Calculate the total mass of the protons and electrons in 19 9F. Use 1.007825 amu as the mass of 11H (mass of a proton and an electron). Express your answer in atomic mass units to 6 significant figures.

Answer: The equilibrium concentration of [tex]PCl_5[/tex] is 1.285 M.

Explanation:

The chemical equation for the decomposition of phosphorus pentachloride follows:

[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)[/tex]

The expression for equilibrium constant is given as:

[tex]K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]

We are given:

[tex]K_c=0.042[/tex]

[tex][PCl_3]=0.18M[/tex]

[tex][Cl_2]=0.30M[/tex]

The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Putting values in above equation, we get:

[tex]0.042=\frac{0.18\times 0.30}{[PCl_5]}[/tex]

[tex][PCl_5]=1.285[/tex]

Hence, the equilibrium concentration of [tex]PCl_5[/tex] is 1.285 M.

The concentration of PCl5 is determined to be 1.29 M.

The student is asking about the concentration of PCl5 at equilibrium in the reaction PCl5(g) ⇌ PC13(g) + Cl2(g) given the equilibrium constant (K) and the concentrations of the products PCl3 and Cl2.

To solve for the concentration of PCl5, we use the equilibrium constant expression:

K = [PCl3][Cl2] / [PCl5]

Given that K = 0.042, [PCl3] = 0.18 M, and [Cl2] = 0.30 M, we can rearrange the expression to solve for [PCl5]:

[PCl5] = [PCl3][Cl2] / K

[PCl5] = (0.18 M * 0.30 M) / 0.042

After carrying out the calculation, the concentration of PCl5 at equilibrium is found to be 1.29 M.

Answer:

Joules

Explanation:

The another ones are units of tempeture (B and D) and unit of electricity that relatione energy and charge. In chemistry the energy es measured in Joules, because the energy is  work done on an object when a force of one newton acts on that object in the direction of the force's motion through a distance of one metre. In other words, J=Nm

Answer:

Option A is true

Explanation:

When you are a scientist conducting an experiment on energy transfer .

The reaction in which you measure a large amount of heat energy transfer.

We have to find the units which you should record them in

Energy: It is defined as the capability of doing the work .

When current I is flowing in ampere A  V is potential in volt  v applied  in the experiment  and R be resistance  in ohm used in experiment for time t in seconds

Then, heat energy =[tex]VI[/tex]=A-volt=Joule

Heat energy=[tex]I^2Rt[/tex]

Heat energy=[tex]A^2ohm sec[/tex]=Joule

The S.I unit of heat energy is Joules.

Hence, option A is true.

Answer : The pressure of the hydrogen gas exert before its volume was decreased will be, 2.28 atm

Explanation :

According to the Boyle's law, the pressure of the gas is inversely proportional to the volume of the gas at constant temperature of the gas and the number of moles of gas.

[tex]P\propto \frac{1}{V}[/tex]

or,

[tex]PV=k[/tex]

or,

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure of the gas = ?

[tex]P_2[/tex] = final pressure of the gas = 5.09 atm

[tex]V_1[/tex] = initial volume of the gas = 12.16 L

[tex]V_2[/tex] = final volume of the gas = 5.45 L

Now put all the given values in this formula, we get the initial pressure of the gas.

[tex]P_1\times 12.16L=5.09atm\times 5.45L[/tex]

[tex]P_1=2.28atm[/tex]

Therefore, the pressure of the hydrogen gas exert before its volume was decreased will be, 2.28 atm

Answer:

b. 9.29 L is the new volume of the gas if the pressure is constant.

Explanation:

As per Charles’s law  

At constant pressure for a given amount of a gas,

Volume is directly proportional to its temperature.

Thus the expression is [tex]V \propto T[/tex]

[tex]\frac {V}{T} = k[/tex] where k is a constant  

When there is a change in the volume and temperature the expression will be

[tex]\frac {V_1}{T_1} = \frac {V_2}{T_2}[/tex]

where [tex]V_1[/tex] and [tex]T_1[/tex] are the initial volume and initial temperature and [tex]V_2[/tex] and [tex]T_2[/tex] are the final volume and temperature.

