An object has a variable volume and shape before it changes state. After the
change, it can still change in shape but has a fixed volume. Which change of
state has occurred

The following compounds create solutions that are listed below

A solution is a homogeneous mixture of one or more solutes dissolved in a solvent.

solvent: the substance in which a solute dissolves to produce a homogeneous mixture

solute: the substance that dissolves in a solvent to produce a homogeneous mixture

1. Nal and H₂O

It will form an ionic solution .

2. KBr and CH4

It will not form a solution

3. CO₂ and CF₄

It will not form a solution

4. LiCI and CCl₄

It will not form a solution

5. NH₃ and H₂O

It will also form a solution and liberate ions of ammonium

6.CO₂ and H₂O

It will form a solution named as carbonic acid

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The molar mass of MgCrO4 is approximately 140.30 g/mol.

To determine the molar mass of MgCrO4 (magnesium chromate), we need to calculate the sum of the atomic masses of each individual element in the compound.

The chemical formula MgCrO4 indicates that the compound consists of one magnesium atom (Mg), one chromium atom (Cr), and four oxygen atoms (O).

The atomic masses of the elements can be found on the periodic table:

Magnesium (Mg) has an atomic mass of approximately 24.31 g/mol.

Chromium (Cr) has an atomic mass of around 51.99 g/mol.

Oxygen (O) has an atomic mass of about 16.00 g/mol.

Now, we can calculate the molar mass of MgCrO4 by summing up the atomic masses of each element, considering the respective subscripts:

Molar mass = (Atomic mass of Mg) + (Atomic mass of Cr) + 4 × (Atomic mass of O)

Molar mass = (24.31 g/mol) + (51.99 g/mol) + 4 × (16.00 g/mol)

Molar mass = 24.31 g/mol + 51.99 g/mol + 64.00 g/mol

Molar mass ≈ 140.30 g/mol

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Answer: 212

Explanation:

1. The % concentration is 20.7% and the molar concentration, Cm, is 1.5 M.

2. 7.8 grams of potassium sulfate will be formed.

3. 10.7 grams of ammonium chloride will be formed.

4. The volume of hydrogen gas that will be produced is 3.86 liters.

5. 21.43 grams of the 70% acetic acid is needed to prepare 500 grams of 3% acetic acid solution.

1. Mass of potassium sulfate = 1.5 moles * (174.26 g/mol) = 261.39 g

Mass of water (H₂O) = 1000 g

% = (mass of solute/mass of solution) x 100

% = (261.39 g / (261.39 g + 1000 g)) x 100

% ≈ 20.7%

Cm = moles of solute / volume of solution

Moles of potassium sulfate (K2SO4) = 1.5 moles

Volume of water (H2O) = 1000 g / (density of water) = 1000 g / 1 g/mL = 1000 mL = 1 L

Cm = 1.5 moles / 1 L

Cm = 1.5 M

2. The balanced equation for the reaction is:

H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O

Molar mass of sulfuric acid (H₂SO₄) = 98.09 g/mol

Moles of sulfuric acid = 10 g / 98.09 g/mol

Moles of sulfuric acid = 0.102 mol

Based on the mole ratio of the reaction, 0.102 moles of sulfuric acid will react to form 0.102 moles of potassium sulfate.

Molar mass of potassium sulfate = 174.26 g/mol

Mass of potassium sulfate = 0.102 mol x 174.26 g/mol

Mass of potassium sulfate ≈ 17.8 g

3. The balanced equation for the reaction is:

Molar mass of hydrochloric acid (HCl) = 36.46 g/mol

Moles of hydrochloric acid (HCl) = 7.3 g / 36.46 g/mol

Moles of hydrochloric acid ≈ 0.2 mol

Based on the mole ratio of the reaction, 0.2 moles of hydrochloric acid will react to form 0.2 moles of ammonium chloride.

