What bonds are broken during the combustion of methane (CH4) to form water and carbon dioxide?


C–H bonds in methane and O–O bonds in oxygen

C–O bonds in carbon dioxide and O–O bonds in oxygen

C–H bonds in methane and H–O bonds in water

H–O bonds in water and C–O bonds in carbon dioxide

The volume percent % (v/v) of an isopropyl alcohol solution made by dissolving 135 ml of isopropyl alcohol in 745 ml of water is 15.34%. It's calculated by dividing the volume of the alcohol by the total volume of the solution, then multiplying by 100%.

The question asks for the volume percent % (v/v) of an isopropyl alcohol solution made by dissolving 135 ml of isopropyl alcohol in 745 ml of water. The volume percentage (v/v) is commonly used to express the concentration of a solution. It's calculated by the volume of the solute divided by the volume of the solution, then multiplied by 100%.

In this case, the solute is the isopropyl alcohol, and the solution is the mixture of isopropyl alcohol and water. The volume of the solute (alcohol) is 135 ml, and the total volume of the solution is 745 ml (water) + 135 ml (alcohol) = 880 ml.

Therefore, the volume percent (v/v) of the isopropyl alcohol is given by (135 ml / 880 ml) * 100% = 15.34%, rounded to two decimal places.

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The number of moles of the gas in the system can be given by the ideal gas equation. There are 3.00 moles of gas in a 25.0 L container.

An ideal gas is a law that gives the relationship between the pressure, volume, moles, and temperature of the hypothetical gas conditions.

Given,

Pressure (P) = 300 kPa = 2.96 atm

Volume (V) = 25 L

Temperature (T) = 300.15 K

Gas constant (R) = 0.0821 atm.L/Kmol

Substituting values in the ideal gas equation, moles are calculated as:

PV = nRT

n = PV ÷ RT

= 2.96 × 25 ÷ 0.0821 × 300.15

= 74 ÷ 24.64

= 3.00 moles

Therefore, 3.00 moles of gas are present in a 25 L container.

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Final answer:

The reaction where Sn⁴⁺(aq) gains electrons to form Sn²⁺(aq) is a reduction process, making the correct answer to the question 'd) reduction' within an oxidation-reduction reaction.

Explanation:

The given chemical equation is Sn⁴⁺(aq) + 2e⁻ → Sn²⁺ (aq), which represents a type of reaction where the oxidation state of tin (Sn) is changing. When an element gains electrons during a reaction, it is undergoing a process known as reduction. Since tin is gaining two electrons, going from a +4 oxidation state to a +2 oxidation state, this reaction is best described as a reduction process within an oxidation-reduction reaction. This transformation signifies a shift towards a lower oxidation state, indicating a reduction in the oxidation number of tin atoms. Such reactions, where oxidation states change, are fundamental in various chemical processes, underpinning diverse phenomena in both natural and synthetic contexts.

Helium and lithium are elements that have two electrons in their outermost energy shell. Helium's both electrons fill its first shell, while lithium has two in the first shell and one in the second shell. Boron also has two electrons in its first shell and the remaining three in the second shell.

The element that has 2 electrons in its outermost energy level is helium (He). Helium is the second element in the periodic table and it has two electrons that fill its first shell, making it stable. Elements like helium that have stable electron configurations in their outermost energy levels are called noble gases and they tend not to react with other elements because they are energetically stable.

Another example is lithium (Li). It has three electrons, two fill its first shell and the third is in the second shell. The first shell aka K shell can hold up to two electrons, and the second shell aka L shell usually has 8 electrons when completely filled, but lithium has only one electron in its second shell. As per the filled-shell electron configuration, which reflects a state of maximum stability, lithium will tend to lose that third electron in chemical reactions, attaining a stable electron configuration like the noble gas helium.

Boron (B) is yet another example which has five electrons with two in its first shell and the remaining three in its second shell. Elements that have similar energy level distributions with their outer shells partially filled tend to form bonds with other elements in order to reach a stable configuration, possibly like the noble gases.

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Final answer:

To find the formula of a hydrate composed of 85.3% barium chloride and 14.7% water, we calculate the molar mass of both compounds, and then determine the molar ratio of barium chloride to water. The simplest whole number ratio gives us the empirical formula of the hydrate, including the number of water molecules associated with each formula unit of barium chloride.

Explanation:

The formula and name of a hydrate that is 85.3% barium chloride and 14.7% water can be determined using stoichiometric calculations. First, one needs to understand that a hydrate is an ionic compound that includes water molecules directly attached to its crystal lattice. The general formula for a hydrate is expressed as XYZ·nH₂O, where XYZ is the ionic compound, and n represents the number of water molecules per formula unit.

