In the redox conversion of Cr(OH)4 to
CrO2, the oxidation number of Cr goes from
(-1, 0, +3) to (-2, +6, +8). Recall that the
oxidation number of oxygen is typically -2.

Solid, Heat of fusion, Liquid, Heat of vaporization, and Gas are the answers for the following questions represented by the graph.

At phase transition temperature is constant so BC & DE represent phase transition.

1)AB represents solid here \(C_{solid}\) constant use to calculate heat absorbed here.

2)BC represents melting solid here heat of fusion is constantly used to calculate heat absorbed here.

3)CD represents liquid here \(C_{liquid}\) constant use to calculate heat absorbed here.

4)DE represents Boiling liquid here Heat of vaporization is constant used to calculate heat absorbed here.

5)EF represents gas here \(C_{gas}\) constant used to calculate heat absorbed here.

An alteration in the state from one phase to another is known as a phase transition. A phase transition is identified by a sudden change in one or more physical attributes accompanied by a minute change in temperature.

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Given the symbol Cl-37, the element is Cl, but there is no information about its charge. When this is like that, it is implicit it is an atom, that is, it is neutral, its charge is 0.

The number indicates is the mass number, so its mass is 37.

So, mass is 37 and charge is 0.

Answer:

2.7 mL

Explanation:

The equation of the reaction is;

2AlBr3 + 3Cl2 -----> 2AlCl3 + 3Br2

Number of moles in 2.12 x 1022 formula units of aluminum bromide

1 mole of AlBr3 = 6.02 * 10^23 formula units

x moles = 2.12 x 1022 formula units

x = 2.12 x 1022 formula units  * 1 mole/ 6.02 * 10^23 formula units

x = 0.0352 moles of AlBr3

According to the reaction equation;

2 moles of AlBr3 produces 3 moles of Br2

0.0352 moles of AlBr3 produces 0.0352 moles  *  3 moles /2 moles

= 0.0528 moles of Br2

Mass of Br2 produced = 0.0528 moles of Br2 * 159.808 g/mol

Mass of Br2 produced = 8.44g

But density = mass/volume

volume = mass/density

volume of Br2 = 8.44 g/ 3.12 g/mL

volume of Br2 =  2.7 mL

Answer:

body-centered cubic

Explanation:

:D

Hydrocid and oxacid acids (formation, nomenclature, properties and application):

Formation: binary compounds formed by HYDROGEN and a NON-METAL (from groups 6A and 7A of the periodic table).

Properties: they are found naturally in a gaseous state. They are called hydrocids because when they dissolve in water and dissociate, they generate acidic solutions.

Applications : In the industry they are mostly used for organic synthesis and for the leather tanning industry.

Formula: HX, H is hydrogen and X is the chemical symbol for halogen.

Systematic: name of the acid + suffix -ide + hydrogen. Example: Hydrogen chloride → HCl

Traditional: acid + non-metal + suffix "-hydric". Example: Hydrogen telluric acid → H₂Te

Formation: Acid compounds formed by HYDROGEN, a NON-METAL and OXYGEN. They are obtained by reaction between an anhydride (acid oxide) and water . Hydrogen acts with oxidation number +1 and oxygen with -2.

Properties: They are ternary compounds (formed by three chemical elements), formed by a non-metallic chemical element, oxygen together with hydrogen.

Applications: In the industry it is used to obtain fertilizers for agriculture . As well as it is used for the synthesis of other acids, sulfates and in the petrochemical industry .

Nomenclature

Formula: HaXbOc, H is hydrogen, X is a non-metal element and O is oxygen

Stock : Acid + prefix indicating the number of oxygens + oxo + prefix for the number of non-metallic atoms + root of that atom ending in "-ico" + valence in Roman numerals (in brackets). Example: Dioxochloric acid (III) → HClO₂

Systematic : prefix that indicates the number of oxygens + oxo + prefix for the number of non-metallic atoms + root of that atom ending in "-ate" + valence in Roman numerals (in parentheses) + hydrogen. Example: Hydrogen tetraoxochlorate (VII) → HIO₄

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Translated Question: Prepare, in your notebook, a graphic organizer in which you synthesize the information obtained in the text. Provides information on the formation of hydroacid and oxacid acids, their nomenclature, their properties and their application in industry. help me

The dividing line between metals and nonmetals is a diagonal line that separates the two categories in the periodic table.

