how are isotopes used to determine average atomic mass?

In the event a particle's size of a solute is decreased, the surface area of the solute gradually increases. This proceeds to an optimum increase in the rate of solution and results in an increase in solubility.

Therefore, this effect is very important when the size goes down to the nanometric range . In many cases, a low dissolution rate is correlated with low solubility.
Solubility is claimed as the ability of a substance, the solute, to create a solution with another substance, the solvent . It is projected as the maximum quantity of a substance that could be dissolved in another . The maximum amount of solute that can be dissolved in a solvent at equilibrium produces a saturated solution .
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The heat content of a system is equal to the enthalpy only for a system that is at constant pressure.

The heat content of a system is also known as its internal energy, which is the sum of the kinetic and potential energies of all the particles in the system. Enthalpy, on the other hand, is the heat content of a system plus the work done by the system on its surroundings at a constant pressure. Therefore, for a system that is at constant pressure, the heat content or internal energy is equal to the enthalpy.

The reason for this is that at constant pressure, the only form of work done by the system on its surroundings is pressure-volume work, which can be expressed as PΔV. In other words, if the volume of the system changes, work is done on the surroundings or by the surroundings, and the enthalpy of the system changes. However, if the volume is constant, then no work is done, and the enthalpy change is equal to the heat content change.

This relationship is used extensively in chemical thermodynamics, where enthalpy changes are often measured calorimetrically at constant pressure. It is also useful in industrial applications, such as in power generation, where the enthalpy of the working fluid is an important parameter in determining the efficiency of the process.

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Answer:

what?

Explanation:

what the hell?

The final product will be CH3-CH2-C≡C-CH2CH3. ( Hex-3-yne)

The reaction between CH3-CH2-C≡C:–Na+ and CH3CH2Br is a nucleophilic substitution reaction. The sodium acetylide will act as a nucleophile and attack the electrophilic carbon of the CH3CH2Br molecule. This will result in the formation of a new carbon-carbon bond and the displacement of the bromine atom.

. In this reaction, the negatively charged carbon in the acetylide ion acts as a nucleophile and attacks the electrophilic carbon in ethyl bromide, resulting in the formation of a new carbon-carbon bond. The bromide ion is then released as a leaving group.

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8.6 mole sample of NO₂ in a 5. 1 L container has 4.6 atmospheric pressure at 33 K.

Given, number of moles of No₂(n) = 8.6 mole

Volume (V) = 5.1 liters

Temperature (T) = 33K

R= 0. 08206 L•atm / (mol•K)

Using ideal gas equation ,

PV=nRT

P (5.1L) = (8.6mol) (0.08206L•atm/(mol•K)) (33K)

P (5.1) = 8.6 × 0.08206 × 33

P (5.1) = 23.3

P = 23.3 ÷ 5.1

P = 4.6 atm

Hence, the pressure of the given gas is 4.6 atmosphere (atm).

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The absolute pressure in the chamber is 149 kPa.

Absolute pressure refers to the total pressure exerted by a fluid, including both atmospheric pressure and any additional pressure from the system itself. To determine the absolute pressure in the chamber, we need to add the vacuum gauge reading to the atmospheric pressure.

Given that the vacuum gauge reads 57 kPa and the atmospheric pressure is 92 kPa, we can calculate the absolute pressure:

Absolute pressure = Vacuum gauge reading + Atmospheric pressure

Absolute pressure = 57 kPa + 92 kPa

Absolute pressure = 149 kPa

Therefore, the absolute pressure in the chamber is 149 kPa.

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The cell potential (Ecell) of the zinc-air battery at 25.0 degrees Celsius and a partial pressure of oxygen of 0.15 atm is 1.65 V.

