The boiling point of a liquid is 64 °c and the enthalpy change for the conversion of this liquid to the gas is 32.21 kj/mole. what is the entropy change for vaporization, δsvap?

Answer:

Species

Explanation:

The acronym, DKPCOFGS; Gives you an explanation of the subgroups highest - lowest.

Did

King

Phil

Come

Over

For

Good

Spagettii

Spagettii is species in the acronym

Enjoy your test :D

-Snooky

Answer : The correct option is, 0.961 g/ml

Solution : Given,

Mass of sample = 35.4 g

Volume of sample = 36.82 ml

Formula used :

[tex]\text{Density of sample}=\frac{\text{Mass of sample}}{\text{Volume of sample}}[/tex]

Now put all the given values in this formula, we get the density of the sample.

[tex]\text{Density of sample}=\frac{35.4g}{36.82ml}=0.961g/ml[/tex]

Therefore, the density of the sample is, 0.961 g/ml

Answer:

The correct option is, 0.961 g/ml

Explanation:

The Complete balanced equation for the reaction of Aqueous iron (iii) nitrate, Fe(NO3)3 and aqueous lithium hydroxide, LiOH, is given by;

Fe(NO₃)₃(aq) + 3LiOH(aq) → 3LiNO₃(aq) + Fe(OH)₃(s)

Keywords: Chemical equations, balancing of chemical equations

Level: high school  

Subject: Chemistry  

Topic: Chemical equations

Sub-topic: Balancing chemical equations  

The balanced chemical equation between aqueous iron(III)nitrate and aqueous lithium hydroxide is Fe(NO₃)₃[tex]_(aq)[/tex] + 3 LiOH [tex]_(aq)[/tex][tex]\rightarrow[/tex]  3 LiNO₂ [tex]_(aq)[/tex] +Fe(OH)₃[tex]_(s)[/tex].

Chemical equation is a symbolic representation of a chemical reaction which is written in the form of symbols and chemical formulas.The reactants are present on the left hand side while the products are present on the right hand side.

A plus sign is present between reactants and products if they are more than one in any case and an arrow is present pointing towards the product side which indicates the direction of the reaction .There are coefficients present next to the chemical symbols and formulas .

The first chemical equation was put forth by Jean Beguin in 1615.By making use of chemical equations the direction of reaction ,state of reactants and products can be stated. In the chemical equations even the temperature to be maintained and catalyst can be mentioned.

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The term that name the water in a chemical reaction that water  decomposes is.

  reactant ( answer D)

Explanation

Reactant is a substance  that take part in and undergoes change  during a reaction.   In other  words reactant  are consumed to form product.

water  is   reactant when it decomposes because  it is consumed  to form  hydrogen and oxygen.

The decomposition  of water  is as below  equation,

2H2O→ 2 H2 + O2

Answer:

A) Metals

Explanation: They mainly explain metals in "periodic table" well it's in it too.

Answer:

Option c, They are not in a fixed position.

Explanation:

As per Bohr's model of atomic atoms,

Electrons revolve around around the nucleus in a fixed path or energy level. Electrons revolving in these energy levels do not radiate energy.

Therefore, electrons are not present in a fixed position. They present outside the nucleus.

Mass of protons and neutrons are same whereas electrons are smaller.

Among the given options, options c best describe the difference between electrons from protons and neutrons.

The molar mass of element X is determined in g/mol by taking into account the mole ratio in the X₂O₃ compound. The mass of X (1.0000 g) is first divided by the moles of X, which is two-thirds of the moles of Oxygen, calculated by dividing the mass of Oxygen in X₂O₃ by its molar mass.

To calculate the molar mass of element X, the mole ratio in the X₂O₃compound needs to be considered. This mole ratio suggests that for every 1 mole of X, 1.5 moles of Oxygen is used given that from the formula of X₂O₃hat would be 2 moles of X for every 3 moles of oxygen (2:3).

The mass of Oxygen in the compound can be calculated by subtracting the mass of X from the total compound weight. Given that 1.1xxx grams of X₂O₃ was obtained from 1.0000 grams of X, the mass of Oxygen in the compound would be 1.1xxx g - 1.0000 g giving a mass of Oxygen in the compound as 0.1xxx grams.

We know that the molar mass of Oxygen (O) is 16.00 g/mol. So, the number of moles of Oxygen in the compound can be calculated as (0.1xxx g)/(16.00 g/mol) which gives the number of moles for Oxygen. Given the mole ratio from the X₂O₃ formula, the number of moles of X will be two-thirds of the moles of Oxygen.

Finally, the molar mass of X can be calculated by dividing the mass of X (1.0000 g) by the number of moles of X calculated. This will give the answer for the molar mass of X in g/mol.

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Answer:

3

Explanation:

Hello,

In this case, as nitrogen is not bearing any formal charge, we apply its mathematical definition to find such oxidation state:

Formal charge= # of valence electrons - # of lone-pair electrons - 1/2 # of bond pair electrons

Since no formal charge is born, no lone-pair electrons are formed and 6 bond pair electrons are always present for the nitrogen to complete the octate, the valence turns out into:

# of valence electrons = 0 + 0 + 1/2 * 6

# of valence electrons = 3

Best regards.

