21.Which of the following is not a sign of a chemical change?Select one:a. Change in color.b. Gas is given off.c. Light is given off.d. Change of state.

The compound has an empirical formula of \(K_2H_2P_2O_8\) and a molecular formula of \(K_2HPO_4\).

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The correct question would be as

The composition of a compound is 28.73% K. 1.48% H, 22.76% P, and 47.03% O. The molar mass of the compound is 136.1 g/mol. What is the Molecular Formula of the compound?

\(KH_2PO_4\\KH_3PO_4\\K_2H_4P_20_{12}\\K_2H_3PO_6\)

Answer:

The second option

Explanation:

The mass of 2.40 × 10²³ molecules of anonymous substance with molar mass of 18.02g/mol is 7.18grams.

The mass of a substance can be calculated by multiplying the number of moles in the substance by its molar mass as follows:

mass = no of moles × molar mass

First, we convert the number of molecules in the anonymous substance to moles as follows:

2.40 × 10²³ ÷ 6.02 × 10²³ = 0.39moles

mass of anonymous substance = 0.39 moles × 18.02g/mol = 7.18grams

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Rutherford's results declined the results of Thomson's model, the correct option is C.

Atomic Models are the scientific theories proposed to determine the structure of an atom.

There are mainly 5 theories proposed for atomic models.

1. John Dalton's Atomic Model: An atom is the basic building block of all physical entities in the universe.

2. Plum Pudding Model, created by J.J. Thomson, uses the comparison of plum pudding, where the positive charge is uniformly dispersed throughout and the negative charge is randomly sprinkled on top, to explain how subatomic particles are structured.

3. Rutherford's model: proved the presence of a nucleus.

4.Niel Bohr's model: Arrangement of electrons in shells.

5. Erwin Schrodinger's model is also called as Quantum Model.

In J.J. Thomson's model, the equal distribution of positive and negative charges is proposed while,

Rutherford's theory declines this arrangement and proposes that the positively charged nucleus occupies a very small part and there is empty space in the atom.

So, Rutherford’s results suggested that the model proposed by Thomson was based on false research and required.

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Based on the information provided, we have a reaction between hydrogen iodide (HI) gas and hydrogen gas (H₂) to form iodine gas (I₂). The equilibrium is represented by the equation:

2HI(g) = H₂(g) + I₂(g)

The concentration values given in the table correspond to the concentrations of H₂ and HI at different times.

a) From t=0 s to t=5 s: Without the specific graph mentioned in Figure 7.10, it is difficult to provide a precise explanation of the physical situation in the container during this time period. However, based on the equilibrium reaction given, we can make some general observations. At the start (t=0 s), the concentrations of H₂ and HI may be high. As time progresses, the reaction proceeds, and the concentrations of H₂ and HI may decrease while the concentration of I₂ increases. The specific behavior will depend on the rate of the forward and reverse reactions.

b) External factor altered at t=5 s: To bring about a change in the shape of the graph at t=5 s, some external factor must have been altered. The most likely factor is the total pressure within the container. Since the reaction involves gases, changes in pressure can affect the equilibrium position. However, according to the information given, a change in pressure will not affect equilibrium in this case since the number of moles of gas is the same on both sides of the equation. Therefore, if the shape of the graph changes at t=5 s, some other external factor, such as temperature or the addition of a catalyst, must have been altered.

c) Calculation of K at t=3 s: The equilibrium constant (K) can be calculated at any given time using the concentrations of the reactants and products. However, the concentrations of H₂ and HI at t=3 s are not provided in the information given. Without the necessary data, it is not possible to calculate K at t=3 s.

Lastly, the statement "1 dm³ COCI, decomposes" seems incomplete. If you provide additional information or clarify the question, I'll be happy to assist you further.

Answer:

a. evaporation .

Explanation:

In this case, given the described situation, we should take into account that there are two types of liquid-gas phase transitions, evaporation and vaporization, which occur in totally different way.

Firstly, evaporation is a superficial phenomena, it means that it occurs at the surface of the liquid only whereas the vaporization is a bulk phenomena, which means that it occurs along the whole volume of liquid.

In such a way, we can infer that cold steam stream flowing over the pond has the capacity to strip or remove liquid water molecules in the pond and take them to the vapor phase, which means that the answer is a. evaporation .

Best regards.

