How many moles are in 5.8 x 1024 atoms of C?

Answer:

B and C

Explanation:

is it multiple choice question??

Van der Waals forces are the interaction between the molecules and the atoms. It increases with an increase in the molecular size and the number of electrons. Thus, options B and C are correct.

Van der Waals forces are the weak forces of attraction and repulsion present between the atoms and the molecules of the compounds. They are also called London Dispersion Forces and dipole-dipole forces.

An increase in the number of electrons and size of the molecule increases the surface area and the electron cloud which in turn, increases the Van der Waals force.

The intermolecular force increases as in large atoms or molecules the valence are held loosely as they are far from the nucleus or the center of the atom.

Therefore, the size and the electron increase the Van der Waals force.

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Answer:

\(\huge\boxed{\sf 39\ L}\)

Explanation:

In the reaction:

for 1 mol of propane, 3 moles of CO₂ are produced.

Volume of propane = 13 L

At STP,

1 mol of propane = 22.4 L/dm³ molar value

So,

Moles in 13 L of propane:

= volume / molar volume

= 13 / 22.4

= 0.58 mol

Since,

1 mol of propane = 3 moles of CO₂

0.58 mol of propane = 3 × 0.58 mol of CO₂

0.58 mol of propane = 1.74 mol of CO₂

Converting 1.74 mol of CO₂ into L:

[Multiply it by 22.4 L/mol]

= 1.74 mol × 22.4 L/mol

= 39 L

\(\rule[225]{225}{2}\)

We can conclude that the reaction will shift to the right in order to reach equilibrium due to equilibrium constant being \(4.63 * 10^(20)\) which is much greater than 1.

In order to determine which way the reaction would shift to reach equilibrium, we need to look at the reaction quotient (Qc) and compare it to the equilibrium constant (Kc).

At standard state conditions, the concentrations of all species are assumed to be 1 M. Therefore, the initial Qc for this reaction would be:

\(Qc = [SO3]^2 / [SO2]^2[O2] = 1^2 / 1^2(1) = 1\)

If the reaction shifts to the right, the concentration of SO3 will increase, while the concentrations of SO2 and O2 will decrease. This would result in a decrease in Qc, as the denominator would become smaller.

If the reaction shifts to the left, the concentration of SO3 will decrease, while the concentrations of SO2 and O2 will increase. This would result in an increase in Qc, as the numerator would become smaller.

Since the equilibrium constant for this reaction is \(Kc = [SO3]^2 / [SO2]^2[O2] = 4.63 * 10^(20)\), which is much greater than 1, we can conclude that the reaction will shift to the right in order to reach equilibrium. This means that the concentration of SO3 will increase, while the concentrations of SO2 and O2 will decrease, until the reaction reaches equilibrium.

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i am NOT gonna read all of that

1) The 7Ti(SO₄)₃ number which represent the coefficient is 7.

2) the expression CaCO₃--> CaO + CO₂ is an example of chemical equation

3) the arrow in a chemical reaction be translated to mean a) yields c) react to form.

1) The 7Ti(SO₄)₃, in the compound the coefficient is 7. the coefficient is the number which is place in front of the chemical symbol or the chemical formula.

2) the chemical reaction is given as :

CaCO₃--> CaO + CO₂

this is an example of the chemical equation.

3)  the arrow in a chemical reaction be translated to mean the reactants will combine to form the products. so the coorect option is a) yields and c) react to form.

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Isotopes are members of an element family that have the same number of protons but differ in the number of neutrons.

Atomic mass of Iodine = 126.86 amu

The elements in nature have several types of isotopes

Atomic mass is the average atomic mass of all its isotopes

MassAtomic mass of Iodine = 126.86 amu atom X = mass isotope 1 . % + mass isotope 2.% + ... mass isotope n.%

Atomic mass of Iodine = 0.8 x 127 + 0.17 x 126 + 0.03 x 128

Atomic mass of Iodine = 101.6 + 21.42 + 3.84

Atomic mass of Iodine = 126.86 amu

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Rutherford said that the atom is mostly empty space, with a small, dense positively charged nucleus located at the center.

Ernest Rutherford conducted the famous gold foil experiment, which involved bombarding a thin sheet of gold with alpha particles. He observed that most of the alpha particles passed straight through the gold foil, but some were deflected at angles, and a few bounced back in the direction they came from.

This led him to conclude that the atom was mostly empty space, with a small, dense nucleus at the center that was positively charged and contained most of the atom's mass. The negatively charged electrons were located in the space surrounding the nucleus, but they were much smaller and lighter than the nucleus itself.

In conclusion, Rutherford's discovery of the atomic nucleus revolutionized our understanding of the structure of the atom. It provided a more accurate picture of the atom than the previous plum pudding model, which suggested that the atom was a uniform blob of positive charge with electrons embedded throughout. Rutherford's model showed that the atom was a complex structure with a small, but incredibly dense nucleus at its center, surrounded by electrons in a cloud-like arrangement.

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Answer:

ocean

Explanation:

because its easy

1) Balance the chemical equation.

