Why is HCI a strong acid and HCIO a weak acid?

The equilibrium constant for the reaction given that the equilibrium concentration of [PH₃] = 0.250 M, [H₂] = 0.580 M, and [P₄] = 0.750 M is 7.3

From the question given above, the following data were obtained:

The equilibrium constant for the reaction can be obtained as shown below:

Equilibrium constant = [Product]ᵐ / [Reactant]ⁿ

Where

Equilibrium constant = [H₂]⁶[P₄] / [PH₃]⁴

Equilibrium constant = [(0.580)⁶ × 0.750] / (0.250)⁴

Equilibrium constant = 7.3

Thus, the equilibrium constant for the reaction is 7.3

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Answer:

.0014 M/s ( (mol*L^-1 / s) )

Explanation:

Since the rate law of a zero order reaction is Rate = k[A]^0, the rate is .0014 * (1.65)^0 = .0014

Answer:it is wrong answer

Explanation:estro man

Scientific theories explain why a phenomenon occurs and a scientific law explains what occurs. The correct option is A.

A scientific theory is an explanation for a phenomenon in the natural world or the universe that has undergone extensive testing and verification using approved procedures for observation, measurement, and result evaluation.

A scientific law often describes an observed phenomenon. It doesn't explain the phenomenon's existence or its origins.

A scientific theory is the explanation for a phenomenon. It is a myth that with enough study, theories can become laws.

Thus, the correct option is A.

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Answer:  Protons

Explanation:

a. 4- bromocyclohexanol is an acidic proton

b. Benzenesulfonic acid deactivates the entire process

c. (2S, 3R)-2,3-Dibromobutane are racemic mixtures

d.  Williamson synthesis cannot be used with tertiary alkyl halides

e. Nucleophiles are not involved in the rate determining step

a. Grignard reagents cannot be made if acidic functional groups are also present in the halogen compound and it is also destroyed by reaction with acidic hydrogen atoms of water, alcohols, phenols, or carboxylic acid groups.

4-bromocyclohexanol is an acidic proton compound

b. Benzenesulfonic acid does not undergo Friedel-Crafts alkylation because the carboxylic group causes deactivation, carboxylic group and the Lewis acid catalyst are also bonded in the process.

c. (2S, 3R)-2,3-Dibromobutane has a specific rotation, [a]o, of 0° are racemic mixtures having equal and opposite specific rotations and a racemic mixture has a specific rotation of zero (0°)

d. This is because the Williamson synthesis cannot be used with tertiary alkyl halides which in this case is t-butyl bromide because they undergo elimination reactions instead of participating in SN2 reactions and form alkenes instead of ethers.

e. Doubling the concentration of a nucleophile has no effect on the rate of a substitution reaction because the nucleophile is not involved in the rate-determining step of the SN-1 reaction.

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Answer:

A logic bomb and a time bomb are both types of malicious software or code that are designed to cause harm to a computer system or network. Here's a brief explanation of each:

Logic Bomb:

A logic bomb is a piece of code or software that is intentionally inserted into a system to execute a malicious action when specific conditions are met. It remains dormant until triggered by a predefined event or circumstance, such as a specific date, time, or user action. Once triggered, the logic bomb may perform various harmful actions, such as deleting files, corrupting data, or disrupting system functionality. The purpose of a logic bomb is often to cause damage or to gain unauthorized access to a system.

Time Bomb:

A time bomb is similar to a logic bomb, but it is specifically designed to activate or execute its malicious payload at a certain date or time. It is usually programmed to remain undetected until the predetermined time arrives. The time bomb can be set to trigger after a specific time period or on a particular date, at which point it may carry out destructive actions. Time bombs can be used by attackers to target specific events or to coordinate an attack to occur simultaneously across multiple systems.

Both logic bombs and time bombs are considered forms of malicious code or malware and are used with malicious intent to disrupt, damage, or compromise computer systems or networks. They can be extremely harmful, and it is important to have strong security measures, such as antivirus software and regular system updates, to protect against such threats.

Explanation:

There are approximately 1.62 x 10^25 water molecules in the 499.8 mL bottle of water.

To determine the number of water molecules in the given volume of water, we need to use the relationship between mass, volume, and molar mass of water.

First, we need to find the mass of water in the bottle:

Mass = Density * Volume

Mass = 0.967 g/mL * 499.8 mL = 483.9 g

Next, we need to convert the mass of water to moles using the molar mass of water. The molar mass of water (H2O) is approximately 18.015 g/mol.