Answer:

The answer is A, B and E

Explanation:

A - carbohydrates

B - Lipids

C- COH

E - Proteins

all except O and NaOH

The mass(in grams) of iron, Fe produced from the 92.5g of iron (ii) oxide, FeO is 71.9 grams

The mass of Fe produced from the 92.5g of iron (ii) oxide, FeO can be obtained as follow:

2FeO → 2Fe + O₂

From the balanced equation above,

143.7 g of FeO reacted to produce 111.7 g of Fe

Therefore,

92.5 g of Fe will react to produce = (92.5 × 111.7) / 143.7 = 71.9 g of Fe

Thus, the mass of Fe produced from the reaction is 71.9 grams

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Complete question:

How many grams of Fe are produced from 92.5 g of FeO given the following reaction 2FeO → 2Fe + O₂

Answer:

E.none

Explanation:

because the bonds join with each other and no lone pair is left

Synthesis. The synthesis of aspirin is classified as an esterification reaction. Salicylic acid is treated with acetic anhydride, an acid derivative, causing a chemical reaction that turns salicylic acid's hydroxyl group into an ester group (R-OH → R-OCOCH3).

hope it's help

2) The heat capacity of aluminum is 219.44 J/mol.°C.

3) a) the experimental ΔHs of ice is -0.154 kJ/mol.

b) too high

Calculate the heat released by the aluminum:

q = mcΔT

where q = heat released, m = mass of aluminum, c = specific heat capacity of water and ΔT = change in temperature.

q = (24.7 g) (0.903 J/g°C) (100.0°C - 23.4°C)

q = 18643.26 J

Next, calculate the heat absorbed by the calorimeter:

q = mcΔT

q = (99.5 g + 24.7 g) (15.8 J/°C) (23.4°C - 19.5°C)

q = 4009.92 J

The heat released by the aluminum is equal to the heat absorbed by the calorimeter and water:

18643.26 J = 4009.92 J + q3

where q3 = heat absorbed by the water.

q3 = 14633.34 J

Calculate the molar heat capacity of aluminum:

Cp,m = q3 / (nΔT)

where Cp,m = molar heat capacity, n = number of moles of aluminum, and ΔT = change in temperature.

n = m / M

where m = mass of aluminum and M = molar mass of aluminum (26.98 g/mol).

n = 24.7 g / 26.98 g/mol

n = 0.916 mol

Cp,m = 14633.34 J / (0.916 mol * 76.6°C)

Cp,m = 219.44 J/mol.°C

Therefore, the heat capacity of aluminum is 219.44 J/mol.°C.

3) (a) To calculate the experimental ΔHs of ice, we first need to calculate the heat gained by the water and the heat lost by the ice during the process.

Heat gained by water = mass of water × specific heat capacity of water × change in temperature

= 100.0 g × 4.184 J/g·°C × (-20.1°C)

= -8,423.84 J

Heat lost by ice = mass of ice × heat of fusion of ice

= 25.6 g × 6.01 kJ/mol

= 154.496 J

Since the process is assumed to be adiabatic (no heat exchange with the surroundings), the heat gained by the water must be equal to the heat lost by the ice.

Thus, -8,423.84 J = 154.496 J = -8,269.344 J

The negative sign indicates that the process is exothermic. Therefore, the experimental ΔHs of ice is:

ΔHs = -154.496 J/mol = -0.154 kJ/mol

(b) If the student forgets to include the calorimeter term in the calculation, the calculated ΔHs of ice will be too high. This is because the heat absorbed by the calorimeter during the process is not accounted for, leading to an overestimation of the heat gained by the water and underestimation of the heat lost by the ice.

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Answer:

12: This is decomposition because nitrogen triiodide is breaking apart.

13: This is double displacement because the elements in the compounds are "switching".

14: This is synthesis because the water and carbon dioxide are combining.

15: This is also synthesis because the hydrogen and oxygen are combining.

16: This is single displacement because the sodium is "switching" the element it's bonding with.

The percent by mass of the carbon is 30.4%.

The term percentage has to do with the ratio of the mass of a particular atom to the total mass of the compound multiplied by one hundred. Thus the first step is to find the total mass or the molar mass of the compound.

