Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you the substance is a pure substance or a mixture, if you can. • Sample A is a solid yellow cube with a total mass of 50.0 g. The cube is put into a beaker filled with 250. mL of water. The cube collapses into a small pile of orange powder at the bottom of the beaker. When this powder is filtered out, dried and weighed, it has a total mass of 29.9 g. If the experiment is repeated with 500. mL of water, the powder that's left over has a mass of 10.0 g. Sample B is 100. g of a coarse grey powder with a faint unpleasant smell. 15. mg of the powder are put into a very thin tube and heated. The powder begins melting at 66.2 °C.The temperature stays constant as the powder slowly melts. After the last of the powder melts, the temperature starts to rise again, eventually reaching 76.0 °C. O pure substance Is sample A made from a pure substance or a mixture? x ? mixture If the description of the substance and the outcome of the experiment isn't enough to decide, choose "can't decide. (can't decide) O pure substance Is sample B made from a pure substance or a mixture? If the description of the substance and the outcome of bstance and the outcome of the experiment isn't enough to decide, choose "can't decide." mixture (can't decide)

Answers

Answer 1

Answer:

Answer and Explanation: Sample A is a mixture. Solubility is characteristics of each substance, which means a substance can be distinguished from other substances and can be useful to separate mixtures.

In Sample A, when is added different volumes of water, the resulting powder has different mass. This means there are more than one substance forming the yellow cube. Therefore, sample A is a mixture.

Sample B is a puresubstance. Each substance has its own melting point. Whe na pure substance reaches its melting point, temperature is constant until all of that substance is melted. In sample B, temperature is stable at 66.2°C and then, after all the powder is melted, it rises again. Therefore, sample B is a pure substance.

Answer 2

Answer:

Final answer:

Sample A is a mixture based on the experiment result, while the nature of Sample B is inconclusive.

Explanation:

The result of the experiment with sample A indicates that it is a mixture. When the solid yellow cube is put into water, it collapses into a small pile of orange powder. The mass of the powder that is left over depends on the amount of water used. This suggests that the cube is composed of different substances that can be separated by filtration.

On the other hand, the result of the experiment with sample B is inconclusive, so we can't decide whether it is a pure substance or a mixture. Heating the coarse grey powder causes it to melt at a constant temperature, but there is a temperature increase after the last of the powder melts. This could indicate that the powder is a pure substance with a melting point range, or it could suggest the presence of impurities.

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Molecular compounds result from covalent bonding which are called _______.This is for high school physical science

Answers

Answer:

Diatomic Molecule

Explanation:

Wich substances is most likely to form in a precipitation reaction?

Answers

Answer:

A precipitation reaction refers to the formation of an insoluble salt when two solutions containing soluble salts are combined. The insoluble salt that falls out of solution is known as the precipitate, hence the reaction's name.

Explanation:

After 56.0 min, 40.0% of a compound has decomposed. What is the half‑life of this reaction assuming first‑order kinetics?

Answers

Answer:

Go ahead and plug in the percentages and time to find the answer.

Explanation:

The amount of a substance with half-life h, that remains after time t is 0.5t/h

Since 26% has decomposed, 74% remains.

So .74 = 0.580/h

ln .74 = (80/h) ln 0.5

h/80 = ln 0.5 / ln .74

h = 80 ln 0.5 / ln .74

h = 184.16 minutes

A student was asked to prepare exactly 250 mL of a 0.500 M aqueous potassium hydroxide solution. What mass of potassium hydroxide (molar mass = 56.10 g/mol) must the student dissolve in the 250 mL of solution? 1. 28.1 g 2. 3.0 g 3. None of these 4. 14.0 g 5. 7.01 g 6. 56.1 g

Answers

Answer:

Option 5 → 7.01 g

Explanation:

Molarity . volume (L) = Moles

This can help us to determine the moles of KOH that are in the solution.

We convert the volume from mL to L → 250 mL . 1L / 1000mL = 0.250 L

0.5 mol /L . 0.250L = 0.125 moles of KOH

Now, we only have to convert the moles to mass, by the molar mass:

Moles . molar mass = mass → 0.125 mol . 56.1 g/mol = 7.01 g

Answer:

We need 7.01 grams of KOH (option 5)

Explanation:

Step 1: Data given

Volume aqueous KOH solution = 250 mL = 0.250 L

Molarity = 0.500 M

Molar mass of KOH = 56.10 g/mol

Step 2: Calculate moles KOH

Moles KOH = molarity * volume

Moles KOH = 0.500 M * 0.250 L

Moles KOH = 0.125 moles

Step 3: Calculate mass of KOH

Mass KOH = moles KOH * molar mass KOH

Mass KOH = 0.125 moles * 56.10 g/mol

Mass KOH = 7.01 grams

We need 7.01 grams of KOH

What is an extensive property that can be calculated?

Answers

Answer: The property which depends on the quantity of the substance is called an extensive property. The free energy change for a reaction (Δ G) depends on the quantity of the substance and is therefore an extensive property. It shows the additive nature. The extensive property Δ G is easily calculated from the formula, ΔG = -nFE cell.

Explanation:

Final answer:

An extensive property is one that changes when the size of the sample changes. One such property that can be calculated is enthalpy. Enthalpy can be calculated using the formula H = E + PV.

Explanation:

An extensive property is a property that changes when the size of the sample changes. Examples include mass, volume, length, and total charge. One extensive property that can be calculated is enthalpy.

The enthalpy of a system can be calculated using the formula H = E + PV, where H represents the enthalpy, E the internal energy of the system, P the pressure, and V the volume. Like other extensive properties, the enthalpy of a system would change with the quantity or size of the sample.

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What is the correct equilibrium constant expression for the following reaction? 3A2 = 2B3 when the reaction started with the initial concentrations of A2 = 3 M and B3 = 2 M and continued until the equilibrium concentrations of A2 = 2.5 M and B3 = 2.5 M

Answers

Answer:

Kc = [B₃]²/[A₂]³ = 0.40

Explanation:

3A₂ ⇄ 2B₃

Given at equilibrium => [A₂] =2.5 and [B₃] = 2.5

Kc = [B₃]²/[A₂]³ = (2.5)²/(2.5)³ = (2.5)⁻¹ = 0.40

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