Moles = Mass / Molar mass

Moles = 483.9 g / 18.015 g/mol = 26.88 mol

Finally, we can calculate the number of water molecules using Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol.

Number of molecules = Moles * Avogadro's number

Number of molecules = 26.88 mol * (6.022 x 10^23 molecules/mol) = 1.62 x 10^25 molecules

Therefore, there are approximately 1.62 x 10^25 water molecules in the 499.8 mL bottle of water.

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the pressure required for 1.35 g of oxygen to dissolve in 1.5 L of water is 3.56 atm.

The first step in solving this problem is to identify the relevant equation.

Henry's law is the formula that relates the pressure of a gas above a liquid to the concentration of the gas that dissolves in the liquid.

In mathematical terms, Henry's law can be expressed as follows:P = kH * Cwhere P is the pressure of the gas, kH is Henry's law constant, and C is the concentration of the gas in the liquid.

To solve the problem, we need to first determine the value of kH using the given data.

kH can be calculated using the following formula:kH = P / CSubstituting the values given in the problem into this formula, we get:kH = 1.65 atm / (0.654 g / 1.5 L) = 3.97 atm/(g/L).

Now that we have the value of kH, we can use Henry's law to calculate the pressure required for 1.35 g of oxygen to dissolve in 1.5 L of water.

To do this, we simply rearrange the formula to solve for P:P = kH * CSubstituting the values of kH and C into this formula, we get:P = 3.97 atm/(g/L) * (1.35 g / 1.5 L) = 3.56 atm

Therefore, the pressure required for 1.35 g of oxygen to dissolve in 1.5 L of water is 3.56 atm.

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Answer:

C and B

Explanation:

1) In order for nitrogen to make a stable electronic configuration, it would have to gain three electrons.

Nitrogen is atomic number 7.  Hence, the electronic configuration would be:

           1s2, 2s2, 2p3

The p orbital is partially filled with 3 unpaired electrons. In order to be stable, the 3 electrons in the p orbital have to be paired. Hence, only by gaining 3 electrons can nitrogen obtain a stable electronic configuration.

2) The quantity the determines how two atoms would bond is the difference in the number of valence electrons in the atoms.

For example, an atom that is lacking 2 electrons will readily form a bond with another atom that has two free electrons.

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Answer:

Fine chemicals are used as starting materials for specialty chemicals, particularly pharmaceuticals, biopharmaceuticals and agrochemicals. ... The term "fine chemicals" is used in distinction to "heavy chemicals", which are produced and handled in large lots and are often in a crude state.

Explanation:

Answer:

The answer is in the attachment below

Explanation:

We have that 70kg of Copper Cu is having a volume of \(V=7.76442liters\)

1 cmCu

=9 g Cu

9.5 x 10^21 atoms Cu

=1 g Cu

1 kg  =1000 g

1 cm

=1 ml

1

=1000 cm3

70.0 kg x  X  _X  II

70.0 kg•L

Generally We First determine the No of Atoms in 70kg of cu

No of Atoms of c\(u=\frac{weight}{atomic mass of cu}*n\)

Where

n=Avogadro's Constant

\(n= 6.02214076 x 10^{23}\)

Therefore

No of Atoms of cu\(=\frac{70*10^3}{63.5}*6.02214076 * 10^{23}\)

No of Atoms of cu=6.63858037x10^{26} atoms of Cu

Hence

We determine the the grams of copper derived from \(6.63858037x10^{26}\)atoms of Cu

No of grams of cu\(=\frac{6.63858037x10^{26}}{9.5 x 10^{21}}*1\)

No of grams of cu=69879.79337 g of Cu

Generally

No of grams of Cu=liters

Where

9 g of Cu= 1cm^3 Cu

Therefore

Volume of cu

\(V=(69879.79337)*\frac{1}{9}*\frac{1}{1000}\)

\(V=7.76442liters\)

In conclusion

70kg of Copper Cu is having a volume of \(V=7.76442liters\)

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We are to drag the given conversion units onto the boxes in the equation in order to make conversions in the boxes that are left empty.

The dimensional analysis for the conversion ratio is given as follows

1cm^3 Cu = 9g Cu        9.5*10^21  atoms of Cu = 1 g Cu

1 kg     =  1000 g                                       1 cm^3 = 1 mL

1 L      =  1000 cm^3

 Here is the box equation

        box         box        box

70 g    ×-----  ×  ------   × -----  =  

        box         box        box

From the given information, the volume of the Cu sample in Liters is:

= 7.765  liters

 From the parameters given:

The first thing we are to do is to calculate the number of atoms of Cu in 70 kg by using Avogadro's constant.

