Oxidation Unit Test
Use the galvanic cell reaction to answer the question.
2Cr(s) + 3Cu2+(aq) → 2Cr3+ (aq) + 3Cu(s)
Which half reaction occurs at the cathode?
(1 point)

Answers

Answer 1

The reduction half equation of the galvanic cell is is;  3Cu2+(aq)  + 3e ------> 3Cu(s).

A galvanic cell is a cell that produces chemical energy by a spontaneous chemical reaction. The equation of the reaction is; 2Cr(s) + 3Cu2+(aq) → 2Cr3+ (aq) + 3Cu(s). We can see that chromium was oxidized and copper was reduced.

Since reduction occurs at the cathode, it follows that the reduction half equation is;  3Cu2+(aq)  + 3e ------> 3Cu(s).

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Related Questions

Use the diagram below to answer 3 questions

Answers

Answer:

The element X is sulfur.

Sulfur will gain 2 electrons to become an ion.

The charge of sulfide ion is (2-).

Explanation:

If 38.6 grams of iron react with an excess of bromine gas, what mass of FeBr2 can form?

Answers

Answer:

›› FeBr2 molecular weight. Molar mass of FeBr2 = 215.653 g/mol. This compound is also known as Iron(II) Bromide. Convert grams FeBr2 to moles or moles FeBr2 to grams. Molecular weight calculation: 55.845 + 79.904*2 ›› Percent composition by element

Explanation:

If 38.6 grams of iron react with an excess of bromine gas, the mass of FeBr2 can form is 149 grams.

What is mass?

Mass is defined as a way to gauge how much matter there is in a substance or thing. The kilogram (kg) is the fundamental SI unit of mass, while lower masses can also be measured in grams (g). Atoms make up everyday matter. A majority of an atom's mass is contained in its nucleus.

Given Fe = 38.6 g.

Fe has a molar mass = 55.845 g/mol.

Given mass/molar mass equals 38.6g/55.845gmol-1, or 0.6912 moles of iron.

The reaction is described as Fe + Br2 FeBr2.

One mole Fe yields 1 mole of FeBr2.

FeBr2 would be produced from 0.6912 moles of Fe.

FeBr2 has a molar mass of 215.65 g/mol.

Moles of FeBr2 x Molar mass of FeBr2

= 215.65 g/mole x 0.6912 mole

= 149.06 g FeBr2 produced is the formula.

Thus, if 38.6 grams of iron react with an excess of bromine gas, the mass of FeBr2 can form is 149 grams.

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Doing Labs at home

I’m a junior and I’m staying home for this semester and I have to take chemistry and a lot of my work is Labs but I don’t know how to do them since I don’t have the materials at home to do the labs. Someone please help!!!

Answers

Answer:

go get the stuff.

Explanation:

the solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm. what is the solubility of nitrogen gas in a deepsea divers blood at a depth of 200 feet and pressure of 7.00 atm

Answers

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

We want to relate the solubility of a gas with its partial pressure.

We can do so using Henry's law.

What does Henry's law state?

Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

C = k × P

where,

C is the concentration of a dissolved gas. k is the Henry's Law constant. P partial pressure of the gas.

The solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm.

Since the solvent is basically water, we can understand that the concentration of nitrogen gas is 1.90 mL/dL at 1.00 atm.

We can use this information to calculate Henry's Law constant.

k = C/P = (1.90 mL/dL)/1.00 atm = 1.90 mL/dL.atm

We want to calculate the solubility of nitrogen gas at a pressure of 7.00 atm.

We will use Henry's law.

C = k × P = (1.90 mL/dL.atm) × 7.00 atm = 13.3 mL/dL

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

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A metal (FW 341.1 g/mol) crystallizes into a body-centered cubic unit cell and has a radius of 1.74 Angstrom. What is the density of this metal in g/cm3

Answers

This  problem provides the molar mass and radius of a metal that has an BCC unit cell and the density is required.

