Can the ph scale be utilized for all acids (Arrhenius, Bronsted-Lowery, and Lewis)? Explain reasoning.

Answer:

3.17 V

Explanation:

The cell is operating under standard conditions. These standard conditions include; that the reaction takes place at 298 Kelvin (room temperature), the pressure of the system is 1 atmosphere (standard pressure), and the solutions have a Molarity of 1.0 M for both the anode and cathode solutions. All these conditions are satisfied in the cell under review in the question.

Hence;

E°anode (magnesium)= -2.37 V

E°cathode (silver) = 0.80 V

E°cell= E°cathode -E°anode

E°cell= 0.80-(-2.37)

E°cell= 0.80 + 2.37

E°cell= 3.17 V

Hence the standard cell potential of this cell at 25°C is 3.17 V

Answer:

The standard potential at 25ºC is 3.17 V.

Explanation:

The anode in  a galvanic cell is the electrode at which oxidation occurs and the cathode is the electrode  at which reduction occurs.

The overall cell reaction  will be the sum of two half-cell reactions. The standard reduction potentials are:

Mg²⁺ (1.0 M) + 2e⁻  →  Mg (s)          Eº = -2.37

Ag⁺ (1.0 M) + e⁻ → Ag (s)                 Eº= +0.80

Since the reactants are in their standard states (1.0 M) and at 25ºC we can write the half-cell reactions as follows:

Anode (oxidation):                        Mg (s) → Mg²⁺ (1.0 M) + 2e⁻

Cathode (reduction):     2Ag⁺ (1.0 M) + 2e⁻ → 2Ag (s)

Overall:            Mg (s) +2 Ag⁺ (1.0 M) + 2e⁻ → 2Ag (s) + Mg²⁺ (1.0 M) + 2e⁻

In order to balance the overall equation we multiply the reduction of  Ag⁺ by 2. We can do so because, as an intensive property, E° is not affected by  this procedure.

The standard emf of the cell, E°cell , which is composed of a contribution  from the anode and a contribution from the cathode, is given by:

[tex] Eº cell = Eº cathode - Eº anode [/tex]

[tex] Eº cell = EºAg⁺/Ag - Eº Mg²⁺/Mg [/tex]

[tex] Eº cell =0.80 V - (-2.37 V) [/tex]

Eº cell = 3.17 V

Answer:

Mass of precipitate (PbSO₄) formed = 0.696 g

Explanation:

The reaction of Lead II Nitrate and Sodium Sulfate is given as

Pb(NO₃)₂ (aq) + Na₂SO₄ (aq) → 2NaNO₃ (aq) + PbSO₄ (s)

The precipitate from this reaction is the PbSO₄

0.76 g of Pb(NO₃)₂ is available for reaction, we convert to number of moles

Number of moles = (Mass)/(Molar Mass)

Mass = 0.76 g

Molar mass of Pb(NO₃)₂ = 331.2 g/mol

Number of moles of Pb(NO₃)₂ = (0.76/331.2) = 0.002294686 moles = 0.002295 moles

25.00 ml of 0.2010M sodium sulfate is available for the reaction, we also conert to number of moles.

Number of moles = (Concentration in mol/L) × (Volume in L)

Concentration of Na₂SO₄ in mol/L = 0.201 M

Volume of Na₂SO₄ in L = (25/1000) = 0.025 L

Number of moles of Na₂SO₄ = 0.201 × 0.025 = 0.005025 moles

From the stoichiometric balance of the reaction, 1 mole of Pb(NO₃)₂ reacts with 1 mole of Na₂SO₄

Hence, this indicates that Pb(NO₃)₂ is the limiting reagent as it is in short supply and it will therefore dictate the amount of precipitate (PbSO₄) formed.

