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

Answer:

D  BLESSS Y

Explanation:

The temperature of an object will increase if the kinetic energy of the particle increases.

Kinetic energy is an energy that is said to be in motion. According to the kinetic molecular theory of ideal gas, the particles of the gas are usually moving in constant random motion and they exert no force on each other.

Also, the temperature varies directly proportional to the average kinetic energy of the gas particles. As a result, when the kinetic energy of the particles increases, the temperature will also increase.

Learn more about the kinetic theory of gas here:

https://brainly.com/question/9949658?referrer=searchResults

Molar mass of Argon, Ar = 39.95 g Mass of 1 atom, m = 39.95 / (6.023 * 10^23) = 6.64 * 10^(-23) g =6.64 * 10^(-26) Kg Now, Kinetic Energy, KE = (1/2) * m * v^2 where velocity,v = 650 m/s Mass of 1 atom, m = 6.64 *...

- BRAINLIEST answerer

Answer:

1.40×10-20

Explanation:

Oh calculate the mass of 1 atom of argon the molar mass of argon 39.95g that is 6.0²×10²³atoms maybe...?

Answer:

[tex]\blue{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

Answer:

go get the stuff.

Explanation:

Answer:

Explanation:

Hydrodistillation is a traditional method for the extraction of bioactive compounds from plants. In this method, plant materials are packed in a still compartment then water is added in sufficient amount and brought to a boil. ... The vapor mixture of water and oil is condensed by indirect cooling with water.

Answer:

D

Explanation:

D Is The Answer Fella, My Head Hurt Really Bad

Answer:

landslide mudflows slump and creep

Explanation:

no explanation

Answer:

Answer D. Picture II shows a chemical change, because the same substance changes form

Explanation:

This is the temperature that water molecules slow down enough to stick to each other and form a solid crystal

Answer:

D Im 90% Sure

Explanation:

If its not right i owe you one I did this one before

Answer:

44

Explanation:

Since nonmetals gain electrons, the correct question to ask about group 16 elements is; "Will group 16 elements gain electrons to bond with group 2 in an XY format?"

Group 16 elements are divalent and they form divalent negative ions. The periodic table is arranged in groups and periods. The elements in the same group have the same number of valence electrons. All elements in group 2 have six valence electrons.

If a wants to form a compound with the non metals of group 16, the correct question to ask is;"Will group 16 elements gain electrons to bond with group 2 in an XY format?"

Learn more: https://brainly.com/question/14281129

Answer:

Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?

Answer: Tristan is most like the Golgi Body

Explanation:

Vectors may be plasmids. An example of one of several useful features of commercially prepared cloning vectors is MULTIPLE CLONING SITES.

The pUC18 vector is a widely used standardized cloning vector for replication in Escherichia coli.

A multiple cloning site can be defined as a short DNA fragment observed in genetically engineered plasmids.

These DNA fragments (multiple cloning sites) contain twenty (20) or more sites where restriction enzymes can cut in order to generate recombinant DNA molecules.

Learn more in:

https://brainly.com/question/14020637

Answer: 1.324L

Explanation: use Boyles law, sorry so late!

Answer:

hydrogen bond is a type of dipole-dipole interaction; it is not a true chemical bond it is a mere electrostatic attraction. These attractions can occur between molecules or within different parts of a single molecule. Intramolecular hydrogen bonds are those which occur within one single molecule.

Explanation:

CAN I GET BRAINLIEST

This is the answer i hope it helps

A sample of gas has Pi = 0.768 ATM, Vi = 10.5 L, and Ti = 300 K. What is the final pressure if VF = 7.85 L and T and f = 250 K?

From the dilution formula, we have that at constant value of the solute, the volume of the solution is inversely proportional to the concentration.

The correct responses are;

Reasons:

The given parameters are;

The concentration of the KCl solution = 4% m/v

Taking the solution as solution of KCl

The volume of the solution in container 1 = Two times less than the volume of KCl solution.

[tex]V_{container \, 1} = \displaystyle \mathbf{ \frac{1}{2} \cdot V_{4\% \, solution}}[/tex]

The volume of the solution in container 2 = Two times larger compared to the volume of KCl solution.

[tex]V_{container \, 2} = \mathbf{\displaystyle 2 \times V_{4\% \, solution}}[/tex]

The volume of the solution in container 3 = Two times larger than the container two solution volume.

