What volume would 4 moles of gas occupy at stp? do this without a calculator. you may need to use longhand calculations. the question is designed to tolerate reasonable rounding?

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.

Answer:

199,927.29 L is the volume air that holds enough air to contain 55 kg of oxygen.

Explanation:

Density of oxygen = 1.31 g/L

Mass of oxygen gas,m = 55 kg = 55,000 g

Volume of the oxygen gas = V

Volume of the air = [tex]V_a[/tex]

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

[tex]1.31 g/L=\frac{55,000 g}{V}[/tex]

V = 41.984.732 L

Air is 21% oxygen by volume.That is in 100 L of air 21 L of oxygen  is present.

So, when oxygen gas is 41.984.732 L the volume of the air will be:

[tex]21\%=\frac{41.984.732 L}{V_a}\times 100[/tex]

[tex]V_a=199,927.29 L[/tex]

199,927.29 L is the volume air that holds enough air to contain 55 kg of oxygen.

The volume of the room that can hold enough air containing 55 Kg of oxygen is 199927.29 L

We'll begin by converting 55 kg of oxygen to grams (g). This can be obtained as follow:

1 Kg = 1000 g

Therefore,

55 kg = 55 × 1000

Next, we shall determine the volume of oxygen. This can be obtained as follow:

Mass of oxygen = 55000 kg

Density of oxygen = 1.31 g/L

[tex]Density = \frac{mass}{volume} \\\\1.31 = \frac{55000 }{volume}[/tex]

Cross multiply

1.31 × Volume = 55000

Divide both side by 1.31

Volume = 55000 / 1.31

Finally, we shall determine the volume of the room by calculating the volume of air. This can be obtained as follow:

Volume of oxygen = 41984.73 L

Percentage of oxygen in air = 21%

[tex]Percentage of oxygen =\frac{volume of oxygen}{Volume of air} * 100[/tex]

21% = [tex]\frac{41984.73}{Volume of air}[/tex]

[tex]0.21 = \frac{41984.73}{Volume of air }[/tex]

Cross multiply

0.21 × Volume of air = 41984.73

Divide both side by 0.21

[tex]Volume of air = \frac{41984.73}{0.21}\\\\[/tex]

Therefore, we can conclude that the volume of the room that can hold enough air containing 55 Kg of oxygen is 199927.29 L

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94.0,                 is the answer for sure

Answer:

A) Metals

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

The valence shell electrons of a molecule are shown in an extremely simplified manner by a Lewis structure.

Butane is an alkane represented by C4H10. All the bonds in this situation are single bonds, since this substance is an alkane. Two shared electrons are present in every bond. Each carbon atom requires 8 electrons around it in order to follow the octet rule.

The octet rule was fulfilled by the hydrogen bonds, eliminating the necessity for the electron dots. Straight-chained butane is the initial structure. The y-chained isobutane is the second structure.

The Lewis structures for a compound with the formula C₄H₁₀ is attached in the image below.

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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 change in energy when the electron drops to the ground state can be calculated. The wavelength of the released photon can be determined using the equation: wavelength = hc / energy.

When an electron drops from a higher energy level to the ground state, it releases energy in the form of a photon. The change in energy can be calculated by taking the difference between the initial energy and the energy of the ground state. The wavelength of the photon can be determined using the equation:

wavelength = hc / energy

where h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3.00 x 10^8 m/s), and energy is the change in energy calculated earlier.

The structure of formaldehyde (CH2O) features a carbon atom double-bonded to an oxygen atom with two lone pairs, and two hydrogen atoms single-bonded to the carbon, resulting in a trigonal planar molecular geometry with bond angles around 120°.

Formaldehyde, with the chemical formula CH2O, is a simple aldehyde commonly used as an embalming agent and preservative. In its Lewis structure, the carbon atom is double-bonded to the oxygen, which also has two lone pairs of electrons, creating a region of high electron density. This double bond contributes to the molecule's trigonal planar geometry, with bond angles close to 120°. Each of the two hydrogen atoms is single-bonded to the carbon atom, completing the tetrahedral geometry around the carbon atom. The molecular structure is depicted as:

:O:

H-C-H

The oxygen atom has two lone pairs of electrons, and there are three regions of bonding around the carbon atom, which has no lone pairs, providing an acceptable Lewis electron structure for formaldehyde.

Explanation:

When an atom gain electrons then it acquires a negative charge whereas when an atom loses electrons then it acquires a positive charge.

For example, atomic number of bromine is 35 and its electronic distribution is 2, 8, 18, 7. So, in order to gain stability bromine accepts one electrons and thus it acquires a negative charge as [tex]Br^{-}[/tex].

Therefore, the symbol for the ion formed when each given element gains electrons and attains a noble-gas electron configuration are as follows.

To determine the number of atoms in 1.70ml of mercury, we have to determine the moles of mercury.

First find the mass in grams by using the formula:

Mass = volume x density

(1.70 mL Hg) x (13.5 g/mL) = 22.95 g Hg 

Now find number of moles;

(22.95 g Hg) x (1 mol Hg/200.59 g/mol) = 0.1144 mol Hg 

For number of atoms multiply with Avogardro's number that is 6.023 x 10²³

(0.1144 mol Hg) x (6.023 x 10²³ atoms/mol) = 6.89 x 10²² atoms Hg 

So, there is 6.89 x 10²² atoms of mercury are present in 1.70ml of mercury.

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

In the d- block of the  periodic table are the majority of radioactive isotopes found.

Periodic table is a tabular arrangement of elements in the form of a table. In the periodic table, elements are arranged according to the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers.

It is called as periodic because properties repeat after regular intervals of atomic numbers . It is a tabular arrangement consisting of seven horizontal rows called periods and eighteen vertical columns called groups.