1 mole of Pb(NO₃)₂ gives 1 mole of the precipitate (PbSO₄)

0.002295 mole of Pb(NO₃)₂ will give 0.002295 mole of the precipitate (PbSO₄)

Mass = (Number of moles) × (Molar mass)

Number of moles of precipitate (PbSO₄) formed = 0.002295 mole

Molar mass of PbSO₄ = 303.26 g/mol

Mass of precipitate (PbSO₄) formed = 0.002295 × 303.26 = 0.6958864734 = 0.696 g

Hope this Helps!!!

Answer: The amount of energy required is 101 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed = ?

m= mass of gold = 30.0 g

c = specific heat capacity of gold = [tex]0.13J/g^0C[/tex]

Initial temperature of the water = [tex]T_i[/tex] = 15°C

Final temperature of the water = [tex]T_f[/tex]  = 41°C

Change in temperature ,[tex]\Delta T=T_f-T_i=(41-15)^0C=26^0C[/tex]

Putting in the values, we get:

[tex]Q=30.0g\times 0.13J/g^0C\times 26^0C[/tex]

[tex]Q=101J[/tex]

The amount of energy required is 101 Joules

Answer:

See attachment.

Explanation:

Mono-substituted cyclohexanes are more stable with their substituents in an equatorial position. However, with poly-substituted cyclohexanes, the situation is more complex since the steric effects of all substituents have to be taken into account. In this case, you can see that the interconversion is shifted towards the conformation in the bottom because there is less tension between the substituents.

Answer:

27.6mL of LiOH 0.250M

Explanation:

The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:

LiOH + HClO₂ → LiClO₂ + H₂O

That means, 1 mole of hydroxide reacts per mole of acid

Moles of  20.0 mL = 0.0200L of 0.345M chlorous acid are:

0.0200L ₓ (0.345mol / L) = 6.90x10⁻³ moles of HClO₂

To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:

6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =

Answer:

NaBr

Explanation:

Bromine's symbol is Br, not B, for B is for Boron

Answer:

23.3mL of 0.150M sodium thiosulfate

Explanation:

The  net reaction of sodium hypochlorite (NaClO) with sodium thiosulfate (Na₂S₂O₃) is:

NaClO + 2 Na₂S₂O₃ + 2H₃O⁺ → NaCl + 3 H₂O  +  Na₂S₄O₆ + 2 Na⁺

2.0mL of the sample of bleach are:

2.0mL ₓ (1.084g / mL) ₓ (6 / 100) ₓ (1 mol / 74.44g) = 1.75x10⁻³ moles of NaClO

As 1 mole of NaClO reacts with 2 moles of thiosulfate:

1.75x10⁻³ moles of NaClO ₓ (2 mol Na₂S₂O₃ / 1 mol NaClO) =

3.50x10⁻³ moles of Na₂S₂O₃

If you have a 0.150M solution of thiosulfate, mL of this solution you need to reach the end point of the titration are:

3.50x10⁻³ moles of Na₂S₂O₃ ₓ (1L / 0.150mol) = 0.0233L =

Answer:

139.98 g to nearest hundredth.

Explanation:

Using Avogadro's Number:

One mole (167.26 g) of Erbium  equates to  6.022141 * 10^23 atoms.

So 5.04 * 10^23 = 167.26 * 5.04/6.022141

= 139,98 g.

Answer:

0.188 moles

Explanation:

moles=mass/molar mass

moles= 19.2/ (32+32+38)

moles= 0.188

Answer:

0.188 moles.

Explanation:

Molar mass of SO2F2

=  32 + 2*16 + 2 * 19

= 102 g.

So 19.2g = 19.2 / 102

= 0.188 moles.

Answer:

45.6 mm

Explanation:

1 centimeter = 10 millimeter

4.56 cm = x mm

x = 4.56 × 10

= 45.6 mm

Answer:

Explanation:I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thnk me...