[tex]V_{container \, 3} = \displaystyle \mathbf{ 2 \times V_{container \, 2}}[/tex]

Therefore;

[tex]V_{container \, 3} = \displaystyle 2 \times 2 \times V_{4\% \, solution} = \mathbf{4 \times V_{4\% \, solution }}[/tex]

Part A Required:

a. To select the container that represent the dilution of the 4% solution to 2%

Solution:

The dilution formula is; C₁·V₁ = C₂·V₂

Therefore;

[tex]\displaystyle V_1 = \mathbf{\frac{C_1 \cdot V_1}{C_2}}[/tex]

C₁ =4%, C₂ = 2%, we get;

[tex]\displaystyle V_1 = \frac{4 \cdot V_1}{2} = 2 \cdot V_1[/tex]

The volume of the container that represents a 2% dilution is container 2

which is filled with a solution of a volume two times larger compared to the

KCl solution.

Part B:

Required:

The container diluted to a 1% m/v KCl solution.

Solution;

Using the dilution formula, we have;

C₁ = 4%, C₂ = 1%

Therefore;

[tex]\displaystyle V_1 = \frac{C_1 \cdot V_1}{C_2}[/tex]

[tex]\displaystyle V_1 = \frac{4 \cdot V_1}{1} = \mathbf{4 \cdot V_1}[/tex]

The volume of the solution is four times the volume of the 4% KCl solution, which is equivalent to the volume in container 3.

Possible parts of the question are;

Select the container that represents the dilution of the 4% (m/v) KCl solution to obtain the solutions that follows;

Part A: a 2% (m/v) KCl solution

Part B: a solution that is a 1% (m/v) KCl solution

Please see attached drawings

Learn more here:

https://brainly.com/question/11493179

Answer:

Explanation:

It''s chemical formula is Na3PO4

[tex]pH = -\log[H^{+}] \\\\\implies \log[H^{+}] = -pH\\\\\implies [H^{+}] = 10^{-pH}\\\\\implies [H^{+}] = 10^{-4.83} = 0.000015[/tex]

Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

To determine which elements are oxidized and reduced,

First, we will define the terms Oxidation and Reduction

Oxidation is simply defined as the loss of electrons. It can also be defined as increase in oxidation number.

Reduction is defined as the gain of electrons. It can also be defined as decrease in oxidation number.

The given chemical equation is

Br₂(l) + 2NaI(aq) → I₂(s) + 2NaBr(aq)

Oxidation number of Bromine decreased from 0 to -1.

Therefore, Bromine is reduced.

Oxidation number of Iodine increased from -1 to 0.

Therefore, Iodine is oxidized.

Oxidation number of sodium did not change.

Therefore, Sodium is neither oxidized nor reduced.

Hence, Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

Learn more here: https://brainly.com/question/12913997

Answer:

D. works with bones to help you move

Explanation:

Your muscles and tendons are what help bones move.

Answer:

KNO2, KBr

Explanation:

Chemical compounds are any substance composed of identical molecules consisting of atoms of two or more chemical elements. So NO2 and KBr are compounds, Br2 and Fe are not.

The time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days

We'll begin by calculating the number of half-lives that has elapsed.

Original amount (N₀) = 0.240 mole

Amount remaining (N) = 1.88×10¯³ mole

N = 1/2ⁿ × N₀

1.88×10¯³ = 1/2ⁿ × 0.240

Cross multiply

1.88×10¯³ × 2ⁿ = 0.240

Divide both side by 1.88×10¯³

2ⁿ = 0.240 / 1.88×10¯³

2ⁿ = 128

2ⁿ = 2⁷

Thus, 7 half-lives has elapsed

Number of half-lives (n) = 7

Half-life (t½) = 11.4 days

t = n × t½

t = 7 × 11.4

Therefore, the time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days

Learn more: https://brainly.com/question/13266270

AnswerExplanation:I finsished d

Answer:

17.55L

Explanation:

0.48mol : 11.7

0.72mol :   x

0.48x = 8.424

x = 17.55

Answer:

qrtyuioplkjhgfdssssssazxcvbn

Answer:

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

leaving the other behind

The heat produced by 4.5 moles of magnesium when burnt is 2700 kJ.

A thermochemical reaction is a reaction in which the amount of heat lost or gained is included in the reaction equation. The thermochemical reaction equation for the combustion of magnesium is shown as follows;

2 Mg(s) + O2(g) → 2 MgO(s) + 1200 kJ

From the reaction equation;

2 moles of magnesium produced 1200 kJ of heat

4.5 moles of magnesium will produce 4.5 moles × 1200 kJ/2 moles

= 2700 kJ

Learn more: https://brainly.com/question/14281129

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.

Learn more: https://brainly.com/question/967776

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

For more information, refer to the link:--

https://brainly.com/question/3523174

Answer:

12 kilograms is larger.

Explanation:

Of the three units, the kilogram is the largest and the milligram is the smallest. The prefix “kilo” means a thousand and “milli” means one-thousandths. A gram is the basic unit of mass.