H2 burning in o2 to form h2o (l) is an example of a system where the entropy of the universe decreases.

1) Vapor pressure. Water exerts vapor pressure. This pressure is produced by the water molecules that change their state from liquid to gas.

2) Setting a similar system and watching what happens.

If we set a system similar to that shown in the image (place the cup of water in the jar and then close the jar), we will see the following

1) The big jar has no molecules of water gas since we just closed it.

2) Some water molecules change from liquid to gas and start filling the big jar.

3) The big jar will have a certain amount of vapor water.

4) The big jar has so much water that it starts condensing on the walls of the big jar.

5) After enough water is condensed on the walls, it runs down the walls and precipitates.

3) The order observed.

1) Water evaporated from the cup

2) water condensed on the walls of the jar.

3) Water precipitated, or ran down the walls of the jar.

Answer: Yes.

Explanation:

the number of protons in the nucleus determines an element's identity. Chemical changes do not affect the nucleus, so chemical changes cannot change one type of atom into another. The number of protons in a nucleus does change sometimes, however. The identity of the atom, therefore, changes.

Ketoses such as fructose are expected to give a positive Tollens test because of their ability to reduce Tollens' reagent.

The Tollens test is a chemical reaction used to test for the presence of aldehydes. The test is named after the German chemist Bernhard Tollens. The Tollens reagent is a solution of ammoniacal silver nitrate, Ag(NH3)2NO3. The test is based on the fact that aldehydes are readily oxidized to carboxylic acids by a solution of silver nitrate in ammonia.

This reaction is exothermic and produces a silver mirror on the inside of the test tube.The test solution is prepared by adding a few drops of a solution of Tollens' reagent to the substance being tested. The test solution is then heated in a water bath for a few minutes.

If the substance being tested is an aldehyde, it will reduce the silver ions in the Tollens' reagent to metallic silver. The metallic silver will form a mirror on the inside of the test tube. If the substance being tested is not an aldehyde, no mirror will form.The aldehydes are more easily oxidized than the ketoses.

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To determine the density of a rock, we can utilize the formula density = mass / volume. In this scenario, the rock's mass is provided as 45.0 grams, while the volume of water displaced by the rock is calculated to be 17.7 milliliters (by subtracting the initial volume of 10.0 milliliters from the final volume of 27.7 milliliters).

By substituting these values into the density formula, we obtain the following calculation: density = 45.0 g / 17.7 ml = 2.54 g/ml. Hence, the density of the rock is determined to be 2.54 grams per milliliter.

In summary, the given rock has a mass of 45.0 grams, and its displacement in water amounts to 17.7 milliliters.

By applying the density formula, which states that density is equal to mass divided by volume, we calculate that the rock's density is 2.54 grams per milliliter.

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The optimal dimensions for the cylindrical cup are approximately a radius of 1.918 cm and a height of 3.836 cm.

To design an optimal cylindrical cup with a volume of 100 ml and minimized surface area, you'll need to apply calculus to find the dimensions. First, let's convert the volume to cubic centimeters (1 ml = 1 cm³) and set up the equations.

Volume (V) of a cylinder: V = πr²h, where r is the radius and h is the height.
Surface area (A) of a cylinder: A = 2πrh + 2πr².

Given V = 100 cm³, we can express h in terms of r: h = 100/(πr²).

Now substitute this expression for h in the surface area equation:
A(r) = 2πr(100/(πr²)) + 2πr² = 200/r + 2πr².

To find the optimal radius, minimize A(r) by taking its derivative with respect to r and setting it equal to 0:
dA/dr = -200/r² + 4πr = 0.

Solve for r:
4πr³ = 200, so r³ = 50/π, and r ≈ 1.918 cm.

Now find the height using h = 100/(πr²):
h ≈ 100/(π(1.918²)) ≈ 3.836 cm.

So, the optimal dimensions for the cylindrical cup are approximately a radius of 1.918 cm and a height of 3.836 cm.

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In Chemistry, we always have to balance the equations, in many topics within the subject of Chemistry, it is required to do that, this is because we have to account for every atom in the reaction, on both sides, reactant and product, the same number of atoms that are on the reactants side must be on the products side. Therefore, the best answer for this question will be letter D

Answer:

In units of measurement of length we use centimeter (cm) to measure. We can use this unit for measuring the length of a pencil, the width of a book etc. but this unit is too big to measure the thicken of a pencil. So we use another unit called millimeter (mm).

Explanation:

A dihalide in which the halogens are attached on adjacent carbons is called a vicinal dihalide.

Dihalides are organic compounds that contain two halogen atoms in a molecule. In particular, they are compounds containing two halogen atoms in the same molecule. They are also known as geminal dihalides because the two halogens are on the same carbon atom. Some examples of dihalides include ethylene dichloride, ethylene dibromide, and carbon tetrachloride. The halogens that make up dihalides are fluorine, chlorine, bromine, and iodine.

