one mole of h2o(g) at 1.00 atm and 100.8c occupies a volume of 30.6 l. when 1 mole of h2o(g) is condensed to 1 mole of h2o(l) at 1.00 atm and 100.8c, 40.66 kj of heat is released. if the density of h2o(l) at this temperature and pressure is 0.996 g/cm3 , calculate de for the condensation of 1 mole of water at 1.00 atm and 100.8c.

Answer:

Chemistry is used almost everywhere you go. from the car you use to get to work (gasoline burning) to the nuclear reactors that power many homes ( uranium becoming unstable to product electricity.

Answer:

Explanation:

Therefore, the pH of the HCl solution is 5.

The correct formula of copper (II) sulfate is CuSO₄ because Copper ion (Cu²⁺) has +2 charge and sulfate ion SO₄²⁻ has -2 charge, while CuSO₃ is copper(II) sulfite.

The CuSO₄ molecule contains an ionic bond between the copper cation (Cu²⁺) and the sulfate anion (SO₄²⁻).

The copper sulfate in its pentahydrate form is given by the chemical formula CuSO₄.5H₂O. This form of copper sulfate is characterized by its bright blue color but the anhydrous form of this salt is a white powder.

The oxidation state of the copper atom in a CuSO₄ molecule is +2. The molar mass of the pentahydrate and anhydrous forms of copper sulfate are 249.685 grams/mole and  159.609 grams/mole respectively.

Anhydrous CuSO₄ has a powdery appearance and a grey-white, while the pentahydrate has a bright blue color.

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1. Using your knowledge of the Brønsted-Lowry theory of acids and bases, complete the following acid-base reactions and indicate each conjugate acid-base pair.

i. OH + HPO₂ → H₂O + H₂PO₄²⁻

The conjugate acid-base pair is OH/H₂O, HPO₂²⁻/H₂PO₄²⁻

2) Identify the conjugate acid-base pairs in the following reactions. Write A, B, CA, and CB below the appropriate substance.

i. HCO₃⁻ + NH₃ → NH₄⁺ + CO₃²⁻

The conjugate acid-base pair is HCO₃⁻/CO₃²⁻, NH₃/NH₄⁺

ii. HCI + H₂O → H₃O⁺ + Cl⁻

The conjugate acid-base pair is H₂O/OH⁻, HCI/Cl⁻, H₃O⁺/H₂O, Cl⁻/HCI

iii. CH₃COOH + H₂O → H₃O⁺ + CH₃COO⁻

The conjugate acid-base pair is CH₃COOH/CH₃COO⁻, H₂O/OH⁻, H₃O⁺/H₂O

iv. HOCI + NH₃ → NH₄⁺ + ClO⁻

The conjugate acid-base pair is HOCI/ClO⁻, NH₃/NH₄⁺

3. Write the formula for conjugate bases formed by the following acids.

i. HPO₄²⁻ → PO₄³⁻

ii. H₂O → OH⁻

iii. CN⁻ → HCN

iv. HOOC-COO⁻ → HOOCCOOH

4) Write the formula for conjugate acids formed by each of the following bases.

i. H₃O⁺ → H₂O

ii. HCN → H₂CN⁺

iii. NH₃ → NH₄⁺

5. Classify each of the following pH values as acidic, basic, or neutral.

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Answer:

ethyl ethanoate

Explanation:

it helps you

Answer:

\(\rho =1.96\frac{g}{L}\)

Explanation:

Hello there!

In this case, since this imaginary gas can be modelled as an ideal gas, we can write:

\(PV=nRT\)

Which can be written in terms of density and molar mass as shown below:

\(\frac{P}{RT} =\frac{n}{V} \\\\\frac{P}{RT} =\frac{m}{MM*V}\\\\\frac{P*MM}{RT} =\frac{m}{V}=\rho\)

Thus, by computing the pressure in atmospheres, the resulting density would be:

\(\rho = \frac{165/760 atm * 314.2 g/mol}{0.08206\frac{atm*L}{mol*K}*425K} \\\\\rho =1.96\frac{g}{L}\)

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The balanced chemical equation for the double-replacement precipitation reaction between lead(II) nitrate and chromium(II) sulfate can be written as: Pb(NO3)2 + CrSO4 → PbSO4 + Cr(NO3)2

In this reaction, lead(II) nitrate and chromium(II) sulfate react to form lead(II) sulfate and chromium(II) nitrate. The lead sulfate is insoluble in water and precipitates out of the solution, while the chromium nitrate remains in the solution as a soluble salt.

