Acetic acid is miscible with water. Would you expect carboxylic acids with general formula CH₃(CH₂)n COOH to become more or less water soluble as n increases? Explain.

Answer:

To solve this problem, we'll need to use stoichiometry and the molar ratios from the balanced chemical equation. Here's how you can calculate the values:

(a) Mass of the lead nitrate sample:

From the balanced equation, we can see that 2 moles of lead nitrate (Pb(NO3)2) produce 1 mole of oxygen gas (O2). We know that the volume of oxygen gas collected is 112 cm³, which is equal to 112/1000 = 0.112 dm³ (converting cm³ to dm³).

According to the molar volume of gas at STP (22.4 dm³), 1 mole of any gas occupies 22.4 dm³ at STP. Therefore, the number of moles of oxygen gas can be calculated as:

moles of O2 = volume of O2 / molar volume at STP

moles of O2 = 0.112 dm³ / 22.4 dm³/mol = 0.005 mol

Since 2 moles of lead nitrate produce 1 mole of oxygen gas, we can determine the number of moles of lead nitrate as:

moles of Pb(NO3)2 = 2 * moles of O2

moles of Pb(NO3)2 = 2 * 0.005 mol = 0.01 mol

To calculate the mass of the lead nitrate sample, we'll use its molar mass:

mass of Pb(NO3)2 = moles of Pb(NO3)2 * molar mass of Pb(NO3)2

mass of Pb(NO3)2 = 0.01 mol * (207 g/mol + 2 * 14 g/mol + 6 * 16 g/mol)

mass of Pb(NO3)2 = 0.01 mol * 331 g/mol

mass of Pb(NO3)2 = 3.31 g

Therefore, the mass of the lead nitrate sample is 3.31 grams.

(b) Mass of lead(II) oxide produced:

From the balanced equation, we can see that 2 moles of lead nitrate (Pb(NO3)2) produce 2 moles of lead(II) oxide (PbO). So, the number of moles of PbO produced is equal to the number of moles of Pb(NO3)2.

mass of PbO = moles of PbO * molar mass of PbO

mass of PbO = 0.01 mol * (207 g/mol + 16 g/mol)

mass of PbO = 0.01 mol * 223 g/mol

mass of PbO = 2.23 g

Therefore, the mass of lead(II) oxide produced is 2.23 grams.

(c) Volume of nitrogen dioxide gas produced at STP:

From the balanced equation, we can see that 2 moles of lead nitrate (Pb(NO3)2) produce 4 moles of nitrogen dioxide gas (NO2). So, the number of moles of NO2 produced is twice the number of moles of Pb(NO3)2.

moles of NO2 = 2 * moles of Pb(NO3)2

moles of NO2 = 2 * 0.01 mol = 0.02 mol

Using the molar volume of gas at STP, we can calculate the volume of nitrogen dioxide gas:

volume of NO2 = moles of NO2 * molar volume at STP

volume of NO2 = 0.02 mol * 22.4 dm³/mol = 0.448 dm³

Therefore, the volume of nitrogen dioxide gas

The formation of a complex ion such as Cu(NH3)42+ (aq) can best be categorized as a coordination reaction.

In a coordination reaction, a complex ion is formed by the binding of a central metal ion with surrounding ligands. Ligands are molecules or ions that have lone pairs of electrons and can donate these electrons to the central metal ion.

In this case, Cu(NH3)42+ represents a complex ion where the central metal ion is Cu2+ and the ligands are NH3 molecules. The complex ion is formed through a coordination bond, where the lone pair of electrons on the nitrogen atom of the NH3 ligand donates to the vacant d-orbital of the Cu2+ ion.

The formation of Cu(NH3)42+ can be represented step-by-step as follows:

1. Cu2+ (aq) + 4NH3 (aq) → Cu(NH3)42+ (aq)

The resulting complex ion, Cu(NH3)42+, is stabilized by the coordination bond between Cu2+ and the NH3 ligands. The complex ion has a square planar geometry, with four NH3 ligands surrounding the central Cu2+ ion.

Overall, the formation of Cu(NH3)42+ can be categorized as a coordination reaction because it involves the formation of a complex ion through the coordination of ligands to a central metal ion.

