Major groups of minerals include _____. oxides and carbonates ions and isotopes inorganics and halides silicates and magnetics

A mixture of iodine and sulphur can be separated by various ways:

1. Add Carbon disulphide that is (CS₂) then stir the solution and filter it . As sulphur dissolved in CS₂, so it come into filter and got separated from the mixture.

2. Take the mixture in a china dish and cover it with a inverted funnel and heat it, on heating iodine will get sublime and produces vapors . These vapors condenses at funnel and produce iodine solid.

Answer: Ik this is super late but for anyone looking for the real answer its this

Answer is C4H8O4

Explanation:

CH2O 12.0g+2(1.0g)+16.0g=30.0g

120.1/30.0=4

CH2O x 4 = C4H8O4

The product: product ratio in the provided chemical equation, which represents the combustion of benzene-acetic acid, is 14CO₂:6H₂O. So, for every two molecules of benzene-acetic acid reacted, 14 molecules of carbon dioxide and 6 molecules of water are produced.

The equation provided is a chemical reaction representing the combustion of benzene-acetic acid (C₆H₅COOH). The correct product: product ratio is determined by the stoichiometric coefficients in front of each product. In this selected reaction, for every two molecules of benzene-acetic acid that react with 15 molecules of oxygen, 14 molecules of carbon dioxide (CO₂) and 6 molecules of water (H₂O) are produced.

Therefore, the product: product ratio is 14CO₂:6H₂O. This means for each reaction, you will get more than twice as much carbon dioxide as water. This ratio helps us to understand how much of each product we can expect when the reaction takes place, assuming there are sufficient reactants to complete the reaction.

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Final answer:

Creating a supersaturated solution of cadmium iodide involves dissolving as much solute as possible in water at an elevated temperature and then slowly cooling the solution to room temperature without disturbance.

Explanation:

To make 1.0 L of a supersaturated solution of cadmium iodide (CdI2), first a saturated solution must be prepared at a temperature higher than room temperature by dissolving as much CdI2 as possible. This can be done by gradually adding CdI2 to a liter of water with continuous stirring while heating the solution to increase the solubility. Once no more salt can dissolve, the solution is allowed to slowly cool to room temperature, making sure to avoid any disturbance that could cause premature crystallization. As the solution cools, it becomes supersaturated because the solubility of CdI2 decreases with temperature, allowing more solute to be contained in solution than is stable at the lower temperature.

If the solution remains undisturbed, the excess solute will stay dissolved, resulting in a supersaturated solution. However, any disturbance or addition of a seed crystal will cause the excess solute to precipitate out rapidly. Therefore, care must be taken to keep the solution free from dust and vibrations that can induce crystallization.

Answer:

The Answer is C. magnesium hydroxide

Explanation:

Final answer:

The osmotic pressure of a solution containing 0.048 g of a hydrocarbon with a molar mass of 340 g/mol is calculated by first finding the molarity and then using the osmotic pressure formula, resulting in approximately 0.00345 atm.

Explanation:

To determine the osmotic pressure of a solution containing a hydrocarbon solute, first, calculate the molarity of the solution. The molarity (M) is the number of moles of solute per liter of solution. Since the molar mass of the hydrocarbon is given as 340 g/mol, you can find the number of moles of solute by dividing the mass of the solute by its molar mass.

Number of moles = Mass (g) / Molar Mass (g/mol) = 0.048 g / 340 g/mol = 0.000141 mol

The volume of the solution is not given, but assuming it's 1 liter for the sake of calculation. Divide the number of moles by the volume to get molarity:

Molarity (M) = Number of moles / Volume (L) = 0.000141 mol / 1 L = 0.000141 M

To find the osmotic pressure, use the formula:

Osmotic Pressure (II) = MRT

Where M is the molarity, R is the gas constant (0.0821 L·atm/(K·mol)), and T is the absolute temperature in Kelvin. Assuming room temperature (25°C or 298 K) for the calculation:

II = (0.000141 M) x (0.0821 L·atm/K·mol) x (298 K)

Therefore, the osmotic pressure is:

II = (0.000141) x (0.0821) x (298) ≈ 0.00345 atm

Answer:

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The formation of a stable diatomic molecule from atoms typically results in a negative ">δH°<", a negative ">δS°<", and a negative ">δG°<". This reflects an exothermic reaction with decreased entropy, leading to spontaneity under the right temperature conditions.

