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What is the standard cell potential of an electrochemical cell?

the standard electrode potential of the anode, relative to a hydrogen cell
the difference between the standard electrode potential of the two half-cells
the standard electrode potential of the cathode, relative to a hydrogen cell
the sum of standard electrode potential of the two half-cells

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:

There will be produced 9L of CH4 and 18 L of H2S. There will remain 3 L of CS2

Explanation:

Step 1: Data given

volume of H2 = 36.00 L

volume of CS2 = 12 L

Step 2 = the balanced equation

4H2(g) + CS2(g) → CH4(g) + 2H2S(g)

Step 3: Calculate number of moles of H2

1 mol = 22.4 L

36 L = 1.607 mol

Step 4: Calculate moles of CS2

1 mol = 22.4 L

12 L = 0.5357 moles

Step 5: Calculate the limiting reactant

For 4 moles of H2 we need 1 mol of CS2 to produce 1 mol of CH4 and 2 moles of H2S

H2 is the limiting reactant. It will completely be consumed. ( 1.607 moles)

CS2 is in excess. There will react 1.607/4 = 0.40175 moles

There will remain 0.5357 - 0.40175 = 0.13395 moles of CS2

0.13395 moles of CS2 = 3 L

Step 6: Calculate products

For 4 moles of H2 we need 1 mol of CS2 to produce 1 mol of CH4 and 2 moles of H2S

For 1.607 moles of H2 we'll have 0.40175 moles of CH4 (= 9L) and 0.8035 moles of H2S =(18L)

There will be produced 9L of CH4 and 18 L of H2S. There will remain 3 L of CS2

Answer:

It will reach to its maximum volume.

Explanation:

Using Ideal gas equation for same mole of gas as :-

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]

Given ,  

V₁ = 65.0 mL

V₂ = ?

P₁ = 745 torr

The conversion of P(torr) to P(atm) is shown below:

[tex]P(torr)=\frac {1}{760}\times P(atm)[/tex]

So,  

Pressure = 745 / 760 atm = 0.9803 atm

P₁ = 0.9803 atm

P₂ = 0.066 atm

T₁ = 25 ºC

T₂ = -5 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

T₂ = (-5 + 273.15) K = 268.15 K  

Using above equation as:

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]

[tex] \frac{{0.9803}\times {65.0}}{298.15}=\frac{{0.066}\times {V_2}}{268.15}[/tex]

Solving for V₂ as:-

[tex]V_2=\frac{0.9803\times \:65.0\times 268.15}{298.15\times 0.066}[/tex]

[tex]V_2=\frac{17086.38392}{19.6779}[/tex]

V₂ = 868 L

Given that:- V max = 835 L

Thus, it will reach to its maximum volume.

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.

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⁻

Answer:

Pt = 0.392 atm

Explanation:

First, we must separate the initial condition for each gas:

- N₂ :

V₁ = 1 Lt , T₁ = 308 K , P₁ = 1 atm

- H₂O :

V₂= 1 Lt , T₂= 308 K, P₂= 24 mm Hg = 0.032 atm

- C₂H₆O:

V₃= 1 Lt, T₂= 308 K, P₃= 102 mm Hg = 0.134 atm

For this case, we consider that 1 atm = 760 mm Hg

We then calculate the number of moles for each case (in gas phase), and we consider that al gasses behave like an ideal gas:

PV = nRT we consider R = 0.082 Lt*atm/K*mol

- N₂ :

n H₂ = PV/RT = 1x1/0.082x308 = 0.04

- H₂O:

n H₂O = PV/RT = 0.032x1/0.082x308 = 1.27x10⁻³

- C₂H₆O:

n C₂H₆O = PV/RT = 0.134x1/0.082x308 = 5.31x10⁻³

Finally, when the stopcocks are open and the 3 gases are mixed together, T will remain as 308 K, total volume will be the volume of the three flasks (3 Lt), so total pressure (Pt) after mixing the 3 gases will be:

