, V1 = 200 mL, P2 = 1200 mmHg, we need to solve for V2. When we plug these values into the equation, we get V2 = (P1 * V1) / P2 = (600 mmHg * 200 mL) / 1200 mmHg = 100 mL.

Therefore, the new volume of the CO2 (g) sample when the pressure is increased to 1200 mmHg at constant temperature is 100 mL.

Answer:

HOFO = (0, 0, +1, -1)

Explanation:

The formal charge (FC) can be calculated using the following equation:

[tex] FC = V - N - \frac{1}{2}B [/tex]

Where:

V: are the valence electrons

N: are the nonbonding electrons

B: are the bonding electrons

The arrange of the atoms in the oxyacid is:

H - O₁ - F - O₂

Hence, the formal charge (FC) on each of the atoms is:

H: FC = 1 - 0 - 1/2*(2) = 0            

O₁: FC = 6 - 4 - 1/2*(4) = 0        

F: FC = 7 - 4 - 1/2*(4) = +1

O₂: FC =  6 - 6 - 1/2*(2) = -1

We can see that the negative charge is in the oxygen instead of the most electronegative element, which is the F. This oxyacid is atypical.  

I hope it helps you!

Antoine Lavoisier divided elements into four categories. Dmitri Mendeleev arranged elements by atomic mass and predicted undiscovered elements. Johann Wolfgang Dobereiner created groups of three elements based on similar properties, whereas John Newlands also arranged elements by atomic weights and predicted some properties of missing elements.

The Periodic Table has been created thanks to several scientists who categorized elements by different methods. Antoine Lavoisier, known as the father of modern chemistry, divided elements into four categories: gases, non-metals, metals, and earths. Consequently, the scientist who divided elements into four categories was Antoine Lavoisier. Dmitri Mendeleev arranged elements according to their atomic mass and used this order to predict the properties of undiscovered elements. Hence, Dmitri Mendeleev is the scientist who used patterns to predict undiscovered elements. Johann Wolfgang Dobereiner, suggested a law of triads, where patterns among groups of three elements were identified based on their properties. Thus, Dobereiner is the scientist who created groups of three elements, each based on similar properties. Lastly, John Newlands arranged elements in order of atomic weights, predicting some properties of missing elements.

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Answer:

oxidation number

Explanation:

Describes the degree of oxidation of an atom in a chemical compound.

Answer:

For the reaction N2 (g) and 2N (g) the energy is required to break the bond in N2 to from N atoms. This means AH is positive, is an endothermic reaction.

Explanation:

Entropy increases that  is positive  because atoms are now free to move.

Answer:

Halogens are better oxidizing agents than their alkali metal counterparts within the same period

Halogens have a higher first ionization energy than their alkali metal counterparts within the same period.

Halogens have a higher electron affinity than their alkali metal counterparts within the same period.

Halogens have a higher effective nuclear charge than their alkali metal counterparts within the same period

Explanation:

Effective nuclear charge increases across a period and accounts for increase in ionization energy across a period. This explains why halogens in the same period with alkali metals have higher effective bucket charge as well as ionization energy than the alkali metals.

Similarly, electron affinity increased across a period, halogens have greater ekctron affinity than corresponding alkali metals in the same period.

Answer:

[tex]Q_{iron} = -32730.083\,J[/tex]

Explanation:

By the First Law of Thermodynamics, the piece of iron and water reach thermal equilibrium when both have the same temperature. The heat released by the piece of iron is received by the water. That is to say:

[tex]Q_{iron} = -Q_{w}[/tex]

[tex]Q_{iron} = (93.64\,g)\cdot \left(4.186\,\frac{J}{kg\cdot ^{\textdegree}C} \right)\cdot (7\,^{\textdegree}C-90.5\,^{\textdegree}C)[/tex]

[tex]Q_{iron} = -32730.083\,J[/tex]

Answer:

Explanation:

Bromine comes in handy with reactivity of organic compounds such as alkenes and arenes( aromatic hydrocarbon). In arenes , it is an electrophilic substitutional reaction where bromine becomes the electrophile to form a substituted aromatic ring. The use of bromine in monobromination (i.e substitution of one bromine atom) is due to the higher electronegative effect in bromine which makes it to be a strong electrophile.

In the diagram; the left hand side shows the arene compound and the right hand side shows the major product after it has undergo monobromination with Br₂/FeBr₃.

Answer:

a

Explanation:

Answer:

Heating is the best method

Explanation:

When a mass of crystals containing Water of crystallization is heated sufiiciently to a high temperature, water vapor may be driven off the crystals by the action of heat alone.

Answer:

Explanation:

If solid/crystals, compound can be dried by applying heat to remove the water of crystallization.

Drying agents are also utile in removing moisture from compounds in a dessicator. Substances which absorb water from air can be used as drying agents for gases. For instance, [tex]CaCl_{2}[/tex], CaO or silica gel is suitable for drying gases, depending on reactivity of the substance.

Answer: 68.6 kPa

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.

[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.

We are given:

[tex]P_1=?\\V_1=13.81L\\P_2=261.1kPa\\V_2=3.63L[/tex]

Putting values in above equation, we get:

[tex]P_1\times 13.81L=261.1\times 3.63L\\\\P_1=68.6kPa[/tex]

Thus pressure  before its volume was decreased was 68.6 kPa

Answer:

waxing crescent phase

Answer:

Waxing crescent

Explanation:

because i got the question right