To calculate the cell potential (Ecell) of the zinc-air battery at 25.0 degrees Celsius, we can use the Nernst equation, which relates the cell potential to the standard cell potential and the concentrations or pressures of the reactants:

\(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \left(\frac{RT}{nF}\right) \ln(Q)\)

Where:

E°cell is the standard cell potential (given as 1.65 V)

R is the ideal gas constant (8.314 J/(mol·K))

T is the temperature in Kelvin (25.0 + 273.15 = 298.15 K)

n is the number of electrons transferred in the balanced cell reaction (2 in this case)

F is Faraday's constant (96485 C/mol)

Q is the reaction quotient

In this case, since we are given the partial pressure of oxygen (O2) in the air diffusing through the cathode (0.15 atm), we can use the partial pressure as a substitute for concentration.

The reaction quotient (Q) can be calculated using the partial pressures of the reactants and products (ZnO):

\(Q = \frac{{(P(\text{ZnO}))^2}}{{P(\text{O}_2)}}\)

Plugging in the values into the Nernst equation, we have:

Ecell = 1.65 V -\(\frac{{8.314 \, \text{J/(mol}\cdot\text{K)}} \times 298.15 \, \text{K}}}{{2 \times 96485 \, \text{C/mol}}} \right) \ln \left( \frac{{(P(\text{ZnO}))^2}}{{P(\text{O}_2)}} \right)\)

Let's calculate the cell potential (Ecell) using the given values:

Ecell = 1.65 V -\({{8.314 \, \text{J/(mol}\cdot\text{K)}} \times 298.15 \, \text{K}}}{{2 \times 96485 \, \text{C/mol}}} \right) \ln \left( \frac{{(0.15 \, \text{atm})^2}}{{0.15 \, \text{atm}}} \right)\)

Simplifying the equation further:

\(E_{\text{cell}} = 1.65 \, \text{V} - (0.0257 \, \text{V}) \ln \left( \frac{{0.15^2}}{{0.15}} \right)\)

Calculating the natural logarithm:

\(E_{\text{cell}} = 1.65 \, \text{V} - (0.0257 \, \text{V}) \ln(1)\)

Since ln(1) is equal to 0:

Ecell = 1.65 V - (0.0257 V) * 0

Ecell = 1.65 V

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The overall cell potential for the given reaction is +3.58 V.

To determine the overall cell potential for the reaction involving these two half-reactions, we need to combine them in a way that allows us to cancel out the electrons. One way to do this is to multiply the first half-reaction by 2 and add it to the second half-reaction:

The overall cell potential can be calculated using the equation:

Ecell = Ecathode - Eanode

where Ecathode is the reduction potential of the cathode (the half-reaction with the higher reduction potential) and Eanode is the reduction potential of the anode (the half-reaction with the lower reduction potential).

In this case, the cathode is the second half-reaction (with MnO₂) and the anode is the first half-reaction (with Mg₂+). Thus, we have:

Ecell = Ecathode - Eanode = (+1.21 V) - (-2.37 V) = +3.58 V

Therefore, the overall cell potential for this reaction is +3.58 V, indicating that it is a spontaneous reaction under standard conditions.

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0.0105 mol of non-electrolyte solute, with 297.14 g/mol molar mass, is dissolved in water to make a 0.0488 M solution.

Given:

Osmotic pressure is expressed mathematically as π=M*R*T. The concentration of the solution "M" should be expressed in molarity, mol/L. Through the use of the formula, let us solve for the concentration of the solution. The osmotic pressure must be converted to atm.

Conversion factor: 1 atm has 760 torr.

913 torr * (1 atm/760 torr) = 1.20131 atm

π = 1.20131 atm = M*(0.08206 L-atm/mol-K)*(300 K)

M = 0.0488 M

From the molar concentration of the solution, we can solve for the number of moles of the solute.

M = mol solute / volume solution

mol solute = M * volume solution

mol solute = 0.0488 M * (215 mL) * (1 L/1000 mL)

mol solute = 0.0105 mol

Then, we can solve for the molar mass of the solute.

molar mass solute = mass solute / mol solute

molar mass solute = 3.12 g / 0.0105 mol

molar mass solute = 297.14 g/mol

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Answer:

b) The amount of energy required to eject an electron from an atom.