Molar mass of ammonium chloride (NH₄Cl) = 53.49 g/mol

Mass of ammonium chloride = 0.2 mol x 53.49 g/mol

Mass of ammonium chloride ≈ 10.7 g

4. The balanced equation for the reaction is:

Molar mass of zinc (Zn) = 65.38 g/mol

Moles of zinc = 13 g / 65.38 g/mol

Moles of zinc ≈ 0.199 mol

Based on the mole ratio of the reaction, 0.199 moles of zinc will react to produce 0.199 moles of hydrogen gas.

Volume of sulfuric acid = 30 g / (density of H₂SO₄ )

The density of H₂SO₄ is 1.84 g/mL

Volume of sulfuric acid  = 30 g / 1.84 g/mL

Volume of sulfuric acid ≈ 16.3 mL or 0.0163 L

Using the ideal gas law, the volume of hydrogen gas produced will be:

V = nRT / P

V = (0.199 mol)(0.0821 L·atm/(mol·K))(273 K) / (1 atm)

V ≈ 3.86 L

5. Assuming that the concentrated original solution of acetic acid is 100% acetic acid (CH₃COOH).

Mass of acetic acid = 500 g x (3/100) = 15 g

The concentrated original solution, however, is 70% acetic acid.

70% acetic acid (mass) = 100% acetic acid (unknown mass)

0.7 * (unknown mass) = 15 g

Solving for the unknown mass:

unknown mass = 15 g / 0.7

unknown mass ≈ 21.43 g

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The relative atomic mass of Boron in trimethyl borate [(CH3O)3B] is; 10.8

According to the question;

The relative abundance of Boron in the spectrum is;

Relative atomic mass = (80% of 11) + (20% of 10)

Hence, the relative atomic mass of Boron in trimethyl borate [(CH3O)3B] is; 10.8

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Niels Bohr's 1913 model depicts that the atom has a small, positively charged nucleus which surrounded by electrons that travel in circular orbits around the nucleus.

The Bohr model, is a representation of the structure of atoms, especially that of hydrogen, The Bohr Model is a universal model in which the negatively charged electrons orbit a small, positively charged nucleus like the planets.

In the Bohr model, electrons are pictured as traveling in circles at dissimilar shells, depending on which element you have. The Bohr model is a proportional primitive model of the hydrogen atom, in contrast to the valence shell mode

So we can conclude that the Bohr Model is a structural model of an atom that shows electrons and the nucleus of an atom.

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1. decreases.

According to Boyle's Law, the relation between pressure and volume cab be stated as PV = k, where P is the pressure inside the container and V is the volume of the container.

So, pressure and volume are inversely proportional. If volume decreases pressure in the container increases and vice-versa. But this law is obeyed until the temperature and amount of gas remain unchanged.

2. Solids

Only solids are the substances which have fixed volume, changing shape and are not compressible.

3. 22.4 liters

At STP (Standard Temperature and Pressure), one mole of any ideal gas occupies a volume of approximately 22.4 liters.

4. D.

According to the kinetic Molecular Theory, solid's molecules move very quickly with high kinetic energy.

5. direct

Temperature and volume have direct relationship according to ideal gas law equation PV = nRT. We can see that temperature (T) and volume (V) are directly related.

6. Indirect

Pressure and volume have direct relationship according to ideal gas law equation PV = nRT. We can see that Pressure (T) and volume (V) are indirectly related.

7. P1 V1/n1 T1 = P2 V2/n2 T2

As the values V = 0.35 L, T = 18. degree celsius and P = 980 mmHg and constant R = 8.314 J/(mol·K), we can easily find V2 easily.

8. PV = nRT

The ideal gas law equation PV = nRT can be used fing V as the values for P, T, n are given in the problem and R value is a constant.

3.21 grams of Aluminum Sulfate are got when 4 g of Aluminum Nitrate reacts chemcially with 3 g of Sodium Sulfate.