Answer:

The pressure at equilibrium of Br₂ is 0.10048 atm

Explanation:

Based on the reaction:

Br₂(g) + 3 F₂(g) ⇄ 2 BrF₃(g)

Kp is defined as:

[tex]Kp = \frac{P_{BrF_3}^2}{P_{Br_2}P_{F_2}^3}[/tex] = 5.4x10⁸

If initial pressures of Br₂ and F₂ are 0.30atm and 0.60atm respectively, the pressures in equilibrium are:

Br₂ = 0.30atm - X

F₂ = 0.60atm - 3X

BrF₃ = 2X

Replacing in Kp formula:

5.4x10⁸ =  [2X]² / [0.30atm - X] [0.60atm - 3X]³

5.4x10⁸ = 4X² / [0.30 - X] [0.216 - 3.24 X + 16.2 X² - 27 X³]

5.4x10⁸ = 4X² / 0.0648 - 1.188 X + 8.1 X² - 24.3 X³ + 27 X⁴

Solving for X:

X = 0.3000 → False answer because produce negative concentrations.

X = 0.19952. Replacing in equation of Br₂:

Br₂ = 0.30atm - 0.19952atm = 0.10048 atm

Answer:

GOOD DAY ON YOUR QUIZ

Explanation:

Answer:

1.25 g

Explanation:

Now we have to use the formula;

N/No = (1/2)^t/t1/2

N= mass of cesium-137 left after a time t (the unknown)

No= mass of cesium-137 present at the beginning = 5.0 g

t= time taken for 5.0 g of cesium-137 to decay =60 years

t1/2= half life of cesium-137= 30 years

Substituting values;

N/5= (1/2)^60/30

N/5= (1/2)^2

N/5= 1/4

4N= 5

N= 5/4

N= 1.25 g

Therefore, 1.25 g of cesium-137 will remain after 60 years.

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.  

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,  

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.  

Answer:

melting

Explanation:

Answer:

65450 in^3

Explanation:

Someone deleted my old answer

Answer:

A. Its temperature is uniform.

Explanation:

Cosmic Microwave Background radiation is a remnant of the Big Bang talks about the universe being very hot thereby causing its expansion with a corresponding cooling of the gases within it. The Cosmic Microwave Background radiation being what is left of the radiation after the Big Bang occurrence.

The temperature was uniform with very little fluctuations which makes the A the right choice.

Answer:

a

Explanation:

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

Where K' = K1 * K2

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just mechanism A and B are consistent with observed rate law

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just mechanism A is consistent with the observation of J. H. Sullivan

Answer:

botany because a maple tree is a plant

Explanation:

1. Octagon (8 sides)

2. Hexagon (6 sides)

3. Pentagon (5 sides)

4. Square/Rectangle (4 sides)

5. Triangle (3 sides)

Hope this helps :)

Answer:

D

Explanation:

Your supervisor is in charge of you so therefore you must always tell your supervisor first.

Answer:

C.

Explanation:

Because you might get hurt, could be dangerous.

Answer:

The correct answer is 116.3 g/mol

Explanation:

When a solute is added to a solvent, the freezing point decreases. That is called freezeing point depression and is given by the following expression:

ΔTf= Kf x m

Where ΔTf is the decrease in freezing point (final temperature - initial temperature), Kf is the freezeing point constant, m is the molality of the solution.

We have the following data:

Kf= 5.12ºC/m

ΔTf= 5.48ºC - 3.77ºC = 1.71ºC

We can calculate the molality of the solution from the mathematical expression for freezeing point as follows:

m = ΔTf/Kf = 1.71ºC/(5.12º C/m)= 0.334 m

The molality is the number of moles of solute in 1 kg of solvent. The solute was added to 500 g of solvent (benzene), so we can multiply the molality by the grams of solvent (500 g = 0.5 kg) to obtain the number of moles:

0.334 mol/kg x 0.5 kg = 0.172 moles

We know that the molar mass is equal to the mass of compound per mole, so we divide the mass into the moles to obtain the molar mass as follows:

MM = mass/moles= 20.0 g/0.172 moles = 116.3 g/mol

Answer:

1) Option A is correct.

The covalent bond is the chemical bond that is established by pooling of electrons.

2) Option B is correct.

The covalent bond is established between two non-metals.

3) Option B is correct.

A covalent bond is established between nitrogen and hydrogen.