In order to find the formula of the hydrate of barium chloride, we need to calculate the molar mass of barium chloride (BaCl₂) and water (H₂O). The molar mass of BaCl₂ is approximately 208.23 g/mol and for H₂O it's about 18.015 g/mol. Since the hydrate consists of 85.3% BaCl₂, we can infer that 100 g of the hydrate would contain 85.3 g of BaCl₂ and 14.7 g of H₂O. We can then calculate the molar ratio of BaCl₂ to H₂O and find the simplest whole number ratio to determine the formula of the hydrate. Using this method, the name and formula of the hydrate would reflect the stoichiometry of barium chloride to water in the compound.

The formula of the hydrate is a. BaCl₂ · 2H₂O.

The formula of the hydrate is BaCl₂ · 2H₂O, and its name is barium chloride dihydrate.

The molar mass of BaCl₂ is approximately 208.23 g/mol, and for H₂O, it is approximately 18.02 g/mol.

Next, we find the ratio of moles of H₂O to moles of BaCl₂: 0.816 mol H₂O / 0.409 mol BaCl₂ ≈ 2.

The name of this compound is barium chloride dihydrate.

Correct question is: What is the formula and name of a hydrate that is 85.3% barium chloride and 14.7% water

a. BaCl₂ · 2H₂O.

b. BaCl₃ . 2H₂O.

c. Ba₂Cl₂ · 2H₂O.

d. BaCl₂ · 3H₂O.

The Arrhenius theory limits acids and bases to those that produce hydrogen or hydroxide ions in aqueous solutions, making ammonia a base only when it reacts with water and not recognizing it as an inherent base due to its structure. The Brønsted-Lowry theory overcomes this limitation by including substances like ammonia that accept hydrogen ions, further expanded by the Lewis model where ammonia can donate an electron pair.

The Arrhenius theory is limited in its definition of acids and bases because it requires the presence of hydroxide ions in the formula of bases and hydrogen ions in the formula of acids. This aspect of the theory does not account for substances like ammonia (NH3) which do not release hydroxide ions directly into solution.

Ammonia becomes a base according to the Arrhenius model only by its reaction with water where it forms ammonium (NH4+) and hydroxide (OH-) ions. Water donates a hydrogen ion to the ammonia, making water the Brønsted-Lowry acid and ammonia the Brønsted-Lowry base. Thus, ammonia is a base in the Arrhenius sense only when dissolved in water, and reveals the limitations because it does not contain hydroxide in its structure.

In the broader Brønsted-Lowry theory, ammonia is clearly a base since it accepts a hydrogen ion from a water molecule. Similarly, the Lewis model recognizes ammonia as a base since it can donate a pair of electrons to a hydrogen ion.

The energy at the beginning is low than at the end of the reaction.

Endothermic reactions gain the energy from surroundings to do the reaction. Since the thermal energy gain, the temperature of the surrounding decreases and the enthalpy for those reactions is a positive value. The products have higher energy than the reactants.

In the endothermic reaction, the total energy in the beginning is less than that in the end.

The endothermic relations are those in which the reactants absorb heat from the surrounding to carry out the reaction and to make product.

These reactions absorb energy, thus lower the temperature of the surrounding.

Examples are melting ice cubes, evaporating water.

Thus, due to these reactions takes energy from surrounding, the energy in the beginning is low.

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Answer:

B. A chemical reaction

Explanation:

A chemical equation describes the sequence and amounts of a chemical reaction

The answer is Condensation

The position of equilibrium for the reaction involving nitric acid will lie far to the right, favoring the production of HNO3 and OH-(aq). This is due to HNO3 being a strong acid and fully ionizing in water, while NO3- is a weak base.

The question you asked is about the position of equilibrium for the reaction NO3-(aq) + H2O(l) ↔ HNO3(aq) + OH-(aq). Nitric acid, HNO3, is known to be a very strong acid that is 100% ionized in water. Given that HNO3 is a strong acid, its conjugate base NO3- is a weak base and has a much lower tendency to gain a proton than water (H2O) has to lose one. This means the reaction will lie very far to the right, favoring the production of HNO3 and OH-(aq).

This behavior is consistent with the concept known as the leveling effect, which indicates that strong acids in water are fully ionized to form hydronium ions (H3O+) and their corresponding anions, and thus cannot exist in aqueous solution as anything stronger than the hydronium ion. In this case, HNO3 will ionize in water to give H3O+ and NO3-, showing that NO3- will not readily recombine with H+ to form HNO3 again. The equilibrium is clearly on the side of dissolving nitric acid (HNO3) into its ions.