The equation given is not balanced. Thus,  when balanced the equation becomes:

2 Al(NO₃)₃ + 3 Na₂SO₄ → Al₂(SO₄)₃ + 6 NaNO₃

The molar mass of Al(NO₃)₃ is:

Al(NO₃)₃ = 1(Al) + 3(N) + 9(O) = 213 g/mol

The molar mass of Na₂SO₄ is:

Na₂SO₄ = 2(Na) + 1(S) + 4(O) = 142 g/mol

From the balanced equation, we can see that 2 moles of Al(NO₃)₃ react with 3 moles of Na2SO4 to produce 1 mole of Al₂(SO₄)₃. Therefore, we can calculate the number of moles of Al(NO₃)₃ and Na₂SO₄ that react:

Number of moles of Al(NO₃)₃ = 4 g / 213 g/mol = 0.0188 mol

Number of moles of Na₂SO₄ = 3 g / 142 g/mol = 0.0211 mol

From the balanced equation, we can see that 2 moles of Al(NO₃)₃ produce 1 mole of Al₂(SO₄)₃. Therefore, the number of moles of Al₂(SO₄)₃ produced is:

Number of moles of Al₂(SO₄)₃ = 0.0188 mol / 2 * 1 = 0.0094 mol

The molar mass of Aluminum Sulfate (Al₂(SO₄)₃) is:

Al₂(SO₄)₃ = 2(Al) + 3(S) + 12(O) = 342 g/mol

Therefore, the mass of Aluminum Sulfate produced is:

Mass of Al₂(SO₄)₃ = Number of moles of Al₂(SO₄)₃ * Molar mass of Al₂(SO₄)₃

= 0.0094 mol * 342 g/mol

= 3.21 g

Hence, 3.21 grams of Aluminum Sulfate are liberated when 4 g of Aluminum Nitrate change state with 3 g of Sodium Sulfate.

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Answer:

Neutrons and protons

Explanation:

The average atomic mass of an element is the sum of the masses of its isotopes, each multiplied by its natural abundance (the decimal associated with percent of atoms of that element that are of a given isotope).

The metal X has an approximate molar mass of 42.36 g/mol and the metal is most likely calcium.

The molar mass of the metal carbonate XCO₃ and identify the metal X, we need to calculate the number of moles of XCO₃ and CO₂ using the given masses and molar masses.

The molar mass of CO₂ (carbon dioxide) is 12.011 g/mol (for carbon) + 2 * 15.999 g/mol (for oxygen) = 44.01 g/mol.

The number of moles of CO₂ can be calculated using the formula:

moles of CO₂ = mass of CO₂ / molar mass of CO₂

moles of CO₂ = 0.855 g / 44.01 g/mol

moles of CO₂ ≈ 0.01944 mol

Since the reaction stoichiometry is 1:1 between XCO₃ and CO₂, the number of moles of XCO₃ is also approximately 0.01944 mol.

molar mass of XCO₃ = mass of XCO₃ / moles of XCO₃

molar mass of XCO₃ = 2.012 g / 0.01944 mol

molar mass of XCO₃ ≈ 103.38 g/mol

The molar mass of XCO₃ is approximately 103.38 g/mol.

To determine the metal X:

molar mass of X = molar mass of XCO3 - molar mass of CO3

molar mass of X = 103.38 g/mol - (12.011 g/mol + 3 * 15.999 g/mol)

molar mass of X ≈ 42.36 g/mol

Metal X is most likely Calcium that has a molar mass of 40 g/mol

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Answer:The balanced equation of Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g) does NOT tell you how many moles of Mg or H2 are present.  It only tells you the MOLE RATIO of the reactants and products.  Thus, 1 mole Mg reacts with 2 moles HCl to produce 1 mole of hydrogen gas.

Explanation:The balanced equation of Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g) does NOT tell you how many moles of Mg or H2 are present.  It only tells you the MOLE RATIO of the reactants and products.  Thus, 1 mole Mg reacts with 2 moles HCl to produce 1 mole of hydrogen gas.