Firstly, we consider the formula that relates molar mass and also includes the Avogadro's number and the volume of the unit cell:

[tex]\rho =\frac{Z*M}{V*N_A}[/tex]

Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since BCC is able to hold 2 atoms and M and NA are given, we calculate the volume of the atom in the unit cell given the radius in meters:

[tex]V=a^3=(\frac{4R}{\sqrt{3} } )^3=(\frac{4*1.74x10^{-10}m}{\sqrt{3} } )^3=6.49x10^{-29}m^3[/tex]

 

And finally the required density in g/cm³:

[tex]\rho =\frac{2*341.1g/mol}{6.49x10^{-29}m^3\frac{m^3}{atom} *6.022x10^{23}\frac{atom}{mol} } =17455257.8g/m^3\\\\\rho=17.5g/cm^3[/tex]

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Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 8.32 percent of its original value

Answers

Answer:

la primera va con la última

Thin-layer chromatography explain ?????​

Answers

Answer:

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. ... After the sample has been applied on the plate, a solvent or solvent mixture (known as the mobile phase) is drawn up the plate via capillary action.

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Water waves in a small tank are 6.0 cm long. They pass at a given point at a rate of 4.8 waves per second. What is the speed of the wave?

Answers

Answer:,   Correct option is    0.288m/s

Explanation:    

The relationship between the velocity of the wave, its wavelength and frequency is given by the formula

Wavelengthλ=

Frequency(ν)

Speed(v)

,

where, v - velocity of the wave

           λ - wavelength of the wave

          f - frequency of the wave.

In the question it is given that the frequency is 4.8 Hz and the wavelength is 6.0 cm, that is, 0.06 meters.

The velocity of the sound is calculated as follows.

v=f×λ=4.8 Hz×0.06 m=0.288 m/s

Hence, the speed of the water wave is 0.288 m/s.

Al2(SO3)3
a. Count the number of Sulfur atom
b. How many total atoms are given in the compound
Please helppp

Answers

Answer:

from the words below underline six example of rhetorical patterns

Suppose you have samples of three unknown solids. Explain how you could use their properties to
determine whether or not they are ionic solids.

Answers

Using melting and boiling temperature, hardness and electric current passing testing.

Ionic solids

Ionic solids are materials that have a strong bond between their ions, thus producing well-defined shapes.

In addition, due to this strong attraction, the boiling and melting temperatures of these materials are very high, in addition to the resistance to breakage presented by them.

Finally, ionic solids are also excellent conductors of electricity.

So, their properties used to determine whether or not they are ionic solids are  melting and boiling temperature, hardness and electric current passing testing.

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The boiling point of a substance is tested. After 10 tests, the result is given as 37+/−3°C. Which conclusion can be drawn from this result? (1 point)

The scientists do not need to collect more data because they have narrowed down the range of the results.

The scientists should not report these results until they have the exact number.

The actual boiling point is either 34°C or 40°C.

The actual boiling point is probably between 34°C and 40°C.

Answers

Answer:

Explanation:

The actual boiling point is probably between 34C and 40C.

Temperature measures the average kinetic energy of particles of the substances. Therefore, the correct option is option D that is the actual boiling point is probably between 34°C and 40°C.

What is temperature?

Temperature is used to measure degree or intensity of heat of a particular substance. Temperature is measured by an instrument called thermometer.

Temperature can be measured in degree Celsius °c, Kelvin k or in Fahrenheit. Temperature is a physical quantity. Heat always flow from higher temperature source to lower temperature source.

We can convert these units of temperature into one another. The relationship between degree Celsius and Fahrenheit can be expressed as:

°C={5(°F-32)}÷9

The actual boiling point is probably between 34°C and 40°C.

Therefore, the correct option is option D.

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How many of sodium (Na) are needed to make 4.5 liters of a 1.5mol/L of Na solution?​

Answers

Answer:

Explanation:

First you will find the mole from the molarity and then the desired mass from the mole.

A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction to separate the amine from the mixture

Answers

There are different ways of extraction. The best way to begin an extraction to separate the amine from the mixture is to extract with dilute NaOH.

An acid-base extraction is often used in the extraction of carboxylic acids from the organic layer and thereafter into the aqueous layer.