When a stable diatomic molecule forms from its atoms, there are typical changes in enthalpy (">">δH°"), entropy (">">δS°"), and Gibbs free energy (">">δG°"). The formation of a stable molecule from atoms is typically exothermic, which means that energy is released, and ">">δH°" would be negative. As atoms come together to form a molecule, the system becomes more ordered, and thus, the change in entropy, ">δS°<", is negative.

According to the Gibbs free energy equation, ">δG° = δH° - TδS°<", and at a temperature (T) where this process is spontaneous, the sign of ">δG°<" would be negative because loss of heat (">">exothermic reaction") and orderliness in system (">">decreased entropy") favor the formation of the molecule.

Answer:

The half-life (in seconds) of a zero-order reaction is 8.15 seconds.

Explanation:

Initial concentration of the of the reactant = [tex][A_o]=0.884 M[/tex]

The value of rate constant = [tex]k=5.42\times 10^{-2} M/s[/tex]

The half life for zero order reaction is given as:

[tex]t-{\frac{1}{2}}=\frac{[A_o]}{2k}=\frac{0.884 M}{2\times 5.42\times 10^{-2} M/s}=8.15 s[/tex]

The half-life (in seconds) of a zero-order reaction is 8.15 seconds.

The correct answer is: 6.173 seconds.

Explanation:

The formula for half life (of reaction) is:

Half-life = [tex]\frac{1}{K*I}[/tex] --- (A)

Where,

K = Rate constant = 0.54 [tex]\frac{1}{ms}[/tex]

I = Initial concentration = 0.30 m

Plug in the values in equation (A):

Half-life = [tex]\frac{1}{0.54*0.30} = 6.173[/tex]

Hence, the half life is 6.173 seconds.

The half-life of a second-order reaction is [tex]\boxed{{\text{6}}{\text{.173 seconds}}}[/tex]  

Further explanation:

Second-order reaction:

A reaction is said to be of second-order if its rate is proportional to the square of the concentration of one reactant. The general example of second-order reaction is,

[tex]2{\text{A}}\to{\text{P}}[/tex]

Here,

A is the reactant.

P is the product.

The rate is calculated by using the following equation:

[tex]{\text{Rate}}={\text{k}}{\left[{\text{A}}\right]^2}[/tex]

Another form of second-order reaction is as follows:

[tex]{\text{A}}+{\text{B}}\to{\text{P}}[/tex]

Here,

A and B are the two different reactants.

P is the product.

The rate is calculated by using the following equation:

[tex]{\text{Rate}}={\text{k}}\left[{\text{A}}\right]\left[{\text{B}}\right][/tex]

Half-life:

Half-life is defined as the time required to reduce the concentration of reactant to half of its initial value. It is denoted by [tex]{{\text{t}}_{1/2}}[/tex] . The general expression to calculate [tex]{{\text{t}}_{1/2}}[/tex]  of second-order reaction is,

[tex]{{\text{t}}_{1/2}}=\frac{1}{{{\text{k}}{{\left[{\text{A}}\right]}_{\text{0}}}}}[/tex]                                                              …… (1)

Here,

[tex]{{\text{t}}_{1/2}}[/tex]  is the half-life of the reaction.

k is the rate constant for the reaction.

[tex]{\left[ {\text{A}}\right]_{\text{0}}}[/tex]  is the initial concentration of the reactant.

The rate constant for the given reaction is [tex]0.54\;{{\text{M}}^{-1}}{{\text{s}}^{-1}}[/tex] .

The initial concentration of the given reaction is 0.30 M.

Substitute these values in equation (1).

[tex]\begin{aligned}{{\text{t}}_{1/2}}&=\frac{1}{{\left({0.54\;{{\text{M}}^{ - 1}}{{\text{s}}^{ - 1}}}\right)\left( {{\text{0}}{\text{.30 M}}}\right)}}\\&=\frac{1}{{0.162\;{{\text{s}}^{ - 1}}}}\\&=6.1728\;{\text{s}}\\&\approx 6.173\;{\text{s}}\\\end{aligned}[/tex]

So, the half-life of a second-order reaction is 6.173 seconds.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603

2. Calculate the moles of ions in the solution: https://brainly.com/question/5950133

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical Kinetics

Keywords: second-order reaction, half-life, initial, half, 6.173 seconds, 0.30 m, k, A0, A, B, P, 2A, reactant, product, t1/2.

Answer : The formula for the compound aluminum sulfate is [tex]Al_2(SO_4)_3[/tex]

Explanation :

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non-metals.

