Which of the following occurs in an endothermic reaction but not in an exothermic reaction?
A. Chemical bonds are broken.
B. Atoms are rearranged.
C. Energy is absorbed.

Answer: Solubility increases so more solute dissolves

Explanation:

The solubility is a measure of the concentration of the dissolved gas particles in the liquid and is a function of the gas pressure. As you increase the pressure of a gas, the collision frequency increases and thus the solubility goes up, as you decrease the pressure, the solubility goes down.

Answer:

The nuclide formed by the β decay of 26Al is 26Mg.


Mark my answer as brainliest! this was a difficult one

Regardless of the reactant B concentration, the reactant rate remains constant. As a result, the response rate is unaffected by increasing B's concentration by a factor of four.

The amount of a substance's solute in a specific amount of solution is how concentrated it is. Molarity, or the amount of molecules of solute in one liter of solution, is a common unit of measurement for concentrations.

Focusing for an extended amount of time on one thing without interruption is meditation. Concentrating the mind on a specific object is the basic exercise of concentration that all meditations begin with. After some practice, concentration can eventually turn into meditation when the mind is calm and not easily diverted by other thoughts.

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The volume of the gas will be 560 mL when the pressure is reduced to 125.0 kPa, assuming the temperature remains constant.

Boyle's law simply states that "the volume of any given quantity of gas is inversely proportional to its pressure as long as temperature remains constant.

Boyle's law is expressed as;

P₁V₁ = P₂V₂

Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.

Substituting the given values, we get:

P₁ = 350.0 kPa (initial pressure)

V₁ = 200 mL (initial volume)

P₂ = 125.0 kPa (final pressure)

V₂ = ?

Solving for V₂, we get:

V₂ = ( P₁ × V₁ ) / P₂

V₂ = (350.0 kPa × 200 mL) / 125.0 kPa

V₂ = 560 mL

Therefore, the final volume of the gas is  560 mL.

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When a protonated epoxide is attacked by water, the nucleophile attacks from the "top" or "front" of the epoxide ring in an "S_N2" substitution nucleophilic bimolecular process.

A protonated epoxide is a molecule that contains an epoxide ring (a three-membered ring consisting of two carbon atoms and one oxygen atom) that has been protonated, or had a hydrogen ion (H+) added to it.

The protonation of an epoxide ring can occur in the presence of an acidic medium, such as a strong acid like sulfuric acid (H2SO4) or hydrochloric acid (HCl).

In acidic conditions, the lone pair of electrons on the oxygen atom of the epoxide ring can interact with the positively charged hydrogen ion, resulting in the formation of a protonated epoxide.

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The volume of the nitrogen oxide gas is 35.2 L

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It is used to determine the amount of reactants needed to produce a certain amount of product, or to determine the amount of product that will be produced from a given amount of reactant.

To apply stoichiometry;

We know that;

Number of moles of Cu = 150/ 63.5g/mol = 2.36 moles

If 3 moles of Cu produced 2 moles of NO

2.36 moles of Cu will produce 2.36 * 2/3

= 1.57 moles

If 1 moles of NO occupies 22.4 L

1.57 moles of NO will occupy 1.57 * 22.4/1

= 35.2 L

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40.5%

Explanation:

Ca×1 = 40

C×4 = 48

H×6 = 6

O×4 = 64

64÷158×100% = 40.5%

Answer: C - The amount of gas

Explanation:

Answer:

d

Explanation:

Nitrogen is the limiting reactant.

42 g N2x (1mole N2/28 g N2) X (2 moles NH3/ 1mole N2) X (17 grams NH3/1mole NH3) = 51 g

13 g H2 x (1 mole H2/2 g H2) X (2 moles NH3/3moles H2) X (17 g NH3/1mole NH3) =73.66g

since 51 is smaller than 73.66 and it started with N2 that is why N2 is the limiting reactant)

Answer:

The density of an object is constant.

Density is a derived unit of measure.

The density of an object determines whether it will sink or float.

Answer:

Absolute zero is the theoretical temperature at which all matter would have zero thermal energy and all molecular motion would stop. This temperature is equal to -273.15 degrees Celsius or -459.67 degrees Fahrenheit. It is the lowest possible temperature in the universe and cannot be reached in practice, as it is impossible to completely eliminate all thermal energy from matter.

Answer:

2300J

Explanation:

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

Where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

In this case, ΔU is what we want to find, Q is 700 J, and W is -400 J (note that the work done by the system is negative because it is done on the surroundings). Substituting these values into the equation:

ΔU = Q - W

ΔU = 700 J - (-400 J)

ΔU = 700 J + 400 J

ΔU = 1100 J

The final internal energy of the system is therefore 1100 J + the initial energy of 1200 J, which equals 2300 J.