We know that the number of atoms of Cu is:

\(\mathbf{= \dfrac{mass \ weight \ of \ Cu}{atomic \ mass \ of \ cu}\times avogadro's \ constant}\)

where;

the number of Cu atoms is:

\(\mathbf{= \dfrac{70*10^3}{63.5}\times 6.023 * 10^{23}}\)

\(\mathbf{= 6.6395 \times 10^{26} \ atoms \ of \ Cu}\)

However, from the given conversion ratios:

If \(\mathbf{9.5\times10^{2}}\)  atoms of Cu = 1 gram of Cu  

∴

atoms of Cu will be:

\(\mathbf{= \dfrac{6.6395 \times 10^{26} \ atoms \ of \ Cu \times 1 \ gram \ of \ Cu}{9.5*10^{21} \ atoms \ of \ Cu }}\)

\(\mathbf{= 0.69889 *10^{5} \ grams \ of \ Cu}\)

Since:

∴

The correct setup of the sample in the box can be computed as follows:

\(\mathbf{= 0.69889 *10^{5} \ gram \ of \ Cu \ \times \dfrac{1 cm^3 \ of \ Cu }{9 \ g \ of \ Cu }\times \dfrac{1 \ Liter}{1000 \ cm^3}}\)

= 7.765  liters

Therefore, we can conclude that the Volume of Cu sample in liters is:

= 7.765 liters

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Answer:So, about 0.093 g of the isotope will remain after 19.8 days.

Explanation:

The first step is to find the number of half-lives that have passed during 19.8 days:

Number of half-lives = time elapsed / half-life

Number of half-lives = 19.8 days / 3.8 days per half-life

Number of half-lives ≈ 5.21

This means that the initial amount of the isotope has been halved 5.21 times. The remaining fraction of the original amount can be calculated using the following formula:

Remaining fraction = (1/2)^(number of half-lives)

Substituting the values, we get:

Remaining fraction = (1/2)^5.21

Remaining fraction ≈ 0.031

Therefore, the amount of the isotope remaining after 19.8 days is:

Remaining amount = Remaining fraction x Initial amount

Remaining amount = 0.031 x 3.00 g

Remaining amount ≈ 0.093 g

So, about 0.093 g of the isotope will remain after 19.8 days.

The isotope in grams will remain after 19.8 days would be 0.081 grams.

The formula to calculate the left mass of a radioactive element can be deduced as -

\( \qquad\star\longrightarrow \underline{\boxed{\sf{m =m_{o} \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{t}{T½}} }}} \\\)

Where-

According to the given specific parameters -

Now that we have all the required values, so we can plug them into the formula and solve for the left mass of a radioactive element-

\( \qquad \longrightarrow \sf \underline{m =m_{o} \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{t}{T½} }} \\\)

\( \qquad\longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{19.8}{3.8} } \\\)

\(\qquad \longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{\cancel{19.8}}{\cancel{3.8}} } \\\)

\( \qquad\longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ 5.21052..... } \\\)

\( \qquad\longrightarrow \sf m =3 \times 0.02700... \\\)

\( \qquad\longrightarrow \sf m =0.081020....\;g \\\)

\( \qquad\longrightarrow \sf \underline{m =\boxed{\sf{0.081\;g}}} \\\)

The grams of AgCl are generated when 85 g of BaCl₂ reacts is 10.86 g.

The balanced equation is given as :

BaCl₂    +    2AgNO₃   ----->   2AgCl   +   Ba(NO₃)₂

the mass of the BaCl₂ = 85 g

molar mass of BaCl₂  = 208.23 g/mol

moles of BaCl₂  = mass / molar mass

                          = 85 / 208.23

                          = 0.408 mol

1 mole of BaCl₂  produce 2 moles of AgCl

0.408 moles of BaCl₂  = 2 × 0.408

                                      = 0.816 mol of AgCl

moles of AgCl = 0.816 mol

mass of AgCl = moles × molar mass

                      = 0.816 g × 143.32 g/mol

                      = 10.86 g

Thus,  The grams of AgCl are generated when 85 g of BaCl₂ reacts is 10.86 g.

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The relationship between the rate of diffusion r, and d is r ∝ 1/√d.