Here, the limiting reactant is Mg. The number of moles of Mg is 0.019 and that of HCl is 0.134. The number of moles of H2 produced is 0.019.

The limiting reactant in a reaction is the reactant which is fewer in moles and thus, determine the yield of the reaction. In the given reaction, two moles of HCl reacts with one mole of magnesium metal.

Given the mass of Mg = 0.475 g

atomic mass = 24 g/mol

no.of moles  = 0.475 /24 = 0.124

volume of HCl = 0.124 L

molarity = 1.08 M

number of moles  = molarity × volume = 0.134

The limiting reactant here is magnesium. One  mole of  Mg reacts with 2 moles of HCl giving 1 mole of hydrogen gas. Then 0.019 moles will give 0.019 moles of hydrogen gas.

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Answer:

\(\Delta H=1962.3J\)

Explanation:

Hello,

In this case, we can compute the change in the solution enthalpy by using the following formula:

\(\Delta H=mC\Delta T\)

Whereas the mass of the solution is 350 g, the specific heat capacity is 4.184 J/g °C and the change in the temperature is 1.34 °C, therefore, we obtain:

\(\Delta H=350g*4.184\frac{J}{g\°C} *1.34\°C\\\\\Delta H=1962.3J\)

It is important to notice that the mass is just 350 g that is the reacting amount and by means of the law of the conservation of mass, the total mass will remain constant, for that reason we compute the change in the enthalpy as shown above, which is positive due to the temperature raise.

Best regards.

Answer:

the answer is copper wire

Difluoroethane (DFE) is a cheap, commonly obtainable volatile chemical that is safe for recreational inhalation.

Thus, It can be found in everyday household items including propellants, refrigerants, and compressed air dusters. When breathed, DFE is a central nervous system (CNS) depressant that causes a momentary feeling of euphoria.

Toxic effects are linked to prolonged or excessive usage, and rapid termination might cause withdrawal.3–5 We describe a DFE misuse case that was accompanied by skeletal fluorosis and withdrawal psychosis.

Difluoroethane is a colourless, odourless gas that is transported under its vapour pressure as a liquefied gas. Ingestion of the liquid can result in frostbite.

Thus, Difluoroethane (DFE) is a cheap, commonly obtainable volatile chemical that is safe for recreational inhalation.

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It's never a good idea to use your credit card when experiencing strong emotions, especially if you tend to steer toward 'retail therapy.

The formula of lithium hydroxide is LiOH.

The molar mass of LiOH = 24 g/mol

The mass of 1 mole of a chemical is indicated by its molar mass.

It provides you with the amount of grams per mole of a substance, to put it another way. Hence, grams/mole are the units for molar mass.

The mass of a specific chemical element or chemical compound (g) divided by the amount of substance is known as the molar mass (mol). By multiplying the standard atomic masses (in g/mol) of the constituent atoms, one can determine the molar mass of a compound.

The molar mass of lithium hydroxide, LiOH is derived as folows:

Molar mass = 7 + 16 + 1

Molar mass of LiOH = 24 g/mol

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Answer:

20000 Mg

Explanation:

Answer:

(CH3)3CCH2Br

Explanation:

This reaction proceeds by the following mechanism;

First, irradiation with light of appropriate frequency leads to the homolytic fission of the bromine molecule to yield two bromine free radicals. This is the initiation step of the reaction.

The free radical now attacks one of the -CH3 groups in neopentane leading to the formation of (CH3)3CCH2Br.

The enthalpy of combustion for 1kg of urea is -1223525.84 J/mol.