Answer:

C

Explanation:

94.0,                 is the answer for sure

The temperature change (ΔT) of the water after sitting in the copper pipe can be found using the formula for heat transfer (q = mcΔT), considering that the water releases 2284 J of energy. Thus, ΔT = 2284 / (190.0 * 4.184).

The temperature change of water in the copper pipe can be determined by using the formula for heat transfer (q = mcΔT), where q is heat in joules, m is the mass in grams, c is the specific heat capacity, and ΔT is the temperature change in degrees Celsius. Here, the initial temperature of the water is not given, but we know that the water releases 2284 J of energy. Plugging in the mass (190.0 g), the specific heat of water (4.184 J/g °C), and the heat (2284 J) into the formula, we get the equation

2284 = 190 * 4.184 * ΔT.

Solving for ΔT, we get ΔT = 2284 / (190.0 * 4.184), which should provide the temperature change of the water after sitting in the copper pipe.

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The given equation is a double-replacement reaction where solid copper reacts with an aqueous solution of silver nitrate to produce a solution of copper(II) nitrate and solid silver.

The given equation is a chemical equation, specifically a double-replacement reaction. In this type of reaction, the cations are swapped between two compounds. In this case, solid copper (Cu) reacts with an aqueous solution of silver nitrate (AgNO3) to produce a solution of copper(II) nitrate (Cu(NO3)2) and solid silver (Ag). The balanced equation for this reaction is:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

297440787 tons of carbon dioxide is produced by the plant after a combustion reaction taking place.

Reaction which takes place in presence of air is combustion reaction.When carbon burns in air it produces carbon dioxide .

C+O₂[tex]\rightarrow[/tex]CO₂

80.7% mass of carbon means 80.7/100×12=9.684 g

12 g C is present in 44 g carbon dioxide

∴9.684 g of C  is present in 9.684×44/12=35.508 g

As 9.6854 g carbon/coal produces 35.508 g carbon dioxide ,

∴35.508×8376726=297440787 tonnes of carbon dioxide.

Combustion reactions are reactions which take place at high temperatures and are exothermic and produce products which are oxidized.In combustion reactions ,chemical equilibrium is difficult to achieve. There are 3 types of combustion:

1) rapid combustion

2) explosive combustion

3) spontaneous combustion

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The value of Henry’s law constant for [tex]{\text{C}}{{\text{O}}_{\text{2}}}[/tex] at [tex]20{\text{ }}^\circ {\text{C}}[/tex] is [tex]\boxed{3.70 \times {{10}^{ - 2}}{\text{ L}} \cdot {\text{atm}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}}[/tex].

Further explanation:

Solubility

It is that property of substance by virtue of which it becomes able to dissolve in other substances. It is measured in terms of the maximum amount of solute that can be dissolved in the given amount of solvent.

Henry’s Law

According to this law, solubility of gas dissolved in the liquid is directly related to the partial pressure of gas. High partial pressure means high solubility and vice-versa.

Mathematically,

[tex]{{\text{S}}_{{\text{gas}}}} \propto {{\text{P}}_{{\text{gas}}}}[/tex]  …… (1)                                                                                    

To remove the proportionality constant in equation (1), constant known as Henry’s constant is incorporated and equation (1) modifies to,

[tex]{{\text{S}}_{{\text{gas}}}} = {{\text{k}}_{\text{H}}}{\mathbf{ \times }}\;{{\text{P}}_{{\text{gas}}}}[/tex]         …… (2)                                                                      

Here,

[tex]{{\text{S}}_{{\text{gas}}}}[/tex] is the solubility of gas.

[tex]{{\text{k}}_{\text{H}}}[/tex] is Henry’s constant.

[tex]{{\text{P}}_{{\text{gas}}}}[/tex] is the pressure of gas.

Equation (2) can be rearranged in order to calculate Henry’s constant [tex]\left( {{{\text{k}}_{\text{H}}}} \right)[/tex] and equation (2) becomes,

[tex]{{\text{k}}_{\text{H}}} = \dfrac{{{{\text{S}}_{{\text{gas}}}}}}{{{{\text{P}}_{{\text{gas}}}}}}\;[/tex]    …… (3)                                                                                  

At [tex]20{\text{ }}^\circ {\text{C}}[/tex] , the value of Henry’s law constant for [tex]{\text{C}}{{\text{O}}_{\text{2}}}[/tex] up to three significant figures is [tex]3.70 \times {10^{ - 2}}{\text{ L}} \cdot {\text{atm}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}[/tex].

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Solutions

Keywords: solubility, gas, Henry’s law, partial pressure, solubility, dissolve, CO2, three significant figures, [tex]3.70*10^-2 L atm/mol,[/tex] high partial pressure, high solubility.

