Which of these orbital designations are invalid?
a. 4s
b. 2d
c. 3f
d. 3p

The answer is A,Today is cloudy, but tomorrow will be clear and sunny.

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Answer:

the mixture is made of one substance

Explanation:

I did the quiz

The weight percent of water in the hydrate before heating is 16.58%.

To find the weight percent of water in the hydrate before heating, we need to calculate the percent by mass of water. This can be done by dividing the mass of water in 1 mole of the hydrate by the molar mass of the hydrate, and then multiplying by 100%. The formula for weight percent is:

Given that the weight of water is 0.349 grams and the weight of the hydrate is 2.107 grams, we can substitute these values into the formula:

Hence, The weight percent of water in the hydrate before heating is 16.58%.

The calculated mass of [tex]Al_2O_3[/tex] formed should be approximately 223.04 g.

Given reaction: [tex]2Al + Fe_2O_3[/tex] → [tex]Al_2O_3 + 2Fe[/tex]

Calculate the moles of Al and [tex]Fe_2O_3[/tex]:

Molar mass of Al = 26.98 g/mol

Molar mass of [tex]Fe_2O_3[/tex] = (55.85 g/mol * 2) + (16.00 g/mol * 3)

= 159.70 g/mol

Moles of Al = 118 g / 26.98 g/mol

≈ 4.37 moles

Moles of [tex]Fe_2O_3[/tex] = 601 g / 159.70 g/mol

≈ 3.76 moles

Determine the limiting reactant:

The balanced equation shows that 2 moles of Al react with 1 mole of [tex]Fe_2O_3[/tex]. So, for the given moles of [tex]Fe_2O_3[/tex], Al is the limiting reactant.

Calculate the moles of [tex]Al_2O_3[/tex] formed:

From the balanced equation, 2 moles of Al produce 1 mole of [tex]Al_2O_3[/tex].

Moles of [tex]Al_2O_3[/tex] formed = 4.37 moles Al / 2

≈ 2.19 moles

Calculate the mass of [tex]Al_2O_3[/tex] formed:

Molar mass of [tex]Al_2O_3[/tex] = (26.98 g/mol * 2) + (16.00 g/mol * 3)

= 101.96 g/mol

Mass of [tex]Al_2O_3[/tex] formed = 2.19 moles * 101.96 g/mol

≈ 223.04 g

Therefore, the calculated mass of [tex]Al_2O_3[/tex] formed should be approximately 223.04 g.

Learn more about limiting reagent, here:

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To calculate the mass of Al2O3 formed and the amount of excess reagent left in the reaction between aluminum and iron(III) oxide, we need to perform stoichiometric calculations that involve converting grams to moles, identifying the limiting reagent, applying stoichiometric ratios to determine product amounts, and converting moles back to grams.

To answer this question, we need to apply the concept of stoichiometry, which is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. As the balanced chemical equation for the reaction between aluminum and iron(III) oxide is given as 2Al + Fe2O3 → Al2O3 + 2Fe, this suggests that 2 moles of aluminum react with 1 mole of iron(III) oxide to produce 1 mole of aluminum oxide and 2 moles of iron.

To find the mass of Al2O3 formed, we would convert the given mass of Al and Fe2O3 to moles using their respective molar masses, see which reagent is the limiting reagent, and then use this information to determine the mass of Al2O3 formed in the reaction. The limiting reagent in a chemical reaction is the reactant that is completely consumed and determines the maximum amount of product that can be formed. Once the limiting reagent is consumed, the reaction stops, leaving the other reactant in excess.

By converting the remaining mass of the excess reagent left at the end of the reaction back to grams, we would calculate the mass of the excess reagent left. This involves a multi-step process encompassing the conversion of mass to moles, comparison of molar amounts, application of stoichiometry to find product amounts, and conversion of moles back to grams.

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Answer:

i think its C

Explanation:

An element existing in nature by itself, such as a neutral atom, will have an overall charge of zero due to the balanced number of protons and electrons.

