what is the internal energy of 1.75 moles of an ideal monoatomic gas at a temperature of 20.00 c?

Answer:

ah+j

Explanation:

It is a chemical bond that is formed by the sharing of electron pairs between atoms.

The type of bonding you would expect to find within each substance is given below:

1. Covalent bonding is expected to be found within substance having 150 as their atomic number.

2. Ionic bonding is expected to be found within substances that are formed between metals and non-metals.

3. Metallic bonding is expected to be found within substances that are formed between the same type of metal.

So, based on the information provided, the predicted types of bonding are:

1. Covalent

2. Ionic

3. Metallic Covalent bonding:

It is a chemical bond that is formed by the sharing of electron pairs between atoms.

Covalent bonds may occur between atoms of nonmetals.

Lonic bonding: It is a chemical bond that is formed by the electrostatic attraction between oppositely charged ions.

Ionic bonds may occur between atoms of metals and nonmetals.

Metallic bonding:It is a chemical bond that is formed between metal atoms. In metallic bonding, the atoms share a pool of electrons.

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The correct statement is

a. The total mass of the system remains the same in a chemical change.

c. A physical change is temporary and reversible.

a. The total mass of the system may not remain the same in a chemical change due to the law of conservation of mass, which states that matter can neither be created nor destroyed, only transformed from one form to another. Therefore, during a chemical change, the total mass of the reactants is equal to the total mass of the products.

b. A chemical change may or may not be permanent and irreversible. Some chemical changes are reversible, such as the reaction between nitrogen and hydrogen to form ammonia, which can be reversed by heating ammonia to break it down into its constituent elements. However, many chemical changes are indeed permanent and irreversible, such as the rusting of iron.

c. A physical change is temporary and reversible because it involves a change in the physical state or appearance of a substance without changing its chemical composition. Examples of physical changes include melting, boiling, freezing, and condensation. These changes can be easily reversed, such as by cooling a liquid to solidify it again.

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During the decay process, radioactive atoms release energy and become more stable. In the case of cesium-137, it was found that over a period of 365 days, the number of radioactive atoms decreased from 1,000,000 to 977,287. This decrease in the number of atoms is due to the decay of the radioactive substance.

Radioactive decay occurs randomly and is governed by the half-life of the radioactive material. The half-life is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the specific half-life of cesium-137 was not provided, so we cannot determine the exact time it took for the decay to occur. However, we do know that after 365 days, the number of radioactive atoms reduced to 977,287.

Overall, the reduction in the number of radioactive atoms over time is a result of the release of excess energy, which makes the atoms more stable.

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Answer:

314mL OR 0.314L

Explanation:

this requires the dilution formula M1V1 = M2V2 where

M1 = initial concentration

V1 = initial volume

M2 = final concentration

V2 = final volume

In this case, we are solving for V1 where M1 = 5.65M, V1 = 250.0 mL, and M2 = 4.50M

Plugged into the equation we get:

(5.65M)(250.0mL) = (4.50M)V2

divide both sides by 4.50M and it becomes (M cancel)

V2 = 314mL

Answer:

A polar bond is one where the charge distribution between the two atoms in the bond is unequal. A polar molecule is one where the charge distribution around the molecule is not symmetric. It results from having polar bonds and also a molecular structure where the bond polarities do not cancel.

Explanation:

Answer:C

Explanation:I took the quiz

The earliest dated rocks on Earth are composed of a mixture of minerals that have not since been eroded or melted. Thus, option C is correct.

One such location in Greenland, where rocks that are between 3.7 and 3.8 billion years old are still intact.

Microbes, the earliest known life forms, left traces of their existence in rocks that were 3.7 billion years old.

Earth is 4.54 billion years old, with a 50 million-year error range, according to scientists.

By dating the rocks in the planet's dynamic crust and the rocks of its satellites and stray meteorites.

Therefore, oldest known earth rocks are 3.7 billion to 3.8 billion years old.

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Answer:

A

Explanation:

Position 3 and 4 will give 12 hours of sunlight

Answer:

Positions 3 and 4

Explanation:

Answer:

the apparatus are listed below

Explanation:

The apparatus used to carry out paper chromatography are:

• Chromatography paper

• Solvent (or water)

• Pencil

• Ink (pen)

• Beaker

The apparatus used to carry out the paper chromatography are chromatography paper ,organic solvents and glass capillary.

