For the following reaction at equilibrium, which gives a change that will shift the position of equilibrium to favor formation of more products? 2NOBr(g) 2NO(g) + Br 2(g), ΔHº rxn = 30 kJ/mol

Answers

Answer 1
Answer:

Answer:

Based on the given reaction, it is evident that the reaction is endothermic as indicated by a positive sign of enthalpy of reaction. Thus, it can be stated that the favoring of the forward reaction will take place by upsurging the temperature of the reaction mixture.  

Apart from this, based on Le Chatelier’s principle, any modification in the quantity of any species is performed at equilibrium and the reaction will move in such an orientation so that the effect of the change gets minimized. Therefore, a slight enhancement in the concentration of the reactant will accelerate the reaction in the forward direction and hence more formation of the product takes place.  


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The movement of which layer of Earth drives the motion of the plates on Earth’s crust? Upper mantle
Lower mantle
Outer core
Inner core

Answers

Final answer:

The movement of the upper mantle and the tectonic plates of the Earth's lithosphere results in the movement of the Earth's crust. The tectonic plates float on the asthenosphere, a semi-fluid part of the upper mantle, and are driven by convection currents.

Explanation:

The movement of Earth's plates is driven by the upper mantle. The Earth's lithosphere, which is the topmost layer consisting of the crust and the rigid upper part of the mantle, is broken up into tectonic plates. These tectonic plates float on the semi-fluid layer of the mantle known as the asthenosphere. Convection currents in the asthenosphere, which is part of the upper mantle, move these plates. Thus, the upper mantle has a significant role in the movement of the Earth's crust.

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1. What kind of intermolecular forces act between a chlorine monofluoride molecule and a hydrogen bromide moleculeNote: If there is more than one type of intermolecular force that acts, be sure to list them all, with a comma between the name of each force.

Answers

Final answer:

The intermolecular forces that act between chlorine monofluoride (ClF) and hydrogen bromide (HBr) are dipole-dipole interactions. These types of forces result from the attraction between polar molecules.

Explanation:

The intermolecular forces that act between a chlorine monofluoride (ClF) molecule and a hydrogen bromide (HBr) molecule are

dipole-dipole interactions

. A

dipole-dipole interaction

is a type of force that results from the attraction between polar molecules. Since ClF and HBr are both polar molecules, they exhibit this kind of interaction. For instance, the positive end of the polar ClF molecule would be attracted to the negative end of the polar HBr molecule, and vice versa, leading to a

dipole-dipole interaction

.

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Final answer:

Between chlorine monofluoride and hydrogen bromide, the intermolecular forces present are dipole-dipole forces and London dispersion forces due to their polar nature and instantaneous polarizations of electron clouds respectively.

Explanation:

The intermolecular forces that act between a chlorine monofluoride molecule and a hydrogen bromide molecule are primarily the dipole-dipole forces. Dipole-dipole forces are attractive forces that occur between the positive end of one polar molecule and the negative end of another polar molecule. Both chlorine monofluoride and hydrogen bromide are polar molecules, and as such, they interact through dipole-dipole forces. Apart from this, there exists London dispersion forces which are weak forces resulting from instantaneous polarizations of electron clouds in molecules. Hence, between chlorine monofluoride and hydrogen bromide, both dipole-dipole forces and London dispersion forces act.

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Identify the sample and analyte in each of the scenarios.(a) Determination of the lead content in paint. lead paint
(b) Analysis of the nitrate content of soil near a local water source. soil nitrate
(c) Measurement of the citric acid found in a lime.

Identify the following as either sample or analyte.
(1) lead
(2) paint chips
(3) soil
(4) nitrate
(5) lime wedge
(6) citric acid

Answers

Answer:

a) Analyte: lead. Sample: paint.

b) Analyte: nitrate. Sample: soil.

c) Analyte: citric acid. Sample: Lime

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

Explanation:

A sample is a portion of material selected from a larger quantity of material while an analyte is the chemical of the system that will be analysed.