The relationship between the rate of diffusion (r) and the density of a gas (d) can be explained using Graham's law of diffusion. According to this law, the rate of diffusion of a gas is inversely proportional to the square root of its density. Mathematically, it can be expressed as:

r ∝ 1/√d

where the symbol '∝' represents 'proportional to'. The constant of proportionality (k) can be introduced to this equation as:

r = k/√d

This equation shows that as the density of a gas increases, its rate of diffusion decreases. This is because denser gases have more molecules per unit volume and thus, they experience greater intermolecular collisions that hinder their movement. Therefore, it requires more energy for them to diffuse through a medium compared to less dense gases.

The relationship between the rate of diffusion and density is particularly important in understanding the behavior of gases in different environments. For instance, in a gas chromatography column, the rate of diffusion of a gas determines how quickly it moves through the column and separates from other components. Similarly, in the Earth's atmosphere, the rate of diffusion of greenhouse gases such as carbon dioxide affects their concentration and hence, their impact on climate change.

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The choice of solution has a concentration of 1.144 mol/kg molality.

Molarity corresponds to the moles of solvent divided by the amount of solution in litres, whereas molality is equal with the moles of solvent divided by the quantity of solvent in kilogrammes.

Since 1 mole of solute is present in 1 litre for the solution, which contains both the solute and the solvent, 1 molar aqueous solutions are more concentrated than one decays aqueous solutions.

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NaOH has ais a liter value for titration molarity of 1 mole.

Molarity-

The amount of a substance in a specific volume of solution is known as its molarity (M). The number of moles of a solute per liter of a solution is known as molarity. The molar concentration of a solution is another term for molarity.

If you know how many grams of NaOH is dissolved in a known amount of solvent, you can use the formula below to calculate its molarity.

The molarity of NaOH is determined, for instance, when 10 gm of NaOH is dissolved in 250 ml of water.

Molarity is calculated as (Weight of NaOH taken*1000)/(250*NaOH's molecular weight).

Molarity equals (10*1000)/(250*40).

NaOH has a molarity of 1 mole.

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The full question-

How do you calculate the molarity of NaOH?

Answer:

We will need two potassium ions to balance the charge on the sulfate ion, so the proper chemical formula is K 2SO 4.

Explanation:

hope this helps

The mass of silicon, Si produced from the reaction is 329.41 g

SiCl₄ + 2Mg —> 2MgCl₂ + Si

Molar mass of SiCl₄ = 28 + (35.5×4) = 170 g/mol

Mass of SiCl₄ from the balanced equation = 1 × 170 = 170 g

Molar mass of Si = 28 g/mol

Mass of Si from the balanced equation = 1 × 28 = 28 g

From the balanced equation above,

170 g of SiCl₄ reacted to produce 28 g of Si.

From the balanced equation above,

170 g of SiCl₄ reacted to produce 28 g of Si.

Therefore,

2 Kg (i.e 2000 g) of SiCl₄ will react to produce = (2000 × 28) / 170 = 329.41 g of Si

Thus, 329.41 g of Si were obtained from the reaction

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The ideal gas law can be reorganized to arrive at: assuming that the volume and molecular weight of the gas remain constant:

P/T = constant indicates that the gas's pressure must rise with its temperature in order to maintain a constant ratio of pressure to temperature.

Consequently, if we heat the gas while maintaining its volume and molecular weight, its pressure will rise.

In other words, it is a measure of how quickly the particles of a substance are moving. Temperature is a fundamental physics concept that can be measured in Kelvin (K), Fahrenheit (°F), and Celsius (°C) scales, among others.

On the Celsius scale, water boils at 100°C and freezes at 0°C at normal atmospheric pressure. On the Fahrenheit scale, water boils at 212°F and freezes at 32°F at normal atmospheric pressure. On the Kelvin scale, 0 K is the theoretical minimum temperature at which all particles have no kinetic energy.

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Answer:

100 rbed  KJ |0| +2k

Explanation:

Answer:

C7H8N4O. The molecular formula C7H8N4O2 (molar mass: 180.16 g/mol) may refer to: Paraxanthine. Theobromine.

Approximately 766.08 grams of oxygen are required to completely react with 240g of C₂H₆.

To determine the amount of oxygen required to completely react with 240g of C₂H₆ (ethane), we need to set up a balanced chemical equation for the combustion of ethane.