4) Option B is correct.

Covalent bonds are of two types; polar covalent bonds and non-polar covalent bonds.

5) Option B is correct.

Select the substances in the molecules of which the atoms are united by ionic bond.

potassium chloride KCl

1) Opțiunea A este corectă.

Legătura covalentă este legătura chimică care se stabilește prin punerea în comun a electronilor.

2) Opțiunea B este corectă.

Legătura covalentă este stabilită între două nemetale.

3) Opțiunea B este corectă.

Se stabilește o legătură covalentă între azot și hidrogen.

4) Opțiunea B este corectă.

Legăturile covalente sunt de două tipuri; legături covalente polare și legături covalente nepolare.

5) Opțiunea B este corectă.

Selectați substanțele din moleculele cărora atomii sunt uniți prin legătură ionică. clorură de potasiu KCl

Explanation:

English Translation

1) The covalent bond is the chemical bond that is established: a) by pooling electrons; b) by electron transfer; c) by pooling or electron transfer.

2) The covalent bond is established between: a) two metals; b) two non-metals; c) a metal and a non-metal.

3) A covalent bond is established: a) between hydrogen and sodium; b) between nitrogen and hydrogen; c) between aluminum and oxygen.

4) Covalent bonds are of two types: a) single covalent bonds and double covalent bonds; b) polar covalent bonds and non-polar covalent bonds; c) no answer is correct.

5) Select the substances in the molecules of which the atoms are united by ionic bond: a) phosphorus oxide P2O5; b) potassium chloride KCl; c) H2O water

Solution

Covalent bonds are bonds that result from the sharing/pooling of electrons by two non-metallic atoms. When the electronegativity difference between the two atoms in a cobalent bond is substantial, a polar covalent bond results (in which the shared electrons are closer to the more electronegative atom of the pair). If the electronegativity difference isn't significant, a non-polar covalent bond ensues, in which the shared electrons are approximately equidistant from both atoms.

1) The covalent bond is the chemical bond that is established by pooling of electrons.

Already explained above.

2) The covalent bond is established between two non-metals.

Already explained too.

3) A covalent bond is established between nitrogen and hydrogen.

Of the three options, only option B contains two elements, nitrogen and hydrogen, which are both non-metals amd this satisfy the condition for the formation of a covalent bond.

4) Covalent bonds are of two types; polar covalent bonds and non-polar covalent bonds.

Already explained above.

5) Select the substances in the molecules of which the atoms are united by ionic bond.

potassium chloride KCl

Ionic bonds result from the transfer of electrons from metals (electropositive elements) to non-metals (electronegative elements). Of the 3 options provided, only potassium chloride is a coming together of a metal and a non metal.

In Romanian/In limba romana

Legăturile covalente sunt legături care rezultă din împărțirea / îmbinarea electronilor de către doi atomi nemetalici. Când diferența de electronegativitate între cei doi atomi dintr-o legătură cobalentă este substanțială, rezultă o legătură covalentă polară (în care electronii partajați sunt mai apropiați de atomul mai electronegativ al perechii). Dacă diferența de electronegativitate nu este semnificativă, apare o legătură covalentă non-polară, în care electronii partajați sunt aproximativ echidistanți de la ambii atomi.

Legăturile ionice rezultă din transferul electronilor de la metale (elemente electropozitive) la nemetalele (elemente electronegative).

Hope this Helps!!!

Sper că acest lucru vă ajută!!!!

Answer:

The correct answer is option C, that is, 1.36 L.