Urea is a compound that is used in fertilizers and in some plastics.The enthalpy of combustion for urea is the amount of energy that is released when urea is burned. In order to calculate the enthalpy of combustion for 1kg of urea, we need to use the information that is provided to us in the question. Let us start by writing down the balanced equation for the combustion of urea: CO(NH2)2 + 3/2 O2 → CO2 + 2H2O + N2
The balanced equation shows that 1 mole of urea reacts with 1.5 moles of oxygen gas to produce 1 mole of carbon dioxide, 2 moles of water, and 1 mole of nitrogen gas. The enthalpy change for this reaction is equal to the amount of energy that is released when 1 mole of urea is burned.
The heat of combustion (ΔHc) of urea is -632.6 kJ/mol. This means that 632.6 kJ of energy is released when 1 mole of urea is burned. We know that 12g of urea raised the temperature of water by 30°C. We can use this information to calculate the amount of energy that was released when 12g of urea was burned.
The specific heat capacity of water is 4.18 J/g°C. This means that it takes 4.18 J of energy to raise the temperature of 1 gram of water by 1°C. Therefore, it takes 4.18 x 1000 = 4180 J of energy to raise the temperature of 1 kg of water by 1°C.
We know that 12g of urea raised the temperature of water by 30°C. Therefore, the amount of energy that was released when 12g of urea was burned is:
Energy = mass x specific heat capacity x temperature change
Energy = 0.012 kg x 4180 J/kg°C x 30°C
Energy = 1497.6 J
We can now use this information to calculate the enthalpy of combustion for 1kg of urea:
Enthalpy of combustion = energy released / moles of urea burned
Enthalpy of combustion = 1497.6 J / (0.012 kg / 60.06 g/mol)
Enthalpy of combustion = - 1223525.84 J/mol
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Diffusion does not require any addition of energy to the molecules, which already have energy in their kinetic motion. Facilitated diffusion is diffusion across a membrane that is aided by transport proteins. This makes the diffusion easier than it would otherwise be.

One form of transport protein is called a channel protein. Just like a key lets you get into a building or a room, channel proteins can let some molecules get across a membrane.

The correct chemical structure that corresponds to 1,1,1-trifluoroethane is (a).

What is  1,1,1-trifluoroethane?

A chemical structure is a spatial arrangement of atoms in a molecule. It determines the molecular geometry and when necessary the electronic chemistry as well .1,1,1-Trifluoroethane or simply known as trifluoroethane is Hydrofluorocarbon (HFC) compound that is colourless and highly inflammable gas with ether like odour. One method of preparation of 1,1,1-Trifluoroethane is by fluorination of 1-chloro-1,1-difluoroethane in the presence of hydrofluoric acid.  The chemical formula for 1,1,1-Trifluoroethane is \(C__{2} } H_{3} F_{3}\). The high stability of it's chemical structure because of being heavier than air makes it a greenhouse gas with high infrared absorbent power. It can be used as a propellant or refrigerant and in cleaning of electrical equipments.

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Answer:

2.7 moles of solute per 1000 mL of solution.

Explanation:

Hello,

In this case, molarity is defined as the ratio of the moles of the solute to the volume of the solution in litres, thus, for a 2.7-M solution, it is clear that we have 2.7 moles of the solute in 1 L of solution, thus the option having the proper relation is 2.7 moles of solute per 1000 mL of solution as 1 L equals 1000 mL.

Best regards.

Answer: .819 Moles of Air

Explanation: To solve this problem, we will use the Ideal Gas Law which states that PV=nRT. P represents pressure or internal pressure, V is volume, T is temperature, n is moles of a gas, and R is the Universal Gas Constant. For the ideal gas law, R is .08206. R is 8.314 for any other calculation. We are solving for the moles of gas. The gas in this case is air which is a mixture of gases but that isn't important.

Our givens are P = 7.3 atm, V = 2.67 L and T = 17.0°C. We convert T to Kelvin because the Ideal Gas Law requires that. We simply add 273 to the value in Celcius to convert it to Kelvin. Our T is now 290 K. We also know R is our Universal Gas Constant. We can now plug into the law.

(7.3 atm)(2.67 L) = n(.08206)(290 K)

n = ((7.3 atm)(2.67 L))/(.08206)(290 K)

n = .819 moles of air

Hope this helps!