Step 1: Assign oxidation numbers

PbO2: Pb (+4) and O (-2)

Mn2+: Mn (+2)

Pb2+: Pb (+2)

MnO4-: Mn (+7) and O (-2)

Step 2: Figure out what's being reduced and what's being oxidized.

Mn has been oxidized. It changed from (+2) to (+7)

Pb has been reduced. It changed from (+4) to (+2)

Step 3: Write half-reaction

Oxidation half-reaction

Reduction half-reaction

Step 4: Balance all elements EXCEPT for hydrogen and oxygen

Oxidation half-reaction

Reduction half-reaction

Step 5: Balance oxygens. We do so by adding water molecules to the half-reactions as needed.

Oxidation half-reaction

Reduction half-reaction

Step 6: Balance hydrogens. We do so by adding protons (H+) to the half-reactions as needed.

Oxidation half-reaction

Reduction half-reaction

Step 7: Balance charges. We do so by adding electrons

Oxidation half-reaction

Reduction half-reaction

Step 8: Multiply half-reactions to make the number of electrons equal.

Oxidation half-reaction.

New oxidation half-reaction.

Reduction half-reaction

New reduction half-reaction

Step 9: Cancel electrons and combine the half-reactions

Overall reaction

Step 10: Balance the chemical equation by reducing the number of water molecules and protons.

Overall reaction

2) The balanced chemical equation

Overall reaction

.

Answer:

semiconductors

Explanation:

Answer: Semiconductors

Explanation:

Answer:

Option d is correct

Explanation:

The basis of the VSEPR model of molecular bonding is that to minimize repulsions, electron domains in the valence shell of an atom arrange on their own.

According to VSEPR theory, if there are two electron domains on a central atom, they will be arranged such that the angles between the domains are 180°

The frequency of light required to remove the valence electron(s) from an atom is known as the ionization energy. The ionization energy is dependent on the atomic structure and is specific to each element.

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It is better to observe the law of conservation of mass in a closed system as it is more accurate.

This is referred to as the law which states that mass can neither be created nor destroyed but can be transformed from one form to another.

In a closed system, matter cannot enter or leave the system which means that when water evaporates inside a sealed container, the container and the water vapor have the same mass after evaporation has occurred which depicts this law and is therefore the reason why a closed system was chosen as the correct choice.

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Answer:

1.3 meters

Explanation: use newton third law equation.

Answer:

B. Metal

Explanation:

The catalyst used in the converter is mostly a precious metal such as platinum, palladium and rhodium. Platinum is used as a reduction catalyst and as an oxidation catalyst. Although platinum is a very active catalyst and widely used, it is very expensive and not suitable for all applications.

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The standard reaction enthalpy for the given reaction is +235.8 kJ/mol.

The standard reaction enthalpy (ΔH°) for the given reaction is determined as follows:

Equation of reaction: 3 Fe₂O₃ (s) → 2 Fe₃O₄ (s) + ½ O₂ (g)

The standard enthalpy of formation values for Fe₂O₃ (s), Fe₃O₄(s), and O₂(g) is used to calculate the standard reaction enthalpy.

ΔH° = [2 × ΔH°f(Fe₂O₃)] + [½ × ΔH°f(O₂)] - [3 × ΔH°f(Fe₃O₄)]

where;

ΔH°f(Fe₂O₃) = -824.2 kJ/mol

ΔH°f(Fe₃O₄) = -1118.4 kJ/mol

ΔH°f(O₂) = 0 kJ/mol

ΔH° = [2 × (-1118.4 kJ/mol)] + [½ × 0 kJ/mol] - [3 × (-824.2 kJ/mol)]

ΔH° = -2236.8 kJ/mol + 0 kJ/mol + 2472.6 kJ/mol

ΔH° = 235.8 kJ/mol

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Answer:

The Internet started in the 1960s as a way for government researchers to share information. ... This eventually led to the formation of the ARPANET (Advanced Research Projects Agency Network), the network that ultimately evolved into what we now know as the Internet.

Structure represents the condensation product formed by the reaction of phenylacetaldehyde with diethyl malonate (CH₂(COOEt)₂) in the presence of sodium hydroxide.

Aldol condensation occurs in aldehydes having α-hydrogen with a dilute base to give β-hydroxy aldehydes called aldols. This reaction is most commonly known as aldol condensation. If the condensation reaction occurs between two different carbonyl compounds it is called crossed aldol condensation.

In the crossed aldol condensation of phenylacetaldehyde (C₆H₅CH₂CHO) with CH₂(COOEt)₂ in the presence of sodium hydroxide, the product formed is 2-(phenylmethylidene)pentanedioic acid diethyl ester.

The structural formula of the product can be represented as follows:

H H H

| | |

H-C-C-C-C-C-C-C-C-C-O-CH₂-CH₃

| | |

H H H

(Phenylmethylidene)

This structure represents the condensation product formed by the reaction of phenylacetaldehyde with diethyl malonate (CH₂(COOEt)₂) in the presence of sodium hydroxide.

The structure provided above represents a general representation of the product formed in the crossed aldol condensation reaction. The exact stereochemistry and arrangement of substituents may vary.