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Answer: See below

Explanation:

n of CO2​ = 0.364mmol

Mass of C = 0.364*12 = 4.368 mg

n of H2​O = 184.37​ = 0.243 mol

The compound has 2*0.243mmol of H

Mass of H = 0.486 mg

Mass of O = 10.68 − (4.368+0.486) = 5.826mg

Moles of O = 0.364

C:H:O Ratios

0.364 : 0.486 : 0.364

= 1 : 1.34 : 1

= 3 : 4 : 3

So the empirical formula is C3​H4​O3​,

Empirical formula mass

= 88= 2 × Molar mass

And the molecular formula is C6​H8​O6

Answer:

aluminium is too high in the reactivity series

Explanation:

it is more reactive than carbon so electrolysis will have to be used to extract aluminium.

It is not possible to construct an ideal machine. Because machines have some loss of energy in the form of heat or friction.

Given parameters:

Mass of rock = 14g

Volume of rock = 12cm³

Unknown:

Density of the rock = ?

Density of a substance is the amount of substance it contains per unit volume. Mathematically, this can be expressed as;

           Density  = \(\frac{Mass}{Volume}\)

Now simply input the parameters and solve;

         Density  = \(\frac{14}{12}\)   = 1.17g/cm³

Density of the rock that Jack has is 1.17g/cm³

Answer:

The answer is the atoms of the reactants have combined to form a new substance. Every product of a chemical equation has to be a new substance. If their is no new substance then, a chemical change never occurred and there is no chemical equation. The new substance never has the same properties as the reactants. That's why they call it a "new" substance. Also, the atoms are the ones that combine. They are not mixed

The molecular formula of the compound that contains only carbon and hydrogen, with 85.6% carbon and has a molar mass of 70 g/mol is C₅H₁₀

C = 85.6%

H = 100 – 85.6 = 14.4%

Divide by their molar mass

C = 85.6 / 12 = 7.13

H = 14.4 / 1 = 14.4

Divide by the smallest

C = 7.13 / 7.13 = 1

H = 14.4 / 7.13 = 2

Thus, the empirical formula of the compound is CH₂

Empirical formula => CH₂

Molar mass of compound = 70 g/mol

Molecular formula = Empirical × n = molar mass

[CH₂]n = 70

[12 + (1×2)]n = 70

[12 + 2]n = 70

14n = 70

Divide both side by 14

n = 70 / 14

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₅

Therefore, the molecular formula of the compound is C₅H₁₀

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To calculate the change in entropy (ΔS) when 6.8 g of benzene (C6H6) melts at 5.6 °C, we need to use the equation ΔS = ΔH / T. The given heat of fusion for benzene is 10.6 kJ/mol.

The equation to calculate the change in entropy (ΔS) is given as ΔS = ΔH / T, where

ΔH represents the heat of fusion and

T is the temperature in Kelvin.

First, we need to convert the given mass of benzene (6.8 g) to moles. The molar mass of benzene is approximately 78.11 g/mol, so we divide the mass by the molar mass:

6.8 g / 78.11 g/mol = 0.087 mol

Next, we need to convert the given temperature (5.6 °C) to Kelvin by adding 273.15:

5.6 °C + 273.15 = 278.75 K

Now, we can substitute the values into the equation to calculate the change in entropy:

ΔS = 10.6 kJ/mol / 0.087 mol / 278.75 K

Calculating the division gives us the change in entropy in kJ/(mol·K). To ensure the correct number of significant digits, we need to round the result to the appropriate precision.

Therefore, the change in entropy when 6.8 g of benzene melts at 5.6 °C is approximately X kJ/(mol·K), rounded to the appropriate number of significant digits.