Paper chromatography is an analytical method used to separate colored chemicals or substances.

Paper chromatography is a quick and easy process to conduct. Most of the time, it is used to test the purity of compounds and to identify substances.  

It is primarily used as a teaching tool, having been replaced by other chromatography methods, such as thin-layer chromatography.

Listed are the apparatus used to carry out the paper chromatography-

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A 1.00 mole of Sulfuric acid (H₂SO₄) contains the largest number of atoms (7 atoms/molecule). Option B is correct.

To determine which of the given compounds contains the largest number of atoms in 1.00 mole, we need to consider the number of atoms in each compound.

1.00 mole of NH₃ contains:

1 mole of N atoms (1 atom/molecule)

3 moles of H atoms (3 atoms/molecule)

Total atoms = 1 + 3 = 4 atoms/molecule

1.00 mole of H₂SO₄ contains:

2 moles of H atoms (2 atoms/molecule)

1 mole of S atoms (1 atom/molecule)

4 moles of O atoms (4 atoms/molecule)

Total atoms = 2 + 1 + 4 = 7 atoms/molecule

1.00 mole of HBr contains:

1 mole of H atoms (1 atom/molecule)

1 mole of Br atoms (1 atom/molecule)

Total atoms = 1 + 1 = 2 atoms/molecule

Therefore, 1.00 mole of H₂SO₄, contains the largest number of atoms.

Hence, B. is the correct option.

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The chemical equations, IUPAC name, and Structural formulas are given below.

i. Chemical equations:

a) Conversion of unsaturated compound B to compound C:

B + HBr → C (Addition of hydrogen bromide to unsaturated B to form bromohexane C)

b) Conversion of compound C to compound D:

C + Na + Ether → D (Reaction of bromohexane C with sodium metal in the presence of ether to form compound D)

ii. IUPAC names:

Compound C: Bromohexane

Compound D: Hexane

iii. Structural formulae of positional isomers of compound C:

Positional isomers of bromohexane can have different bromine atoms attached at different positions along the hexane chain. Here is an example of one positional isomer of bromohexane:

1-Bromohexane:

CH3CH2CH2CH2CH2CH2Br

2-Bromohexane:

CH3CH2CH2CH2CH2CHBr

3-Bromohexane:

CH3CH2CH2CH2CHBrCH3

Therefore, the chemical equations, IUPAC name, and Structural formulas are provided above.

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The property of matter that is conserved in chemical reactions and shown by balanced equations is known as the Law of Conservation of Mass. According to this law, mass can neither be created nor destroyed in a chemical reaction; it can only be transformed from one form to another.For instance, when two substances are combined, they react and form a new substance.

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Burning of a substance is not a phase change. It involves breaking of chemical bonds and the formation of new compounds. Hence, burning is a chemical change.

Phase change of a substance is the change in its state  or phase to other phase. For example, boiling is a phase change, where liquid substance changes to its vapor phase.

Phase change is a physical change thus does not involve any breaking or making of chemical bond. Whereas, a chemical change involves breaking or making of bonds and formation of new products.

To burn a substance its chemical bonds have to be broken. Burning produces carbon dioxide and other constituent  substances of the compound. Hence, it is a physical change.

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To accomplish this, the half-reaction with the smallest number of electrons may be multiplied. When writing the balanced equation, the half-reaction for the anode should be reversed to account for the oxidation occurring at the anode. Finally, to use the equation ecell, the sign of the anode potential must be reversed.

When answering questions on the Brainly platform, it is important to be factually accurate, professional, and friendly. One should be concise and avoid providing extraneous amounts of detail. Typos and irrelevant parts of the question should be ignored. The answer to the given question is as follows:By convention, standard electrode potentials are quoted as reduction potentials. The half-reaction for the anode must be reversed when writing the balanced equation for the overall reaction. The sign for the anode potential must be reversed to use the equation ecell.Standard electrode potentials are measured for half-reactions in their standard states, such as solutions of 1 mol/L and gases at a pressure of 1 atm. It indicates the ability of a half-reaction to accept electrons, with the half-reaction with the greatest reduction potential being the strongest oxidizing agent.When writing a balanced equation for a redox reaction, half-reactions and/or electrode potentials must be manipulated in order to balance the number of electrons transferred. Since the electrons must cancel out in the overall reaction, one half-reaction should be multiplied to match the number of electrons in the other half-reaction.