When an element exists in nature by itself, it is in the form of a neutral atom, which means it must have an overall charge of zero. This neutrality is achieved because the atom has an equal number of protons and electrons, and since protons and electrons carry equal but opposite charges, their charges cancel each other out. The proton carries a positive charge, and the electron carries a negative charge, both with a magnitude of 1.60×10-19 coulombs (C). Neutral atoms may form molecules by sharing electrons via covalent bonds but still retain no net charge unless they lose or gain electrons to become ions. Therefore, in its most stable form, an element on its own will exhibit a charge of zero.

Intramolecular forces such as the ionic bond strength in a CrCl2 molecule tend to be stronger than the intermolecular forces, which include the attractive forces between water molecules and chromium chloride ions.

The attractive forces between water molecules and chromium (Cr) and chloride (Cl) ions, or the ionic bond strength in a CrCl2 molecule, can be compared by analyzing intermolecular forces and intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, like the ionic bonds in a CrCl2 molecule. These forces are typically much stronger compared to intermolecular forces. Intermolecular forces, on the other hand, are the attractions between molecules, such as the attractive forces between water and Cr and Cl ions. Using water as an example, it has strong intermolecular forces of hydrogen bonding leading to high surface tension. However, the energy required to overcome these forces (around 17 kilojoules for one mole of water) is significantly less than the energy required to break the covalent bonds in the water molecule itself (about 430 kilojoules). Therefore, intramolecular forces like the ionic bond strength of CrCl2 are generally stronger than intermolecular forces between water molecules and Cr and Cl ions.

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To find the number of moles of gas gained or lost, we can use the ideal gas law equation. By substituting the given values into the equation and simplifying, we find that the system gained or lost 0.921 moles of gas.

To find the number of moles of gas gained or lost, we will use the ideal gas law equation:

P1V1 / T1 = P2V2 / T2

Using the given information:

P1 = 1 atm, V1 = 22.4 L (at STP), T1 = 273 K

P2 = 2.00 atm, V2 = 14.5 L, T2 = 100 °C = 373 K

Substituting the values into the equation:

(1 atm)(22.4 L) / (273 K) = (2.00 atm)(14.5 L) / (373 K)

Simplifying the equation:

Moles gained or lost = (1 atm)(22.4 L)(373 K) / (273 K)(2.00 atm)(14.5 L)

Calculating the result:

Moles gained or lost = 0.921 moles

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Stab pull is the process by which the Earth's tectonic plates move apart at their edges due to the force of gravity at divergent plate boundaries. It leads to the formation of mountains and volcanic activity.

Stab pull refers to the process by which the Earth's tectonic plates move apart at their edges due to the force of gravity. This occurs at divergent plate boundaries, where the edges of the leaves become denser and sink downwards and away from each other. As a result, new crust is created from magma upwelling beneath the leaves, leading to the formation of mountains and volcanic activity.

For example, the East African Rift System is a prime location for stab pull. The African Plate is splitting apart along this boundary, causing the formation of the Great Rift Valley and the creation of a new crust.

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To prepare 1.80 L of a 0.215 M H2SO4 solution, you will need to use 21.5 mL of 18 M sulfuric acid.

To calculate the volume of 18 M sulfuric acid needed, we can use the formula:

M1V1 = M2V2

Where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Plugging in the given values:

(0.215 M)(1.80 L) = (18 M)V2

Solving for V2:

V2 = (0.215 M)(1.80 L) / (18 M) = 0.0215 L = 21.5 mL

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To prepare 1.80 L of 0.215 M H2SO4, you would need to use 21.5 mL of 18 M sulfuric acid. This is calculated using the formula for dilution: C1V1 = C2V2.

To answer this question, you need to use the formula for dilution of solutions: C1V1 = C2V2. Here, C1 is the initial concentration(18 M), V1 is the volume of this concentrated solution that we're trying to find, C2 is the final concentration(0.215 M), and V2 is the final volume(1.80 L).