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We would need stock solutions of 20 ml of 1 M NaCl and 50 ml of 2 M MgCl₂ to prepare 500 mL of buffer A (100 mM NaCl and 250 mM MgCl₂).

The amount of NaCl and MgCl2 needed can be calculated using the formula for buffer concentration:

C1 × V1 + C2 × V2 = C_buffer × V_buffer

where C1 and C2 are the stock concentrations of NaCl and MgCl₂, V1 and V2 are the volumes of NaCl and MgCl₂ stock solutions needed to prepare the buffer, and C_buffer and V_buffer are the final concentration and volume of the buffer.

We know the final concentration of the buffer (100 mM NaCl and 250 mM MgCl₂) and the volume of the buffer (500 mL), so we can solve for the volumes of the stock solutions needed:

V1 = (C_buffer × V_buffer - C2 × V2) / C1

V1 = (100 × 0.5 - 250 × V2) / 1000

V2 = (C_buffer × V_buffer - C1 × V1) / C2

V2 = (100 × 0.5 - 1000 × V1) / 250

V1 = (100 × 0.5 - 250 × (100 × 0.5 - 1000 × V1) / 250) / 1000

V1 = (100 × 0.5 - 250 × 100 × 0.5 / 250 + 250 × 1000 × V1 / 250) / 1000

V1 = (100 × 0.5 - 100 × 0.5 + 1000 × V1) / 1000

V1 × 1000 = 100 × 0.5 / 1000

V1 = 100 × 0.5 / 1000

V1 = 0.05 L = 50 mL

So the volume of the stock solution of 2 M MgCl₂ is 50 ml.

V2 = (100 × 0.5 - 1000 × 50 mL) / 250

V2 = (100 × 0.5 - 1000 × 0.05 L) / 250

V2 = (100 × 0.5 - 50) / 250

V2 = 0.2 L = 20 mL

So the volume of the stock solution of 1 M NaCl is 20 ml.

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Your hand gets warmer as soon as it touches a warm picnic table. Describe the relevance of energy conservation in this case.The method by which heat transmission between things that come into contact is known as conduction.

Heat transmission via conduction occurs when two items are in contact.The heat is transferred to a object with lower temperature when two things touch and one of them has a higher heat capacity than the other.Because the heat is transferred from the item to your hand when you contact anything hot, it feels hot.

The radiator's exterior feels warm to the touch.Through direct contact, its heat is transmitted from of the liquid to the metal and finally to your hand.The term for this is THERMAL CONDUCTION.

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The given reaction N2 (g) + 3H2 (g) à 2NH3 (g) is an example of combination reaction.

As a result, it is a combination reaction.

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In terms of the stream survey and partial pressure of CH₄, the physical variable that is most strongly correlated with CH₄ is likely water temperature

This is because CH₄ production and release is greatly influenced by temperature, with warmer waters often leading to higher levels of CH₄.

Other physical variables that may have some correlation with CH₄include water flow rate and dissolved oxygen levels.

In terms of chemical variables, pH and nutrient levels (such as nitrogen and phosphorus) can also impact CH4 levels, as they influence the growth of methane-producing bacteria in the water.

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Answer:

diffusion is the movement of gases from a higher concentration area to a lower concentration area

osmosis is the movement of water molecules from a higher concentration area to a lower concentration area

Answer:

The claim is: Therefore Mitosis requires less energy than sexual reproduction does.

Explanation:

Answer:N-1

             H-3

             C-1

             H-4

Explanation:Nitrogen-1,Hydrogen-3(Ammonia),Carbon-1,Hydrogen-4(Methane).

Under aerobic conditions, the carbon in decaying vegetation in a lake ends up primarily as carbon dioxide. This excess carbon dioxide in the lake can even damage the ecosystem and create an imbalance.

The process of conversion of carbon into carbon dioxide is given below:
1. Aerobic decomposition occurs in the presence of oxygen.
2. In this process, microorganisms break down the decaying vegetation.
3. The microorganisms consume oxygen and release carbon dioxide as they metabolize the organic matter.
4. As a result, the carbon in the decaying vegetation is primarily converted to carbon dioxide (CO2), which is released into the atmosphere or dissolved in the water.

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Red gold is a gold-copper alloy that is commonly used to make jewelry. To calculate the percentage by mass of each metal in the jewelry, you need to determine the mass of gold and copper in the jewelry. The mass of gold can be calculated using the density of gold and the volume of gold in the jewelry:

mass of gold = density of gold * volume of gold

mass of gold = 19.3 g/cm³ * 0.559 cm³

mass of gold = 10.8 g

The mass of copper can be calculated using the density of copper and the volume of copper in the jewelry:

mass of copper = density of copper * volume of copper

mass of copper = 8.96 g/cm³ * (0.559 cm³ - 0.559 cm³)

mass of copper = 0 g

The total mass of the jewelry is 10.8 g + 0 g = 10.8 g.

Therefore, the percentage by mass of gold in the jewelry is 10.8 g/10.8 g = 100%. The percentage by mass of copper in the jewelry is 0 g/10.8 g = 0%.

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