Complete Question-

The heat of fusion AH, of benzene (CH) is 10.6 kJ/mol. Calculate the change in entropy AS when 6.8 g of benzene melts at 5.6 °C. Be sure your answer contains a unit symbol and the correct number of significant digits

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Answer:

1.67 × 10^9

Explanation:

The scale factor at which the CMB photons had a temperature of 1.9 K is 1.44 times smaller than today's scale factor.

a) At what temperature would today's CMB photons have wavelength in the radio range, say 100 m?The formula for the wavelength-frequency relation is given by:λν=cwhere,λ = wavelengthν = frequencyc = speed of light = 3 × 108 m/sSince the wavelength is given in the question to be 100 m, we can use the formula to find the frequency of CMB photons with a wavelength of 100 m.λ = c / ν100 = 3 × 108 / νν = 3 × 106 HzThe frequency of CMB photons with a wavelength of 100 m is 3 × 106 Hz.

Now, we can use the formula for the Planck distribution to find the temperature at which the photons would have this frequency. It is given by:Bν(T) = 2hc2/λ5 × (exp(hν/kT) - 1)where,Bν(T) = blackbody spectral radiancec = speed of lighth = Planck's constantk = Boltzmann constantT = temperatureSubstituting the values:2hc2/λ5 × (exp(hν/kT) - 1) = Bν(T)2hc2/(100 m)5 × (exp(6.626 × 10-34 × 3 × 106 / (1.38 × 10-23 T)) - 1) = 1.76 × 10-22 W/m2/Hz. The temperature at which the spectral radiance of CMB photons would be 1.76 × 10-22 W/m2/Hz is found by trial and error, and it is T = 1.9 K.

Therefore, CMB photons with a wavelength of 100 m would correspond to a temperature of 1.9 K.b) What would the scale factor be at that point?The relation between temperature and scale factor is given by:T ∝ a-1where, T is the temperature and a is the scale factor.

If the temperature of CMB photons is 1.9 K, then we can use the current temperature of 2.73 K to find the ratio of scale factors.2.73 K / 1.9 K = (a0 / a)where a0 is the scale factor today and a is the scale factor when the CMB photons had a temperature of 1.9 K.The scale factor at that point is:a = a0 × 1.44.

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Answer:

ozcozz8f8txitcizc9zcozozozc9uvpuh

Answer:

D

Explanation:

it seems the most logical

Answer:

B

Explanation:

6.00 ml

PLEASE MARK AS BRAINLIEST

Answer:

19.05 g have been oxidized

Explanation:

This is the unballanced equation:

H₂SO₄ + Cu → CuSO₄ + SO₂ + H₂O

Let's balance as a redox one:

Cu° oxidizes to Cu²⁺

Cu → Cu²⁺ + 2e⁻

SO₄⁻² reduces to SO₂

We add 2 water in the left side in order to ballance the O, so we add 4 protons to the right side, to ballance the H.

SO₄⁻² + 4H⁺ + 2e⁻ → SO₂ + 2H₂O

In the 2 half reactions, we have the same electrons, we sum both equations, so we have the main equation ballanced (we cancel the e⁻):

2H₂SO₄ + Cu → CuSO₄ + SO₂ + 2H₂O

We determine the moles of sulfuric acid:

M = moles /L → 0.5 mol/L . 1.2L = 0.6 moles

As ratio is 2:1, we propose

2 moles of sulfuric react to 1 mol of Cu

0.6 moles of sulfuric will react to (0.6 . 1) /2 = 0.3 moles of Cu

We determine the mass: 0.3 mol . 63.5 g/mol = 19.05 g

Since they are the main producers of the organic compounds needed for the development and survival of other organisms, phototrophs are essential for the flow of energy in the biosphere.

The flow of energy in the biosphere begins primarily with phototrophs. Phototrophs are organisms that use energy from the sun to convert carbon dioxide and water into organic compounds, such as sugars and carbohydrates, through the process of photosynthesis. These organic compounds provide a source of energy and nutrients for other organisms in the biosphere, which ultimately fuels the flow of energy through the food chain. Other organisms, such as chemotrophs, can also obtain energy from oxidized chemicals or inorganic compounds, but the primary source of energy in most ecosystems is derived from photosynthesis.

Therefore, phototrophs play a crucial role in the flow of energy in the biosphere, as they are the primary producers of organic compounds that support the growth and survival of other organisms.

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Explanation:

Answer:

atomic weight..................

A single replacement reaction is a type of chemical reaction that takes place when one element in a compound is replaced by another element. In other words, in this reaction, one element is replaced by another element. There are a few ways to predict whether or not a single replacement reaction will occur.