When we plug these numbers into the formula, we get the equation 18 M × V1 = 0.215 M × 1.80 L. Solving for V1 gives us V1 = (0.215 M × 1.80 L) / (18 M) = 0.0215 L or 21.5 mL.

So, to prepare 1.80 L of 0.215 M H2SO4, you would need to use 21.5 mL of 18 M sulfuric acid solution.

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The partial pressure of Hydrogen gas can directly be calculated by simply taking the difference of the overall pressure and the vapour pressure of water. That is:

P (H2 gas) = 759.2 torr – 23.8 torr

P (H2 gas) = 735.4 torr

The enzyme α-amylase increases the rate of starch breakdown by lowering the activation energy of the reaction, which is option (e). Enzymes act as catalysts in metabolic processes, and their activity can be influenced by environmental factors such as pH and temperature.

The enzyme α-amylase increases the rate at which starch is broken down into smaller oligosaccharides by lowering the activation energy of the reaction. This process does not require altering the equilibrium constant or the change in free energy; rather, it involves facilitating the reaction so that the energy barrier is lower, allowing the reactants to convert into products more easily. In the options provided, the correct answer to how α-amylase increases the rate of the reaction is (e) lowering the activation energy of the reaction.

An enzyme's role is to act as a catalyst, speeding up chemical reactions without being consumed or permanently changed by the reaction. Enzymes are crucial for metabolic processes, such as the hydrolysis of starch into sugars like glucose and maltose, which are used by cells as a source of energy and carbon. Parameters that can influence the activity of α-amylase include pH and temperature, which if not optimal, can reduce the rate of the reaction.

[tex]\boxed{{\text{3}}\;{\text{covalent bonds}}}[/tex] are formed by nitrogen if each of its unpaired electrons participates in one bond.

Further Explanation:

The bond that is formed by the mutual sharing of electrons between the bonded atoms is called a covalent bond. It is also known as a molecular bond. Covalent compounds are those compounds which are formed by the electron sharing between two or more non-metals.

The octet rule is the rule in accordance to which the elements have the tendency to bond with other elements and acquire eight electrons in their valence shells. This results in achieving a stable noble gas configuration.

For example, the formation of [tex]{\text{NaCl}}[/tex] occurs according to the octet rule. The electronic configuration of sodium is [tex]{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{p}}^{\mathbf{6}}}{\mathbf{3}}{{\mathbf{s}}^{\mathbf{1}}}[/tex] while that of chlorine is [tex]{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{p}}^{\mathbf{6}}}{\mathbf{3}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{3}}{{\mathbf{p}}^{\mathbf{5}}}[/tex] .

Chlorine is one electron short of the stable noble gas configuration and sodium can achieve stable configuration by losing an electron. So sodium loses an electron and forms cation and chlorine gains that electron to complete its octet.

In case of nitrogen:

The atomic number of nitrogen is 7. Its ground state electronic configuration is [tex]{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{p}}^{\mathbf{3}}}[/tex] . The partial orbital diagram is the diagrammatic representation of the distribution of electrons in the valence shell only. In case of nitrogen, the valence shell is 2s and 2p.(Refer to the attached image).

Nitrogen atom consists of five electrons in its valence shell. Out of these five electrons, two are paired and present in 2s while three remain unpaired in 2p. If all three unpaired electrons are to be used, it will accept three electrons from the neighboring atoms to make three covalent bonds. The octet of nitrogen is completed with the formation of three covalent bonds with neighboring atoms.

Learn more:

1. Identification of ionic bonding: https://brainly.com/question/1603987

2. Chemical bonds in NaCl: https://brainly.com/question/5008811

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Ionic and covalent compounds

Keywords: covalent bonds, nitrogen, unpaired electrons, bond, 3, paired electrons, covalent compounds, molecular bond, covalent bond, mutual sharing of electrons, five electrons, partial orbital diagram, distribution of electrons.