In part a, we saw that the theoretical yield of aluminum oxide is 1.40 mol . calculate the percent yield if the actual yield of aluminum oxide is 0.938 mol .

The redox reaction between chromium and copper(II) in acidic solution is balanced by adjusting the half-reactions to make the electrons lost equal the electrons gained, combining them into the overall equation, and finally balancing the hydrogen with water and hydrogen ions.

To complete and balance the equation for the redox reaction where chromium (Cr) is oxidized to chromate (CrO42−) and copper(II) (Cu2+) is reduced to copper (Cu), we need to balance each half-reaction and then combine them to make the overall balanced equation. The two half-reactions given in acidic solution are:

First, to balance the number of electrons, we multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2, obtaining:

Next, we combine these balanced half-reactions to form the overall equation:

2Cr(s) + 3Cu2+ (aq) → 2Cr3+ (aq) + 3Cu(s)

Finally, in an acidic solution, we need to balance the hydrogens by adding 7H2O to the left side and 14H+ to the right side, resulting in:

2Cr(s) + 3Cu2+ (aq) + 7H2O(l) → 2CrO42− (aq) + 3Cu(s) + 14H+ (aq)

Answer:

Flowering plant

Explanation:

Flowering plants :

Flowering plants are the dominant plant form on land which reproduce by sexual and asexual means.

Their most distinguishing feature is their reproductive organs, which are  flowers. Sexual reproduction in flowering plants includes

i) Production of male and female gametes,

2) Transfer of the male gametes to the female ovules in a process called pollination.

After pollination occurs, automatically fertilization happens.

Since orange tree follows the above procedure it is a

Flowering plant

One part of evidence that the color of the flame created is from the metal ion and not from the chemical is that not any of the flames with dissimilar metals had the similar color (for each metal had its own flame color). Even if most of the metals tested had chloride, the colors of the flames were all dissimilar. The two flames that both had copper (one had copper (II) chloride and the other had copper (II) sulfate) were exactly close in color. The one was green-blue, and the other was a bright green. This displays that they were nearly the same, and the minor difference could be credited to error.

The observed colors of compounds are typically due to the metal ions, which absorb specific wavelengths of visible light due to electron transitions in their d orbitals. An additional test to confirm this would be to change the ligand, which should result in a change of color if the metal ion is responsible.

The characteristic color observed for each compound is typically due to the metal ion rather than the non-metal anion. This is because the color in transition metal compounds arises from the d-orbital electrons absorbing certain wavelengths of visible light. When this light is absorbed, the electrons transition between different d orbitals, and the remaining light that is not absorbed gives the compound its distinct color.

An additional test that could be conducted to confirm that the color is due to the metal ion involves using a different ligand to form a new compound with the same metal ion. If the color changes with the different ligand, this further confirms that it's the metal ion interacting with the ligand field, rather than the anion, that causes the color of the compound.

Many factors affect the exact color, such as the metal's oxidation state and the types of ligands attached to the metal. For instance, different oxidation states of vanadium exhibit different colors in solution. This variability of color is a unique property of transition metal ions, not observed in compounds with non-transition metal ions.

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A) On a molecular scale, a crystal of alum differs from a crystal of potassium aluminum sulfate in terms of its water content. Alum contains 12 water molecules, whereas potassium aluminum sulfate does not contain any water molecules.

B) When performing stoichiometric calculations with alum compared to potassium aluminum sulfate, the main difference lies in accounting for the water molecules present in alum.

When calculating the number of moles of alum, consider the molar mass of the compound, including the water molecules. For anhydrous potassium aluminum sulfate, consider the molar mass of the anhydrous compound without any water molecules.

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Answer:

The answer is B I took the test

Explanation:

Final answer:

The chemical formula for calcium chloride is CaCl2. It is composed of calcium ions (Ca2+) and chloride ions (Cl-).

Explanation:

Calcium chloride is an ionic compound, composed of calcium ions (Ca2+) and chloride ions (Cl-). The chemical formula for calcium chloride is CaCl2.

When calcium donates two electrons to two chlorine atoms, it forms Ca2+ and two Cl- ions. The oppositely charged ions attract each other, resulting in the formation of calcium chloride.

According to VSEPR theory, the electron pair geometry around the central carbon atom in acetic acid is tetrahedral, while the molecular structure is trigonal planar.

According to VSEPR theory (Valence Shell Electron Pair Repulsion theory), the electron pair geometry and molecular structure of acetic acid can be determined.

The electron pair geometry around the central carbon atom in acetic acid is tetrahedral, while the molecular structure is trigonal planar.