6.4875 moles of nitrogen gas would be produced if 8.65 moles of copper(II) oxide were reacted with excess ammonia.

Given the following balanced equation, 2 NH₃(g) + 3 CuO (s) → 3 Cu(s) + N₂(g) + 3 H₂O(g). We are required to determine the number of moles of nitrogen gas that would be produced if 8.65 moles of copper(II) oxide were reacted with excess ammonia. We can use stoichiometry to solve this problem. Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. It allows us to make predictions about the amount of product produced or reactant required in a chemical reaction. Stoichiometry relies on the balanced chemical equation for the reaction. In this case, the balanced chemical equation is 2 NH₃(g) + 3 CuO (s) → 3 Cu(s) + N₂(g) + 3 H₂O(g).

The balanced chemical equation shows that 2 moles of ammonia react with 3 moles of copper(II) oxide to produce 1 mole of nitrogen gas and 3 moles of water. This means that the mole ratio of ammonia to nitrogen gas is 2:1. We can use this mole ratio to determine the number of moles of nitrogen gas produced in the reaction. We know that 8.65 moles of copper(II) oxide is reacted with excess ammonia. Since copper(II) oxide is the limiting reagent, we can use it to calculate the number of moles of ammonia used in the reaction. The molar ratio of copper(II) oxide to ammonia is 3:2. Therefore, we can calculate the number of moles of ammonia used in the reaction as follows:Number of moles of ammonia = (3/2) × number of moles of copper(II) oxideNumber of moles of ammonia = (3/2) × 8.65Number of moles of ammonia = 12.975 molesWe know that the mole ratio of ammonia to nitrogen gas is 2:1. Therefore, the number of moles of nitrogen gas produced in the reaction is half the number of moles of ammonia used.Number of moles of nitrogen gas produced = (1/2) × number of moles of ammoniaNumber of moles of nitrogen gas produced = (1/2) × 12.975Number of moles of nitrogen gas produced = 6.4875 moles.

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Answer:

79.4 %

Explanation:

From the question given above, the following data were obtained:

Theoretical yield = 3.4 kg

Actual yield = 2.7 kg

Percentage yield =?

Percentage yield is simply defined as the ratio of the actual yield to that of the theoretical yield multiplied by 100 i.e

Percentage yield = Actual yield/Theoretical yield × 100

With the above formula, we can obtain the percentage yield as follow:

Theoretical yield = 3.4 kg

Actual yield = 2.7 kg

Percentage yield =?

Percentage yield = Actual yield/Theoretical yield × 100

Percentage yield = 2.7/3.4 × 100

Percentage yield = 79.4 %

Thus the percentage yield is 79.4 %

The percent yield is 79.4%

From the question,

We are to determine the percent yield.

Percent yield is given by the formula,

\(Percent\ yield = \frac{Actual\ yield}{Theoretical yield } \times 100\% \)

From the given information,

Actual yield = 2.7 kg

Theoretical yield = 3.4 kg

Putting the parameters into the formula,

\(Percent\ yield = \frac{2.7}{3.4} \times 100\% \)

Then,

\(Percent\ yield = 0.7941176\times 100\% \)

\(Percent\ yield = 79.4\% \)

Hence, the percent yield is 79.4%

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There are various kind of elements that are present in periodic table. Elements that are metals are kept on left side of periodic table and elements which are non metals are kept on right side of periodic table.

Periodic table is a table in which we find elements with properties like metals, non metals and metalloids element arranges in increasing atomic number.

The first 20 elements of periodic table are Hydrogen, Helium, Lithium, Beryllium, Boron, Carbon, Nitrogen, Oxygen, Fluorine, Neon, Sodium, Magnesium, Aluminum, Silicon, Phosphorus, Sulfur, Chlorine, Argon, Potassium and Calcium. Out of these Lithium, Beryllium, Sodium, Magnesium. Potassium and Calcium are metals so they are kept on the left side of the periodic table while others are non metals so they are kept on right side of periodic table.

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