As radioactive isotopes are between atomic numbers 84 and 118 the majority of them are found in the d-block of the periodic table.

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Radioactive isotopes are largely found in heavier elements of the periodic table, specifically in the uranium, actinide, and thorium series. Elements after atomic number 83 are inherently unstable and tend to be radioactive due to additional neutrons in their nuclei enforcing instability and resulting in radioactive decay.

The radioactive isotopes are primarily found in the heavier elements of the periodic table. Notably, they occur in the uranium series, the actinide series, and the thorium series. Any elements beyond atomic number 83 (Bismuth) on the periodic table are inherently unstable and tend to be radioactive. For instance, Technetium (element 43) and Promethium (element 61), and most of the elements with atomic number 84 (Polonium) and higher consist entirely of unstable, radioactive isotopes. This instability is due to the additional neutrons they possess that tend to make their nuclei unstable and thereby undergo radioactive decay. Radioactive decay is characterized by the loss of one or more neutrons and the release of energy.

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Answer : Amongst the given choices a weak beam of ultraviolet light is considered to be the worst as compared to others.

The wavelength of UV light is from 10 nm to 400 nm, and its energy varies from 3 eV to 124 eV.

As a single beam of UV light will have much energy in each photon which is emitted. The damage done by UV rays is observed to be very harmful in human cells and may cause many diseases. UV rays can also damage the eyes as more than 99% of UV radiation is absorbed by the front of the eyes. It can also lead to ultimate blindness. Therefore, it is considered to be much worse than an intense beam of infrared light.

To prepare a 6L solution of 170 mM NaCl, you need to calculate the moles required and use the molecular weight of NaCl to convert to grams. Multiply the molarity by the volume to get moles, and then multiply by the molecular weight to get 59.6088 grams.

To prepare a 170 mM NaCl solution, you will need to calculate the number of grams of NaCl required. First, you need to understand the concentration units. Here, 170 mM (millimolar) indicates 170 millimoles of NaCl are needed per liter of solution. To find out how many millimoles you need for 6 liters, you multiply 170 mmol/L by 6 L, yielding 1020 mmol, or 1.020 mol since 1000 mmol is equivalent to 1 mol.

Next, you use the molecular weight of NaCl, which is 58.44 g/mol, to convert moles to grams. The calculation is:

Therefore, you would need to weigh out 59.6088 grams of NaCl and dissolve it in enough water to make the total volume up to 6 liters.

Entropy change of vaporization is simply the ratio of enthalpy change and the temperature in Kelvin.

Temperature = 64 + 273.15 = 337.15 K

Hence,

δsvap = (32.21 kJ / mole) / 337.15 K

δsvap = 0.0955 kJ / mole K = 95.5 J / mole K

Answer:

[tex]\Delta S_{vap}=0.096\frac{kJ}{mol*K} =96\frac{J}{mol*K}[/tex]

Explanation:

Hello,

In this case, the entropy of vaporization (conversion from liquid to gas) is mathematically defined in terms of enthalpy and the boiling temperature in K as shown below:

[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T_b}[/tex]

Thus, for the given data we obtain:

[tex]\Delta S_{vap}=\frac{32.21kJ/mol}{(64+273.15)K} \\\\\Delta S_{vap}=0.096\frac{kJ}{mol*K} =96\frac{J}{mol*K}[/tex]

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Taking into account the definition of percentage in volume, 0.55 mL of active ingredient are in the bottle.

In first place, every solution consists of two classes of components: the solvent and the solute.

The solvent of a solution is the component that is capable of dissolving others and is usually the component present in greater quantity.

The solute is the component that is dissolved in the solvent and is usually present in smaller quantities. A solution can contain more than one solute.

On the other side, the concentration of a solution is the amount of solute in a given amount of solution.

The percentage in volume (% v / v) is a measure of the concentration that indicates the volume of solute per 100 units of volume of the solution and can be calculated mathematically by means of the expression:

[tex]percentage in volume=\frac{solute volume}{solution volume}x100[/tex]

In this case:

Replacing in the expresion for percentage in volume:

[tex]5=\frac{solute volume}{11 mL}x100[/tex]

Solving:

5× 11 mL= solute volume×100

55 mL=solute volume×100

55 mL÷100= solute volume

0.55 mL= solute volume

Finally, 0.55 mL of active ingredient are in the bottle.

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To find the amount of active ingredient in the cough syrup, you multiply the percentage concentration (5.00%) by the total volume of the syrup (11.0 mL) and divide by 100, yielding 0.55 mL of active ingredient.

To calculate the amount of active ingredient in a cough syrup solution, we use the percentage concentration by volume, which in this case is 5.00% v/v. This means that for every 100 mL of the total solution, there are 5 mL of the active ingredient. Given that the total volume of the cough syrup bottle is 11.0 mL, we can determine the volume of active ingredient using the following calculation:

Volume of active ingredient = Percentage concentration × Total volume / 100

Volume of active ingredient = 5.00% × 11.0 mL / 100 = 0.55 mL

Therefore, there are 0.55 milliliters of active ingredient in the 11.0 mL bottle of cough syrup.

Answer:

Mg (Magnesium)

Explanation:

Pearson Chemistry Question (Ions and the Periodic Table)

The mass of helium that it would take to fill the balloon is 1.1 grams.

Relation between the mass of any substance and their density will be represented as:

Density = Mass / Volume

Given that,

Density of helium gas = 1.79 × 10⁻⁴ g/mL

Volume of balloon in which gas is present = 6.3L = 6300 mL

On putting all these values, we get

Mass = (1.79 × 10⁻⁴ g/mL)(6300mL) = 1.12 grams

Hence required mass of helium is 1.1 g.

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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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