Vicinal dihalides: A vicinal dihalide is a type of organic compound with the molecular formula CnH2n-2X2, where X is a halogen atom and n is an integer. It is a compound with two halogen atoms attached to adjacent carbons. As a result, it is known as a vicinal dihalide. This type of compound is also known as 1,2-dihalides, as the halogen atoms are on the first and second carbons. Examples of vicinal dihalides include 1,2-dichloroethane, 1,2-dibromobutane, and 1,2-difluoroethylene. Therefore, option A is the correct answer.

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Mothballs are composed of naphthalene, C10H8. Naphthalene has a total of 3 resonance structures. The C−C bond lengths in the molecule are expected to be intermediate between C−C single and C=C double bonds. Based on the resonance structures, we can expect that 4 out of the 10 C−C bonds in naphthalene will be shorter than the others.

Naphthalene has a resonance structure due to the delocalization of electrons within the two aromatic rings. The incomplete Lewis structure indicates the presence of two resonance structures for naphthalene. These resonance structures can be obtained by shifting the double bonds within the rings.

In terms of bond lengths, C−C single bonds are longer than C=C double bonds due to the overlapping of orbitals. Since the resonance in naphthalene spreads the electron density across the molecule, the C−C bond lengths are expected to be shorter than those in C−C single bonds but longer than those in C=C double bonds. The delocalization of electrons results in a partial double bond character in the C−C bonds, making them intermediate in length.

As for the variation in bond lengths, not all of the C−C bonds in naphthalene are equivalent due to the presence of resonance structures. The delocalization of electrons causes a redistribution of electron density, leading to a difference in bond lengths. The bonds adjacent to the double bonds in the resonance structures are expected to be shorter than the other C−C bonds.

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Answer:

The burette is contaminated with an acid solution- The concentration is too high

The burette contains a large air bubble in the tip, which disappears during the titration- The concentration of the acid is too low.

A small volume of the acid is spilled when you transfer it into the Erlenmeyer flask- The concentration of the acid is too low.

Explanation:

The molality of a solution is the number of moles of solute per liter of solution. The concentration of an unknown solution is obtained by titration against a standard solution of acid or base whose concentration  is known. It is a volumetric method of analysis.

If the burette is contaminated with the acid, then the concentration of the base that reacts with the acid is less than the stated amount and consequently the concentration of the acid calculated is higher than it should be.

If the tip of the burette contains air bubbles, the volume of the acid reported is less than the volume of acid actually delivered during the titration hence the calculated concentration of the acid is too low.

If some volume of acid is spilled when you transfer the acid into the Erlenmeyer flask, the volume of the acid reacted is decreased and consequently the calculated concentration of the acid is too low.

The molar mass of the unknown gas is 142.53 g/mol.

Graham's law states that the square root of a gas's molar mass is inversely proportional to the rate of effusion of that gas (r) (M).

rH₂/rX = √[M(X)/ M(H₂)]

(rH₂/rX)² = M(X)/ M(H₂)

M(X) = (rH₂/rX)² × M(H₂)

M(X) = (8.4)² × 2.02 g/mol

M(X) = 164 g/mol

Therefore, molar mass of the unknown gas is 142.53 g/mol.

According to Graham's Law, a gas's rate of effusion is inversely related to the square-root of its molecular mass. The gas whose molecular weight is lowest will effuse first.

Based on the molar weights of the two gases, the rate of effusion calculator determines the rate of diffusion or effusion of one gas into another. Depending on the rate of diffusion, it can also provide you with the molar weight of a gas.

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Answer: A: Explaining natural events using facts and data AND B:Developing theories using many lines of evidence

Explanation: trust me.

Answer:

your ear detecteds by cells with shorter hair bundles, located closest to where sound enters the ear; lower-pitched sounds are detected by cells with taller hair bundles located further in the ear!

Explanation:Hope this helps!

Answer:

High-pitched sounds are detected by cells with shorter hair bundles, located closest to where sound enters the ear; lower-pitched sounds are detected by cells with taller hair bundles located further in.

Explanation:

that pattern progresses through the several thousand hair cells that are essential for hearing

Answer:

The combustion will release -3,321 KJ/mol of energy

Explanation:

What we need to do here is to first calculate the number of moles of methane burnt.