Thus, the balanced chemical equation for the double-replacement precipitation reaction between lead(II) nitrate and chromium(II) sulfate can be written as:

Pb(NO3)2 + CrSO4 → PbSO4 + Cr(NO3)2

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Answer:

[Ar]4s2

Explanation:

Here are the electron configurations for calcium (Ca) and promethium (Pm) in noble gas notation:

Calcium (Ca):

1s2 2s2 2p6 3s2 3p6 4s2

Or in noble gas notation:

[Ar] 4s2

The electron configuration starts with a full 1s orbital, then a full 2s orbital and 2p subshell.

Then the 3s orbital is full and 3p orbital is full, matching the electron configuration of argon. Thus we write [Ar].

After the noble gas core, the next electron go into the 4s orbital, so we write 4s2.

Promethium (Pm):

1s2 2s2 2p6 3s2 3p6 4s2 3d1 4p6 5s2 4d1 6s2

Or in noble gas notation:

[Xe] 6s2 4f4

The electron configuration is similar up to argon.

After the argon core, the electrons fill the 4s and 3d orbitals, then the 4p.

The electron configuration then matches that of xenon, so we write the xenon core as [Xe].

The remaining electrons go into the 6s and 4f orbitals, shown after the noble gas core.

In summary:

Calcium electron configuration in noble gas notation:

[Ar] 4s2

Promethium electron configuration in noble gas notation:

[Xe] 6s2 4f4

The electron configuration in noble-gas notation for Ca (Calcium) is [Ar] 4s2, while for Pm (Promethium), it's [Xe] 6s2 4f5.

The electron configuration of an atom presents the distribution of electrons in atomic orbitals. It's helpful when understanding its atomic structure and chemical behavior.

For the element Ca (Calcium), which is atomic number 20, the noble gas prior to it on the periodic table is [Ar] (Argon). So, the electron configuration in noble-gas notation for Ca is [Ar] 4s2.

For Pm (Promethium), atomic number 61, the closest noble gas is [Xe] (Xenon), thus the notation starts with [Xe]. Pm falls in the 6th period, f-block, therefore following [Xe], the notation should be [Xe] 6s2 4f5.

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The IUPAC name for 2,3,7,9-tetramethyl-5-(1-methylbutyl)decanoic acid is 2,3,7,9-tetramethyl-5-(1-methylbutyl)decanoic acid. It is a carboxylic acid with a molecular formula of C20H40O2 and a molecular mass of 312.54 g/mol. The compound has a melting point range of 70-75°C.

2,3,7,9-tetramethyl-5-(1-methylbutyl)decanoic acid contains a carboxyl group (-COOH), which is the functional group responsible for its acidic properties and solubility in water. The presence of the carboxyl group makes the compound more polar than similar molecules, enhancing its water solubility. Carboxylic acids are also known for their reactivity with bases and ability to undergo esterification reactions.

In summary, 2,3,7,9-tetramethyl-5-(1-methylbutyl)decanoic acid is a carboxylic acid with a unique molecular structure. Its IUPAC name describes the positions and nature of the substituent groups. The presence of the carboxyl group contributes to its polarity, water solubility, acidity, and reactivity with bases and esterification reactions. The compound's characteristics are significant in understanding its properties and potential applications in various chemical reactions and processes.

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Answer:

It is difficult to cut the steam of water because of the unique properties of water and steam.Water and steam are two different states of matter, but they have a common property - they are both molecules of H2O. Steam is formed when water is heated, and the molecules of H2O begin to move faster and further apart from one another.

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The mass of the asteroid is C. 1.2 x \(10^{12}\) Kg

To find the mass of the other asteroid, we can rearrange the equation for the gravitational force between two objects:

F = (G * m1 * m2) / \(r^{2}\)

where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two asteroids, and r is the distance between them.

Given that the distance between the asteroids is 75000 m, the force of gravity between them is 1.14 N, and one asteroid has a mass of 8 x \(10^{7}\) kg, we can substitute these values into the equation and solve for the mass of the other asteroid (m2):

1.14 N = (6.67430 × \(10^{-11}\) N \(m^{2}\)/\(Kg^{2}\) * 8 x \(10^{7}\) kg * \(m2\)) / \((75000 m)^{2}\)

Simplifying and solving the equation, we find that the mass of the other asteroid (m2) is approximately 1.2 x \(10^{12}\) kg. Therefore, Option C is correct.

The question was incomplete. find the full content below:

Two asteroids are 75000 m apart one has a mass of 8 x \(10^{7}\) kg if the force of gravity between them is 1.14 what is the mass of the asteroid

A. 3.4 x \(10^{11}\) kg

B. 8.3 x \(10^{12}\) kg

C. 1.2 x \(10^{12}\) kg

D. 1.2 x \(10^{10}\) kg

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Answer:

The second picture

Explanation:

The first picture shows a solid because the particles of a solid are organized and compact. They can only vibrate in place

The third picture represents a gas because of how spaced out each particle is. They can move fast and erratically.

The second picture is the liquid because the particles aren't extremely packed, so they have space to move.