Answer:

B. The elements in the substance are mixtures of their isotopes.

Charge flowed : 240 Coulombs

Given

t = time = 60 s

I = current = 4 A

Required

Charge

Solution

General formula :

Q = I x t

Q = charge (electricity), C

Input the value :

Q = 4 x 60

Q = 240 Coulombs

Answer:

put a test tube over the opening, remove it and quickly put a lit splint near the mout or in the tube. if you hear a squeaky pop it is hydrogen.

Explanation:

hydrogen ignites in air.

The complete catabolism of a reduced organic energy source to CO2, using glycolytic pathways and the TCA cycle, with oxygen as the terminal electron acceptor for electron transport, is called aerobic respiration.

Aerobic respiration is the process by which living organisms convert organic compounds such as glucose into carbon dioxide, water, and energy in the form of ATP. The process begins with glycolysis, which occurs in the cytoplasm of the cell and converts glucose into pyruvate.

Pyruvate then enters the TCA cycle in the mitochondria, where it is further broken down into CO2 and water, releasing energy in the form of ATP. The final step is electron transport, where electrons are transferred to oxygen, producing water and ATP. This process is known as oxidative phosphorylation, and it generates most of the ATP in aerobic organisms.

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Answer:

47

Explanation:

False, Atoms are not charge-neutral and some have permanent dipole moments(1).False, Molecules are not always charge neutral and can have permanent dipole moments(2).

Both atoms and molecules have some charge because of the electrons.

The atoms themselves have no net charge, but the electrons are negatively charged. In other words, atoms are charge-neutral but their electrons are not.

Furthermore, atoms can have permanent dipole moments as well. For instance, a molecule of HCl has a permanent dipole moment because chlorine has a higher electronegativity than hydrogen, which means that the shared electron pair is drawn closer to the chlorine than to the hydrogen.Molecules can also have a charge imbalance or permanent dipole moment.

When the molecule has an uneven distribution of charge, it becomes polar. A permanent dipole moment exists when the electrons within a covalent bond are not equally shared between the atoms.

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Answer:

B

Explanation:

The statement false about solution has been that all solutions are mixtures. Thus, option A is correct.

The solution can be defined as the dissolution of two or more substances. Thus, all solutions are been considered as dissolution of the solute and solvent molecules.

The solution has generally been the homogenous mixture as there has been the complete dissolution of the substances. However, the mixture can be defined as the combination and not the complete dissolution of the substances.

Thus, all solutions have not been considered as mixtures. Thus, option A is correct.

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Place the atomic number subscript and mass number superscript to the left of the atomic symbol.

The chemical properties of an atom are determined by its atomic number and are represented by the symbol Z. The total number of nucleons in an atom is the atomic mass number. This value is represented by the symbol A.

Isotopes are members of a family of elements that have the same number of protons and a different number of neutrons. The number of protons in the nucleus determines the atomic number of an element on the periodic table. Isotopes are atoms of the same element that contain the same number of protons but different numbers of neutrons.

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Answer: pic

Explanation:

plz

Answer:

4. matter is neither created nor destroyed

Explanation:

The Law of Conservation of Mass dates from Antoine Lavoisier's 1789 discovery that mass is neither created nor destroyed in chemical reactions. In other words, the mass of any one element at the beginning of a reaction will equal the mass of that element at the end of the reaction.

Answer:

the answer is

191 mL

hope this will help you ❤️

Answer:

Based on the search results [1], [3], and [6], the calculation to determine the empirical formula of butane gas if it consists of 81.82% carbon and 18.18% hydrogen is:

Step 1: Assume 100 g of the compound, so there are 81.82 g of carbon and 18.18 g of hydrogen.

Step 2: Convert the mass of each element to moles using their molar masses.

- Carbon: 81.82 g C × (1 mol C/12.01 g C) = 6.82 mol C

- Hydrogen: 18.18 g H × (1 mol H/1.01 g H) = 18.0 mol H

Step 3: Divide each amount of moles by the smallest number of moles to get the simplest whole-number ratio.

- Carbon: 6.82 mol C ÷ 6.82 = 1 mol C

- Hydrogen: 18.0 mol H ÷ 6.82 = 2.64 mol H ≈ 3 mol H

Step 4: Write the empirical formula using the integers found in the previous step.