NaOH is known to be the most common compound that is used to convert a carboxylic acid into its more water-soluble ionic carboxylate form.

But if  the mixture has a compound that you want, and that can react with NaOH, another milder base such as sodium bicarbonate is preferably used.

See full question below

A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction in order to separate the carboxylic acid from the mixture?

A) Extract with dilute NaOH

B) Extract with dilute HCl

C) Extract with dichloromethane

D) Extract with water

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At 298 K, the reaction 2 HF (g) ⇌ H2 (g) + F2 (g) has an equilibrium constant Kc of 8.70x10-3. If the equlibrium concentrations of H2 and F2 gas are both 1.33x10-3 M, determine the initial concentration of HF gas assuming you only started with HF gas and no products initially.

Answers

This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.

Then, we can write the following equilibrium expression for hydrofluoric acid once the change, [tex]x[/tex], has taken place:

[tex][HF]=[HF]_0-2x[/tex]

Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):

[tex]8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }[/tex]

[tex][HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M[/tex]

Finally, the initial concentration of HF is calculated as follows:

[tex][HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M[/tex]

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Which stage of a fire can be described as follows: The flames are not visible and the combustible item no longer generates heat or combustion products

Answers

Fire extinguishment is referred to the stage in which flames are not visible

and the combustible item no longer generates heat.

This stage refers to when the fire has been put out through a fire

extinguisher or other compounds.   When fire is put out, there is no longer

flames present.

The absence of flames also means that there is no heat generation or

combustion products due to the absence of a heat source.

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Help !

Student A uses 3N of force to move a cart 5 meters in 10 seconds. Student B uses 6N of force to move the same cart the same distance in 5 seconds. Which student did more work? Which student used more power? Use evidence and explain your answersss

Work = force x distance

Power = work/time

Answers

Work is said to be done if the force applied to a body causes the body to move through a distance.

Student B used more work and power

Workdone = Force * Distance

For student A:

Force = 3N

distance = 5meters

Time taken = 10 secs

Workdone by Student A = 3 * 5

Workdone by student A = 15Nm

Power used up = workdone/time

Power used up = 15/10

Power used up = 1.5 Watts

For student B:

Force = 6N

distance = 5meters

Time taken = 5 secs

Workdone by Student A = 6 * 5

Workdone by student A = 30Nm

Power used up = workdone/time

Power used up = 30/5

Power used up = 6 Watts

This shows that student B used more work and power

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Oil does not dissolve in water because

Answers

Explanation:

Oils are made up of long hydrocarbon chains which are non polar. Since water is a polar solvent oil doesn't dissolve in water well.( Polar solutes dissolve in polar solvents and non polar solutes dissolve in non polar solvents) The hydrocarbon chains are hydrophobic.

he hybridization of carbon in diamond is _________. Enter your answer with no superscripts or subscripts, i.e., ab3.

Answers

Diamond is composed of hexagonal rings in which sp3 hybridized carbon atoms are linked together.

Hybridization refers to the mixing of atomic orbitals to yield hybrid orbitals that are suitable for bonding. The energy of orbitals that combine to form hybrid orbitals must be close enough for such combination to take place.

Diamond is composed of hexagonal rings in which sp3 hbridized carbon atoms are linked together. Each carbon atom in diamond is tetrahedral.

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The mixing of the two different orbitals to form a compound is called hybridization. For example mixing of s and p orbits.

The correct answer is sp3.

The arrangement of the elements in a different manner to form a new compound is called allotropes. For example, diamond and graphite are the allotropes of carbon.

The valence electrons are in p orbitals and p orbit mixed after the s orbitals.

Therefore, the correct answer is sp3

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The sum of the number of proteins and neutrons in an atoms nucleus is its __________ ___________.

Answers

Answer:

Mass Number

Explanation:

In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.

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Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)

Answers

The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s)   (1)

We can write the above reaction in two reactions, one for oxidation and the other for reduction:

Oxidation reaction

Li⁰(s) → Li⁺(aq) + e⁻   (2)

Reduction reaction

Fe²⁺(aq) + 2e⁻ → Fe⁰(s)    (3)

We can see that Li⁰ is oxidizing to Li⁺ (by losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).  