Answer:

D

Explanation:

because replusive force is much more

To solve for Δ2, we need to reverse the reaction and change the sign of Δ1:

2Al2O3(s)⟶4Al(s)+3O2(g)Δ2 = -Δ1

Therefore, Δ2 = -Δ1.

To solve for Δ3, we need to add the reactions for the formation of two moles of Al2O3 from aluminum and oxygen:

4Al(s)+3O2(g)⟶2Al2O3(s)Δ1

2Al2O3(s)⟶4Al(s)+3O2(g)Δ3

Adding these equations gives:

12Al(s)+9O2(g)⟶6Al2O3(s)Δ3

Therefore, Δ3 = 2Δ1.

To solve for Δ4, we need to divide the reaction for the formation of two moles of Al2O3 by two:

2Al(s)+3O2(g)⟶Al2O3(s)Δ1/2

Multiplying this equation by 16 gives:

32Al(s)+48O2(g)⟶16Al2O3(s)8Δ1/2

We can then cancel out the formation of 14 moles of Al2O3:

2Al(s)+32O2(g)⟶Al2O3(s)Δ4 = 8Δ1/2 - 7Δ1

Therefore, Δ4 = 8Δ1/2 - 7Δ1.

Explanation:

The balanced chemical equation for the reaction between sulfuric acid and potassium hydroxide is:

H₂SO4(aq) + 2KOH(aq) → K₂SO4(aq) + 2H₂O(l)

From the equation, we can see that 1 mole of H₂SO4 reacts with 2 moles of KOH. We can use this information to determine the number of moles of H₂SO4 and KOH present in the mixture before neutralization:

moles of H₂SO4 = 0.850 L x 0.400 mol/L = 0.34 mol

moles of KOH = 0.800 L x 0.250 mol/L = 0.20 mol

Since KOH is the limiting reagent, it will be completely consumed in the reaction. The number of moles of H₂SO4 that reacts with the KOH is given by:

moles of H₂SO4 reacted = 2 x moles of KOH = 0.40 mol

The remaining moles of H₂SO4 after neutralization is:

moles of H₂SO4 remaining = moles of H₂SO4 - moles of H₂SO4 reacted

moles of H₂SO4 remaining = 0.34 mol - 0.40 mol

moles of H₂SO4 remaining = -0.06 mol

Since the moles of H₂SO4 remaining is negative, it means that all of the H₂SO4 has reacted with the KOH and there is an excess of KOH. Therefore, the concentration of sulfuric acid remaining after neutralization is 0 M.

The IUPAC name of the compound is 3-Ethyl-2,2-dimethylhexane.

IUPAC naming is a systematic method of naming chemical compounds according to a set of rules established by the International Union of Pure and Applied Chemistry. It ensures that each compound has a unique and unambiguous name based on its molecular structure.

From the image:

Thus, the IUPAC name of the compound is 3-Ethyl-2,2-dimethylhexane.

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Answer:

To solve this problem, we can use the Boyle's Law equation, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

Boyle's Law: P1V1 = P2V2

where P1 and V1 are the initial pressure and volume, and P2 and V2 are the new pressure and volume.

Using the given values:

P1 = 30.0 atm

V1 = 2 L

P2 = 10.0 atm

Substituting these values into the Boyle's Law equation, we get:

30.0 atm x 2 L = 10.0 atm x V2

Simplifying and solving for V2, we get:

V2 = (30.0 atm x 2 L) / 10.0 atm

V2 = 6 L

Therefore, the new volume of the gas will be 6 L if the pressure is reduced to 10.0 atm, assuming the temperature remains constant.

I Hope This Helps!

O:C:H ratio is 34.8/16/52.2/12/13/0/1 = 2.17/4.35/13/0 = 1:2/6. Hence, C2H6O is the empirical formula (option D).

Fuel cells may produce heat and energy from hydrogen. Although transportation and utilities are expanding businesses, fertilizer manufacturing and petroleum refining still use hydrogen most frequently today.

According to the 1992 Energy Policy Act, hydrogen qualifies as an alternative fuel. The ability of hydrogen to power fuel cell technology in zero-emission vehicles, the potential for home consumption, and the high efficiency and quick filling time of fuel cells all contribute to the interest in hydrogen as such an alternative transportation fuel.

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Answer:2.115 L

Explanation:

4.5 x .47 = 2.115 L

4.5 pints are approximately equal to 2.129292 liters. The answer closes to the result is 2.215 liters.

This question is concerned with conversion between pints and liters. 1 pint is equal to approximately 0.473176 liters. Therefore, to find out how many liters that 4.5 pints is equal to, we multiply 4.5 by 0.473176.

The calculation is as follows: 4.5 pints * 0.473176 liters/pint = 2.129292 liters

Considering the options that was given, the closest answer is 2.215 liters.