Plugging in the values given

[tex]\frac {(9.3L)}{(279.5K)}=\frac {V_2}{(279.1K)}[/tex]

[tex]V_2=\frac {(9.3L\times279.1K)}{279.5K} \\\\=9.29L[/tex]

(Answer)

Answer:

b. the rate of destruction is equal to the rate of carbonate input.

Explanation:

At the Calcite Compensation Depth(CCD) the rate of addition of calcite and dissolution of the mineral is the same. Below this depth, all calcite minerals are dissolved.

The CCD occurs at a depth of 3000-4000m. It varies from places to places within the ocean. Other factors play important roles in determining the CCD. Some of the factors are temperature and pressure.

Answer: pH = 4.11

Explanation: pH of the buffer solution is calculated using Handerson equation:

[tex]pH=pKa+log(\frac{base}{acid})[/tex]

pKa is calculated from the given Ka value as:

pKa = - log Ka

[tex]pKa=-log1.34*10^-^5[/tex]

pKa = 4.87

pH of the solution before adding HCl to it:

[tex]pH=4.87+log(\frac{0.112}{0.322})[/tex]

pH = 4.87 - 0.46

pH = 4.41

Now, 0.069 moles of HCl are added to the buffer solution. This added HCl react with base(sodium propanoate) to produce acid(propanoic acid).

Initial moles of acid = 0.322*1.44 = 0.464

initial moles of base = 0.112*1.44 = 0.161

moles of base after reacting with HCl = 0.161 - 0.069 = 0.092

moles of acid after addition of HCl = 0.464 + 0.069 = 0.533

Let's plug in the values in Handerson equation to calculate the pH:

[tex]pH=4.87+log(\frac{0.092}{0.533})[/tex]

pH = 4.87 - 0.76

pH = 4.11

So, the original pH of the buffer solution is 4.41 and after addition of HCl the pH is 4.11 .

The pH of solution following the addition of 0.069 moles of HCl is 4.11.

The pH of a buffer solution can be calculated using Handerson equation as follows:

pH = pka + log (base/acid)

pKa of the acid is calculated from the given Ka value as follows:

pKa = - log Ka

pKa = - log 1.34 × 10-⁵

pKa = 4.87

The pH of the solution before adding HCl to it is as follows:

pH = 4.87 + log(0.112/0.322)

pH = 4.87 - 0.46 = 4.41

According to this question, 0.069 moles of HCl are added to the buffer solution.

Therefore, the pH of the buffer after adding HCl is:

pH = 4.87 + log(0.092/0.533)

pH = 4.87 - 0.76 = 4.11

Therefore, the pH of solution following the addition of 0.069 moles of HCl is 4.11.

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Using Hess's law, the enthalpy change for the reaction can be calculated by manipulating and adding together the enthalpy changes of the given reactions to match the sought reaction.

The enthalpy change for the reaction: 2 C(graphite) + 3H₂(g) → C₂H₆(g) can be calculated using Hess's law. According to

, the enthalpy change for a reaction is equivalent to the sum of the enthalpy changes for each step that makes up that reaction.

Looking at the given equations, we can manipulate them to establish a path to our desired equation. We can reverse the first equation and multiply it by 2, keep the second equation as it is and also multiply it by 3, and then reverse and divide the third equation by 2. Once these are added together, they should form our sought reaction. The overall enthalpy change for the reaction is then found by summing the enthalpy changes of each of these steps multiplied by their corresponding coefficients.

This approach utilizes the fact that enthalpy is a state function, meaning it only depends on the initial and final states of the system, not the path taken.

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The enthalpy change for the reaction 2 C(graphite) + 3H2(g) → C2H6(g) can be calculated using Hess's Law and the given enthalpy changes for other reactions. The reaction is broken down into steps and the total enthalpy change is the sum of the changes for each step. The calculated enthalpy change for the reaction is -941.8 kJ/mol.

To calculate the enthalpy change for the reaction 2 C(graphite) + 3H2(g) → C2H6(g), we use the values given for three known reactions and apply Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction process, regardless of the number of steps.

First, reverse the equation C(graphite) + O2(g) → CO2(g)ΔH o rxn = −393.5 kJ/mol and multiply it by 2 to get 2C(graphite) + 2O2(g) → 2CO2(g), ΔH = 2*(-393.5) kJ/mol = -787 kJ/mol.

Second, multiply the equation H2(g) + 1/2 O2(g) → H2O(l)ΔH o rxn = −285.8 kJ/mol by 6 to get 6H2(g) + 3 O2(g) → 6H2O(l), ΔH = 6*(-285.8) kJ/mol = -1714.8 kJ/mol.

Finally, add the values obtained to the equation for the formation of ethane, 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)ΔH o rxn = −3119.6 kJ/mol, but reverse it to get C2H6(g) → 2C(graphite) + 3H2(g), ΔH = -(-3119.6/2) kJ/mol = 1560 kJ/mol.

The total enthalpy change for the reaction is then found by adding these values together: (-787) + (-1714.8) + 1560 = -941.8 kJ/mol. Thus, the enthalpy change for the reaction 2 C(graphite) + 3H2(g) → C2H6(g) is -941.8 kJ/mol.

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Answer:

Explanation:

When a chemical reaction happens, the net change of enthalpy of the products and the reactants is the heat of reaction:

Then, when ∑ ΔHproducts > ∑ ΔA reactants, ΔH rxn > 0 and heat is released. This is what is called an exothermic reaction.

In an exothermic reaction, heat is released, ΔH > 0, and the surroundings will get hotter.

In and endothermic reaction heat is absorved, ΔH < 0, and the surroundings will get cooler.

Answer:

I think it's the Zeroth law of thermodynamics.

Answer:

the second law of thermodynamics.

Explanation:

Answer : The number of moles of gas added to the container will be, 20.89 mole

Explanation :

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

[tex]V\propto n[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas  = 8.15 L

[tex]V_2[/tex] = final volume of gas  = 17.9 L

[tex]n_1[/tex] = initial number of moles of gas  = 9.51 mole

[tex]n_2[/tex] = final number of moles of gas  = ?

Now we put all the given values in this formula, we get  the final moles of gas.

[tex]\frac{8.15L}{17.9L}=\frac{9.51mole}{n_2}[/tex]

[tex]n_2=20.89mole[/tex]

Therefore, the number of moles of gas added to the container will be, 20.89 mole

The solute that dissolves in ethanol depends on the polarity of the solute and the solvent. If both the solute and the solvent are polar, then the solute will dissolve in the solvent. Similarly, if both the solute and the solvent are nonpolar, then the solute will also dissolve in the solvent. However, if the solute and the solvent have different polarities, they will not dissolve in one another.

A substance can dissolve in a solvent, and form a solution, if the solute and solvent are attracted to each other. For example, water molecules that are held together by hydrogen bonding will dissolve solutes that can also hydrogen bond, like ethanol (CH3CH₂OH). The new hydrogen bonds between the water and the ethanol molecules (solvent-solute attractions) are nearly as strong as the hydrogen bonds in water (solvent-solvent) and ethanol (solute-solute) alone, making the process of solution formation (also called dissolution or dissolving) favorable.

When water mixes with other polar substances, like ethanol, some of the hydrogen bonding between water molecules replace with similar hydrogen bonding with ethanol molecules. Since the electrostatic potential energy is similar, the natural tendency to go towards more dispersion drives the dispersion of ethanol molecules uniformly in water resulting in the solution.

Nonpolar compounds do not dissolve in water. The attractive forces that operate between the particles in a nonpolar compound are weak dispersion forces. However, the nonpolar molecules are more attracted to themselves than they are to the polar water molecules. When a nonpolar liquid such as oil is mixed with water, two separate layers form because the liquids will not dissolve into each other. When another polar liquid such as ethanol is mixed with water, they completely blend and dissolve into one another. Liquids that dissolve in one another in all proportions are said to be miscible. Liquids that do not dissolve in one another are deemed immiscible. The general rule for deciding if one substance is capable of dissolving another is 'like dissolves like'. A nonpolar solid such as iodine will dissolve in nonpolar lighter fluid, but will not dissolve in polar water.

Answer:  Hypothesis

Explanation:  Hypothesis is a kind of idea that has been put forward prior to experimentation and whose results are not out yet.

The laws makes the theory then there are some rules which are needed to be followed in order to make the most out of the experiments. Then prior to the experiments , there comes the hypothesis whis the explanation of the work and which would be tested through the experiments.

Answer:

option b

Explanation:

When the energy is released the process is called exothermic reaction. This happens when the bonds are broken in the reactants and the system release energy.

Answer:

a. forming bonds

Explanation:

The energy required to break a bond is endothermic that is energy is absorbed to break a bond.

The energy is released in the formation of a bond that is energy is released when a bond is formed.

The formula to find the ∆H of the reaction is  

∆H (reaction) = ∆H (bonds Broken) - ∆H (bonds formed)

For example  

[tex]N_2+3H_2< >2NH_3[/tex]

[tex]N_2[/tex] contains one N≡N triple bond (Bond breaking 946 KJ per mol)

[tex]H_2[/tex] contains a single H-H bond (bond breaking 436 KJ per mol)

[tex]NH_3[/tex] contains 3, N - H single bonds(389 KJ per mol)

So ∆H (bonds broken) = 946 + (3 × 436) = 2254KJ

∆H (Bonds formed ) = (2 × 3 × 389) = 2334KJ

So

∆H (reaction) = 2254 KJ - 2334 KJ

= - 80KJ and the reaction is Exothermic

In this example we see energy required to break the bond is lesser than energy released in forming the bond.

A main goal of most environmental scientists is to achieve biodiversity.

Answer : The value of [tex]\Delta E[/tex] of the reaction is, 479.958 KJ/mole

Explanation :

The relation between the internal energy and enthalpy of reaction is:

[tex]\Delta E=\Delta H-\Delta n_g\times RT[/tex]

where,

[tex]\Delta E[/tex] = internal energy of the reaction = ?

[tex]\Delta H[/tex] = enthalpy of the reaction = 483.6 KJ/mole = 483600 J/mole

From the balanced reaction we conclude that,

[tex]\Delta n_g[/tex] = change in the moles of the reaction = Moles of product - Moles of reactant = 3 - 2 = 1 mole

R = gas constant = 8.314 J/mole.K

T = temperature = [tex]165^oC=273+165=438K[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta E=483600J/mole-(1mole\times 8.314J/mole.K\times 438K)[/tex]

[tex]\Delta E=479958.468J/mole[/tex]

[tex]\Delta E=479.958KJ/mole[/tex]

Therefore, the value of [tex]\Delta E[/tex] of the reaction is, 479.958 KJ/mole

Final answer:

To calculate the change in internal energy for the given reaction, we can use the equation ∆U = q - P∆V, where q is the heat transferred and P∆V is the work done on or by the system. Given the enthalpy change for the reaction, we can calculate the heat transferred and therefore determine the change in internal energy.

Explanation:

The question asks for the change in internal energy (∆U) for the reaction 2H2O(g) → 2H2(g) + O2(g) at a pressure of 1.0 atm and a temperature of 165°C. To find ∆U, we need to use the equation ∆U = q - P∆V, where q is the heat transferred and P∆V is the work done on or by the system. Since the pressure is constant, P∆V is zero, so we only need to calculate the heat transferred, q.

Given that the enthalpy change (ΔH) for the reaction is 483.6 kJ/mol, we can use the equation q = nΔH, where n is the number of moles. Since 2.0 moles of H2O(g) are being converted, the heat transferred can be calculated as q = 2.0 mol × 483.6 kJ/mol = 967.2 kJ.

Therefore, the change in internal energy for this reaction, ∆U, is 967.2 kJ.

Answers:

A. Disrupts hydrogen bonding in proteins, linearizing the protein

B. Provides an overall negative charge on proteins, making the migration distance on gel a function of only protein size

Final answer:

SDS in gel electrophoresis A. linearizes proteins into a rod-like shape and coats them with a uniform negative charge, allowing separation by molecular weight when an electric current is applied.

Explanation:

The sodium dodecyl sulfate (SDS) in gel electrophoresis serves primarily two functions:

SDS binds to proteins roughly in proportion to the number of amino acids, which correlates with the protein's mass. Thus, SDS-PAGE allows for the separation of proteins primarily by their molecular weight once an electric current is applied. The molecular weight of the proteins is estimated by comparing their migration distance to that of known standards.

Answer:

4.083 * 10^20 atoms.

Explanation:

One Mole of phosphorus  contains 6.022 * 10^23 atoms (Avogadros number)'

Since 1 mole of Phosphorus  has a mass of  30.974 grams, 21 milligrams has

6.022 * 10^23  * 0.021 / 30.974

= 0.004083 * 10^23

= 4.083 * 10^20

Answer:

The order of increasing entropy will be:

1 mol of NaCl (s) at 25 ° C < 1 mol of HCl (g) at 25 ° C<  1 mol of HCl (g) at 50 ° C < 2 mol of HCl (g) at 50 ° C

Explanation:

The entropy increases with

a) increase in temperature

b) state of matter (the entropy order of matter is gas > liquid > solid)

So here

i) 1 mol of HCl (g) at 50 ° C : it is gas at high temperature as compared to HCl gas at lower temperature.

ii. 1 mol of NaCl (s) at 25 ° C : this is solid so will have lower entropy than gas.

iii. 2 mol of HCl (g) at 50 ° C: the moles are more so will have more entropy than 1 mol of HCl (g) at 50 ° C

iv. 1 mol of HCl (g) at 25 ° C : will have lower entropy than HCl gas at higher temperature but will have higher entropy than solid NaCl.

The correct order of increasing entropy is given as:

1 mol of NaCl (s) at 25 ° C < mol of HCl (g) at 50 ° C < 1 mol of HCl (g) at 25 ° C < 2 mol of HCl (g) at 50 ° C

Entropy can be defined as that measure of thermal energy of a system per unit temperature that is unavailable for doing work.

Mathematically, the entropy change in a chemical reaction is given by:

The sum of the entropies of the products - the sum of the entropies of the reactants.

In conclusion, the correct order of increasing entropy is given as

1 mol of NaCl (s) at 25 ° C < mol of HCl (g) at 50 ° C < 1 mol of HCl (g) at 25 ° C < 2 mol of HCl (g) at 50 ° C

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Final answer:

The epicenter of an earthquake can be located using the arrival times of S- and P-waves at three or more seismographic stations, allowing for triangulation and precise determination of the distance to the earthquake source.

Explanation:

To locate the epicenter of an earthquake, seismographic stations use the arrival times of S- and P-waves. The S-wave is slower than the P-wave, so the difference in arrival times at three or more seismographic stations can be used to determine the distance to the epicenter. By triangulating the distances from multiple stations, the epicenter can be pinpointed.

For example, if the S-wave and P-wave travel at speeds of 4.00 km/s and 7.20 km/s respectively, the difference in arrival times can be used to calculate the distance. The precision of the seismographs in measuring the arrival times allows for a precise determination of the distance to the earthquake source.

It is important to note that uncertainties in the propagation speeds of the waves can introduce greater uncertainty in determining the distance to the epicenter.

Answer:

option B

Explanation:

A spontaneous reaction drive favorable when enthalpy is decreasing and the entropy is increasing on the system. If that happens the reaction occurs spontaneously

Answer:

b. decreasing enthalpy and increasing entropy

Explanation:

∆H stands for enthalpy change and ∆S stands for entropy change

Spontaneity depends on the enthalpy and entropy changes of the reaction

∆G = ∆H - T∆S

When ∆H is negative and ∆S is positive  

∆G will be negative

For a spontaneous reaction ∆G is negative

If ∆G = 0 then the reaction will be at equilibrium  

If ∆G is positive the reaction is non spontaneous.

Decreasing enthalpy (negative) and increasing entropy (positive) will give a negative number for [tex]\Delta G[/tex]

Answer:

lowers activation energy

Answer:

Explanation:

An enzyme reduces the free energy of activation (EA) of the reaction it catalyzes.

Answer : The percent composition by volume of mixture of [tex]CO[/tex] and [tex]CO_2[/tex] are, 18.94 % and 81.06 % respectively.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

[tex]R\propto \sqrt{\frac{1}{M}}[/tex]

And the relation between the rate of effusion and volume is :

[tex]R=\frac{V}{t}[/tex]

or, from the above we conclude that,

[tex](\frac{V_1}{V_2})^2=\frac{M_2}{M_1}[/tex]            ..........(1)

where,

[tex]V_1[/tex] = volume of helium gas = 29.7 ml

[tex]V_2[/tex] = volume of mixture = 9.28 ml

[tex]M_1[/tex] = molar mass of helium gas  = 4 g/mole

[tex]M_2[/tex] = molar mass of mixture = ?

Now put all the given values in the above formula 1, we get the molar mass of mixture.

[tex](\frac{29.8ml}{9.28ml})^2=\frac{M_2}{4g/mole}[/tex]

[tex]M_2=40.97g/mole[/tex]

The average molar mass of mixture = 40.97 g/mole

Now we have to calculate the percent composition by volume of the mixture.

Let the mole fraction of [tex]CO[/tex] be, 'x' and the mole fraction of [tex]CO_2[/tex] will be, (1 - x).

As we know that,

[tex]\text{Average molar mass of mixture}=\text{Mole fraction of }CO[/tex]

[tex]\text{Average molar mass of mixture}=(\text{Mole fraction of }CO\times \text{Molar mass of } CO)+(\text{Mole fraction of }CO_2\times \text{Molar mass of } CO_2)[/tex]

Now put all the given values in this expression, we get:

[tex]40.94g/mole=((x)\times 28g/mole)+((1-x)\times 44g/mole)[/tex]

[tex]x=0.1894[/tex]

The mole fraction of [tex]CO[/tex] = x = 0.1894

The mole fraction of [tex]CO_2[/tex] = 1 - x = 1 - 0.1894 = 0.8106

The percent composition by volume of mixture of [tex]CO[/tex] = [tex]0.1894\times 100=18.94\%[/tex]

The percent composition by volume of mixture of [tex]CO_2[/tex] = [tex]0.8106\times 100=81.06\%[/tex]

Therefore, the percent composition by volume of mixture of [tex]CO[/tex] and [tex]CO_2[/tex] are, 18.94 % and 81.06 % respectively.

Answer:

Acetylcholine causes an end-plate potential by triggering the opening of sodium channels.

Acetylcholine (ACh), a neurotransmitter released by motor neurons, causes an end-plate potential by binding to receptors in the motor end plate and triggering a depolarization process that eventually leads to an action potential.

Acetylcholine (ACh) is a neurotransmitter that plays a pivotal role in triggering an end-plate potential within the neuromuscular system. The process starts when an action potential travels down the motor neuron's axon, triggering the release of these neurotransmitters. The ACh then binds to receptors in the motor end plate and initiates a series of events that lead to changes in ion permeability, the influx of sodium ions into the muscles cells, and ultimately a reduction in the voltage difference between the inside and outside of the cell - a process called depolarization. When ACh binds at the motor end plate, this depolarization is called an end-plate potential.

This depolarization then spreads along the muscle fiber membrane, the sarcolemma, creating an action potential, which moves across the entire cell, creating a wave of depolarization. Therefore, the acetylcholine indeed causes the end-plate potential by triggering the release of these neurotransmitters and the subsequent sequence of events.

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To calculate the standard enthalpy of formation of carbon disulfide (CS2), we can use Hess's Law and the given enthalpy changes for the reactions involving carbon, sulfur, and oxygen. The standard enthalpy of formation of CS2 from its elements is -310.3 kJ/mol.

To calculate the standard enthalpy of formation of carbon disulfide (CS2) from its elements, we can use Hess's Law and the given enthalpy changes for the reactions involving carbon (C), sulfur (S), and oxygen (O).

First, we can use the given reaction for the formation of carbon dioxide (CO2) to find the enthalpy change for the reaction involving carbon:

C(graphite) + O2(g) → CO2(g) ΔH o rxn = −393.5 kJ/mol

Next, we can use the given reaction for the formation of sulfur dioxide (SO2) to find the enthalpy change for the reaction involving sulfur:

S(rhombic) + O2(g) → SO2(g) ΔH o rxn = −296.4 kJ/mol

Now, we can use the given reaction for the formation of carbon disulfide (CS2) to find the enthalpy change for this reaction:

CS2 + 3O2(g) → CO2(g) + 2SO2(g) ΔH o rxn = −1073.6 kJ/mol

By rearranging these equations and manipulating the enthalpy changes, we can find the standard enthalpy of formation of carbon disulfide:

CS2 = CO2 - C(graphite) - 2SO2 = -1073.6 kJ/mol - (-393.5 kJ/mol) - 2(-296.4 kJ/mol) = -1073.6 kJ/mol + 393.5 kJ/mol - 2(-296.4 kJ/mol) = -310.3 kJ/mol

Therefore, the standard enthalpy of formation of carbon disulfide (CS2) from its elements is -310.3 kJ/mol.

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The standard enthalpy of formation of carbon disulfide (CS2) is calculated using Hess's Law and given reactions. After rearranging and combining reactions to form CS2 from its elements, the standard enthalpy of formation for CS2 is found to be -87.3 kJ/mol.

To calculate the standard enthalpy of formation (ΔHfo) of carbon disulfide (CS2), we use Hess's Law and the given reactions:

The standard enthalpy of formation of CS2 is calculated by rearranging the reactions to derive the formation reaction for CS2 from its elements:

In doing so, we get:

C(graphite) + 2S(rhombic) + O2(g) → CS2(l), ΔHfo = -1073.6 kJ/mol + (1 x 393.5 kJ/mol) + (2 x 296.4 kJ/mol)

We sum the enthalpy changes of these steps to find the enthalpy of formation for CS2 which is:

ΔHfo(CS2) = -1073.6 kJ/mol + 393.5 kJ/mol + 592.8 kJ/mol = -87.3 kJ/mol

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Answer:

Hygroscopic

Explanation:

An hygroscopic substance is one that absorbs moisture from the atmosphere and becomes wet. Their ability to remove water from air is less than that of deliquescent substances. Most of the solid hygroscopic substances forms pasty substances and not solutions like the deliquescent compounds.

Examples are sodium trioxonitrate(v), copper(ii) oxide e.t.c

Efflorescence compounds gives off their water of crystallization to the atmosphere.

Hydrates capable of removing moisture from the air due to their low vapor pressure are known as hygroscopic.

Hydrates that have a low vapor pressure and can remove moisture from air are hygroscopic. Substances such as anhydrous calcium chloride and magnesium chloride exhibit hygroscopic properties due to their ability to absorb moisture, ultimately becoming hydrates in the process.

For example, anhydrous calcium chloride mixed with cobalt chloride serves as both a drying agent and an indicator; cobalt chloride is blue when anhydrous and pink when hydrated, thus revealing the condition of the desiccant.

Furthermore, the presence of nonvolatile solutes, such as these hydrates, can lower the vapor pressure of a solution by preventing the evaporation of solvent molecules. The waters of hydration in compounds are loosely bound water molecules that can often be removed through heating, turning hydrates back into their anhydrous form.

Answer:

because energy is released

Explanation:

freesing is an exothermic process

C. Because energy is released, freezing is an endothermic process.

In thermochemistry, an endothermic system is any process with an increase in the enthalpy H of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transferred into the system.

Endothermic reactions are chemical reactions wherein the reactants absorb heat energy from the surroundings to form products. these reactions lower the temperature in their surrounding region, thereby creating a cooling effect.

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Acid rain is the process that involves  heavy chemicals falling to the Earth

as dry particles.

Acid rain is common in areas which have a lot of industries which release

chemicals into the atmosphere. With time, these chemicals accumulate and

falls to earth as dry particles through precipitation.

The acid rain normally contains water which pushes the chemicals down and the water is usually  acidic with a pH between 4.2 and 4.4 as a result of the chemicals.

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Answer:

Explanation:

1) Data:

2) Formula:

Combined law of gases:

3) Solution:

T₂ = P₂ V₂ T₁ / (P₁ V₁)

T₂ = 872.15 mmHg × 7.63 liter × 261.2 K / ( 424.9 mmHg × 0.66 liter)

T₂ = 6198 K

Answer:

=0.5 moles

Explanation:

Let us assume that the sodium chloride solution has a density of 1g/cm³.

Therefore the volume of the 0.5 kg of solution will be calculated as follows.

0.5kg into grams=0.5 kg×1000g/kg

=500g

volume= mass/density

=500g/1g/cm³

=500cm³

The solution is 1.0 M which means that 1.0 moles are in 1000 cm³

500cm³ will have:

(500 cm³×1.0 moles)/1000 cm³

=0.5 moles

Answer:

B. Peer review

Explanation:

Peer review ensures that the results of an experimental procedures are consistent are reliable and they meet their objective statement.

When peers which are professionals in a field of study subjects the results from an experiment into a test, they can give their own verdict as to wether such findings are consistent and reliable with the problem in view.

Answer:

the answer would be peer review

Explanation:

founders education chemistry