We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.

In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).  

Therefore, the reducing agent in reaction (1) is lithium (Li).  

 

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2 nitrogen atoms and five chlorine atoms what compound does that make

Answers

Answer:

dinitrogen pentachloride

In experiment 9, in one operation, we heat up the alcohol with acid and do a concurrent distillation. What was the purpose of doing this

Answers

Answer:

we heat up because the component with lower boiling evaporates first,

leaving the other behind

To what pH should you adjust a standard hydrogen electrode to get an electrode potential of -0.128 V ? (Assume that the partial pressure of hydrogen gas remains at 1 atm.) Express your answer using two decimal places.\

Answers

The pH of the standard hydrogen electrode that has electrode potential of -0.128 V  is 4.3.

The equation of the hydrogen electrode is;

2H^+(aq) + 2e ⇄ H2(g)

The standard electrode potential of hydrogen is 0.00 V

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Now;

E°cell = 0.00 V

n = 2

Q = 1/[H^+]

-0.128 = 0.00 - 0.0592/2 log  1/[H^+]

-0.128 = 0.00 - 0.0296 log 1/[H^+]

 -0.128 =  - 0.0296 log 1/[H^+]

-0.128/  - 0.0296  =  log 1/[H^+]

1/[H^+] = Antilog (4.32)

[H^+] = 4.79 × 10^-5

Now;

pH = -log[H^+]

pH = -log (4.79 × 10^-5)

pH = 4.3

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Pleeeeasee someone who’s good at chemistry?! 10 grade

ASAP
I’ll give points, just help please

Answers

Answer:

where is the question????????????

solvent extraction explain ???​

Answers

Answer:

this the partial removal of a substance from a solution or mixture by dissolving it in another immiscible solvent in which it is more soluble.

g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The height of the cylinder is 30 cm. The outside pressure is 105 Pa. The temperature of the gas is kept at 250 K throughout the experiment. The volume filled by the gas is 2.0 l. Now assume that solid cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. Cylinder and piston have the same diameter. Assume that the kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Calculate the change of entropy of the gas and of the environment. Please read this text very carefully

Answers

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

[tex]v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s[/tex]

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

[tex]K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J[/tex]

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

[tex]\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K[/tex]

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, [tex]\Delta S_{env} = -1.18 J/K[/tex]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

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Convert 1.36x10 to standard form​

Answers

Answer:

13.6 is the correct answer written in standard form.

Explanation:

1.36, move the decimal once to the right to get 13.6

Answer:

13.6

Explanation:

The standard form is 13.6

Using the Kf value of 1.2×109 calculate the concentration of Ni2+(aq) and Ni(NH3)62+ that are present at equilibrium after dissolving 1.47 gNiCl2 in 100.0 mL of NH3(aq) solution such that the equilibrium concentration of NH3 is equal to 0.20 M.
Express your answers in moles per liter to two significant figures separated by a comma.

Answers

This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.

Firstly, we can write out the chemical equation to be considered:

[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]

Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:

[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]

Afterwards, we set up an equilibrium expression for this chemical reaction:

[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]

Which can be written in terms of the reaction extent, [tex]x[/tex]:

[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]

Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:

[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]

Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.

Thus, the required concentrations at equilibrium are about:

[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]

[tex][Ni^{2+}]=0M[/tex]

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HELP!! what are the usual products of combustion reactions?

Answers

Explanation:

Carbon dioxide and water

I hope it helps

Answer:

The usual products of combustion reactions are carbon dioxide and water.

Explanation:

Combustion reaction is when a substance reacts with oxygen gas, resulting in a release of energy in the form of light and heat. Combustion reactions must have oxygen (O2) as one of the reactants.

A pan containing 40 grams of water was allowed to cool from a temperature of 91.0 °C. If the amount of heat released is 1,300 joules, what is the approximate final temperature of the water? 74 °C 78 °C 81 °C 83 °C

Answers

Answer:

d. 83 °c

Explanation:

Answer:

D. 83 c

Explanation:

took the test

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