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Acetic acid has the moleculer formula CH3COOH, which consists of a methyl group and a carboxyl group. The geometry around each interior atom varies: the carbon atom in the methyl group is tetrahedral; the carbon atom in the carboxyl group is trigonal planar; and the oxygen atoms in the carboxyl group have bent geometries.

Acetic acid, represented by the formula CH3COOH, can be divided into three sections for purpose of geometry prediction. These sections are CH3, C atom of COOH and the O atoms in COOH.

CH3: Carbon atom in methyl group (CH3) forms four single bonds (with three hydrogens and the carbon atom of the carboxyl group), thus it is sp3 hybridized and its geometry is tetrahedral.

C (in COOH): The carbon atom in the carboxyl group (COOH), forms a single bond with the methyl group, a double bond with one oxygen atom, and a single bond with the OH group. This indicates sp2 hybridization and it has a trigonal planar geometry.

O (in COOH): Both the oxygen atoms in COOH are also sp2 hybridized. The oxygen atom which forms a double bond with the C atom has a bent (or V shaped) geometry. The oxygen atom of the hydroxyl group is also bent, however, it forms one single bond with the carbon atom and one single bond with the hydrogen atom.

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Answer:

The main feature of an oxidation-reduction reaction is electron exchange.

Explanation:

Every oxeduction reaction is related to an electron transfer between the atoms and / or ions of the reactant substances.

An oxeduction reaction is characterized as a simultaneous process of electron loss and gain, as electrons lost by one atom, ion or molecule are immediately received by others.

For example, a copper sulfate (CuSO4 (aq)) solution is blue due to the presence of Cu2 + ions dissolved in it. If we place a metal zinc plate (Zn (s)) in this solution, over time we may notice two modifications: the color of the solution will be colorless and a metallic copper deposit will appear on the zinc plate.

Therefore, the reaction that occurs in this case is as follows: Zn (s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq). Note that there was an electron transfer from zinc to copper. Analyzing separately the transformation that occurred in each of these elements, we have: Zn (s) → Zn2 + (aq). Zinc has lost 2 electrons from metal zinc to cation. In this case, zinc has oxidized. Cu2 + (aq) → Cu (s)  Already with the copper the opposite happened, it gained 2 electrons, from cation copper II to metallic copper. Copper has been reduced.

This explains the two changes observed, as the solution became colorless because the copper ions turned to metallic copper, which settled on the zinc plate. Since there was a simultaneous loss and gain of electrons, this reaction is an example of a redox reaction.

When sodium hydroxide (NaOH) is used to treat 3-bromo-2,4-dimethylpentane, an elimination reaction occurs that produces an alkene. In this instance, a dihydrohalogenation reaction specifically occurs.

In this process, both a bromine atom (Br) from the 3-position and a hydrogen atom (H) from the -carbon (the carbon next to the bromine atom) are removed. As a result, the -carbon and the carbon previously bonded to the bromine form a double bond (alkene).

2,4-dimethylpent-2-ene is the alkene that is produced when 3-bromo-2,4-dimethylpentane is reacted with sodium hydroxide. Depending on the reaction conditions, solvent and other variables, this reaction can also occur via E1 or E2 mechanism. The unique conditions of the reaction can have an effect on the selectivity of the reaction.

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Final answer:

3-Bromo-2,4-dimethylpentane will form only one alkene product upon treatment with sodium hydroxide due to the presence of only one β-hydrogen in the correct position for an E2 elimination reaction.

Explanation:

When 3-bromo-2,4-dimethylpentane is treated with sodium hydroxide, the reaction is an elimination reaction, specifically an E2 mechanism. In an E2 mechanism, the base removes a hydrogen atom that is anti to the leaving group, in this case bromide, resulting in the formation of a double bond and thus creating an alkene. For 3-bromo-2,4-dimethylpentane, there is only one β-hydrogen available that is anti to the leaving group, which is on the third carbon. This means that only one alkene can form, as there are no other β-hydrogens that can be removed to form a different alkene.

The optically inactive stereoisomer of 1,3-cyclopentanediol is the meso isomer, where both -OH groups are on the same side of the cyclopentane ring, creating a plane of symmetry.

The question asks us to draw the optically inactive stereoisomer(s) of 1,3-cyclopentanediol. To be optically inactive, the molecule must not be chiral. A molecule is chiral if it cannot be superimposed on its mirror image, meaning it has at least one chiral center. 1,3-cyclopentanediol can have chiral centers at the two carbon atoms bearing the OH groups (C1 and C3).