The periodic table is a table of the chemical elements in which the elements are arranged according to their atomic number, electronic configurations, and chemical properties. It is a standard tool used in chemistry to predict an element's behavior and to determine its relationships to other elements. It is divided into groups, which share common properties, and periods, which reflect the electron structure of the elements. The periodic table is generally divided into four categories, including metals, nonmetals, metalloids, and noble gases. Metals are generally characterized by their ability to conduct electricity and heat, their ductility and malleability, their luster and shine, and their reactivity. Nonmetals are generally characterized by their lack of metallic properties, including their lack of ability to conduct electricity and heat, their brittleness, and their lack of luster and shine. The dividing line between metals and nonmetals lies along a diagonal line that separates the two categories. The elements that fall along this line, including boron, silicon, germanium, arsenic, antimony, tellurium, and polonium, are considered to be metalloids, as they possess properties of both metals and nonmetals.

The dividing line between metals and nonmetals is a diagonal line that separates the two categories in the periodic table. Elements that possess metallic and nonmetallic properties, such as boron, silicon, germanium, arsenic, antimony, tellurium, and polonium, fall along this line and are known as metalloids. The periodic table is a fundamental tool that helps predict the behavior of elements and determine their relationships to one another.

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When the enthalpy is positive (H > 0), the reaction is endothermic, that is, it absorbs energy.

When the enthalpy is negative (H< 0), the reaction is exothermic, that is, it releases energy.

Therefore, the correct matches are

Endothermic:

• Positive H Value.

• Absorbs Heat.

• Products have higher energy.

Exothermic:

• Negative H Value.

• Releases Heat.

• Products have lower energy.

The illustration of Thomson's atom represents a neutral atom. In this case, the total amount of positive charge and the total amount of negative charge must be equal. This means that there are equal numbers of protons and electrons in the atom. This is what makes the atom neutral.

A neutral atom is an atom that has no electrical charge. An atom is neutral because it has the same amount of positively charged protons and negatively charged electrons. The nucleus of an atom contains protons, which are positively charged particles. Electrons, which are negatively charged particles, are located in the atom's electron cloud around the nucleus.

Electrons, protons, and neutrons are the three components of atoms. Electrons are negatively charged, protons are positively charged, and neutrons have no charge. Electrons are found outside the nucleus of the atom and are continually moving at high speeds.

In summary, if the illustration of Thomson's atom represents a neutral atom, then the total amount of positive charge and the total amount of negative charge must be equal. This means that there are equal numbers of protons and electrons in the atom. This is what makes the atom neutral.

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Answer:

t = 42.4 years

Explanation:

To find the amount of time needed for the sample to decay, you need to use the half-life equation:

\(N(t) = N_0(\frac{1}{2})^{t/h}\)

In this equation,

-----> N(t) = final mass (mg)

-----> N₀ = initial mass (mg)

-----> t = time passed (yrs)

-----> h = half-life (yrs)

You can find how much time passed by plugging the given variables into the equation and solving for "t". The final answer should have 3 sig figs like the given values.