In the given question, the aluminum metal forms an ionic compound with polyatomic anion (sulfate) [tex]SO_4^{2-}[/tex].

Here, aluminum is having an oxidation state of +3 called as [tex]Al^{3+}[/tex] cation and sulfate [tex]SO_4^{2-}[/tex] is an anion with oxidation state of -2. The charges are not balanced. So, the charges are balanced by the cris-cross method. Thus, the compound formed will be, [tex]Al_2(SO_4)_3[/tex]

Hence, the formula for the compound aluminum sulfate is [tex]Al_2(SO_4)_3[/tex]

The formula for aluminum sulfate is Al₂(SO₄)₃. This is because two Al³+ ions balance three SO₄²⁻ ions to achieve electrical neutrality.

The compound aluminum sulfate is formed when an aluminum cation (Al³+) combines with a sulfate ion (SO₄²⁻). Given that atoms strive to reach electrical neutrality, the formula for this ionic compound reflects the simplest ratio that can balance the positive and negative charges. In this case, two aluminum ions (2 x +3 for a total charge of +6) are needed to balance three sulfate ions (3 x -2 for a total charge of -6), resulting in the chemical formula Al₂(SO₄)₃.

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Answer:

A solution is another name for homogeneous mixture.

Explanation

Solution or homogeneous mixture are names for the classification of matter.

Matter is classified in 2 big groups: pure substances and mixtures.

Pure substances: are those that cannot be separated into more simple ones by physics methods, and are classified in compounds and elements.

Mixtures: are those that can be separated into more simple ones by physics methods, and are classified in homogeneous and heterogeneous.

Answer:

D. Fe2(SO4)3

is the answer

Explanation:

Answer:

The answer is 450 g

Explanation:

When the solid sublimes, it gives the same mass of gas because sublimation is a physical change, and it is valid the energy and mass conservation law (mass cannot be destroyed or created in a chemical or physical process). If sublimation is complete, all the molecules changes its physical state from solid to gas.

Sublimation is the process where a solid changes directly into a gas without passing through the liquid phase. The mass of the substance remains unchanged, so a solid with a mass of 450 g will yield a gas with a mass of 450 g.

The process that changes a solid directly into a gas is called sublimation. In this process, a solid transitions to a gas phase without passing through the liquid phase. Importantly, the mass of the substance remains constant during this phase change. Therefore, if you start with a solid that has a mass of 450 g, the resulting gas will also have a mass of 450 g. This is in accordance with the law of conservation of mass which suggests that mass can neither be created nor be destroyed.

An example of sublimation occurs with dry ice (solid CO₂), which transitions directly from a solid to a gas at standard temperature and pressure.

THE ANSWER IS

the number of perching birds present

Final answer:

The thiazoline ring in thiamine hydrochloride and thiamine contains two p electrons, one from each of the two carbon atoms in the ring.

Explanation:

The student has asked about the number of p electrons in the thiazoline ring of thiamine hydrochloride and thiamine. Thiazoline is a heterocyclic compound made up of a five-membered ring containing both sulfur and nitrogen atoms. In thiamine, the thiazoline ring contributes to the structure of thiamine pyrophosphate (TPP), which is the active co-enzyme form of thiamine.

In the thiazoline ring, there are two carbon atoms that can have p electrons in their p orbitals. Each carbon atom has one p electron that is not involved in sigma bonding (single covalent bonds). Therefore, the thiazoline ring contains a total of two p electrons from the carbon atoms. The nitrogen and sulfur atoms do not contribute p electrons that are relevant to the aromaticity or reactive properties of the thiazoline ring in thiamine. The thiazoline ring does not have a completely conjugated pi system like the thiazole ring, which is a closely related structure.

The answer is D. Two tectonic plates are colliding and one is sliding under the other

The two chair conformations of cis 1 chloro 2 methylcyclohexane are shown in the image attached.

Due to the cyclohexane ring's capacity for ring flipping, isopropyl chloro-2-methylcyclohexane displays two chair conformations. The methyl group is positioned equatorially in the first conformation, while the chlorine atom is located in the axial position.

The chlorine atom is in the equatorial position and the methyl group is in the axial position in the second conformation, which has the substituents arranged in reverse order. These conformations show the dynamic character of cyclohexane rings, where substituents can be positioned in either an axial or an equatorial orientation to minimize steric hindrance between them by ring flipping.