Answer:

The second one is correct

Answer:

silver (III) phosphate

Explanation:

the limiting reagent is nitrogen gas.

steps:

write out balanced equation: 3H2 + N2 = 2NH3

convert to moles to find how much NH3 each yields

find molar mass of each

10 g H2 (1 mol/2.016g) (2NH3/3H2) = 3.31 mol NH3

26g N2 (1mol/28.02g)(2NH3/1N2)=1.86 mol NH3

since N2 yields less NH3, it is the limiting reagent.

1.12 M is the dextrose solution's final concentration.

The following formula can be used to determine the concentration of the diluted dextrose solution:

C₁V₁ = C₂V₂

The starting concentration and volume are denoted by C1 and V1, while the final concentration and volume are denoted by C2 and V2.

Given: 2.0 M initial concentration (C1)

14 mL is the initial volume (V1).

25 mL is the final volume (V2).

When the values are substituted into the formula, we get:

C2 = (2.0 M) (14 mL) (25 mL)

To make the calculation easier:

28 = 25C₂

Calculating C2: C2 = 28 / 25 = 1.12 M

As a result, 1.12 M is the dextrose solution's final concentration.

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The enthalpy of formation (ΔHf​) of \(Br_2\)​(g) at 1.00 atm and 298 K is 0 kJ/mol. Hence, option A is correct.

The standard enthalpy of formation is defined as the enthalpy change for creating one mole of a compound from elements in their standard states.

The enthalpy of formation (ΔHf​) of \(Br_2\) ​(g) at 1.00 atm and 298 K is 0 kJ/mol.

It is because the enthalpy of formation of the molecule is only 0 in its standard state and for bromine, the standard state at 298K is the liquid state.

Hence, option A is correct.

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Answer:

A: The Δ.Hf for Br2(I) is 0 kJ/mol.

C: The Δ.Hf for Hg(I) is 0 kJ/mol.

D: The Δ.Hf for Cl2(g) is 0 kJ/mol.

Explanation:

Xeriscaping is a landscaping approach that focuses on conserving water by using drought-tolerant plants and efficient irrigation techniques. The goal is to create a visually appealing and sustainable garden while minimizing water usage.

Successful xeriscaping projects involve several key elements. Firstly, careful plant selection is crucial, opting for species that can thrive in arid conditions without excessive watering. Mulching is used to reduce evaporation and retain soil moisture.

Proper soil preparation, such as improving drainage and adding organic matter, promotes healthier plant growth. Efficient irrigation systems, like drip irrigation or soaker hoses, deliver water directly to plant roots, minimizing wastage.

Additionally, controlling erosion through the use of retaining walls or terracing is important. Lastly, regular maintenance, including appropriate pruning and weed control, ensures the longevity and vitality of the xeriscape garden. Overall, a successful xeriscaping project harmonizes sustainable practices with a beautiful outdoor environment.

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Jeremiah can arrange the materials in the following way to create a model that shows the processes by which water is cycled from a lake into the atmosphere and back to the lake

The following can be a representation of the water cycle;

Fill the glass jar with water to resemble the lake.

Put the lamp next to the jar to symbolize the sun.

Wrap the jar in plastic sheet to imitate the atmosphere.

Turn on the bulb to represent the sun warming the water.

When the water in the jar warms up and evaporates into water vapor, moisture will condense on the plastic wrap.

The water vapor will ascend and collect on the plastic wrap to represent the water vapor rising into the atmosphere.

Water vapor cools as it rises and condenses back into liquid form, as shown by the water droplets gathering on the plastic wrap.

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Answer:

among these which is not a base is CaSO4

An atom of 51V has 51 protons, 51 electrons, and approximately 49 neutrons.

The number of protons, electrons, and neutrons in an atom of 51V (vanadium) can be determined based on its atomic number, which is 51.

The atomic number of an element indicates the number of protons in its nucleus. Therefore, an atom of 51V has 51 protons.

Since atoms are electrically neutral, they must have an equal number of electrons as protons. Therefore, an atom of 51V has 51 electrons.

The number of neutrons in an atom can be calculated by subtracting the atomic number from the atomic mass, which is typically given in atomic mass units (amu). For 51V, the atomic mass is approximately 50.94 amu.

Therefore, the number of neutrons can be calculated as:

atomic mass - atomic number = number of neutrons

50.94 amu - 51 = approximately 49 neutrons

So, an atom of 51V has 51 protons, 51 electrons, and approximately 49 neutrons.