Pt = (n N₂ + n H₂O + n C₂H₆O)xRT/V = (0.04658x0.082x308)/3

Pt = 0.392 atm

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:

[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. 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:

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:

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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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:

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:

30.25°C

Explanation:

The calorimeter is an equipment used to measure the combustion enthalpy of a substance. The heat loss in the reaction is used to heat the water and the equipment. By the conservation of energy:

Qcombustion + Qcalorimeter + Qwater = 0

Because there is no phase change:

Qcalorimeter = C*ΔT, where C is the heat capacity, and ΔT the variation in temperature (final - initial)

Qwater = m*c*ΔT, where m is the mass, and c is the specific heat (4.184 J/g°C).

The molar mass of lactose is 342.3 g/mol, so the number of moles in 2.50 g is:

n = mass/molar mass

n = 2.50/342.3

n = 0.0073 mol

Qcombustion = -5652 kJ/mol * 0.0073 mol

Qcombustion = -41.28 kJ

Qcombustion = - 41280 J

Thus,

-41280 + 1630*(T - 24.58) + 1350*4.184*(T - 24.58) = 0

(T - 24.58) * (1630 + 5648.4) = 41280

7278.4(T - 24.58) = 41280

T - 24.58 = 5.67

T = 30.25°C

The student needs to calculate the final temperature of water in a calorimeter after lactose combustion by using the initial temperature, lactose's enthalpy of combustion, and heat released by the combustion is absorbed by the water and calorimeter, raising their temperature.

The student wants to calculate the final temperature of water in a calorimeter after burning a 2.50 g sample of lactose with an enthalpy of combustion of -5652 kJ/mol. To solve this, we need to find the amount of heat absorbed by the water and calorimeter, which can be calculated using the specific heat capacities of water and the calorimeter and the mass of the water.

First, the number of moles of lactose combusted is found by dividing the mass in grams (2.50 g) by the molar mass of lactose ($C_{12}H_{22}O_{11}$). Then, the total heat released by the combustion of lactose is calculated by multiplying the number of moles by the enthalpy of combustion. Since the heat of combustion is expressed in kJ/mol, and the heat capacity of the calorimeter is in J/°C, we must ensure our units are consistent when calculating the energy absorbed by the water and calorimeter. The final temperature is obtained by adding the temperature change to the initial temperature of 24.58°C.

Given the initial temperature and the enthalpy of combustion of lactose, the final temperature can be found after determining the heat transferred to the water and calorimeter.

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:

[tex]1s^22s^22p^63s^23p^63d^{4}[/tex]

Explanation:

Manganese is the element of group 7 and forth period. The atomic number of Manganese is 25 and the symbol of the element is Mn.

The electronic configuration of the element, manganese is -  

[tex]1s^22s^22p^63s^23p^63d^{5}4s^2[/tex]

To form [tex]Mn^{3+}[/tex], it will lose 3 electrons from the valence electrons and thus the configuration of the ion is:-

[tex]1s^22s^22p^63s^23p^63d^{4}[/tex]

The ground state electron configuration for the Mn3+ ion is 1s2 2s2 2p6 3s2 3p6 3d4 after removing three electrons from the highest energy orbitals, which are the 4s and 3d in the case of manganese.

The ground state electron configuration for the Mn3+ ion starts with the neutral manganese atom, which has an atomic number of 25. The neutral Mn atom would have an electron configuration of 1s2 2s2 2p6 3s2 3p6 3d5 4s2. However, for Mn3+, three electrons are removed to account for the +3 charge, typically from the outermost orbitals first which in this case are the 4s and 3d orbitals. Since the 4s electrons are generally removed before the 3d electrons, after losing two electrons from 4s and one electron from 3d, the configuration will be 1s2 2s2 2p6 3s2 3p6 3d4.

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:

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:

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.

The mechanisms for producing concentrated urine are countercurrent multiplication, ADH secretion, and osmoregulation. Excretion of excess water is not involved.