- The empirical formula of butane gas with 81.82% carbon and 18.18% hydrogen is CH3.

Therefore, the empirical formula of butane gas when it contains 81.82% carbon and 18.18% hydrogen is CH3.

Explanation:

the answer is boiling

Explanation:

as you boil it evaporates,equal to external atmospheric pressure

Explanation:

the answer is boiling cuz if heat is added to a liquid it will start to boil

Answer:

Potassium Oxide is KO.

Calcium Oxide added to Potassium Oxide is Potassium dioxide.

potassium oxide: K₂O

calcium oxide: CaO

When a compound is oxidized, one of its elements may experience a change in oxidation number.

Oxidation is the loss of electrons, so if an element loses electrons, its oxidation number increases. For example, in the compound H2O, the oxidation number of oxygen is -2 and the oxidation number of hydrogen is +1. If H2O is oxidized, the oxygen may gain electrons and its oxidation number will decrease.

At the same time, the hydrogen may lose electrons and its oxidation number will increase. This change in oxidation numbers is important in understanding chemical reactions and the transfer of electrons between molecules.

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Answer:

Looks like 2 hydrogen and sulfuric oxide 4

Answer:

four atoms of sulfuric oxide

Explanation:

Given only the chemical formula of a compound, it is difficult to determine the charge of the cation without additional information. However, there are several ways to infer the charge of cations.

Look for polyatomic ions.

Many cations are composed of metal ions and polyatomic ions. Polyatomic ions often exhibit a cationic charge. For example, if the chemical formula is NaNO3, the cation is Na+ because nitrate (NO3) is always -1.

Note the presence of transition metals:

transition metals can have multiple oxidation states, and the oxidation state can be inferred by the presence of other elements in the compound.

Look at the periodic table:

Some elements on the periodic table are the alkali metals (group 1), which are always +1, and the alkali metals, which are always +2.

However, it is important to note that even using these methods it can be difficult to determine the charge of the cation without additional information. In many cases, it is necessary to refer to references such as textbooks or chemistry databases to ascertain the load.

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74.5 grams of lead (II) nitrate were reacted to produce 62.6 grams of lead (II) chloride.

To determine the mass of lead (II) nitrate that was reacted when 62.6 grams of lead (II) chloride is produced, we need to use the stoichiometry of the balanced chemical equation and calculate the molar masses of the compounds involved.

The balanced chemical equation for the reaction is:

2Pb(NO3)2 + 2NaCl → 2PbCl2 + 2NaNO3

From the equation, we can see that 2 moles of Pb(NO3)2 react to produce 2 moles of PbCl2. Therefore, the molar ratio of Pb(NO3)2 to PbCl2 is 1:1.

First, let's calculate the molar mass of PbCl2 and Pb(NO3)2:

Molar mass of PbCl2 = Atomic mass of Pb + 2 × Atomic mass of Cl

= 207.2 g/mol + 2 × 35.45 g/mol

= 278.1 g/mol

Molar mass of Pb(NO3)2 = Atomic mass of Pb + 2 × (Atomic mass of N + 3 × Atomic mass of O)

= 207.2 g/mol + 2 × (14.01 g/mol + 3 × 16.00 g/mol)

= 331.2 g/mol

Next, we can calculate the moles of PbCl2 produced:

Moles of PbCl2 = Mass of PbCl2 / Molar mass of PbCl2

= 62.6 g / 278.1 g/mol

≈ 0.225 mol

Since the molar ratio of Pb(NO3)2 to PbCl2 is 1:1, the moles of Pb(NO3)2 reacted will also be 0.225 mol.

Finally, to find the mass of Pb(NO3)2 that was reacted, we can use the moles and molar mass:

Mass of Pb(NO3)2 = Moles of Pb(NO3)2 × Molar mass of Pb(NO3)2

= 0.225 mol × 331.2 g/mol

≈ 74.5 g

Therefore, approximately 74.5 grams of lead (II) nitrate were reacted to produce 62.6 grams of lead (II) chloride.

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