Thus:

a) Analyte is lead while you must take a sample of paint to analyze this lead.

b) Analyte is the nitrate while sample must be soil.

c) Analyte is citric acid and lime is the sample

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

g For a given arrangement of ions, the lattice energy increases as ionic radius ________ and as ionic charge ________.

Answers

For a given arrangement of ions, the lattice energy increases as ionic radius decreases and as ionic charge increases.

What is ion?

An atom or molecule is said to be an ion if one or more of whose valence electrons have been acquired or lost, providing it a net negative or positive electrical charge.

Faraday knew that metals disintegrated together into solution place at a single electrode and that a second metal was placed first from solution at the opposite electrode, as such matter had to be trying to move underneath the impact of an electrical current even though he was unable to identify the particles trying to move between the electrodes.  For a given arrangement of ions, the lattice energy increases as ionic radius decreases and as ionic charge increases.

Therefore, for a given arrangement of ions, the lattice energy increases as ionic radius decreases and as ionic charge increases.

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Answer:

as the charge of the ions increases, the lattice energy increases. as the size of the ions increases, the lattice energy decreases.

A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half-life for 238U  206Pb is 4.5  109 yr.)

Answers

Half-life of a radioactive element

The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay

Calculating the original amount of U-238

  • The number of moles present in each element is first determined:
  • Number of moles = mass/molar mass

For Pb-206:

mass = 0.37 mg = 0.37 * 10⁻³ g

molar mass = 206 g/mol

number of moles = 0.37 * 10⁻³ g/206 g/mol

number of moles of Pb-206 = 1.79 * 10⁻⁶ moles

For U-238:

mass = 0.95 mg = 0.95 * 10⁻³ g

molar mass = 238 g/mol

number of moles = 0.95 * 10⁻³ g/238 g/mol

number of moles = 3.99 * 10⁻⁶ moles

Assuming that all the Pb-206 were formed from U-238

  • Initial moles of U-238 = Moles of present U-238 + molesw of present Pb-208

Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles

Initial moles of U-238 = 5.78 * 10⁻⁶ moles

One mole of U-238 contains = 6.02 * 10²³ atoms

5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms

Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms

Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms

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Iron(II) chloride is added to sodium carbonate. The reaction is described by the balanced equation below. (7 points) FeCl2(aq) + Na2CO3(aq) FeCO3(s) + 2NaCl(aq) i. Write the complete ionic equation from the balanced equation. (3 points) ii. What are the spectator ions in the reaction? (1 point) iii. Write the net ionic equation from the complete ionic equation. (2 points)

Answers

Answer:

Explanation:

FeCl₂(aq) + Na₂CO₃(aq) ⇒ FeCO₃(s) + 2NaCl(aq)

complete ionic equation

Fe⁺²(aq) + 2 Cl⁻(aq) + 2 Na⁺(aq) + CO₃⁻² (aq) ⇒ FeCO₃(s)+ 2 Na⁺(aq) +2Cl⁻(aq)

Spectator ions =  Cl⁻(aq) , Na⁺(aq)

Net ionic equation

Fe⁺²(aq) + CO₃⁻² (aq) ⇒ FeCO₃(s).

The Complete ionic equation:

Fe²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + CO₃²⁻(aq) → FeCO₃(s) + 2Na⁺(aq) + 2Cl⁻(aq)

ii. Spectator ions: Na⁺(aq) and Cl⁻(aq)

iii. Net ionic equation:

Fe²⁺(aq) + CO₃²⁻(aq) → FeCO₃(s)

An ionic equation is a chemical equation that shows the dissociation of all soluble ionic compounds into their respective ions in a solution. It represents the species present in their ionic form rather than as complete compounds.

In an ionic equation, only the ions involved in the chemical reaction are shown, while the spectator ions, which do not participate in the reaction, are omitted. This type of equation provides a more focused representation of the chemical reaction occurring between the ions.

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