Molar mass = 40 + 2(2(12) + 3(1) + 2(16))

= 40 + 2(24 + 3 + 32)

= 40 + 2(59)

= 40 +118

= 158

Thus the mass of carbon is;

4(12)/158 * 100/1

= 30.4%

Thus carbon is only about 30.4% by mass of the compound.

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The statement "increasing the force will increase the moment" is true.

This is because the moment is a measure of the turning effect of a force on an object about a pivot point. It is defined as the product of the force and the perpendicular distance between the force and the pivot point. The unit of moment is the newton-meter (Nm) or the joule (J).When a force is applied to an object, it will produce a moment about the pivot point if the force is not acting along the same line as the pivot point. The magnitude of the moment depends on the force applied and the distance of the force from the pivot point. As the force increases, the moment also increases, provided that the distance from the pivot point remains constant. Conversely, if the force remains constant, but the distance from the pivot point increases, the moment also increases. This is because the perpendicular distance is directly proportional to the moment, meaning that a longer distance results in a larger moment.Therefore, it can be concluded that increasing the force applied to an object will increase the moment produced about a pivot point.

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Based on the given reactions, the compounds are as follows:

A: The specific product formed from the reaction between prop-1-yne and either 2HBr or H2O2.

B: The product formed when compound A reacts with 2H2O.

C: The product formed when compound B reacts with K2CO3(aq).

D: The product formed from the heat-induced reaction of compound C.

E: The product formed when compound D reacts with HBr.

Based on the given reactions, let's analyze the compounds involved:

Reaction 1: prop-1-yne + 2HBr/H2O2 = A

The reactant prop-1-yne reacts with either 2HBr or H2O2 to form compound A. The specific product formed will depend on the reaction conditions.

Reaction 2: A + 2H2O = B

Compound A reacts with 2H2O (water) to form compound B.

Reaction 3: B + K2CO3(aq) = C

Compound B reacts with K2CO3(aq) (potassium carbonate dissolved in water) to form compound C.

Reaction 4: C + heat = D

Compound C undergoes a heat-induced reaction to form compound D.

Reaction 5: D + HBr = E

Compound D reacts with HBr (hydrobromic acid) to form compound E.

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Clofibrate is a lipid-lowering pharmaceutical that has been utilized to treat hyperlipidemia. Its metabolism system includes different pathways such as hydrolysis, oxidation, conjugation, and others.

Metabolites are little molecules that are produced during the process of  metabolism in living living beings.

Clofibrate hydrolysis metabolite:

Clofibrate oxidation metabolites:

Clofibrate alkylation metabolites:

Clofibrate conjugation metabolites:

Conclusively, the relative abundance and significance of each metabolite can vary among individuals.

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CuI2 is not a known compound. Copper compounds typically have different oxidation states for copper, resulting in various compound names.

Copper(II) oxide (CuO): It is a black solid compound where copper is in the +2 oxidation state. It is commonly used as a pigment and in catalytic reactions.

Copper(II) sulfate (CuSO4): It is a blue crystalline compound in which copper is in the +2 oxidation state. It is used in various applications such as agriculture, electroplating, and as a laboratory reagent.

Copper(I) oxide (Cu2O): It is a red crystalline compound in which copper is in the +1 oxidation state. It is used as a pigment, in solar cells, and as a catalyst.

Copper(II) chloride (CuCl2): It is a greenish-brown solid compound in which copper is in the +2 oxidation state. It is utilized in various chemical processes, including etching and catalyst synthesis.

Copper(II) nitrate (Cu(NO3)2): It is a blue crystalline compound where copper is in the +2 oxidation state. It is commonly used in the production of catalysts, as a coloring agent, and in electroplating.

These are just a few examples of copper compounds with different oxidation states and properties. It's important to note that the compound CuI2 mentioned in the question, if it exists, would be an exception to the typical nomenclature for copper compounds.

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Ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)) mixture could be a useful buffer in a solution. Option C

A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base or a weak base and its conjugate acid. The buffer system works by the principle of Le Chatelier's principle, where the equilibrium is shifted to counteract the changes caused by the addition of an acid or a base.

In option A, acetic acid (\(CH_3CO_2H\)) is a weak acid, but hydrochloric acid (HCl) is a strong acid. This combination does not form a buffer because HCl is completely dissociated in water and cannot provide a significant concentration of its conjugate base.