In the production of concentrated urine, all of the following mechanisms are involved:

The mechanism that is not involved in producing concentrated urine is the excretion of excess water.

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In producing a concentrated urine, the kidney uses several mechanisms including high NaCl concentration, ADH secretion, aquaporin insertion and facultative water reabsorption, but not obligatory water reabsorption in the proximal convoluted tubule. therefore, option D is correct

The key mechanism for producing concentrated urine in the kidney involves several steps, but the option D, obligatory water reabsorption in the proximal convoluted tubule, does not directly contribute to the urine's concentration. This obligatory water reabsorption happens regardless of your body's state of hydration, and it reabsorbs about 70% of the water from the urine before it reaches the section of the kidney where the urine's concentration is adjusted. The other options - a high concentration of NaCl in the interstitial fluid, secretion of ADH, insertion of aquaporins, and increase in facultative water reabsorption - are all steps which allow the regulation of the urine's concentration depending on the body's need to retain or expel water.

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the complete Question is given below

The mechanism which the kidney uses to produce a concentrated urine involves all of the following EXCEPT

A. a high concentration of NaCl in the interstitial fluid that surrounds the collecting ducts.

B. secretion of ADH by the posterior pituitary gland.

C, aquaporins being inserted into the membranes of the collecting duct cells.

D. obligatory water reabsorption in the proximal convoluted tubule.

E. an increase in facultative water reabsorption.

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

Explanation:

The two figures are missing so I can't tell which reactions will proceed faster, but the answer will follow this logic:

If the two samples of Mg(s) are equal, the two amounts of HCl(aq) are equal, the operation conditions are the same (Temperature, pressure,etc.), the difference in the speed of reaction will only depend difference in the quality of mixing of the Mg(s).

Being that said:

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 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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The element with the correct symbol for the ion, and the name of the ion has been fluorine, [tex]\rm F^-[/tex],  Fluoride. ion. The correct option is a.

The elements tend to attain the stable electronic configuration and acquire the noble gas configuration. The elements gain electrons and acquire a -ve charge while losing electrons attain a +ve charge onto the element.

The IUPAC nomenclature gives the addition '-ide' as a suffix to the metal losing electrons.

Thus, the element with the correct symbol for the ion, and the name of the ion has been fluorine, [tex]\rm F^-[/tex],  Fluoride. ion. The correct option is a.

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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:

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:

In order to determine the rate at which the sediment layers were deposited

Explanation:

Iridium is an important element that belongs to the Platinum group and they are dominantly present in the asteroids and comets. They are the key evidence that suggests the occurrence of an asteroidal impact or a mass extinction event that has taken place in the geological past.

The presence of Iridium mixed with the clay sediments in the boundary between the Cretaceous and Tertiary (K-T boundary) suggested the mass extinction event that wiped out numerous life forms from the earth.

The famous scientist Dr. Luis Walter Alvarez suggested measuring the concentration of Iridium in this K-T boundary in order to determine the rate at which these sediment layers were deposited. They were able to determine that these Iridium elements were present due to the asteroid impact on earth, that wiped out the dinosaurs from the earth. They also considered that this element can be produced from the eruptions of volcanoes.

Answer:

The presence of Iridium mixed with the clay sediments in the boundary between the Cretaceous and Tertiary (K-T boundary) suggested the mass extinction event that wiped out numerous life forms from the earth.

The famous scientist Dr. Luis Walter Alvarez suggested measuring the concentration of Iridium in this K-T boundary in order to determine the rate at which these sediment layers were deposited. They were able to determine that these Iridium elements were present due to the asteroid impact on earth, that wiped out the dinosaurs from the earth. They also considered that this element can be produced from the eruptions of volcanoes.

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Explanation:Iridium is a component of cosmic dust that rains down upon the earth at a constant rate. Why did Luis Alvarez suggest measuring iridium levels in the K-T boundary?

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.

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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