The Henry's law constant for CO₂ at 20°C is 3.91 × 10⁻² M/atm. The pressure required to achieve a CO₂ concentration of 6.90 × 10⁻² M at 20°C is approximately 23.4 atm.

Part A - To answer the question, we need to understand that Henry's Law states that the concentration of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. The law is represented by the equation:

C = kH * P

where C is the concentration of the gas, kH is the Henry's law constant, and P is the partial pressure of the gas.

Given that the Henry's law constant for CO₂ in water at 25°C is 3.4 × 10⁻² M/atm, we can use interpolated values to find the constant at 20°C. Experimentally, at 20°C, the Henry's law constant for CO₂ is found to be 3.91 × 10⁻² M/atm.

Therefore, the Henry's law constant for CO₂ at 20°C is 3.91 × 10⁻² M/atm.

Part B: Pressure Required for CO₂ Concentration

To find the pressure required to achieve a CO₂ concentration of 6.90 × 10⁻² M at 20°C, we can use Henry's Law:

pCO₂ = KH × [CO₂]

where pCO₂ is the partial pressure of CO₂, KH is the Henry's Law constant, and [CO₂] is the concentration of CO₂ in moles per liter.

Rearranging the equation to solve for pCO2:

pCO₂ = KH × [CO₂] = 3.39 × 10² L·atm/mol × 6.90 × 10⁻² mol/L ≈ 23.4 atm

So, the pressure required to achieve a CO₂ concentration of 6.90 × 10⁻² M at 20°C is approximately 23.4 atm.

Complete Question - Part A What is the Henry's law constant for CO₂ at 20°C? Express your answer to three significant figures and include the appropriate units. Part B What pressure is required to achieve CO₂ concentration of 6.90x10⁻² M at 20°C? Express your answer to three significant figures and include the appropriate units.

The room is approximately 0.0212 hectometers long.

To convert the length of the room from meters to hectometers, we need to know that 1 hectometer (hm) is equal to 100 meters (m). Therefore, to convert meters to hectometers, we divide the length in meters by 100.

Given that the room is 212 meters long, we perform the following calculation:

[tex]\[ \text{Length in hectometers} = \frac{\text{Length in meters}}{100} \][/tex]

[tex]\[ \text{Length in hectometers} = \frac{212 \text{ m}}{100} \][/tex]

[tex]\[ \text{Length in hectometers} = 2.12 \text{ hm} \][/tex]

However, to express this length in a more standard form, we can write 2.12 hectometers as 212/10000 hectometers, which simplifies to 212 hm divided by 100, or 2.12 hm. To be more precise, we can express this as a decimal:

[tex]\[ \text{Length in hectometers} = 0.0212 \text{ hm} \][/tex]

Thus, the room is approximately 0.0212 hectometers long.

The combination of a single polar group and dominant non-polar regions prevents effective hydrogen bonding with water, leading to the immiscibility of cinnamaldehyde in water.

**Structure:**

       O

       |

   C=CH-CH=CH-CH2-CH2-CH3

       |

       H

**Analysis:**

a) **Non-polar regions:**

* **Double bonds:** The double bonds (C=C) have evenly distributed electron density, making them non-polar. Circle both double bonds.

* **Alkyl chains:** The aliphatic chains (CH2-CH2-CH3) also have similar electron densities across their C-H bonds, making them non-polar. Circle both alkyl chains.

b) **Polar region:**

* **Carbonyl group (C=O):** The carbonyl group has a significant dipole moment due to the electronegativity difference between oxygen and carbon. This creates a partial positive charge on the carbon and a partial negative charge on the oxygen. Circle the carbonyl group (C=O).

c) **Water's preferred intermolecular force:**

* **Hydrogen bonding:** Water molecules are highly polar due to their lone pairs and O-H bonds, enabling strong hydrogen bonding with other polar molecules.

d) **Cinnamaldehyde's immiscibility with water:**

* Although the carbonyl group is polar, its interaction with water is significantly weakened by the extensive non-polar regions (double bonds and alkyl chains) dominating the molecule.

* These non-polar regions prefer London dispersion forces (weak attractive forces between all molecules due to temporary fluctuations in electron density) over the stronger hydrogen bonding with water.

* The weak London dispersion forces cannot overcome the stronger intermolecular hydrogen bonding network of water, resulting in immiscibility.

Therefore, the combination of a single polar group and dominant non-polar regions prevents effective hydrogen bonding with water, leading to the immiscibility of cinnamaldehyde in water.

The probable question may be:

1. Re-draw   the   chemical   structure   of   cinnamaldehyde and use it to answer the following solubility-based questions:

a) Circle and label all regions of the molecule that exhibit non-polar behavior.

b) Circle and label the region of the molecule that exhibits polar behavior.

c) Water   prefers   to   interact   with   solute   molecules   with   what   type   of intermolecular   force   (choose   one   answer):   hydrogen   bonding,   dipole- induced dipole, or induced dipole-induced dipole (London dispersion)?

d) Based on the answers to parts (a)-(c), explain why cinnamaldehyde is NOT miscible with water. Incorporate relevant  intermolecular forces in your answer.