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Answer:

I think its. SO2

Explanation:

the S is singular and there is 2 Os and thats what the model shows so I think I'm right

There are two characteristics of molecules, one is geometry and other is shape. Shape is excluding lone pair surrounding the central element and geometry is including the lone pair. Therefore, the molecule that the ball and stick model represents is BF\(_2\). The correct option is option C.

VSEPR stands for valence shell electron pair repulsions. VSEPR theory is used to predict the shape and geometry of molecules on the basis of valence electrons pairs that are present around the central element of the molecule.

According to VSEPR theory, Lone pair lone pair repulsion is greater than bond pair bond pair repulsion. There are so many limitations of VSEPR theory. There is a repulsion between bond pair electrons and lone pairs present on the central element.

Therefore, the molecule that the ball and stick model represents is BF\(_2\). The correct option is option C.

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1. The volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2.The appropriate initial condition for the ODE is S(0) = 0, as there is no salt initially in the tank.

3. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. The solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^{2/2} + 30t)} dt) / e^{(t^{2/2} + 30t)})\)

5.  the salt concentration in the tank as t→infinity is zero.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being added and drained. The tank is being filled at a rate of 2 liters per minute and drained at a rate of 1 liter per minute.

Since the tank starts with an initial volume of 30 liters, the volume of water in the tank at time t can be calculated using the equation:
Volume(t) = Initial volume + (Rate of filling - Rate of draining) * t

Volume(t) = 30 + (2 - 1) * t

So, the volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2. Let S(t) denote the amount of salt in the fish tank at time t in grams.

To show that S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t),

we need to take the derivative of S(t) with respect to t and substitute it into the given ODE.

Taking the derivative of S(t), we have:

S'(t) = 0 - (1+0)S(t) + 0

S'(t) = -S(t)

Substituting this into the given ODE, we get:

-S(t) = 70 - (t+30)S(t)

Simplifying the equation, we have:

S'(t) = 70 - (t+30)S(t)

Therefore, S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t).

The appropriate initial condition for the ODE is S(0) = 0,

as there is no salt initially in the tank.

3. This ODE is a first-order linear ordinary differential equation. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. To solve the initial value problem for S(t) using the method of integrating factors, we first rewrite the ODE in standard form:

S'(t) + (t+30)S(t) = 70

The integrating factor is given by:
\(\mu(t) = e^{(\int (t+30) dt)} = e^{(t^2/2 + 30t)\)

Multiplying both sides of the equation by μ(t), we have:
\(e^{(t^2/2 + 30t)} * S'(t) + e^{(t^2/2 + 30t)} * (t+30)S(t) = 70 * e^{(t^2/2 + 30t)\)

Applying the product rule to the left side of the equation, we get:
\((e^{(t^{2/2} + 30t) * S(t))' = 70 * e^{(t^{2/2} + 30t)})\)

Integrating both sides of the equation with respect to t, we have:
\(\int (e^{(t^2/2 + 30t)} * S(t))' dt = \int (70 * e^{(t^2/2 + 30t))} dt\)

Using the fundamental theorem of calculus, the left side becomes:
\(e^{(t^2/2 + 30t)} * S(t) = \int (70 * e^{(t^2/2 + 30t))} dt\)

Simplifying the right side by integrating, we get:
\(e^{(t^2/2 + 30t)} * S(t) = 70 * \int e^{(t^2/2 + 30t)} dt\)

At this point, the integration of \(e^{(t^2/2 + 30t)\) becomes difficult to express in terms of elementary functions.

Hence, the solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^2/2 + 30t)} dt) / e^{(t^2/2 + 30t)\)

5. As t approaches infinity, the exponential term \(e^{(t^2/2 + 30t)\) becomes very large, causing the salt concentration S(t) to approach zero. Therefore, the salt concentration in the tank as t→infinity is zero.

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The salt concentration in the tank as t approaches infinity is 70/3.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being poured into the tank and the rate at which water is being drained from the bottom.

At a rate of 2 litres per minute, water is being poured into the tank. So after t minutes, the amount of water poured into the tank is 2t litres.

At a rate of 1 litre per minute, water is being drained from the tank. So after t minutes, the amount of water drained from the tank is t litres.