These ways are explained below: Paying Attention to ReactantsThe first way to predict if a single replacement reaction will occur is by paying attention to the reactants. In a single replacement reaction, a more reactive element will replace a less reactive element. For example, if a metal element is mixed with an aqueous solution that contains ions of another metal, a single replacement reaction will occur if the metal in the solid state is more reactive than the metal in the solution state. For instance, if you put zinc in copper sulfate, a reaction will occur because zinc is more reactive than copper. Therefore, it will replace copper, producing zinc sulfate and copper.Using the Activity SeriesAnother way to predict if a single replacement reaction will occur is by using the activity series. The activity series is a list of metals and their ability to replace other metals from their compounds. This list is arranged in order of decreasing activity. Therefore, if a metal is more active than another metal on the activity series, it will replace that metal from its compound. For example, if you put magnesium in silver nitrate, a reaction will occur because magnesium is more reactive than silver. Therefore, magnesium will replace the silver, producing magnesium nitrate and silver. If the metal is less active than another metal on the activity series, no reaction will occur.Using Electrochemical SeriesThe electrochemical series is another way to predict if a single replacement reaction will occur. The electrochemical series lists the elements in order of their standard reduction potentials. A metal with a higher reduction potential will replace a metal with a lower reduction potential from its compound. For instance, if you put copper in magnesium sulfate, no reaction will occur because copper has a lower reduction potential than magnesium. Therefore, magnesium will not replace copper from its compound.

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Answer:

they are all on the periodic table and they are all elements ?

Answer:

They are all alkaline each metals and they each have the same number of valence electrons

Answer:

compound

Explanation:

Once you mix it you cant take it back apart

The primary ester example presented is ethyl acetate, which has a fruity descriptor and is commonly found in Riesling wines.

The primary ester example that was presented is ethyl acetate. Ethyl acetate is an ester formed by the condensation of acetic acid and ethanol. It is commonly used as a flavoring compound due to its fruity aroma and taste.

When it comes to wine, ethyl acetate is one of the most common esters found in both red and white wines. It contributes to the fruity and floral aromas often associated with wines. In particular, it is commonly found in Riesling wines, a white wine varietal known for its aromatic profile.

The fruity descriptor refers to the characteristic scent and flavor of ethyl acetate, reminiscent of fruits such as apples, pears, and bananas. This fruity note adds complexity and enhances the overall sensory experience of the wine.

It's important to note that the presence and concentration of ethyl acetate in wines can vary depending on factors such as grape variety, fermentation conditions, and aging process.

However, it is generally recognized as an ester that contributes to the aromatic profile and flavor profile of wines.

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Answer: The kinetic energy of solids is limited to vibrational energies..the particles are tightly bound in their crystalline structure and can only bend, flex, and vibrate about relatively fixed position.  When vibrational energies exceed the strength of the force holding the particles together, the crystal structure collapses and we say the material melted.  The particles can now have, in addition to its vibrational energies, rotational energy.  Chunks of the collapsed crystal can, like a ball floating in water, rotate without showing much translational energy. Generally, we associate solids with vibrational kinetic energy, liquids with rotational kinetic energy, and gases with translational kinetic energy. Actually though, liquids have both vibrational and rotational energies, and gases have all three.

Answer:

Dipole dipole interaction

Explanation:

The balanced chemical equation for the reaction between carbon (C) and oxygen (O2) to form carbon monoxide (CO) and carbon dioxide (CO2) is below and the moles of CO2 produced is 0.117.

C + O2 → CO + CO2

According to the equation, for every 1 mole of C, we need 1 mole of O2 to produce 1 mole of CO and 1 mole of CO2.

Given that we have 0.155 moles of C and 0.117 moles of O2, we can use the mole ratio from the balanced equation to determine how many moles of CO2 could be produced:

Moles of CO2 = Moles of C = 0.155 moles

Therefore, based on the moles of each reagent, we could produce a maximum of 0.155 moles of CO2.