N(t) = 9.38 mg                       t = ? yrs

N₀ = 25.0 mg                        h = 30.0 yrs

\(N(t) = N_0(\frac{1}{2})^{t/h}\)                                <----- Half-life equation

\(9.38mg = 25.0mg(\frac{1}{2})^{t/30.0yrs}\)              <----- Insert variables

\(0.3752 = (\frac{1}{2})^{t/30.0yrs}\)                          <----- Divide both sides by 25.0 mg

\(ln(0.3752) = ln((\frac{1}{2})^{t/30.0yrs})\)              <----- Take the natural log of both sides

\(ln(0.3752) = \frac{t}{30.0yrs} ln(\frac{1}{2})\)                   <----- Rearrange the exponent

\(-9.803 = \frac{t}{30.0yrs} (-0.6931)\)                 <----- Solve the natural logs

\(1.1414= \frac{t}{30.0yrs}\)                                  <----- Divide both sides by -0.6931

\(42.4 yrs= t\)                                        <----- Multiply both sides by 30.0 yrs

Answer: just want points

Explanation: POINTS!! ;)

Answer: the volume increases

Explanation: It increasses because the volume is how tall it is

Answer: mass of oxygen is needed to react with 10g of hydrogen is

Explanation:Calculation:

- The reaction of oxygen with hydrogen is depicted as:

H2 + 1/2 O2 → H20

- From the above reaction, we can calculate the mass of oxygen as:

Mass of oxygen required to react with 2 gm hydrogen = 16 gm

⇒ Mass of oxygen required to react with 10 gm hydrogen = (16/2) × 10

⇒ Mass of oxygen = 80 gm

- According to the given amounts, mass of water formed = 10 + 80 = 90 gm

- No of moles of water in 90 gm = 90/18 = 5 moles

- The volume of 5 moles of water = 22.4 × 5

⇒ The volume of 5 moles of water = 112 Litre

- Hence, 80 gm of oxygen is required to react with 10 gm of hydrogen and 112 L of water is formed on this reaction.

5 million

Explanation:

trust me bro

1. The longer distance for each choice is:

2. The complete statements are as follows;

3. The basic unit of length in the metric system is the meter and is represented by a lowercase m.

4. The meter is defined as the distance traveled by light in absolute vacuum in 1⁄299,792,458 of a second.

5. The values that complete each statement is given below:

6. The larger value for each option is:

7. The number of millimeters in 1 centimeter is 10 mm

8. Using the ruler and line, the answers are:

The unit for measuring distance in the metric system is the meter. Smaller and larger values of the meter are also used such as millimeters, centimeters, kilometers, etc.

Other units for measuring distance include yards, miles, and inches.

The various units for measuring distance can also be interconverted using their conversion factors.

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Answer:

Chemical bonds hold molecules together and create temporary connections that are essential to life.

Explanation:

Chemical Bonding refers to the formation of a chemical bond between two or more atoms, molecules, or ions to give rise to a chemical compound. These chemical bonds are what keep the atoms together in the resulting compound.

Answer:

Chemical bonds hold molecules together and create temporary connections that are essential to life.

Explanation :

A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.

Answer:

thx for giving points and in reaturn i won't answer qustion because i don't know the answer thx for th points agian

Explanation:

The number of years it will take for 570.7 mg ³H to decay to 0.56 mg ³H is approximately 103.1 years.

To determine the time it takes for 570.7 mg of hydrogen-3 (³H) to decay to 0.56 mg, we'll use the half-life formula:

N = N₀ * (1/2)^(t/T)
where:
N = remaining amount of ³H (0.56 mg)
N₀ = initial amount of ³H (570.7 mg)
t = time in years (unknown)
T = half-life (12.3 years)

Rearrange the formula to solve for t:

t = T * (log(N/N₀) / log(1/2))

Plugging in the values:

t = 12.3 * (log(0.56/570.7) / log(1/2))
t ≈ 103.1 years

It will take approximately 103.1 years for 570.7 mg of hydrogen-3 to decay to 0.56 mg.

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The element being oxidized in this reaction is  ["C"].

The element being reduced in this reaction is  ["Mn"].

In the given reaction, MnO4⁻ is being reduced to Mn2+, indicating that Mn is undergoing a reduction process and gaining electrons.

Therefore, Mn is the element being reduced.

On the other hand, \(H_{2} C_{2} O_{4}\)  is being oxidized to \(CO_{2}\), which means that carbon (C) is losing electrons and undergoing oxidation.

Therefore, C is the element being oxidized.

To determine which element is oxidized and which is reduced, we look at the change in oxidation states.

Mn goes from +7 to +2, indicating a reduction (a decrease in oxidation state), while C goes from +3 to +4, indicating an oxidation (an increase in oxidation state).

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To make 1 liter of a 10% NaCl solution from a solid stock, you will require the following materials and containers.MaterialsSolid NaClDistilled water1-Liter volumetric flask250-mL volumetric flask 2-beakersProcedureTo prepare 1 liter of a 10% NaCl solution, the following procedure should be followed:Measure out 100g of NaCl using a balance.

Measure the weight of an empty 250-mL volumetric flask.Add the NaCl to a 250-mL beaker and add a small amount of distilled water to it to dissolve the NaCl.Carefully pour the dissolved NaCl solution into the 250-mL volumetric flask. Add distilled water to the mark on the flask to make up the volume. Stopper the flask and invert it several times to mix the solution.Measure the weight of the 1-Liter volumetric flask.Add the 250-mL volumetric flask solution to a 1-Liter volumetric flask.Add distilled water to the mark on the flask to make up the volume.

Stopper the flask and invert it several times to mix the solution.The final volume of the solution will be 1 liter of a 10% NaCl solution.PrecautionsEnsure the NaCl has completely dissolved before adding more water to avoid making a less concentrated solution.Measure the weight of the volumetric flask before and after adding the solution to calculate the volume of solution that was added.Use distilled water to prepare the solution.

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In benzene (C6H6), all of the carbon-carbon bonds are equal to one and one-half bonds. The bond length of these bonds is in between that of a single bond and a double bond all of the bond angles at 120 degrees.

Benzene (C6H6) is an aromatic hydrocarbon consisting of a ring of six carbon atoms with alternating single and double bonds. The structure of benzene is often represented as a hexagon with a circle inside, indicating the delocalization of electrons. In reality, benzene is a planar molecule with all carbon atoms in the same plane.

In benzene, each carbon atom forms three sigma bonds with its neighboring carbon atoms, resulting in a total of six sigma bonds in the ring. These sigma bonds are considered one and one-half bonds, also known as delocalized pi bonds. The delocalization of electrons creates a resonance structure where the double bonds continuously shift between adjacent carbon atoms.

The bond length of the carbon-carbon bonds in benzene is intermediate between that of a single bond and a double bond. This is due to the electron delocalization and the partial double bond character of the bonds. The bond angles in benzene are approximately 120 degrees, which is consistent with the trigonal planar geometry around each carbon atom.

Overall, the unique structure of benzene with its delocalized pi bonds and bond lengths in between those of single and double bonds contribute to its stability and aromaticity.

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The density of the liquid is 0.98 g/mL

The density of a substance is defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as:

Density = mass / volume

With the above formula, we can obtain the density of the liquid.

Density = mass / volume

Density of liquid = 30.8 / 31.5

Density of liquid = 0.98 g/mL

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Answer:

1/32

Explanation:

Answer:

En la mejora de la investigación y la medicina.

Explicación:

La aparición del microscopio ha permitido la creación de nuevas áreas de estudio, tanto en la mejora de la investigación como en la medicina. La invención del microscopio nos permite crecer y desarrollarnos en el campo de la investigación y el estudio. Con este microscopio, los científicos pudieron descubrir la estructura de la célula, así como las partículas subatómicas que están presentes dentro del átomo. Gracias a este microscopio, los científicos pudieron crecer y desarrollarse en el campo de la creación de nuevos medicamentos.

The intermolecular force for BrF is dipole-dipole.

Dipole-dipole is the intermolecular force for BrF. The molecules of a substance in a specific state of matter are subject to intermolecular forces.

The difference in electronegativity between the atoms that make up a substance determines the type of intermolecular forces that are present.