The balanced equation for the combustion of ethane is as follows:

C₂H₆ + O₂ → CO₂ + H₂O

From the balanced equation, we can see that the stoichiometric ratio between C₂H₆ and O₂ is 1:3. This means that for every one mole of C₂H₆, three moles of O₂ are required for complete combustion.

To calculate the amount of oxygen required, we need to convert the given mass of C₂H₆ to moles using its molar mass, and then use the stoichiometric ratio to determine the moles of O₂ required. Finally, we can convert the moles of O₂ back to grams using the molar mass of oxygen.

The molar mass of C₂H₆ is calculated as follows:

(2 x atomic mass of carbon) + (6 x atomic mass of hydrogen)

(2 x 12.01 g/mol) + (6 x 1.01 g/mol) = 30.07 g/mol

Now, we can proceed with the calculation:

Calculate the moles of C₂H₆:

moles of C₂H₆ = mass of C₂H₆ / molar mass of C₂H₆

moles of C₂H₆ = 240 g / 30.07 g/mol ≈ 7.98 mol

Determine the moles of O₂ using the stoichiometric ratio:

moles of O₂ = moles of C₂H₆ x (3 moles of O₂ / 1 mole of C₂H₆)

moles of O₂ = 7.98 mol x 3 ≈ 23.94 mol

Convert moles of O₂ to grams:

mass of O₂ = moles of O₂ x molar mass of O₂

mass of O₂ = 23.94 mol x 32.00 g/mol = 766.08 g

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The dicotyledons, also known as dicots, are one of the two groups into which all the flowering plants or angiosperms were formerly divided. The name refers to one of the typical characteristics of the group, namely that the seed has two embryonic leaves or cotyledons.

The reaction that would most likely require the use of an inert electrode is a reaction involving highly reactive substances or species that can react with or be oxidized by the electrode material itself.

One such reaction that often requires the use of an inert electrode is the electrolysis of aqueous solutions containing halide ions (Cl-, Br-, I-). When halide ions are electrolyzed, they can undergo oxidation at the anode, resulting in the formation of halogen gas (Cl2, Br2, I2). However, these halogens are highly reactive and can react with many common electrode materials, such as metals, leading to unwanted side reactions.

To prevent these undesired reactions, an inert electrode, usually made of materials such as platinum, gold, or graphite, is employed. Inert electrodes do not react with the halogen gases formed during electrolysis, allowing the desired reaction to take place without interference from the electrode material itself.

For example, in the electrolysis of a sodium chloride (NaCl) solution, chloride ions (Cl-) can be oxidized at the anode to form chlorine gas (Cl2). To ensure that the chlorine gas is produced without any reactions involving the anode material, an inert electrode like platinum or graphite is used.

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Answer:

65.5J

Explanation:

ML=Q

ML=MC(change in temperature)

0.75 X 23000 =0.75 X 351 X C

C= 65.5J

The energy supplied to the lead to melt from 21°c to its melting point is 51521 Joules.

Specific heat is the amount of heat energy supplied to change the temperature of one unit mass of a substance by 1 °C. The SI unit of the specific heat capacity of a substance is J/Kg.

The mathematical expression for the specific heat capacity can be written as:

Q = mCΔT      Where C is the specific heat of the substance.

The specific heat capacity depends upon the starting temperature and is an intensive characteristic of the material.

Given, the melting point of the lead T₂ = 327.5° C

The initial temperature of the lead, T₁ = 21° C

The latent heat of the lead given, L = 230000 J/Kg K

The specific heat of the lead, C = 130 J/Kg K

The heat required to melt the lead from 12°C to 327.5 °C is :

Q = m× [C × (T₂ - T₁) + L ]

Q = 0.75 × [0.130 (327.5 - 21)  + 23000]

Q = 51521 J

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Answer:

Metal C

Explanation:

Density = mass (g)/ volume (mL, which is the same thing as cm cubed)

Divide mass by volume for each metal, the metal with the lowest value (metal C) is your answer.

Answer:

IT IS

NOT. I REPEAT - IT IS NOT> D

Explanation:

i got it wrong :(

The volume of copper (II) sulfate in a container is equivalent to 0.650L.

Volume is the three-dimensional measure of space that comprises a length, a width and a height. It is measured in units of cubic centimeters (cm³) in metric, cubic inches or cubic feet in English measurement.

According to this question, a container seems to have about 650 mL of copper(II) sulfate solution in volume. The volume can be converted to litres as follows:

1 millilitre = 0.001 litre

650 millilitres = 0.650 litres

Therefore, 0.650L is the volume of copper II sulfate in litres.

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