Explanation:

The reaction mentioned in the question is:  

Zn (s) + 2HCl (aq) ⇒ H2 (g) + ZnCl2 (aq)

It is clear that one mole of zinc is generating one mole of hydrogen gas as seen in the reaction. The mass of zinc mentioned in the question is 3.566 grams, the no of moles can be determined by using the formula,  

n = mass / molecular mass

The molecular mass of zinc is 65.39 g/mol, now putting the values in the formula we get,  

n = 3.566 g/ 65.39 g/mol

= 0.0545 mol

Based on the question, the partial pressure of water vapor is 18.65 mmHg and the atmospheric pressure is 752 mmHg. Therefore, the pressure of hydrogen gas will be,  

Pressure of hydrogen gas (H2) = 752 mmHg - 18.65 mmHg

= 733.35 mmHg

The liters of hydrogen gas produced can be calculated by using the equation, PV =nRT

R is the gas constant, having the value 62.36 L mmHg/K/mol and T is the temperature (273 + 21 = 294 K).  

Now putting the values in the equation we get,  

733.35 mmHg * V = 0.0545 mol * 62.36 mmHg/K/mol * 294 K

= 999.19 L / 733.35

= 1.36 L

Hence, the volume of hydrogen gas is 1.36 L.  

Answer:

a. Impurities

c. Recrystallizing solvent

Explanation:

In this type of reaction the products are never considered totally pure, that is why as a final product it must always be taken into account that it is proportional, it will be a recrystallization solvent and other impurities with which the products were mixed.

Answer:

[tex]m_{NaCl}=33g[/tex]

Explanation:

Hello,

In this case, in order to compute the grams of sodium chloride starting by the molecules, the first step is to compute the moles contained in the given amount of molecules by using the Avogadro's number:

[tex]n_{NaCl}=3.4x10^{23}molecules*\frac{1mol}{6.022x10^{23}molecules} =0.56mol[/tex]

Then, by using the molar mass of sodium chloride (58.45 g/mol) we can directly compute the grams:

[tex]m_{NaCl}=0.56mol*\frac{58.45g}{1mol} \\\\m_{NaCl}=33g[/tex]

Regards.

Answer:  The property is called Luster

Explanation:

Luster is a term used to describe the appearance of  light from the surface of the mineral when it reflects light. The two types of classification of Luster is Mettalic or Non Mettalic.

some examples of minerals with mettalic lustre are --- Galena, graphite, Gold, Silver, Copper etc

Some examples of minerals with non mettalic lustre are--- Barite, Serpentine, Fluorite etc.

Answer:

Lustrous property

Explanation:

Materials like metals that have lustre usually glow or display sheen when light comes in contact with the surface of the material.

It is derived from the Latin word "lux" meaning "light", which is indicative of radiance or gloss.

Lustre is a common property of metals, including sonourousity, malleability and ductility amongst others.

Answer:

The reaction is a double displacement reaction

Explanation:

Let us consider the reaction equation of the reaction between ammonium oxalate and lithium acetate.

(NH4)2C2O4(aq) +2 CH3COOLi(aq) -------> 2NH4CH3CO2(aq) + Li2C2O4(s)

This is a displacement reaction. A double displacement reaction is a type of reaction in which two reactants exchange their ions to form two new compounds. Double displacement reactions usually lead to the formation of a solid product which is also called a precipitate.

The general form of a Double displacement reaction is of the format:

AB + CD → AD + CB

Where A,B,C and D represents different ions respectively.

A double displacement reaction can also be referred to as salt metathesis reaction, double replacement reaction, exchange reaction, or a double decomposition reaction, although the latter term is more strictly used when one or more of the reactants does not dissolve in the solvent.

Answer:

198.8mL

Explanation:

Step 1:

Data obtained from the question:

Molarity of base (Mb) = 3.5M

Volume of base (Vb) =..?

Molarity of acid (Va) = 2.75M

Volume of acid (Va) = 253mL

Step 2:

The balanced equation for the reaction.

HF + LiOH —> LiF + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 3:

Determination of the volume of the base, LiOH needed for the reaction.

The volume of the base needed for the reaction can be obtained as follow:

MaVa /MbVb = nA/nB

2.75 x 253 / 3.5 x Vb = 1

Cross multiply

3.5 x Vb = 2.75 x 253

Divide both side by 3.5

Vb = 2.75 x 253 / 3.5

Vb = 198.8mL

Therefore, the volume of the base needed for the reaction is 198.8mL