To make it optically inactive, we need to eliminate any chiral centers by arranging the substituents so that the molecule is symmetrical.

Therefore, the optically inactive isomer of 1,3-cyclopentanediol must have both OH groups on the same side of the cyclopentane ring (either both above or both below the plane), making it a meso compound and achiral. This structure is known as the meso form.

Answer:

The Nazca plate forms the southeastern part of the Pacific plate. The Nazca and the Pacific plate share both divergent and transform type of plate boundary. The Pacific and the Nazca plate are separating at an increasing rate of about 122-142mm/year. These are the regions where the rate of tectonic activity is very high.

The Nazca Plate and the South American Plate are also converging at a rate of about 62mm/year, being the fastest subduction rate on earth.

Explanation :

Extraction : It is a separation technique in which one compound is separated from a mixture of compounds.

In  a chemical process, the solution may not contain only the desired product but there maybe a mixture of products present.

So, for obtaining the desired product we need to get rid of the other products in a mixture like catalyst (if used), other by products or unreacted starting material.

For a compound to be extracted it has to be soluble in the extracting solvent.

Therefore, this process is based on the miscibility of solute and solvent in a mixture.

Extraction is a useful method of “pre-purification” of organic compounds in order to remove impurities. It is based on different solubilities of components in the mixture.

Further Explanation:

Purification of organic compounds:

The method of purification depends on the nature of substances and the type of impurities present in them. There are several methods for purification of compounds as follows:

1. Chromatography

2. Steam distillation

3. Fractional distillation

4. Simple distillation

5. Simple crystallization

6. Fractional crystallization

7. Sublimation

8. Azeotropic distillation

Extraction is the process to separate the desired substance mixed with some impurities. The mixture is added with a solvent such that the desired substance is soluble in it whereas the impurities are insoluble in it. Any substance is never so pure that it is free from impurities. So the undesirable substances are to necessary to be removed from the mixture and therefore extraction is an essential procedure in order to remove impure substances from the mixtures.

Extraction makes use of two different phases that are immiscible with each other. It is carried out on the basis of relative solubilities of different substances present in the mixture.

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Answer details:

Grade: College

Subject: Chemistry

Chapter: Solvent extraction

Keywords: extraction, purification, organic compounds, sublimation, crystallization, distillation, mixtures, impurities, chromatography, different phases, undesirable substances.

Heat is applied to the cover and empty crucible to remove moisture, because the reading and test results of the chemical are affected by the presence of water.

Heat is defined as a type of energy that moves through systems or items with varying temperatures. Increased kinetic energy  of a substance's constituent particles is another effect of heat. Conduction, convection, and radiation are the three mechanisms through which thermal energy is transferred.

A crucible will probably display a weight that differs from what it would if it were at the same temperature as the balance compartment if it is not at that temperature. Due to the creation of convection currents that impact the apparent mass, the weight will vary depending on the temperature, being slightly less or higher in the case of heat.

Thus, heat is applied to the cover and empty crucible to remove moisture, because the reading and test results of the chemical are affected by the presence of water.

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The -23.0 kJ energy is released or absorbed when 40.0 g of steam at 250.0 c is converted to water at 30.0 c

To find the total energy change when 40.0 g of steam at 250.0°C is converted to water at 30.0°C, we need to consider the following steps:

1. **Heating steam to 100.0°C:**

[tex]\[ q_1 = m \cdot c_{\text{steam}} \cdot \Delta T_1 \][/tex]

2. **Phase change from steam at 100.0°C to water at 100.0°C:**

[tex]\[ q_2 = m \cdot \Delta H_{\text{vap}} \][/tex]

3. **Heating water from 100.0°C to 30.0°C:**

[tex]\[ q_3 = m \cdot c_{\text{water}} \cdot \Delta T_2 \][/tex]

The total energy change (\(q_{\text{total}}\)) is the sum of these contributions:

[tex]\[ q_{\text{total}} = q_1 + q_2 + q_3 \][/tex]

Substitute the given values and constants, taking into account the correct sign conventions for energy release (negative) or absorption (positive).

Calculate the result:

[tex]\[ q_{\text{total}} = (40.0 \, \text{g} \cdot 2.03 \, \text{J/g°C} \cdot (100.0 - 250.0)) + (40.0 \, \text{g} \cdot 2260 \, \text{J/g})[/tex][tex]+ (40.0 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot (30.0 - 100.0)) \][/tex]

The calculated value should be -23.0 kJ.

Therefore, the correct answer is (b) -23.0 kJ.

This negative sign indicates that energy is released during the process.