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Answer: Titanium dioxide

Answer: 55.5 oC is 0.014 atm (3rd decimal point)

Explanation:

The Clausius-Clapeyron equation is given as:

ln(P2/P1) = -(ΔH_vap/R) * (1/T2 - 1/T1)

where:

P1 = vapor pressure at temperature T1

P2 = vapor pressure at temperature T2

ΔH_vap = enthalpy of vaporization

R = gas constant = 8.314 J/(mol*K)

Converting the enthalpy of vaporization to J/mol:

ΔH_vap = 35.2 kJ/mol = 35,200 J/mol

Converting temperatures to Kelvin:

T1 = 64.7 + 273.15 = 337.85 K

T2 = 55.5 + 273.15 = 328.65 K

Substituting the values into the equation and solving for P2:

ln(P2/1 atm) = -(35,200 J/mol / 8.314 J/(mol*K)) * (1/328.65 K - 1/337.85 K)

ln(P2/1 atm) = -4.231

P2/1 atm = e^(-4.231)

P2 = 0.014 atm

Therefore, the vapor pressure for methanol at 55.5 oC is 0.014 atm, to the third decimal point.

The theoretical yield of the titanium tetra fluoride is  2.3 g.

We know that we can be able to obtain the theoretical yield from the stoichiometry of the reaction. We already have the balanced chemical reaction equation as shown in the question.

Number of moles of Titanium = 2.6 g/49 g/mol = 0.053 moles

Number of moles of fluorine = 1.4 g/38 g/mol = 0.037 moles

Given that 1 mole of Titanium reacts with 2 moles of fluorine

0.053 moles reacts with  0.053 moles * 2 moles/ 1 mole

= 0.106 moles

Thus fluorine is the limiting reactant

Now;

2 moles of fluorine produces 1 mole of titanium fluoride

0.037 moles of fluorine produces 0.037 moles * 1 mole/ 2 moles

= 0.0185 moles

Theoretical yield  of the product = 0.0185 moles * 124 g/mol

=  2.3 g

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The percent by mass of the carbon is 30.4%.

The term percentage has to do with the ratio of the mass of a particular atom to the total mass of the compound multiplied by one hundred. Thus the first step is to find the total mass or the molar mass of the compound.

Molar mass = 40 + 2(2(12) + 3(1) + 2(16))

= 40 + 2(24 + 3 + 32)

= 40 + 2(59)

= 40 +118

= 158

Thus the mass of carbon is;

4(12)/158 * 100/1

= 30.4%

Thus carbon is only about 30.4% by mass of the compound.

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Answer:

none of the above

Explanation:

A system is said to have attained dynamic equilibrium when the forward and reverse reactions proceed at the same rate. That is;

Rate of forward reaction = Rate of reverse reaction

The implication of this is that the concentrations of reactants and products remain constant when dynamic equilibrium is attained in a system. This does not mean that the reactant and product concentrations become equal; it rather means that their concentrations do not significantly change once dynamic equilibrium has been attained.

Answer: Pentane C5H12

Explanation:

The boiling point of a substance is simply defined as the temperature whereby a liquid's vapor pressure is equal to the pressure that is surrounding the liquid and hence, the liquid will changes into vapor.

The likely molecular formula for this compound is Pentane i.e C5H12 due to the fact that its boiling point is between Butane with formula C4H10 and Hexane with formula C6H14 boiling points.

The molecular formula of the of the substance is \(\bold {C_5H_1_0}\) (pentane).

Boiling point:

It is the temperature at which liquid's vapor pressure is equal to the pressure that is surrounding the liquid and hence, the liquid start to into vaporize.

The boiling point of the given compound is 309 K which is between Butane  and Hexane.

Therefore, the molecular formula of the of the substance is \(\bold {C_5H_1_0}\) (pentane).

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Approximately 4.37 x 10^23 molecules of HCl would be required to form 34.52 g of MgCl₂.

To determine the number of molecules of HCl required to form 34.52 g of MgCl₂, we need to use the molar mass and stoichiometry of the balanced equation:

Mg(OH)₂(s) + 2 HCl(aq) → 2 H₂O(l) + MgCl₂(aq)

The molar mass of MgCl₂ is 95.21 g/mol.

First, we need to calculate the number of moles of MgCl₂ formed:

Moles of MgCl₂ = mass of MgCl₂ / molar mass of MgCl₂

Moles of MgCl₂ = 34.52 g / 95.21 g/mol

Moles of MgCl₂ = 0.363 mol

According to the balanced equation, the stoichiometric ratio between HCl and MgCl₂ is 2:1. Therefore, the moles of HCl required can be calculated as follows:

Moles of HCl = 2 * Moles of MgCl₂

Moles of HCl = 2 * 0.363 mol

Moles of HCl = 0.726 mol

To calculate the number of molecules, we need to use Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol.

Number of molecules of HCl = Moles of HCl * Avogadro's number

Number of molecules of HCl = 0.726 mol * 6.022 x 10^23 molecules/mol

Number of molecules of HCl = 4.37 x 10^23 molecules

Therefore, approximately 4.37 x 10^23 molecules of HCl would be required to form 34.52 g of MgCl₂.

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