Because of the non-zero difference in electronegativity between Br and F, the molecule of BrF is polar and the intermolecular forces are dipole-dipole forces.

Given that BrF has a high boiling point and a known dipole moment of 1.4D, the dipole-dipole force is the greatest interparticle force acting on it. Dispersion force is also present.

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When each trial is started with 10.0 g of tert-butanol the freezing points would be higher. Hence option A is correct.

Generally the term freezing point is described as the, temperature at which a liquid becomes a solid. Basically with the melting point, increase in the pressure usually increases the normal freezing point. Basically the freezing point is always lower than the melting point in the case of mixtures and also for certain organic compounds such as fats.

When each trial is started with 10.0 g of tert-butanol the freezing points would be higher because there is more tert-butanol present in order to balance out the additives added to the solution. Therefore, when each trial is started with 10.0 g of tert-butanol, the freezing points would be higher. Hence, option A is correct.

The given question is incomplete the complete question is given as,

If each trial started with 10.0 g of tert-butanol, what would have changed? explain your reasoning.

a) the freezing points would be higher

b) the freezing points would be lower

c) the freezing points doesn't change

d) None of the above

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The group number 16 on the periodic table is represented by the description .

The valency of group number 16 is - 2 . Hence , two other atom ( like aluminum ion ) is required .

Similarly , the valency of aluminium ion is + 3 . Hence , three of other atom( like  oxygen ) is required .

Oxygen ions have a -2 charge with the formula \(O^{-2}\) . Aluminum ions have a 3+ charge with the formula \(Al^{+3}\) .

An ionic compound must be neutral , which means the positive and negative charges must be equal. So we need three oxygen ions to bond with two aluminum ion , and the formula for aluminum oxide is \(Al_{2} O_{3}\) .

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Changing the mass of the acid used in the experiment would not affect the Ka value. it's worth noting that if the experimental conditions were altered along with the increase in acid mass.

If you weighed out more than 1.2 g of your unknown acid, the final calculated Ka (acid dissociation constant) would likely be the same as the value determined in the experiment, assuming the conditions and methodology remain constant.

The Ka value represents the equilibrium constant for the dissociation of the acid in water, which is a characteristic property of the acid itself. It is independent of the amount of acid present in the solution. Therefore, changing the mass of the acid used in the experiment would not affect the Ka value.

In acid-base experiments, the concentration of the acid is typically used to calculate the Ka value. Concentration is defined as the amount of substance per unit volume. In this case, if you weigh out more than 1.2 g of the acid, you would dissolve it in a larger volume of solvent to maintain the same concentration. This would ensure that the ratio of acid to water remains constant, and the equilibrium constant, Ka, would not change.

However, it's worth noting that if the experimental conditions were altered along with the increase in acid mass, such as using a different volume of water or altering the temperature, it could potentially impact the calculated Ka value. In such cases, it would be important to consider the specific details of the experiment to determine the potential effects on the final calculated Ka.

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The pH of the base is 4.42.

No of moles of CH₃COOH = 47.0mL x L/1000L x 0.313 mol/L = 0.0147 mol

No of moles of OH⁻ added = 4.97 mL x L/1000 mL x (0.481 mol Ba(OH)₂)/L × (2 mol OH⁻)/(1 mol Ba(OH)₂) = 0.00478 mol

CH₃COOH + OH⁻ → CH₃C00⁻ + H₂O

Total volume = 47.0 mL + 4.97 mL = 51.97 mL = 0.05197 L

Concentration of CH₃COOH, [CH₃COOH] = 0.00992 mol/0.05197 L = 0.191M

Concentration of CH3C00⁻, [CH3C00⁻] = 0.00478 mol/0.05197 L = 0.0920 M

pKᵇ, of CH₃COOH = -log Kₐ = -log (1.8x10⁻⁵) = 4.74

According to Henderson equation,

pH = pKₐ + log [CH3C00⁻] / [CH₃COOH]

     = 4.74 + log 0.0920 / 0.191 = 4.42

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