Option B consists of sodium hydroxide (NaOH), which is a strong base, and elemental sodium (Na), which is a metal. This combination does not form a buffer as there is no weak acid-base pair involved.

Option D contains acetic acid (\(CH_3CO_2H\)), a weak acid, and ammonia (\(NH_3\)), a weak base. Although they are weak acid and base, they do not form a buffer system together as they are both weak acids or bases and lack the required conjugate acid-base pair.

Option C, ammonia (\(NH_3\)), is a weak base, and ammonium chloride (\(NH_4Cl\)) is its conjugate acid. This combination can form a buffer system. When ammonia reacts with water, it forms ammonium ions (NH4+) and hydroxide ions (OH-).

The ammonium ions act as the weak acid, while the ammonia acts as the weak base. The addition of a small amount of acid will be counteracted by the ammonium ions, and the addition of a small amount of base will be counteracted by the ammonia, thus maintaining the pH of the solution relatively stable.

Therefore, option C, consisting of ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)), is the suitable mixture that could be a useful buffer in a solution.

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Answer:

it should be rinsed to clean the glassware

Explanation:

this is answer.

15.6 grams of water are formed when 500.0 kJ of energy is released in the synthesis of water from its elements.

The balanced chemical equation for the synthesis of water from its elements is:

O2(g) + 2H2(g) → 2H₂O(l) + 572 kJ

The given energy change is -500.0 kJ, which means that energy is released during the reaction.

We can use the relationship between the amount of energy released and the amount of substance involved in the reaction to calculate the mass of water formed.

The molar enthalpy change for the reaction can be calculated as follows:

ΔH = -500.0 kJ/mol H2O

The molar enthalpy change tells us how much energy is released when 1 mole of water is formed.

The molar mass of water (H2O) is 18.015 g/mol.

To calculate the mass of water formed, we can use the following equation:

mass of water = (ΔH / molar enthalpy change) x molar mass of water

mass of water = (-500.0 kJ / mol H2O / -572 kJ/mol) x 18.015 g/mol

mass of water = 15.6 g

Therefore, 15.6 grams of water are formed when 500.0 kJ of energy is released in the synthesis of water from its elements.

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Answer:

hope it helps

Explanation:

The balanced chemical equation for the formation of ammonia gas by the reaction between nitrogen gas an hydrogen gas is given. N2+3H→2NH3.

Type #1 Flame symbols are among the WHMIS emblems.

Type 2: Symbols with a flame above a circle.

Exploding bomb symbols are of type 3.

Compressed gas symbols are of type 4.

Corrosion symbols are type #5.

Skull and water the water symbols are type #6.

Exclamation mark symbols are type #7.

Health hazard symbols are type #8.

Because workplaces require a defined technique to detect hazardous items, WHMIS labels are crucial.

The national ’s hazard standard for Canada is the Health And Safety At work System (WHMIS). Hazard categorization, cautionary container labeling, the distribution of safety data sheets, and worker information and training programs are the system's main components.

The U.S. Ohs Hazard Identification Standard and WHMIS are quite similar.

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Answer:

the volume increases

Explanation:

The growth of the new plant depicts the asexual reproduction characteristic. The characteristic that describes the growth of the new plant in Stephan's mother cutting a twig from a rose bush and planting it in the soil is asexual reproduction.

Asexual reproduction is the mode of reproduction by which organisms generate offspring that are identical to the parent's without the fusion of gametes. Asexual reproduction is a type of reproduction in which the offspring is produced from a single parent.

The offspring created are clones of the parent plant, meaning they are identical to the parent.The new plant in Stephan’s mother cutting a twig from a rose bush and planting it in the soil depicts the process of asexual reproduction, which is the ability of a plant to reproduce without seeds. In asexual reproduction, plants can reproduce vegetatively by cloning themselves using their roots, bulbs, or stems.

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The functional groups that are present the compound is shown in the image is;

The functional group is the atom, group of atoms or bond that is responsible for the chemical reactivity of the members of a particular family of organic compounds. We have to note that there are several families of organic compounds and the reactivity of these families is determined mostly by the kind of functional group that we have in that particular molecule.

Now we have the structure that has been shown in the image attached. We now have to look at the image and determine the kind of functional groups that we have in the molecule. In this case, we are actually looking out for the very points along the chain where a chemical reaction is possible.

We can see that the functional groups that we have here are;

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