The probable question may be:

How much energy is released or absorbed when 40.0 g of steam at 250.0 c is converted to water at 30.0 c?

sp.ht. of water 4.18 J/gC                            ΔH_{fus} of water is 333 J/g.

sp.ht. of steam 2.03 J/gC                          ΔH_{vap} of water is 2260 J/g

a) -24.0 kJ

b) -23.0 kJ

c)-32.9 kJ

d)-114 kJ

e) -122 kJ

To calculate the pH of a buffer solution, you can use the Henderson-Hasselbalch equation. In this case, you can calculate the initial concentration of the conjugate base and acid using the equation. Then, find the new pH of the buffer solution after adding NaOH.

To calculate the pH of a buffer solution, you can use the Henderson-Hasselbalch equation:

pH = pKa + log([conjugate base]/[acid])

In this case, the initial pH of the buffer is 4.19, so you can calculate the initial concentration of the conjugate base and acid using the equation.

Then, you can calculate the new concentrations of the conjugate base and acid after adding 0.010 mol of NaOH and use the Henderson-Hasselbalch equation again to find the new pH of the buffer solution.

Using the given Ka value of 6.5 x 10^-5 for benzoic acid and the initial concentrations of the buffer solution, you can calculate the initial pH and the new pH after adding NaOH.

The final pH of the buffer solution after the addition of NaOH is approximately 4.37.

The final pH of the buffer solution after the addition of NaOH can be calculated using the Henderson-Hasselbalch equation:

[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]

where [tex]\([\text{A}^-]\)[/tex] is the concentration of the benzoate ion and [tex]\([\text{HA}]\)[/tex] is the concentration of benzoic acid.

Given that the initial pH is 4.19, we can find the [tex]\(\text{pKa}\)[/tex] of benzoic acid:

[tex]\[ \text{pKa} = \text{pH} - \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]

Since the initial concentrations of benzoic acid and sodium benzoate are both 0.10 M, the ratio [tex]\(\frac{[\text{A}^-]}{[\text{HA}]}\)[/tex] is 1, and [tex]\(\log(1) = 0\)[/tex], so:

[tex]\[ \text{pKa} = 4.19 \][/tex]

Now, upon the addition of 0.010 mol of NaOH to 500.0 ml of buffer solution, we need to calculate the new concentrations of benzoic acid and sodium benzoate. The NaOH will react with benzoic acid to produce sodium benzoate:

[tex]\[ \text{C}_6\text{H}_5\text{COOH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{COONa} + \text{H}_2\text{O} \][/tex]

The amount of benzoic acid that reacts is 0.010 mol, which will produce the same amount of sodium benzoate. The initial moles of benzoic acid and sodium benzoate are both 0.050 mol (0.10 M * 0.500 L). After the reaction, the moles of benzoic acid will be 0.040 mol, and the moles of sodium benzoate will be 0.060 mol. The new concentrations are:

[tex]\[ [\text{HA}]' = \frac{0.040 \text{ mol}}{0.500 \text{ L}} = 0.080 \text{ M} \][/tex]

[tex]\[ [\text{A}^-]}' = \frac{0.060 \text{ mol}}{0.500 \text{ L}} = 0.120 \text{ M} \][/tex]

Now we can use the Henderson-Hasselbalch equation to find the new pH:

[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}'}{[\text{HA}]'} \right) \][/tex]

[tex]\[ \text{pH} = 4.19 + \log \left( \frac{0.120}{0.080} \right) \][/tex]

[tex]\[ \text{pH} = 4.19 + \log(1.5) \][/tex]

[tex]\[ \text{pH} = 4.19 + 0.176 \][/tex]

[tex]\[ \text{pH} = 4.366 \][/tex]

Final answer:

To find the mass percent of KCl in the solution, add the mass of KCl to the mass of water to get the total mass of the solution, then divide the mass of KCl by this total mass and multiply by 100% to get a mass percent of 12.11%.

Explanation:

To calculate the mass percent of KCl in a solution where 31.0 g of KCl is dissolved in 225 g of water, you should first determine the total mass of the solution. This is done by adding the mass of the solute (KCl) to the mass of the solvent (water). Therefore, the mass of the solution is:

mass of solution = mass of solute + mass solvent

mass of solution = 31.0 g KCl + 225 g water = 256.0 g

Next, to find the mass percent of KCl, divide the mass of KCl by the total mass of the solution and then multiply by 100%:

Percent by mass = (mass of solute / mass of solution) x 100%

Percent by mass = (31.0 g / 256.0 g) x 100% = 12.11%

Therefore, the mass/mass percent concentration of KCl in the solution is 12.11%.