Mathematically, the number of moles of methane burnt = mass of methane/molar mass of methane

Molar mass of methane = 16g /mol

Thus the

number of moles will be ;

59.7/16 = 3.73 moles

Now, 1 mole of methane produces -890 KJ/mol if energy

Then the energy produced by by 3.73 moles of methane would be 3.73 * 890 = -3,320.8125

which is approximately -3,321 KJ/mol

Kindly note that the amount of heat is negative because it is an exothermic reaction

The volume of oxygen at STP is approximately 295.03 cm³

To calculate the volume of oxygen at standard temperature and pressure (STP), we can use the ideal gas law equation:

PV = nRT

Where:

P = Pressure

V = Volume

n = Number of moles

R = Gas constant

T = Temperature

First, let's convert the given temperature of 47°C to Kelvin:

T1 = 47°C + 273 = 320K

We can rearrange the ideal gas law equation to solve for the volume at STP:

V2 = (P1 * V1 * T2) / (P2 * T1)

Where:

V1 = Initial volume (320 cm³)

P1 = Initial pressure (1.05 * 10⁵ Nm-²)

T2 = STP temperature (273K)

P2 = STP pressure (1.01 * 10⁵  Nm-²)

Now we can plug in the values:

V2 = (1.05 * 10⁵ Nm-² * 320 cm³ * 273K) / (1.01 * 10⁵ Nm-² * 320K)

Canceling out the units and performing the calculation, we get:

V2 = 295.03 cm³

Therefore, the volume of oxygen at STP is approximately 295.03 cm³.

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Answer:

Plants : Cactus, tumbleweed

Animals : camel, fennec fox

Answer:

Desert plant - Cactus & Tumbleweed

Desert animal - Camel & Scorpion

Answer:

That would be fracture

Explanation:

A way that a mineral breaks apart curved at  the edges or split into uneven pieces

Answer:

Fracture

Explanation:

Got it right!

The pH of the solution is 2.55 and the concentrations of C₂H₅COOH and C₂H₅COO⁻ are 0.594 M and 0.00282 M respectively.

The following chemical equation describes how propanoic acid, a weak acid, will ionise in water.

C₂H₅COOH(aq) + H₂O(l) ↔ C₂H₅COO⁻(aq) + H₃O⁺(aq)

The concentration of the initial solution of propanoic acid is 0.597 M.

The Ka of propanoic acid is 1.34 × 10⁻⁵.

Let the concentration of each of the products is x.

Now, We have Ka = [C₂H₅COO⁻] [H₃O⁺] / [C₂H₅COOH] [H₂O]

or, 1.34 × 10⁻⁵ = x × x / 0.597

or, x² = 0.8 × 10⁻⁵

or, x = 2.82 × 10⁻³ M

or, x = 0.00282 M

So, concentration of C₂H₅COO⁻ = x = 0.00282 M

Concentration of C₂H₅COOH = 0.597 - 0.00282 = 0.594 M

Now, pH of solution = -log[H₃O⁺] = -log[x] = -log(0.00282) = 2.55

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Answer:

If the water was turned off i would go look for some in a river and filter it out and bring for my family too.

Explain:

The required mass of water vaporized is 15.5 grams, from a 250 gram sample of water at the boiling point had 35.0 kj

Given: Mass of water (m) = 250 gHeat added (q) = 35.0 kJHeat of vaporization (ΔHvap) = 40.6 kJ/mole

To find:Mass of water vaporized (x) Formula:q = ΔHvap × nx = (q / ΔHvap) × nMass = moles × molar mass

We know that molar mass of water (H2O) = 18 g/molMoles of water vaporized (n) = (35.0 kJ / 40.6 kJ/mol) = 0.861 mol

Therefore,Mass of water vaporized (x) = 0.861 mol × 18 g/mol= 15.5  

Detailed Solution: According to the given statement,250g of water was taken at its boiling point and 35.0 kJ of heat was added to it, we need to find how many grams of water were vaporized. To solve this question, first, we need to know the heat of vaporization for water, which is 40.6 kJ/mole. It means to vaporize 1 mole of water, 40.6 kJ of heat is required.

Mass of water (m) = 250 g Heat added (q) = 35.0 kJHeat of vaporization (ΔHvap) = 40.6 kJ/molen = q / ΔHvapn = (35.0 kJ / 40.6 kJ/mol) = 0.861 molMoles of water vaporized (n) = 0.861 mol

Therefore, Mass of water vaporized (x) = 0.861 mol × 18 g/mol= 15.5 g Hence, the required mass of water vaporized is 15.5 grams.

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Answer: A FILTERING

Explanation:

WHEN YOU FILTER YOU SEPARATE (sorry for caps) btw can you check out my recent question plz

Yes, a metal will always be produced at one electrode during electrolysis.

An electrode is an electrical conductor used to make contact with a nonmetallic part of a circuit.

An electrode can be anode or cathode. During electrolysis, that is , the decomposition of a chemical compound by electricity, a metal will be deposited at the cathode.

Thus, a metal will always be produced at one electrode during electrolysis.

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Answer: FALSE

Explanation: Often, gases rather than metals form at electrodes, especially if it involves the electrolysis of a solution.

Answer:

Explanation:

E