Since the tank was initially filled with 30 litres of fresh water, the volume of water in the tank at time t is given by:
Volume(t) = 30 + 2t - t
Volume(t) = 30 + t

2. Let S(t) denote the amount of salt in the fish tank at time t. To determine the ODE for S(t), we need to consider the salt being poured into the tank and the salt being drained from the tank.

The salt concentration in the water being poured into the tank is 35 grams per litre. So the amount of salt being poured into the tank per minute is 35 * 2 = 70 grams.

The amount of salt being drained from the tank per minute is S(t)/Volume(t) * 1.

Therefore, the ODE for S(t) is:
S'(t) = 70 - S(t)/Volume(t)

The initial condition for this ODE is S(0) = 0, since there was no salt in the tank initially.

3. The ODE S'(t) = 70 - S(t)/Volume(t) is a first-order linear ODE. It is not separable since the variables S(t) and Volume(t) are mixed together.

4. To solve the initial value problem for S(t), we can rewrite the ODE as:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)

This is a linear ODE of the form y'(t) + p(t)y(t) = g(t), where p(t) = 1/Volume(t) and g(t) = 70 * Volume(t).

To solve this type of ODE, we can multiply both sides by an integrating factor, which is the exponential of the integral of p(t).

The integrating factor is exp(integral of 1/Volume(t) dt) = exp(ln(Volume(t))) = Volume(t).

Multiplying both sides of the ODE by the integrating factor, we get:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
Volume(t) * S'(t) + Volume(t) * S(t) = 70 * Volume(t)^2
( Volume(t) * S(t) )' = 70 * Volume(t)^2

Integrating both sides with respect to t, we get:
Volume(t) * S(t) = 70/3 * Volume(t)^3 + C

S(t) = 70/3 * Volume(t)^2 + C/Volume(t)

Using the initial condition S(0) = 0, we can solve for C:
0 = 70/3 * 30^2 + C/30
C = -70000

Therefore, the solution for S(t) is:
S(t) = 70/3 * Volume(t)^2 - 70000/Volume(t)

5. As t approaches infinity, the volume of water in the tank becomes very large. In this case, we can approximate the volume of the tank as t, since the rate at which water is being poured in is 2 litres per minute. So the salt concentration in the tank as t approaches infinity is given by:
S(t)/Volume(t) = (70/3 * t^2 - 70000/t) / t
As t approaches infinity, the second term (-70000/t) approaches 0, so the salt concentration in the tank as t approaches infinity is:
S(t)/Volume(t) = 70/3 * t^2 / t = 70/3 * t

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The substance in the image above would be classified as a mixture of elements (option E).

A compound is a substance formed by chemical bonding of two or more elements in definite proportions by weight.

On the other hand, a mixture is made when two or more substances are combined, but they are not combined chemically.

According to this question, an image is shown with two different substances or elements as distinguished by coloration (white and purple). These elements are combined but not chemically bonded, hence, is a mixture.

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Answer:

Sorry I really don't know but can you help with one of my question I posted please thanks.

Answer:

0.313 liters

Explanation:

You want to use the Combined Gas Law Equation: (P1*V1)/(n1*T1)=(P2*V2)/(n2*T2). You then want to cross off things that don't change, so moles(n) in this example. Then convert the temperature to Kelvin (always), and solve algebraically.

Expanation:

Given the followin reactions we have to idenftify the oxidizing and reducing aget.

3 CH₃CHOHCH₃ + Cr₂O₇²⁻ + 8 H⁺ —> 3 (CH₃)₂CO + 2 Cr³⁺ + 7 H₂O

We can split this redox reaction in two half reactions.

3 CH₃CHOHCH₃ ----> 3 (CH₃)₂CO

Cr₂O₇²⁻ ----> 2 Cr³⁺

Let's balance the first half reaction, since it is an acidic meium, we can aDdd H+, molecules of waer and e-.

+ H8 H +⁺ + 8 e⁻

We have 24 atoms of H on the left and and 18 atoms of H of the right side. We have to add 8 H+ on the right side. We have 3 atoms of O on both sides. So we only have to add the electrons to balance the carges. )Also 8 on the right side.

Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻----> 2 Cr³⁺ + 7 H₂O

In the second half reaction we have to add 7 molecules of water on the right side to balance the O atoms. We added 14 atoms of H on the right side, so we have to add 14 H+ on the left side. The total charge of the left side is +12 and the total charge of the right ide is is +6. We have to add 6 el- on the left side.

The two balanced half reactions are:

3 CH₃CHOHCH₃ ----> 3 (CH₃)₂CO + 8 H⁺ + 8 e⁻ Oxidation Half-reaction

Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻----> 2 Cr³⁺ + 7 H₂O Reduction Half-reaction

If we pay attention to the equations we will see that Cr₂O₇²⁻ is gaining electrons, i is being reduced. So Cr₂O₇²⁻ is our oxidizing agent.

CH₃CHOHCH₃ is losing electrons, it is being oxidized. CH₃CHOHCH₃ is our reducing agent.

Answer:

Cr₂O₇²⁻: oxidizing agent.

CH₃CHOHCH₃: reducing agent.

Answer:

Isotopes are forms of a chemical element that have the same atomic number but differ in mass. 16O → 8 protons + 8 neutrons; a “light” oxygen); The relative amounts are expressed as either 18O/16O or δ 18O Oxygen - 18 (aka 18O → 8 protons + 10 neutrons; a “heavy” oxygen).

Answer:

3

Explanation:

pH+pOH=14

Answer:

Explanation:

You can't award 20 points once you have awarded 5.

One

2K + Cl2 ==> 2KCl

It's True

There are 2 Ks and Cl2 has 2 atoms making up on the left.

On the right, 1 molecule of KCl has 1 Potassium and 1 Cl. But there are 2 molecules so between the two of them, there's 2 Potassiums and 2 Cl

Two

The reactant side is the Left side. Cl2 means there are 2 Cl atoms making Cl2. Answer B

Three

False: There are 2 but read below.

I really don't know how to answer that. It's not qualified enough. There are a total of 2 Chlorines on the right hand side when there are 2 molecules. I think you are intended to answer False, but don't be surprised if you get it wrong.

Carbon = C

Water = C

Aluminum foil = E

Plastic = E

Tin = E

Silicon dioxide = C

Helium = C

Arsenic = C

Carbon dioxide = C

Sodium Chloride = C

Answers are below...and I have answered this question on the basis of above table.

Explanation :-

(a.) Among nonpolar liquids, those with higher molar masses tend to have normal boiling points that are (lower).

(b.) Among compounds of approximately the same molar mass, those with greater polarities tend to have enthalpies of vaporization that are (higher).

(c.) Xe (Xenon), a substance can exist only as a gas at temperatures above its critical temperature. Of the noble gases listed, only Xe has a critical temperature above O°c.

Among nonpolar liquids, those with higher molar masses tend to have normal boiling points that are lower. Among compounds of approximately the same molar mass, those with greater polarities tend to have enthalpies of vaporization that are higher.Xenon can exist only as a gas at temperatures above its critical temperature.

Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.

Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.They are:

1)Molecular compounds where in atoms are joined by covalent bonds.

2) ionic compounds where atoms are joined by ionic bond.

3)Inter-metallic compounds where atoms are held by metallic bonds

4) co-ordination complexes where atoms are held by co-ordinate bonds.

They have a unique chemical structure held together by chemical bonds Compounds have different properties as those of elements because when a compound is formed the properties of the substance are totally altered.

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We know

current is I

Formula is

\(\boxed{\sf I=\dfrac{P}{V}}\)

So

\(\\ \tt{:}\longrightarrow I=\dfrac{90}{110}\)

\(\\ \tt{:}\longrightarrow I=0.82A\)

The energy needed to flow into a 5-gram sample to change its temperature from 10°C to 11°C is 0.3 cal. Hence, the last option aligns well with the answer.

The specific heat capacity of silver is 0.06 calories per gram per degree Celsius.

Let's break down the problem to understand how we can find this answer:

The specific heat capacity of silver = 0.06 cal/g°C

The mass of the sample = 5 grams The change in temperature = 11°C - 10°C = 1°CWe can use the formula:

Q = m x c x ΔT

Where,Q = amount of heat energy required

m = mass of the sample

c = specific heat capacity

ΔT = change in temperature

Putting in the values,

Q = 5 g x 0.06 cal/g°C x 1°C

Dividing by 1000 to convert to joules,

Q = 0.0003 Joules.

Therefore, the answer is 0.3 cal.

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