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Answer:

many milliliters of 1.50 M HCl(aq) are required to react with 5.05 g Zn(s)?

volume: 16 mL

First, we need to calculate the number of moles of Zn:

m(Zn) = 5.05 g

M(Zn) = 65.38 g/mol (molar mass of Zn)

n(Zn) = m(Zn) / M(Zn) = 5.05 g / 65.38 g/mol = 0.0773 mol

According to the balanced chemical equation, 1 mole of Zn reacts with 2 moles of HCl. Therefore, the number of moles of HCl needed is:

n(HCl) = 2 × n(Zn) = 2 × 0.0773 mol = 0.1546 mol

Now we can use the molarity of the HCl solution to calculate the volume needed:

M(HCl) = 1.50 mol/L

n(HCl) = V(HCl) × M(HCl)

V(HCl) = n(HCl) / M(HCl) = 0.1546 mol / 1.50 mol/L = 0.103 L = 103 mL

Therefore, we need 103 mL of 1.50 M HCl(aq) to react with 5.05 g Zn(s).

The mole fraction of HCl in the given solution is 0.051.

To calculate the mole fraction of HCl in the given solution, we need to first find the mass of HCl and water present in the solution.

Let's assume we have 100 g of the solution, so 9.8 g of it is HCl, and 90.2 g is water.

Next, we need to find the moles of HCl present in the solution.

To do this, we divide the mass of HCl by its molar mass.

The molar mass of HCl is 36.46 g/mol (1.01 g/mol for hydrogen + 35.45 g/mol for chlorine).

moles of HCl = 9.8 g / 36.46 g/mol = 0.269 mol

The moles of water can be calculated using its molar mass which is 18.015 g/mol.

moles of water = 90.2 g / 18.015 g/mol = 5.005 mol

The total number of moles in the solution is the sum of the moles of HCl and water.

total moles = 0.269 mol + 5.005 mol = 5.274 mol

The mole fraction of HCl can now be calculated by dividing the moles of HCl by the total number of moles.

mole fraction of HCl = 0.269 mol / 5.274 mol = 0.051

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Answer:

24.31 g/mol.

Explanation:

moles =mass/molar mass

n=w/m

Answer: The amount of heat absorbed for the reaction of 25 mL of 1.0 M H₂SO4 and 50 mL of 1.0 M NaOH, resulting in a temperature increase from 25°C to 33.9°C, is 10.14 kJ.

Explanation:

Answer:

To find the value of y, we need to use the concept of molar ratios and the mole concept.

First, we need to calculate the number of moles of aluminum and oxygen in the given sample of ore. We can do this by dividing the mass of each element by its respective atomic mass:

Number of moles of aluminum = 5.4 kg / 27 g/mol = 200 moles

Number of moles of oxygen = 4.8 kg / 16 g/mol = 300 moles

Next, we can determine the ratio of the number of moles of aluminum to oxygen in the sample. This ratio is:

Aluminum : Oxygen = 200 : 300

Simplifying this ratio by dividing both sides by 100, we get:

Aluminum : Oxygen = 2 : 3

According to the chemical formula of aluminum oxide (Al2O3), it contains 2 atoms of aluminum for every 3 atoms of oxygen. Therefore, the sample of ore must contain a whole number of units of this chemical formula. Let the number of units of Al2O3 be y.

Then, we can set up the following equation to solve for y:

2 moles of aluminum * y = 200 moles of aluminum

3 moles of oxygen * y = 300 moles of oxygen

Simplifying each equation, we get:

y = 100

y = 100

Since both equations give the same value for y, we can conclude that the sample of ore contains 100 units of Al2O3. Therefore, the value of y is 100.

Answer:

The density of a substance is given by dividing the mass by the volume.

The S.I units of density are kg/m³, but can be measured by other units such as g/cm³

The mass of the iron is 29 kg or 29000 g

Therefore; Density = 29000 g/ 3680 cm³

                              = 7.880 g/cm³

Hence, the density of the iron plate is 7.880 g/cm³

Explanation:

It is a synthetic statin, an dihydroxy monocarboxylic acids, a pyrimidine, a sulfonamide, and a monofluorobenzene. It shares a functional connection with hept-6-enoic acid.

20 mg of rosuvastatin are contained in each film-coated tablet (as rosuvastatin calcium). Each 20 mg tablet also includes 0.025 milligrammes Sunset yellow FCF, 0.029 mg Allura red AC, and 91.755 mg lactose monohydrate.

Yet, because cholesterol has a steroid nucleus, it will behave differently. Aldehyde, ketone, ether, and amide groups don't exist in cholesterol. It only possesses one hydroxyl group, which, like carbohydrates, contains the functional group alcohol.

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