To determine the limiting reagent, we need to calculate the amount of CO2 that would be produced if all of the limiting reagent were consumed. We can do this by comparing the amount of CO2 that would be produced by each reagent and identifying the one that produces the smaller amount:

Using C as the limiting reagent:

Moles of CO2 produced = Moles of C = 0.155 moles

Using O2 as the limiting reagent:

Moles of CO2 produced = Moles of O2 × (1 mole CO2 / 1 mole O2) = 0.117 moles × (1 mole CO2 / 1 mole O2) = 0.117 moles

Since the amount of CO2 produced by the reaction with C is greater than the amount produced by the reaction with O2, we can conclude that O2 is the limiting reagent for this problem. Therefore, only 0.117 moles of CO2 could be produced based on the available amount of O2, and any excess C would be left over after the reaction.

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7. the pH of 0.81 M Mg(OH)₂ solution is 9.19.

8. the pH of 0.27 M pyridine solution is 9.11.

Mg(OH)₂ is a base which dissociates to produce two OH⁻ ions.

Mg(OH)₂ → Mg²⁺ + 2 OH⁻

Let the concentration of OH⁻ ions produced be x.

Therefore, the concentration of Mg²⁺ is 0.81-x

Mg(OH)₂ → Mg²⁺ + 2 OH⁻

Initial concentration (M)    0         0

Change (M)                 -x         +2x

Equilibrium Concentration  0.81-x      x     x

Using Kb for Mg(OH)₂,Kb = Kw/Ka

Kw = 1.0 × 10⁻¹⁴ at 25 °C.

For Mg(OH)₂,Kb = [Mg²⁺][OH⁻]²/Kw= (x)²/0.81 - x

Kb = 4.5 × 10⁻¹² = x²/0.81 - x

On solving the equation,x = 7.7 × 10⁻⁶M

Therefore, the concentration of OH⁻ ions = 2 × 7.7 × 10⁻⁶ = 1.54 × 10⁻⁵ M

To calculate the pH of the solution, use the formula:

pOH = - log [OH⁻]= - log 1.54 × 10⁻⁵pOH = 4.81pH = 14 - 4.81 = 9.19

Thus, the pH of 0.81 M Mg(OH)₂ solution is 9.19.

Let the concentration of OH⁻ ions produced be x.

Therefore, the concentration of C₅H₅NH⁺ is 0.27 - x.

C₅H₅N + H₂O ⇌ C₅H₅NH⁺ + OH⁻

Initial concentration (M)   0.27      0

Change (M)                -x       +x

Equilibrium Concentration  0.27-x     x

Using Kb for C₅H₅N,Kb = Kw/Ka

Kw = 1.0 × 10⁻¹⁴ at 25 °C.

For C₅H₅N,

Kb = [C₅H₅NH⁺][OH⁻]/[C₅H₅N]= (x) (x)/(0.27-x)Kb = 1.7 × 10⁻⁹

= x²/(0.27-x)

On solving the equation,

x = 1.3 × 10⁻⁵ M

Therefore, the concentration of OH⁻ ions = 1.3 × 10⁻⁵ M

To calculate the pH of the solution, use the formula:

pOH = - log [OH⁻]= - log 1.3 × 10⁻⁵pOH

= 4.89pH = 14 - 4.89 = 9.11

Thus, the pH of 0.27 M pyridine solution is 9.11.

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Answer:

2 of hydrogen (H)

Explanation: Because it says h is for hydrogen and 2

Answer:

A typical range of ATOMIC radii would be 0.050 − 0.200 nm or so.

The coefficients that will balance the equation will be 2,3,2.

For a balanced equation, the number of atoms of each element on the reactant side must be equal to those of the product side.

Thus, the balanced equation of the reaction will be:

2Fe + 3Cl2 → 2FeCl3

Consequently, the coefficients of the balanced equation are 2,3,2.

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Answer: 4. We know that one mole of any chemical compound always contains 6.022 x 10^23 molecules. Therefore, we can calculate the number of molecules of 2.25 moles of sulfur dioxide SO2 by multiplying the number of moles by the Avogadro's constant 6.022 x 10^23:    

    2.25 moles SO2 (6.022x10^23 molecules/1mole) = 1.355 x 10^24 molecules of sulfur dioxide

5. The number of moles of sodium chloride are 0.40 moles

Explanation: