Consider a hypothetical experiment in which the left beaker contains 4 mM NaCl, 9 mM glucose and 10 mM albumin. The right beaker contains 10 mM NaCl, 10 mM glucose and 40 mM albumin. The dialysis membrane is permeable to all substances except albumin. In which direction will glucose move?

Answer:

644.2918 g of turpentine are consumed.

Explanation:

Calculation of the moles of [tex]C[/tex] as:-

Mass = 569 g

Molar mass of carbon = 12.0107 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{569\ g}{12.0107\ g/mol}[/tex]

[tex]Moles_{C}= 47.3744\ mol[/tex]

According to the given reaction:-

[tex]C_{10}H_{16}+8Cl_2\rightarrow 10C+16HCl[/tex]

10 moles of C are produced when 1 mole of turpentine undergoes reaction

1 mole of C are produced when [tex]\frac{1}{10}[/tex] mole of turpentine undergoes reaction

47.3744 moles of C are produced when [tex]\frac{1}{10}\times 47.3744[/tex] moles of turpentine undergoes reaction

Moles of turpentine = 4.73744 moles

Molar mass of turpentine = 136 g/mol

[tex]Mass=Moles\times Molar\ mass=4.73744\ moles\times 136\ g/mol=644.2918\ g[/tex]

644.2918 g of turpentine are consumed.

Answer: Saturated Fatty Acids

Explanation: Saturated fats are one's in which fatty Acids chains

Largely or totally contains a single bond. Most animal fats contains high proportion of Saturated Fatty acid such as milk, cheese and butter. Foods like pizza, sausage also have high proportion of Saturated Fatty acid.

Answer:

The final temperature of the water is 28.0 °C

Explanation:

Step 1: Data given

Mass of liquid water = 10.0 grams

Temperature = 23.0 °C

Heat absorbed = 209 Joules

Since heat was absorbed by the water, you must have a positive value for  

Δ T

Step 2: Calculate final temperature

q = m*c* ΔT

⇒ with m = the mass of the water = 10.0 grams

⇒ with c = the specific heat of water = 4.184 J/g°C

⇒ with ΔT = The change in temperature = T2 -  T1 = T2 - 23.0 °C

⇒ with q = the heat absorbed = 209 Joule

209 = 10.0 * 4.184 * ΔT

ΔT = 5

ΔT = 5 = T2 - 23

T2 = 28 °C

The final temperature of the water is 28.0 °C

The final temperature of the liquid water sample is 28.0°C.

The specific heat capacity of water is 4.184 J/(g°C). To find the final temperature of the water sample, we can use the equation:

q = mcΔT

Where q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

Given that m = 10.0 grams, ΔT = final temperature - 23.0°C, and q = 209 joules, we can rearrange the equation to solve for the final temperature:

209 joules = (10.0 grams)(4.184 J/(g°C))(final temperature - 23.0°C)

Simplifying the equation gives:

final temperature - 23.0°C = 209 joules / (10.0 grams)(4.184 J/(g°C))

final temperature - 23.0°C = 4.9886854°C

final temperature = 28.0°C

Answer: The dentist and at least one other dental auxiliary must be present, in the treatment room, during the administration of nitrous oxide. True.

Explanation: there is a minimum standard of care while performing nitrous oxide inhalation by a medical personnel and in addition shall maintain under continuous direct supervision auxiliary personnel who shall be capable of reasonably assisting in procedures, problems and emergencies incident to the used of nitrous oxide inhalation sedation.

Answer:

85g

Explanation:

Firstly, from the balanced equation, we can see that one mole of silver nitrate yielded one mole of silver chloride. This is the theoretical relation. We need to get the actual relation. To do this, we will first need to get the number of moles of silver nitrate.

The number of moles is the mass of silver nitrate divided by the molar mass of silver nitrate. The molar mass of silver nitrate is 108 + 14 + 3(16) = 108 + 14 + 48 = 170g/mol

The number of moles is thus 100/170 = 0.59 moles

Since the mole ratio is 1 to 1, the number of moles of silver chloride produced too is 0.59 moles.

To get the mass of silver chloride produced, we simply multiply the number of moles of silver chloride by the molar mass of silver chloride. The molar mass of silver chloride is 108 + 35.5 = 143.5g/mol

The mass is thus = 143.5 * 0.59 = 84.67g

Answer:

E. Evaporation to dryness

Explanation:

E. - Evaporation to dryness is the best method for the recovery of solid KNO3 from an aqueous solution of KNO3.

(KNO3 is very soluble, and will violently decompose if overheated.)

Paper chromatography is for separation of different weight molecules in solution.

B. Filtration  won't work on a solution

C.Titration would contaminate the salt with something else and is used  

to determine concentrations

D. Electrolysis would destroy the salt

Answer:

E

Explanation:

Answer & Explanation:

The law of multiple proportion states that two elements A and B can react with each other to form multiple compounds while one of the two elements remain fixed.

For Example

Hydrogen reacts with oxygen to form:

Water (H2O)

Hydrogen peroxide (H2O2)

The Law of Multiple Proportions states that the mass ratio of elements in different compounds is a simple whole number ratio. In the case of water, the mass ratio of hydrogen to oxygen is always 1:8.

The Law of Multiple Proportions is a principle in chemistry that states that when two elements combine to form different compounds, the ratio of their masses will be a simple whole number ratio. In the case of water, the ratio of hydrogen to oxygen by mass is always 2:16 or 1:8. This means that for every 1 gram of hydrogen in water, there are 8 grams of oxygen. The mass ratio of hydrogen to oxygen in water is therefore constant, not variable.

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Answer:

Smog

Explanation:

Smog is formed when fossil fuels or turpentine are incompletely burned in the presence of sunlight.

Smog is type of highly dirty air that contain various impurity in it. A form of visible air pollution consists of oxides of ammonia, sulfur , asbestos, haze, and other particulate matter. Man-made smog is derived from emissions from coal burning, vehicle emissions, industrial emissions, forest and agricultural fires, as well as photochemical reactions from these emissions.

Answer:

smog

Explanation:

Answer:

1. 351.62L

2. 15.05 * 10^23 molecules

3. 54.6g

4. 8.29 * 10^21 molecules

Explanation:

1. First we will need to calculate the number of moles present in such number of molecules. We know that one mole contains 6.02 * 10^23 molecules. Hence to calculate the number of moles in 9.45 * 10^24 carbon iv oxide Miley, we simply divide. And that is (9.45 * 10^24) / ( 6.02 * 10^23) = 15.7 moles

At s.t.p, a mole of a gas will occupy a volume of 22.L. Hence 15.7 moles will occupy a volume of 15.7 * 22.4 = 351.62L

2. We first need to calculate the number of moles of calcium metal present in 100g. To do this we simply divide the mass by the atomic mass of calcium. The atomic mass of calcium is 40 amu. The number of moles is thus 100/40 = 2.5 moles

One mole contains 6.02 * 10^23 number of molecules. Hence 2.5 moles will contain 2.5 * 6.02 * 10^23 = 15.05 * 10^23 molecules

3. Firstly we need to calculate the number of moles contained in those number of molecules. What we simply need do is to divide by 6.02 * 10^23

Since the power of 10 is same, we simply divide 5.06 by 6.02 = 0.84 moles

To get the mass, we simply multiply by the atomic mass of zinc. The atomic mass of zinc is 65a.m.u

The mass is thus 0.84 * 65 = 54.6g

4. Firstly we will need to calculate the number of moles here. To do this we divide the mass by the molar mass. The molar mass of lead nitrate is 331g/mol

The number of moles is thus 4.56/331 = 0.014 moles

1 mole contains 6.02 * 10^23 molecules, then 0.014 moles will contain 0.014 * 6.02 * 10^23 molecules = 8.29 * 10^21 molecules

Answer:

1. 0.178 moles ; 2. 8x10²³ atoms ; 3. 7.22x10²³ molecules ; 4. 89.6 g ; 5. 1.34x10²² atoms ; 6. 1.67x10²⁵ molecules

Explanation:

1. Mass / Molar mass = Mol

5g / 28 g/m = 0.178 moles

2. 1 molecule of N₂ has 2 atoms, it is a dyatomic molecule.

4x10²³  x2 = 8x10²³ atoms

3. 1 mol of anything, has 6.02x10²³ particles

6.02x10²³ molecules . 1.2 mol = 7.22x10²³

4. 1 atom of C weighs 12 amu.

4.5x10²⁴ weigh ( 4.5x10²⁴ . 12) = 5.24x10²⁵ amu

1 amu = 1.66054x10⁻²⁴g

5.24x10²⁵ amu = (5.24x10²⁵ . 1.66054x10⁻²⁴) = 89.6 g

5. Molar mass NaCl = 58.45 g/m

1.3 g /  58.45 g/m = 0.0222 moles

1 mol has 6.02x10²³ atoms

0.0222 moles → ( 0.0222 . 6.02x10²³) = 1.34x10²²

6. Density of water is 1 g/mL, so 500 mL are contained in 500 g of water

Molar mass H₂O = 18 g/m

500 g / 18 g/m = 27.8 moles

6.02x10²³ molecules . 27.8 moles = 1.67x10²⁵

Answer:

The reaction proceeds with a net release of free energy.

Explanation:

Exergonic reactions: It is known as the chemical reaction where the change in the free energy is occur negative or there is a net release of free energy, and indicating a spontaneous reaction. For the processes this takes place under constant temperature, and pressure conditions.

The Gibbs free energy is used whereas the processes which takes place under constant volume, and temperature conditions, and their Helmholtz energy is used. Cellular respiration is the example of an exergonic reaction.

Answer:

A

Explanation:

Pico(p) has a value of 10^-12.

Nano(n) has a value of 10^-9.

A. 374ps= 374×10^-12 s

B. 3.74ps= 374×10^-14 s

C. 374ns= 374×10^-9 s

Answer:

19.5g

Explanation:

Firstly, we can use the ideal gas equation to calculate the number of moles of nitrogen gas needed.

We use the following formula:

PV = nRT

n = PV/RT

In the question, we were given:

P = 1 atm

V = 10L

T = 273.15k

R = molar gas constant= 0.082L.atm/mol.k

Let us insert these values into the ideal gas equation.

n = (1 * 10)/(273.15 * 0.082) = 0.45moles

The number of moles of nitrogen required is 0.45 moles

From the balanced reaction equation, we can see that 2 moles of sodium azide yielded 3 moles of Nitrogen.

Hence the number of moles of sodium azide produced from 0.45 moles of nitrogen would be : (0.45 * 2)/3 = 0.3mole

To get the mass of sodium azide needed, we simply multiply the number of moles of sodium azide by the molar mass of sodium azide. We know the number of moles but we do not know the molar mass. The molar mass of sodium azide is 23 + 3(14) = 23 + 42 = 65g/mol

The mass = 65 * 0.3 = 19.5g

The mass of azide required is 19.3 g.

The equation of the reaction is given as; 2NaN₃(s)⟶3N₂(g)+2Na(s)

We have to obtain the number of moles of the nitrogen gas from;

P = 1.000 atm

V = 10.00L

T= 273.15 K

R = 0.082 atm L K-1mol-1

n = PV/RT

n =  1.000 atm × 10.00L/0.082 atm L K-1mol-1 × 273.15 K

n = 0.446 moles

Now;

2 moles of azide yields 3 moles of nitrogen

x moles of azide yieilds 0.446 moles of nitrogen

x = 2 moles × 0.446 moles/3 moles

x = 0.297 moles

Mass of azide required=  0.297 moles × 65 g/mol = 19.3 g

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The substance with the highest Rf value in thin-layer chromatography using a silica plate would be CH3CH2CH2CH2CH3 (A), because it is the least polar, and thus has lesser degree of interaction with the silica plate, allowing it to travel furthest. Option A

The substance with the highest Rf value in thin-layer chromatography using a silica plate would be the least polar. Rf value depends on the polarity of the substance and the solvent; the less polar the substance, the higher its Rf value. Thin-layer chromatography, or TLC, is a method used in chemistry to separate mixtures.

It works because different compounds in a mixture have different affinities for the stationary phase (silica in this case) and the mobile phase (the solvent).

In this case, we have (A) CH3CH2CH2CH2CH3, (B) HOCH2CH2CH2CH3, (C) HOCH2CH2CH2OH, and (D) HOOCCH2CH2CH3. Molecules B, C, and D all contain a polar -OH group, which would lead to hydrogen bonding with the silica plate, slowing their movement along the plate and decreasing their Rf values.

Molecule A, however, is nonpolar, as it is composed entirely of carbon and hydrogen atoms which have similar electronegativities resulting in minimal polarity. Thus, molecule A will travel furthest and hence have the highest Rf value when using a silica plate in TLC.

Option A

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Answer:

8 shared electrons

Explanation:

When you are looking for the number of shared electrons via the equation:

S = N-A

Where:

S = means the shared electrons

N = Needed electrons

A = available electrons

"Needed electrons" means how many electrons does it need to have a noble gas configuration, in this case, to complete the octet rule.

"Available electrons" means how many valence electrons is actually available considering the compound or the elements involved in the compound.

To get the needed electrons, treat the elements involved separately.  We have a silicon (Si) atom and 4 chlorine (chlorine) atoms in this compound. Let's list it down first:

              Number of atoms

Si                      1

Cl                     4

Next step is to determine how many electrons should it have in its outer shell to achieve the octet rule. Both of them in this case would be 8. Multiply that by the number of atoms and add up the needed electrons to determine how many you will need for this particular compound.

             Number of atoms             Electrons to achieve Octet       Needed

Si                      1                      x                         8                           =       8

Cl                     4                      x                         8                           =     32

                                                                     TOTAL:                              40

This is now our N. N = 40 electrons

Next step is to determine how many we actually have. Your clue in determining how many valence electrons the atom has is the group. Silicon is in Group 4A, this means it has 4 valence electrons. Chlorine is in Group7A, so ths means it has 7 valence electrons.

So first we write the number of atoms again, then in the next column, you write down the actual number of valence electrons and multiply them. Sum it up to see how many electrons available  in this particular compound.

             Number of atoms             Valence electrons                 Available

Si                      1                      x                         4                           =      4

Cl                     4                      x                         7                          =     28

                                                                     TOTAL:                              32

This is now our A. A = 32 electrons

Now we apply this:

S = N - A

N = 40 electrons

A =  32 electrons

S = 40 - 32 = 8

Number of shared electrons is 8

The oxidation numbers for the elements Co, Al, C, N, Cl, and Cr in the given compounds are +3, -3, -2, -3, +1, and +6 respectively.

The oxidation numbers for the elements in the compounds are determined as follows:

(a) In LiCoO2, Co has an oxidation number of +3, as lithium contributes +1 and each of the two oxygen atoms contribute -2.
(b) In NaAlH4, Al has an oxidation number of -3, as sodium contributes +1, while hydrogen as a metal hydride contributes -1 each adding up to -4.
(c) In CH3OH, C has an oxidation number of -2, as hydrogen contributes +1 each from the three hydrogen atoms and the fourth hydrogen attached to oxygen contributes -1 and oxygen contributes -2.
(d) In GaN, N has an oxidation number of -3, as gallium contributes +3.
(e) In HClO2, Cl has an oxidation number of +1, as hydrogen contributes +1, and each of the two oxygen atoms contribute -2.
(f) In BaCrO4, Cr has an oxidation number of +6, as barium contributes +2 and each of the four oxygen atoms contribute -2.

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The oxidation numbers for the given elements in their compounds are as follows: Co in LiCoO₂ is +3, Al in NaAlH₄ is +3, C in CH₃OH is -2, N in GaN is -3, Cl in HClO₂ is +3, and Cr in BaCrO₄ is +6.

(a) Co in LiCoO₂: The oxidation number of Co is +3. This is because Lithium (Li) has an oxidation number of +1, and Oxygen (O) has an oxidation number of -2. Since the molecule LiCoO₂ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Co, we get: 1(+1) + x + 2(-2) = 0, which simplifies to x = +3.

(b) Al in NaAlH₄: The oxidation number of Al is +3. This is because Sodium (Na) has an oxidation number of +1, and Hydrogen (H) has an oxidation number of -1. Since the molecule NaAlH₄ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Al, we get: 1(+1) + x + 4(-1) = 0, which simplifies to x = +3.

(c) C in CH₃OH (methanol): The oxidation number of C is -2. This is because Hydrogen (H) has an oxidation number of +1, and Oxygen (O) has an oxidation number of -2. Since the molecule CH₃OH is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of C, we get: 4(+1) + x + 1(-2) = 0, which simplifies to x = -2.

(d) N in GaN: The oxidation number of N is -3. This is because Gallium (Ga) has an oxidation number of +3. Since the molecule GaN is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of N, we get: 1(+3) + x = 0, which simplifies to x = -3.

(e) Cl in HClO₂: The oxidation number of Cl is +3. This is because Hydrogen (H) has an oxidation number of +1, and Oxygen (O) has an oxidation number of -2. Since the molecule HClO₂ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Cl, we get: 1(+1) + x + 2(-2) = 0, which simplifies to x = +3.

(f) Cr in BaCrO₄: The oxidation number of Cr is +6. This is because Barium (Ba) has an oxidation number of +2, and Oxygen (O) has an oxidation number of -2. Since the molecule BaCrO₄ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Cr, we get: 1(+2) + x + 4(-2) = 0, which simplifies to x = +6.

Answer:

Group 3-12 transition metals has 5 occupied principal energy levels

Explanation:

The group 3-12 transition metal includes rubidium, strontium, yttrium, zirconium, niobium, molybdenum, technetium, ruthenium, rhodium, palladium, silver, cadmium, indium, tin, antimony, tellurium, iodine, xenon. The elements can also go beyond the 5th principal energy level. The filling of the orbital always depends on the atomic number of the atom. The orbitals filled with the electrons in the ascending order of the energy of the orbitals. The starting element of the period of the the periodic table shows the new principal energy level.

samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness. In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.

Answer:

[H+] = 1.74 x 10⁻⁵

Explanation:

By definition pH = -log  [H+]

Therefore, given the pH,  all we have to do is  solve algebraically for  [H+] :

[H+]  = antilog ( -pH ) =  10^-4.76 = 1.74 x 10⁻⁵

Final answer:

The concentration of [H+] in a 0.1 M solution of acetic acid at a pH of 4.76 is 1.32 × 10^-3 M.

Explanation:

The concentration of [H+] in a 0.1 M solution of acetic acid at a pH of 4.76 can be calculated using the equation Ka = [H+][CH3COO-]/[CH3COOH]. Since the pK of acetic acid is 4.76, it means that at this pH, [H+] will be equal to [CH3COO-]. Substituting the values, we get [H+] = [CH3COO-] = 1.32 × 10^-3 M. Therefore, the concentration of [H+] in the solution is 1.32 × 10^-3 M.

Answer:

A) 36 g

Explanation:

From the source, some values have been correct.

Given that;-

Mass of NaCl formed = 116 grams

Molar mass of NaCl = 58 g/mol

Moles of NaCl = [tex]\frac{Mass}{Molar\ mass}=\frac{116}{58}\ mol=2\ mol[/tex]

According to the reaction shown below:-

[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]

1 mole of NaCl and 1 mole of [tex]H_2O[/tex] are produced upon reaction.

So,

2 moles of NaCl and 2 moles of [tex]H_2O[/tex] are produced upon reaction.

Moles of water = 2 moles

Molar mass of water = 18 g/mol

Mass = [tex]Moles\times Molar\ mass=2\times 18\ g=36\ g[/tex]

36 g of [tex]H_2O[/tex] is produced.

Answer:

B

Explanation:

The species on the right are the more stable acids and bases. A weak acid always has a strong conjugate base as in HCOO- and HCOOH. The acid is a weak acid because it has a strong conjugate base. Similarly, a strong acid has a weak conjugate base as in ClO2- and HClO2. ClO2- is a weak base hence it quickly abstracts a proton to form the acid.

HClO₂ is a stronger acid than HCOOH, and ClO₂⁻ is a weaker base than HCOO⁻. Therefore, the correct answer is B.

To determine the relative strengths of acids and bases in the given reaction:

ClO₂⁻ + HCOOH → HClO₂ + HCOO⁻

We use the Brønsted-Lowry definition of acids and bases. According to the reaction and knowing that equilibrium constants (Keq) of less than one (<1) favor reactants:

Thus, the correct answer is: B. HClO₂ > HCOOH, ClO₂⁻ < HCOO⁻

Beer's law states that the absorbance of a solution is directly proportional to its concentration. Given the molar absorptivity and absorbance for a Cu2+ solution, we can calculate the concentration to be approximately 0.00172 M.

The subject of this question is Beer's Law in the field of chemistry. Beer's Law states that the absorbance of a solution is directly proportional to its concentration, and mathematically it can be expressed as A = εcl. In this formula, A is the absorbance of the solution, ε is the molar absorptivity (the slope in the Beer's law plot), c is the concentration of the solution, and l is the length of the light path through the solution.

In your case, you have been given the slope (ε) from the Beer's law plot, which is 435 L/mol, and the absorbance (A) of the Cu2+ solution, which is 0.75. You need to find the concentration (c), so you can rearrange the formula to c = A / ε.

Substituting the given values into this equation gives: c = 0.75 / 435 = 0.0017241379310344827586206896551724 M

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Answer:

E

Explanation:

As the number of covalent bonds increase, atoms are drawn closer together hence the distance between the atoms decreases. However, the strength of the bonds increases. The bond energy of nitrogen gas (N2) is high because of the strong triple bond between its atoms hence the gas is inert.

As the number of covalent bonds between two atoms increases, the distance between the atoms decreases and the strength of the bond between them increases.

As the number of covalent bonds between two atoms increases, the distance between the atoms decreases and the strength of the bond between them increases. Covalent bonds are formed when atoms share electrons. The stronger the bond, the shorter the distance between the atoms.

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Answer:

Partial pressure of H₂ is 0.375 atm

Explanation:

We apply the mole fraction to solve this.

Standard pressure is 1 atm

Mole fraction of a gas = Moles of gas / Total moles

Mole fraction of pressure = Partial pressure of gas / Total pressure

Both values are the same

Total moles = 4 moles of O₂  +  3 moles of H₂ and 1 mol of N₂ = 8 moles

3 moles H₂ / 8 moles = Partial pressure H₂ / 1 atm

(3 / 8 ) .1 = 0.375  atm → Partial pressure of H₂

The partial pressure exerted by the hydrogen gas, H₂ in the container is 0.375 atm

We'll begin by calculating the mole fraction of H₂ in the gas mixture. this can be obtained as follow:

Mole of O₂ = 4 moles

Mole of H₂ = 3 moles

Mole of N₂ = 1 mole

Total mole = 4 + 3 + 1 = 8 moles

[tex]mole \: fraction \: = \frac{mole \: of \: gas}{total \: mole} \\ \\ = \frac{3}{8} \\ \\ [/tex]

Mole fraction of H₂ = 0.375

Total pressure = STP = 1 atm

Partial pressure = mole fraction × Total Pressure

Partial pressure of H₂ = 0.375 × 1

Therefore, the partial pressure of H₂ in the container is 0.375 atm

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Answer:

The partial pressure of hydrogen gas = 1.64 atm

Explanation:

Step 1: Data given

partial pressure of nitrogen = 0.88 atm

Nitrogen occupies 15% of the volume (Let's assume the volume is 1L)

Oxygen gas occupies 57 % of the volume

Step 2: Calculate % volume of hydrogen

100 % - 15% - 57 % = 28 %

Step 2:

P1/V1 = P2/V2

⇒ with p1 = the partial pressure of nitrogen = 0.88 atm

⇒ with V1 = the volme occupied by nitrogen = 15% ( assume 0.15 L)

⇒ with P2 = the partial pressure of hydrogen = TO BE DETERMINED

⇒ with V2 = the volume occupied by hydrogen = 28 % = 0.28 L

0.88/ 0.15 = X/0.28

X = 1.64 atm

The partial pressure of hydrogen gas = 1.64 atm

Answer:

a. P = 182 atm

b. due to daviation from ideal gas behavior

Explanation:

Data Given:

amount of CO = 122g

Volume of CO = .400 L

Temperature of CO = -71.2°C

Convert the temperature to Kelvin

T = °C + 273

T = -71.2 + 273

T = 201.8 K

a. Calculate the pressure exerted by the CO(g) in this system using the ideal gas equation (P) = ?

Solution:

To calculate Pressure by using ideal gas formula

                        PV = nRT

Rearrange the equation for Pressure

                       P = nRT / V . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant = 0.08206 L.atm / mol. K

For this we have to know the mole of the gas and the following formula will be used

                no. of moles = mass in grams / molar mass . . . . . . (2)

Molar mass of CO = 12 + 16 = 28 g/mol

Put values in equation 2

                  no. of moles = 122 g / 28 g/mol

                  no. of moles = 4.4 mol

Now put the value in formula (1) to calculate Pressure for CO

        P = 4.4 x 201.8 K x 0.08206 (L.atm/mol. K) / 0.400 L

          P = 182 atm

So the pressure will be 182 atm

__________

b. Data Given:

Actual pressure exerted by CO = 145 atm

expected pressure exerted by CO = 182 atm

why the actual pressure is less than what would be expected = ?

Explanation:

This is because of the deviation from ideal behavior of real gases.

The real gases approaches to ideal behavior under very high temperature and very low pressure.

But CO deviate from ideal behavior to give expected value for pressure, because it behave at high pressure and low temperature.

This non-ideal behavior is due to two postulate of ideal behavior

• gas molecules have negligible volume

• Gas molecules have negligible inter-molecular interaction

but these postulates not obeyed under real condition. so we calculated the pressure using ideal condition values for gas and obtained the expected value for pressure but the actual pressure value was detected under normal condition.

Answer:

Ionotophoresis

Explanation:

Ionotophoresis  is the process when "chemical ions are transported through intact skin using electrical current"and is usually used to treat skin infections or in order to create other benefits to the skin.

In order to use this method we need to use a low voltage current direct and we need to ensure a continuous mode with a long pulse duration in order to the ions can flow through the surface of interest.

So then the answer would be:

Ionotophoresis is the process where chemical ions are transported along the intact skin by an electrical current.

Explanation:

In order for water to condense from vapor to liquid, the temperature of the air must be at or below the dew point, and there must be condensation nuclei present. The seeding provides the nuclei, but the air temperature must also be below the temperature at which the air is saturated with water vapor.

Answer:

Balanced chemical equation:

CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)

Overall ionic equation:

Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)

Net ionic equation:

2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)

Amount of precipitate:

24.72 g

Explanation:

First, let's identify the compounds and the products of the reaction. Copper(II) chloride is CuCl₂, and lead(II) nitrate is Pb(NO₃)₂, after the reaction the products will be PbCl₂ and Cu(NO₃)₂, the first one is an insoluble salt, which will precipitate, and the second one is a soluble salt. So, the balanced chemical equation will be:

CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)

The ionic equation is done by putting the ions that are presented when the compound ionization at the aqueous solution. The metals have their charged expressed in the name of the compound, and chloride and nitrate have charge -1:

Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)

The net ionic equation is the simplified ionic equation. So let's eliminate the ions that are presented on both sides of the equation:

2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)

The stoichiometry of the reaction is 1 mol of CuCl₂ for 1 mol of PbCl₂(the precipitate). The molar mass of the compounds are:

CuCl₂ = 134.452 g/mol

PbCl₂ = 278.106 g/mol

1 mol of CuCl₂ ------------ 1 mol of PbCl₂

Transforming in mass:

134.452 g of CuCl₂ ----------------- 278.106 g of PbCl₂

11.95 g of CuCl₂ ---------------- x

By a simple direct three rule:

134.452x = 3323.3667

x = 24.72 g of PbCl₂

Balanced chemical equation:

CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)

Overall ionic equation:

Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)

Net ionic equation:

2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)

The maximum amount of precipitate that could be formed is 24.72 g.

Copper(II) chloride is CuCl₂, and lead(II) nitrate is Pb(NO₃)₂, these react together with to form PbCl₂ and Cu(NO₃)₂. The balanced chemical equation will be:

CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)

The metals have their charged expressed in the name of the compound, and chloride and nitrate have charge -1. The ionic equation will be:

Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)

Net-ionic equation:

2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)

The stoichiometry of the reaction is 1 mol of CuCl₂ for 1 mol of PbCl₂(the precipitate). The molar mass of the compounds are:

CuCl₂ = 134.452 g/mol

PbCl₂ = 278.106 g/mol

1 mol of CuCl₂ ------------> 1 mol of PbCl₂

Converting to mass:

134.452 g of CuCl₂ ----------------- 278.106 g of PbCl₂

11.95 g of CuCl₂ ---------------- x g

134.452x = 3323.3667

x = 24.72 g of PbCl₂

The maximum amount of precipitate that could be formed is 24.72 g.

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Answer:

The final pressure in the container at 0°C is 2.49 atm

Explanation:

We apply the Ideal Gases law to know the global pressure.

We need to know, the moles of each:

P He . V He = moles of He . R . 273K

(1atm . 4L) / R . 273K = moles of He  → 0.178 moles

P N₂ . V N₂ = moles of N₂ . R . 273K

(1atm . 6L) / R . 273K = moles of N₂ → 0.268 moles

P Ar . V Ar = moles of Ar . R . 273K

(1atm . 10L) / R . 273K = moles of Ar → 0.446 moles

Total moles: 0.892 moles

P . 8L = 0.892 mol . R . 273K

P = ( 0.892 . R . 273K) / 8L = 2.49 atm

R = 0.082 L.atm/mol.K

The final pressure in the evacuation container at 0 °C is 2.499 atm

To solve this question, we'll begin by calculating the number of mole of each gas. This can be obtained as follow:

Volume (V) = 4 L

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

1 × 4 = n × 0.0821 × 273

4 = n × 22.4133

Divide both side by 22.4133

n = 4 / 22.4133

Volume (V) = 6 L

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

1 × 6 = n × 0.0821 × 273

6 = n × 22.4133

Divide both side by 22.4133

n = 6 / 22.4133

Volume (V) = 10 L

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

PV = nRT

1 × 10 = n × 0.0821 × 273

10 = n × 22.4133

Divide both side by 22.4133

n = 10 / 22.4133

Next, we shall determine the total moles of the gas in the container.

Mole of He = 0.178 mole  

Mole of N₂ = 0.268 mole

Mole of Ar = 0.446 mole

Total mole = 0.178 + 0.268 + 0.446

Finally, we shall determine the pressure in evacuation container. This can be obtained as follow:

Volume (V) = 8 L

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

Number of mole (n) = 0.892 mole

P × 8 = 0.892 × 0.0821 × 273

P × 8 = 19.993

Divide both side by 8

P = 19.993 / 8

Therefore, the final pressure in the evacuation container at 0 °C is 2.499 atm

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