The oxidation state of the lead and chromium metal in the PbCrO₄ compound is +2 and +6 respectively.

Oxidation state of any atom tells about the number of exchangeable electrons from the outermost shell of that atom.

As we know that in the compound PbCrO₄, CrO₄ is an anion and charge on this is -2 i.e. it is represented as CrO₄²⁻. We know that oxidation state of each oxygen atom is -2 and let us assume the oxidation state of chromium is y, so we calculate the value of y for CrO₄²⁻ as:

y + 4(-2) = -2

y = -2 + 8 = 6

Let the oxidation state of lead is x and calculation for that from PbCrO₄ will be done as:

x + 6 + 4(-2) = 0

x = 8 - 6 = 2

Oxidation state of Pb is +2 and of Cr is +6.

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Answer:

No, potassium chloride will not dissolve completely. Only 28.15 g of potassium chloride will get dissolved.

Explanation:

Amount od potassium chloride added to 50 g of boiling water = 250.0 g

The solubility of potassium chloride in boiling water is 56.3 g/100 g water.

Amount of potassium chloride soluble in 100 g of boiling water = 56.3 g

Amount of potassium chloride soluble in 1 g of boiling water = [tex]\frac{56.3 g}{100}[/tex]

Amount of potassium chloride soluble in 50.0 g of boiling water :

[tex]\frac{56.3 g}{100}\times 50.0 g=28.15 g[/tex]

Amount of potassium chloride left undissolved = 250.0 g - 28.15 g = 221.85 g

28.15 g of potassium chloride will dissolve and remaining 221.85 g og potassium chloride will not.

The question is about the combined gas law in Chemistry and using it to solve for the new temperature of a gas given changes in its volume and pressure.

The subject of this question revolves around understanding how variables of a gas such as pressure, temperature and volume change in relation to each other. This falls under the concept of gas laws in Chemistry, particularly the combined gas law.

In this scenario, we are given the initial volume, pressure, and temperature of a gas, and we are asked to find the new temperature after the volume and pressure have been changed. To do this, we have to apply the combined gas law formula, which is (P1V1/T1 = P2V2/T2).

To start, we should convert all our pressures to the same unit- atm, since our given value is in atm. 625 torr is equivalent to 0.821 atm. Substituting all the given and converted values into the combined gas law will allow us to solve for T2, the new temperature of the gas.

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A student whose arrows land close together but not on the bullseye is precise but not accurate. Precision implies consistent results, while accuracy means hitting the intended target.

When a student fires a bow and arrow in gym class and all his arrows land close to each other, but not on the bullseye, the student could be said to be precise but not accurate.

Precision refers to the consistency of the student's shots, as the arrows land near each other, indicating that the student is able to repeat the same motion with consistency. However, accuracy relates to how close the shots are to the intended target, the bullseye, which in this case, the student's arrows are not; thus, the student is not shooting accurately.

To help illustrate this concept, imagine three scenarios represented in Figure 1.27:

(a) Arrows that are close to both the bull's eye and one another are both accurate and precise.

(b) Arrows that are close to one another but not on target are precise but not accurate.

(c) Arrows that are neither on target nor close to one another are neither accurate nor precise.

The student's arrows are precise but not accurate, meaning they consistently hit a similar spot but not the desired target. Accuracy refers to how close a measurement or result is to the true or desired value, while precision refers to how close multiple measurements or results are to one another. With practice and adjustment of aim, the student can improve both accuracy and precision in archery.

In this scenario, the student's arrows are close to each other but not on the bullseye. This means that the student's arrows are precise because they consistently hit a similar spot, but they are not accurate because they are not hitting the desired target.

Accuracy refers to how close a measurement or result is to the true or desired value, while precision refers to how close multiple measurements or results are to one another. In this case, the student's arrows are precise but not accurate.

It is important to strive for both accuracy and precision in any endeavor, including archery. With practice and adjustment of aim, the student can improve both the accuracy and precision of their shots.

Answer:

4.014 g of water

Explanation:

An internet search for your question tells me that the chemical equation is:

So first we use PV=nRT to calculate the moles of hydrogen gas produced:

0.993 atm * 5.50 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

Now we convert mol H₂ to mol H₂O and finally to grams of water:

0.223 mol H₂ * [tex]\frac{1molH_{2}O}{1molH_{2}}*\frac{18g}{1molH_{2}O}[/tex] = 4.014 g H₂O

Answer:

1.99 x 10^(-23) g

Explanation: