what are structure for each of the compounds listed. Explain why the name given is incorrect. And give a correct name in each case
(1) 1-methylbutane
(2) 1,1,3-trimethylhexane
(3) 5octyne
(4)2-ethyl-1-propanol
(5)2,2-dimethyl1-3-butanol​

Answers

Answer 1

Answer:

Explanation:

(1) 1-methylbutane

(2) 1,1,3-trimethylhexane

(3) 5octyne

(4)2-ethyl-1-propanol

(5)2,2-dimethyl1-3-butanol


Related Questions

What volume (mL) of a concentrated solution of sodium hydroxide (6.00 M) must be diluted to 187 mL to make a 1.53 M solution of sodium hydroxide

Answers

Answer:

47.68 mL

Explanation:

In this case, we have a dilution problem. So, we have to start with the dilution equation:

[tex]C_1*V_1=C_2*V_2[/tex]

We have to remember that in a dilution procedure we go from a higher concentration to a lower one. With this in mind, We have to identify the concentration values:

[tex]C_1~=~6.00~M[/tex]

[tex]C_2~=~1.53~M[/tex]

The higher concentration is C1 and the lower concentration is C2. Now, we can identify the volume values:

[tex]V_1~=~X[/tex]

[tex]V_2~=~187~mL[/tex]

The V2 value has "mL" units, so V1 would have "mL" units also. Now, we can include all the values into the equation and solve for "V1", so:

[tex]6.00~M*V_1=1.53~M*187~mL[/tex]

[tex]V_1=\frac{1.53~M*187~mL}{6.00~M}=47.68~mL[/tex]

So, we have to take 47.68 mL of the 6 M and add 139.31 mL of water (187-47.68) to obtain a solution with a final concentration of 1.53 M.

I hope it helps!

What is T2, if T1= 500 k, v1=10L, V2=8L,P1=600 torr,P2=200 torr?

Answers

Answer:

T2 = 133.333°K

Explanation:

Using Combined Gas Laws:

(600 torr)(10L)/500°K = (200 torr)(8L)/x°K

[tex]\frac{600 torr(10L)}{500K} =\frac{200 torr(8L)}{xK}[/tex]

Cross multiply:

x°K (600 torr)(10L) = 500°K(200 torr)(8L)

Divide:

x°K = (500°K(200 torr)(8L))/(600 torr)(10L)

[tex]xK = \frac{500K(200 torr)(8L)}{600 torr(10L)}[/tex]

x = 400/3°K or 133.333°K

We are all familiar with the general principles of operation of an internal combustion engine: the combustion of fuel drives out the piston. It is possible to imagine engines that use reactions other than combustions, and we need to assess the work they can do. A chemical reaction takes place in a container of cross-sectional area 100.0 cm^2; the container has a piston at one end. As a result of the reaction, the piston is pushed out through 10 cm against a constant external pressure of 100 kPa. Calculate the work done by the system in Joules.

Answers

Answer:

The work done by the system is 100 J

Explanation:

Given details

The cross sectional area of the of the container is A = 100.0 cm^2 = 0.01m²

The total distance pushed by the piston is d = 10 cm = 0.10m

The total external pressure by which piston pushed is P = 100 kPa

From above data, the following relation can be used to determine the change in volume of the container

∆V = A * d

∆V = 0.01 * 0.10 = 0.001 m³

By using the following relation, the work done by the system is calculated as;

Work done W = P * ∆V

W = 100 * 0.001 = 0.1 kJ = 100 J

The work done by the system is 100 J

Question 4
2 pts
A careless chemistry student performed a chemical reaction where his theoretical yield of
Magnesium oxide was 57.82 grams, but he actually produced 12.89 grams. What is his percent yield
for this experiment? (include the number with 4 significant figures but no units)

Answers

Answer:

22.29%

Explanation:

Percent yield = experimental yield / theoretical yield * 100

= 12.89 / 57.82 * 100 = 22.29%

You've just synthesized a new molecule and need to purify it by recrystallization. You find that it is poorly soluble in water and highly soluble in ethanol, even when cooled in ice. What solvent should work in this situation

Answers

Answer:

Water is used as a solvent.

Explanation:

In order to purify the new molecule, recrystallization occurs in which the new molecule should be added in water and heated the water in order to increase the solubility of the solution. The new molecule dissolve in water while the impurity remains undissolved which can be removed from the solution and then remain the solution to be cold down and the new molecule will again undissolved and the molecule can be extracted without any impurities. We use water instead of ethanol due to lower solubility.

What is the rate constant of a reaction if rate = 1.5 (mol/L)/s, [A] is 1 M, [B] is
3 M, m = 2, and n = 1?
k=
rate
[A]" [B]"
A. 0.17
B. 13.5
C. 0.5
D. 4.5

Answers

[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the Rate Constant.

Here, the "K" is the Rate Constant.

so the ANSWER IS C.) 0.5

The rate of constant is 0.5.

The answer is option C.

How do find the rate of constant?

To determine the fee regulation from a desk, you have to mathematically calculate how differences in molar concentrations of reactants affect the response charge to parent out the order of every reactant. Then, plug in values of the response charge and reactant concentrations to discover the particular rate constant.

What's the rate of constant?

The particular rate constant is the proportionality steady touching on the rate of the reaction to the concentrations of reactants. The fee regulation and the unique charge constant for any chemical response need to be decided experimentally. The price of the fee regular is temperature-established.

Learn more about the rate of constant here: brainly.com/question/8813467

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Aqueous calcium chloride reacts with aqueous potassium carbonate in a double-displacement reaction. Write a balanced equation to describe this reaction. Include states of matter in your answer. Click in the answer box to open the symbol palett

Answers

Answer: [tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The balanced reaction between aqueous calcium chloride reacts with aqueous potassium carbonate is shown as:

[tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]

As lead chemist for a pharmaceutical manufacturing company, you need to inform the purchasing office of a supply order for the next batch of cisplatin, PtCl2(NH3)2. If you intend to make a 500kg batch, how many kg chlorine gas do you need to order?

Answers

Answer:

mass of chlorine gas required is 118 kg.

Explanation:

Total mass of the drug (Cisplatin) required = 500 kg

For the drug PtCl2(NH3)2, we first find the molar mass of the compound.

The molar mass of the drug is the total of all the molar mass of the elements in the drug

molar mass of Pt (platinum) in the drug = 195.078 g/mol

molar mass of chlorine (Cl) in the drug = 2 x (35.453 g/mol) = 70.908 g/mol

molar mass of ammonia (NH3) in the drug = 2 x (17.031 g/mol) = 34.062 g/mol

Total molar mass of the drug = 195.078 g/mol + 70.908 g/mol + 34.062 g/mol = 300.048 g/mol

fractional composition of chlorine in the drug = 70.908/300.048 = 0.236

mass of chlorine required for 500 kg of the drug = 0.236 x 500 = 118 kg

What would have happened to your % Transmittance reading and to your calculations of Keq if the spectrophotometer had been set at 520 nm rather than 447nm

Answers

Answer:

On the off chance that the wavelength(λ) maximum worth has been changed to 520 nm from 470 nm on the spectrophotometer, less light would be absorbed and in this way %T would be higher than the one found at 470 nm. The wavelength utilized at 520 nm isn't adequate for the excitation and consequently lesser light is absorbed by the arrangement.

Explanation:

A spectrophotometer is an analytical equipment used to quantitatively gauge the transmission(passage) or impression of visible light, UV light or infrared light through a medium.

Transmittance (τ) is the ratio of the brilliant or luminous flux at a given wavelength that is transmitted to that of the incident radiation.

where, Keq is the equilibrium constant.

On the off chance that the wavelength(λ) has been changed to 520 nm from 470 nm on the spectrophotometer, less light would be absorbed and in this way %T would be higher than the one found at 470 nm.

What happens to Transmittance?

A spectrophotometer is an analytical equipment used to quantitatively gauge the transmission(passage) or impression of visible light, UV light or infrared light through a medium. Transmittance (τ) is the ratio of the brilliant or luminous flux at a given wavelength that is transmitted to that of the incident radiation. The wavelength utilized at 520 nm isn't adequate for the excitation and consequently lesser light is absorbed by the arrangement. As the concentration goes up, more radiation is absorbed and the absorbance goes up. Therefore, the absorbance is directly proportional to the concentration.

Find more information about Transmittance here: brainly.com/question/14919298

A chemist dissolves 867. mg of pure barium hydroxide in enough water to make up 170. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 degree C.) Be sure your answer has the correct number of significant digits.

Answers

Answer: The pH of the solution is 11.2

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n = moles of solute

[tex]V_s[/tex] = volume of solution in ml

moles of [tex]Ba(OH)_2[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.867g}{171g/mol}=0.00507mol[/tex]         (1g=1000mg)

Now put all the given values in the formula of molality, we get

[tex]Molarity=\frac{0.00507\times 1000}{170}[/tex]

[tex]Molarity=0.0298[/tex]

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

[tex]pOH=-\log [OH^-][/tex]

[tex]Ba(OH)_2\rightarrow Ba^{2+}+2OH^{-}[/tex]

According to stoichiometry,

1 mole of [tex]Ba(OH)_2[/tex] gives 2 mole of [tex]OH^-[/tex]

Thus 0.0298 moles of [tex]Ba(OH)_2[/tex] gives =[tex]\frac{2}{1}\times 0.0298=0.0596[/tex] moles of [tex]OH^-[/tex]

Putting in the values:

[tex]pOH=-\log[0.0596]=2.82[/tex]

[tex]pH+pOH=14[/tex]

[tex]pH=14-2.82[/tex]

[tex]pH=11.2[/tex]

Thus the pH of the solution is 11.2

The two reactions above, show routes for conversion of an alkene into an oxirane. If the starting alkene is cis-3-hexene the configurations of the oxirane products, A and B are Product A: _______ Product B: _______ Will either of these two oxirane products rotate the plane of polarization of plane polarized light? _____

Answers

Answer:

Product A: cis; no

Product B: cis: no  

Explanation:

Two common methods of forming oxiranes from alkenes are:

Reaction with peroxyacids Formation of a halohydrin followed by reaction with base

1. Reaction with peroxyacids

(a) Stereochemistry

The reaction with a peroxyacid is a syn addition, so the product has the same stereochemistry as the alkene.

The starting alkene is cis, so the product is cis-2,3-diethyloxirane.

(b) Configuration

The product is optically inactive because it has an internal plane of symmetry.

It will not rotate the plane of polarized light.

2. Halohydrin formation

(a) Stereochemistry

The halogenation of the alkene proceeds via a cyclic halonium ion.

The backside displacement of halide ion by alkoxide is also stereospecific, so a cis alkene gives a cis epoxide.

The product is cis-2,3-diethyloxirane.

(b) Configuration

The cyclic halonium ion has an internal plane of symmetry, as does the product (meso).

The oxirane will not rotate the plane of polarized light.

 

Drag the description to the category

Answers

Answer:

ok

Explanation:

How many carbon atoms are represented by the model below

A) 0
B) 3
C) 2
D) 1

Answers

Answer:

Option (B) 3.

Explanation:

In the model represented above, the two extreme represent carbon atoms since no other group are attached to it. The joint at the middle also represent carbon atom.

Thus, we can write a more simplify illustration for the model above as

C—C—C

From the above illustration, we can see that the model contains 3 carbon atom.

How many grams of LiNO3 must be added to 25.0 g of water to prepare a 5.00% (m/m) solution of LiNO3?

Answers

Answer:

1.25 g.

Explanation:

5% or 25 g

= 0.05 * 25

= 1.25 g (answer).

If 175mL of oxygen is produced at STP, how many grams of hydrogen peroxide, H2O2
were decomposed? At STP, 1 mole of gas occupies 22.4L. Be sure to balance first.
2 H202 > H202 + O2
what's the
Mass of H2O2​

Answers

Answer:

0.53g

Explanation:

We'll begin by converting 175mL to L. This is illustrated below:

1000mL = 1L

Therefore 175mL = 175/1000 = 0.175L

Next, we shall calculate the number of mole of O2 that occupy 0.175L. This is illustrated below:

1 mole of O2 occupy 22.4L at stp.

Therefore, Xmol of O2 will occupy 0.175L i.e

Xmol of O2 = 0.175/22.4

Xmol of O2 = 7.81×10¯³ mole

Therefore, 7.81×10¯³ mole of O2 occupy 175mL.

Next, we shall determine the number of mole of H2O2 that decomposed to produce 7.81×10¯³ mole of O2. This is illustrated below:

2H2O2 —> 2H2O + O2

From the balanced equation above,

2 moles of H2O2 decomposed to produce 1 mole of O2.

Therefore, Xmol of H2O2 will decompose to produce 7.81×10¯³ mole of O2 i.e

Xmol of H2O2 = 2 x 7.81×10¯³

Xmol of H2O2 = 1.562×10¯² mole

Therefore, 1.562×10¯² mole of H2O2 decomposed in the reaction.

Finally, we shall convert 1.562×10¯² mole of H2O2 to grams. This is illustrated below:

Molar mass of H2O2 = (2x1) + (16x2) = 34g/mol

Mole of H2O2 = 1.562×10¯² mole

Mass of H2O2 =..?

Mole = mass /Molar mass

1.562×10¯² = mass /34

Cross multiply

Mass of H2O2 = 1.562×10¯² x 34

Mass of H2O2 = 0.53g

Therefore, 0.53g of Hydrogen peroxide, H2O2 were decomposition in the reaction.

Two stereoisomers are obtained from the reaction of HBr with (S)-4-bromo-1-pentene. One of the stereoisomers is optically active, and the other is not. Draw the structure of the optically active stereoisomer.

Answers

Answer:

See explanation

Explanation:

In this case, we have an addition reaction. Additionally, this is a marknovnikov addition, therefore the "Br" atom would be added in the most substituted carbon (in this case carbon a). And we are going to have 2 enantiomers (2S,4S)-2,4-dibromopentane and  (2R,4S)-2,4-dibromopentane. In the case of (2R,4S)-2,4-dibromopentane we will have a symmetry plane (a point in the molecule in which we can divide the molecule into two equal parts). When this happens we will have a mesocompound and we will not have optical activity.

See figure 1

I hope it helps!

How many moles of aqueous magnesium ions and chloride ions are formed when 0.240 mol of magnesium chloride dissolves in water

Answers

Answer:

1. 0.240 mole of magnesium ion, Mg^2+

2. 0.48 mole of chloride ion, Cl^-

Explanation:

We'll begin by writing the balanced dissociation equation for magnesium chloride. This is given below:

MgCl2(aq) —> Mg^2+(aq) +2Cl^-(aq)

1. Determination of the number of mole of magnesium ion, Mg^2+ obtained from the dissolution 0.240 mol of magnesium chloride, MgCl2 in water. This is illustrated below:

From the balanced equation above,

1 mole of MgCl2 produced 1 mole of Mg^2+.

Therefore, 0.240 mole of MgCl2 will also produce 0.240 mole of Mg^2+.

Therefore, 0.240 mole of magnesium ion, Mg^2+ is produced.

2. Determination of the number of mole of chloride ion, Cl^- obtained from the dissolution 0.240 mol of magnesium chloride, MgCl2 in water. This is illustrated below:

From the balanced equation above,

1 mole of MgCl2 produced 2 moles of Cl^-.

Therefore, 0.240 mole of MgCl2 will produce = (0.240 x 2) = 0.48 mole of Cl^-.

Therefore, 0.48 mole of chloride ion, Cl^- is produced.

Lewis structure of methyl metcaptain

Answers

Answer:

The lewis structure of the compounds can be drawn by making the skeleton of the molecule first. Then the different atoms are arranged and the electrons are arranged in their bonding pattern. The lone pair of the atoms, which are not involved in the bonding are represented by the dots.

So the lewis structures of both the compound methyl mercaptan has been in the attached image:

Spelling of methyl metcaptain is wrong, the correct spelling is methyl mercaptan.

Answer:

Methyl mercaptan is also known as Methanethiol with the chemical formula  CH3SH and it is an organosulfur compound.

For lewis structure of methyl mercaptan (CH3SH), there are total 14 valence electrons. Four hydrogen atoms has one valence electron each, carbon has four valence electrons and sulfur has six valence electrons. Carbon form one bond with three hydrogen atoms by sharing one electron with each, carbon form one single bond with sulfur atom by sharing one electron with it and sulfur form one single bond with hydrogen. Sulfur left with four unpair electrons.

A sample of magnesium ribbon is ignited in a crucible to form magnesium oxide. Determine the empirical formula of magnesium oxide from the following data:

mass of crucible and cover + magnesium metal
33.741 g

mass of crucible and cover
33.500 g

mass of crucible and cover + magnesium oxide
33.899 g

Answers

Answer:

MgO

Explanation:

The following data were obtained from the question:

mass of crucible and cover + magnesium metal = 33.741 g

mass of crucible and cover = 33.5 g

mass of crucible and cover + magnesium oxide = 33.899 g

Next, we shall determine the mass of magnesium metal. This can be obtained as follow:

mass of crucible and cover + magnesium metal = 33.741 g

mass of crucible and cover = 33.5 g

Mass of magnesium metal =..?

Mass of magnesium metal = (mass of crucible and cover + magnesium metal) – (mass of crucible and cover)

Mass of magnesium metal = 33.741 – 33.5

Mass of magnesium metal = 0.241g

Next, we shall determine the mass of magnesium oxide. This can be obtained as follow:

mass of crucible and cover + magnesium oxide = 33.899 g

mass of crucible and cover = 33.5 g

Mass of magnesium oxide =?

Mass of magnesium oxide = (mass of crucible and cover + magnesium oxide) – (mass of crucible and cover)

Mass of magnesium oxide = 33.899 –. 33.5

Mass of magnesium oxide = 0.399g

Next, we shall determine the mass of oxygen. This can be obtained as follow:

Mass of magnesium oxide = 0.399g

Mass of magnesium metal = 0.241g

Mass of oxygen =..?

Mass of oxygen = (Mass of magnesium oxide) – (Mass of magnesium metal)

Mass of oxygen = 0.399 – 0241

Mass of oxygen = 0.158g

Now, we can obtain the empirical formula for the magnesium oxide as follow:

Mg = 0.241g

O = 0.158g

Divide by their molar mass

Mg = 0.241 / 24 = 0.01

O = 0.158 / 16 = 0.0099

Divide by the smallest

Mg = 0.01 / 0.0099 = 1

O = 0.0099 / 0.0099 = 1

Therefore, the empirical formula for the magnesium oxide is MgO

(a) How many stereoisomers are possible for 4-methyl-1,2-cyclohexanediol? ___ (b) Name the stereoisomers formed by oxidation of (S)-4-methylcyclohexene with osmium tetroxide. If there is only one stereoisomer formed, leave the second space blank. Isomer #1: Isomer #2: (c) Is the product formed in step (b) optically active? _____

Answers

Answer:

See explanation

Explanation:

For the first part of the question, we have to check the chiral carbons in 4-methyl-1,2-cyclohexanediol. In this case carbons, 1 and 2 are chiral, if we have 2 chiral carbons we will have 4 isomers. We have to remember that formula 2^n in which "n" is the number of chiral carbons, so:

2^n = 2^2 = 4 isomers

And the isomers that we can have are:

1) (1R,2S)-4-methylcyclohexane-1,2-diol

2) (1S,2S)-4-methylcyclohexane-1,2-diol

3) (1S,2S)-4-methylcyclohexane-1,2-diol

4) (1S,2R)-4-methylcyclohexane-1,2-diol

See figure 1

For the second part of the question, we have to remember that the oxidation with [tex]OsO_4[/tex] is a syn addition. In other words, the "OHs" are added in the same plane. In this case, we have the methyl group with a wedge bond, so the "OH" groups will have a dashed bond due to the steric hindrance. Due to this we only can have 1 isomer ((1S,2R,4S)-4-methylcyclohexane-1,2-diol). Finally, on this molecule, we dont have any symmetry planes (this characteristic will cancel out the optical activity), so the product of this reaction has optical activity.

See figure 2

I hope it helps!

Identify the precipitation reaction in the set?

Answers

Answer:

The third reaction

(2NaOH + NiCL2 ---> 2NaCl + Ni(OH)2)

Explanation:

By definition, a precipitation reaction refers to the formation of an insoluble salt when two solutions containing soluble salts are combined.

(Source: lumenlearning)

From the 4 options, we can eliminate the first and second one immediately because there is no formation of an insoluble salt.

Then, the last one can also be eliminated because even though there is insoluble solid formed, but it is not a salt, and, the reactants are not solutions too. In fact, the last one is a displacement reaction. A more reactive metal displaces a less reactive metal to form an ion.

Since the third reaction matches the definition of precipitation reaction, this is the answer.

what is the modification of the Dalton's atomic theory​

Answers

Part two of Dalton's theory had to be modified after mass spectrometry experiments demonstrated that atoms of the same element can have different masses because the number of neutrons can vary for different isotopes of the same element.

Answer:

Dalton's theory had to be modified after mass spectrometry experiments demonstrated that atoms of the same element can have different masses because the number of neutrons can vary for different isotopes of the same element

Calculate the theoretical percentage of water for the following hydrates.
(a) manganese(II) monohydrate, MnSO4 H2O
(b) manganese(II) tetrahydrate, MnSO4 4H2O

Answers

Answer:

(a) [tex]\% H_2O=10.65\%[/tex]

(b) [tex]\% H_2O=32.2[/tex]

Explanation:

Hello.

For this questions we must consider the ratio of the molar mass of water to  hydrated compound molar mass as shown below:

(a) In this case, we can consider that inside the manganese (II) sulfate monohydrate, whose molar mass is 169.02 g/mol, there is one water molecule that has a molar mass of 18 g/mol, for which the theoretical percentage of water is:

[tex]\% H_2O=\frac{18g/mol}{169.0g/mol} *100\%\\\\\% H_2O=10.65\%[/tex]

(b) In this case, we can consider that inside the manganese (II) sulfate tetrahydrate, whose molar mass is 223.1 g/mol, there are four water molecules that have a molar mass of 4*18 g/mol, for which the theoretical percentage of water is:

[tex]\% H_2O=\frac{4*18g/mol}{223.1g/mol} *100\%\\\\\% H_2O=32.27\%[/tex]

Best regards.

Select the chemical equation that represents an acid base reaction??? Please help

Answers

Answer:

HCl + AgNO3 ------ AgCl + HNO3

Explanation:

HCl is an acid called hydrochloric acid, while AgNO3 is a base called Silver Nitrate.

Hope it helps.

At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.00317sâ1: 2N2O5(g) â2N2O4(g) + O29(g) Suppose a vessel contains SO3 at a concentration of 1.44M . Calculate the concentration of SO3 in the vessel 0.240 seconds later. You may assume no other reaction is important.Round your answer to 2 significant digits.

Answers

Answer:

[A] = 1.438M = 1.4M (Two s.f)

Explanation:

Rate constant, k =  0.00317

Initial Concentration, [A]o = 1.44M

Final Concentration, [A] = ?

Time, t = 0.240 s

Since this is a second order reaction, the formula for this is given as;

1 / [A] = 1 / [A]o + kt

1 / [A] = 1 / 1.44 + (0.00317 * 0.240)

1 / [A] = 0.6944 + 0.0007608

1 / [A] = 0.6952

[A] = 1.438M = 1.4M (Two s.f)

g Which statement is incorrect regarding oxidation? Oxidation is a "gain" of electrons. Oxidation is the combination with O atoms. Oxidation is an increase in oxidation state. Oxidation is always accompanied by reduction. none of these

Answers

Answer:

The incorrect statement from the options is OXIDATION IS A "GAIN" OF ELECTRONS

Explanation:

Oxidation in a redox reaction is the loss of electrons. It is also the increase in the oxidation states of an atom or ion or atoms in a molecule. A redox reaction is a type of chemical reaction in which there is a transfer of electrons from an atom or ion to another resulting in a change in oxidation states of the substances involved. The reducing agent in the reaction is undergoes oxidation by losing electrons while the oxidating agent is reduced that is it gains electrons at the end of the reaction. The atom or ion from which electron is lost is said to be oxidized while the other atom or ion involved in the reaction is reduced.

Oxidation is also the combination with O atoms and it is always accompanied by reduction because oxidation forms a half of the whole redox reaction. A substance cannot be oxidized except it has reduced another substance by losing electrons to it.

oxidation number of Se in Se8

Answers

Answer:

0.

Explanation:

All elements existing on their own in their free states has an oxidation number of 0.

The oxidation number of Se in Se8 is 0. The element Se in its elemental state has an oxidation number of 0.

Chemistry uses the idea of oxidation number, also known as oxidation state, to describe the charge that an atom seems to have when it forms a compound or ion. It is a notion that aids in our comprehension of how electrons are distributed within a molecule or ion. Oxidation numbers can be used to balance chemical equations and identify the many kinds of chemical reactions that may take place. For a better understanding of the reactivity and bonding of various elements in compounds, they offer insights into the transfer or sharing of electrons between atoms. Se in Se8 has an oxidation number of 0. Se has an oxidation number of 0 when it is in its purest form.

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. Explain with examples following characteristics of chemical reactions:

  a. Change of colour  b. Evolution of gas  c. Change of smell  d. Change of state

Answers

Answer:

See explanation.

Explanation:

Hello,

a. In this case, the change of color is evident for instance when copper reacts with nitric acid to form hydrogen and copper (II) nitrate since copper orange-like and nitric acid is colorless, but copper (II) nitrate is green (dry) or blue (hydrated).

b. In this case, when we make react hydrochloric acid and magnesium, we notice a gas giving off while the magnesium chloride remains aqueous, due to the fact that magnesium displaces hydrogen which is given off as a gas.

c. In this case, we can consider an egg since when it is edible it has a tasty smell but when it decomposes to rotten egg, hydrogen sulfide is given off due to the action of specific bacteria, causing a change in smell to a quite stinky one.

d. In this case, a reaction by which a change of state is exhibited is for instance when aqueous lead (II) nitrate reacts with aqueous potassium iodide to yield potassium nitrate which remains aqueous whereas the lead (II) iodide precipitates out as a solid due to its tiny solubility as a yellow solid.

Best regards.

Helium (He) is the lightest noble gas component of air, and xenon (Xe) is the heaviest. Perform the following calculations, using R = 8.314 J/(mol·K) and ℳ in kg/mol. (a) Find the rms speed of He in winter (0.°C) and in summer (30.°C). Enter your answers in scientific notation. × 10 m/s (winter) × 10 m/s (summer) (b) Find the ratio of the rms speed of He to that of Xe at 30.°C. (rate He)/(rate Xe) (c) Find the average kinetic energy per mole of He and of Xe at 30.°C. Enter your answers in scientific notation. × 10 J/mol for He × 10 J/mol for Xe (d) Find the average kinetic energy per molecule of He at 30.°C. Enter your answer in scientific notation. × 10 J/He atom

Answers

Answer:

Explanation:

Hello,

Data;

R = 8.314J/(mol.K)

Temp (winter) = 0°C = (0 + 273.15)K = 274.15K

Temp.(summer) = 30°C = (30 + 273.15)K = 303.15K

Molar mass of He = 4g/mol = 0.004kg/mol

Molar mass of Xe = 131.29g/mol = 0.131kg/mol

a) rms speed in winter and summer

Vrms = √(3RT/M)

R = gas constant

T = temperature of the gas

M = molar mass of the gas

In winter,

Vrms = √(3×8.314×273.15) / 0.004

Vrms = 1.30×10³m/s

In summer

Vrms = √(3×8.314×303.15) / 0.004

Vrms = 1.37×10³m/s

b) Vrms of Xe at 30°C

Vrms = √(3 × 8.314 × 303.15) / 0.131

Vrms of Xe = 240.24m/s = 2.40×10²m/s

Vrms of He = 1.37×10³m/s

Rate of He / rate of Xe = 1.37×10³ / 2.40×10²

Rate of He / rate of Xe = 5.7

c) K.E per mole

At 30°C

K.E of He = (3/2) × 8.314 × 303.15

K.E of He = 3.78×10³J/mol

K.E of Xe = (3/2) × 8.314 × 303.15

K.E of Xe = 3.78×10³J/mol

d) K.E per molecule = ½mv²

K.E of He molecule = ½ × 0.004 × 1.37×10³

K.E = 2.74J

The rms speed of He in winter and in summer are 1.30×10³m/s & 1.37×10³m/s respectively, ratio of the rms speed of He to that of Xe at 30 degree celsius is 5.7, average kinetic energy per mole of He and of Xe and average kinetic energy per molecule of He are discussed below.

How do we calculate the root mean square velocity?

Root mean square velocity of the gases will be calculated as:

Vrms = √(3RT/M), where

R = universal gas constant = 8.314J/(mol.K)

M = molar mass of gas in kg/mol

T = temperature in K

Root mean square velocity of Helium at 0°C or 273.15K and 30°C or 303.15K will be calculated as:

At 273.15K-

Vrms = √(3×8.314×273.15) / 0.004 = 1.30×10³m/s

At 303.15K-

Vrms = √(3×8.314×303.15) / 0.004 = 1.37×10³m/s

Vrms (winter×summer) = 1.30×10³m/s × 1.37×10³m/s = 1.78×10⁶m/s

Vrms of Xe at 303.15K

Vrms = √(3 × 8.314 × 303.15) / 0.131

Vrms of Xe = 240.24m/s = 2.40×10²m/s

Vrms of He = 1.37×10³m/s

Rate of He / rate of Xe = 1.37×10³ / 2.40×10²

Rate of He / rate of Xe = 5.7

Average kinetic energy will be calculated as:

K.E = 3RT / 2N, where

N = avgadros number = 6.022×10²³ atoms/mole

Average kinetic energy per mole of He & Xe at 303.15K is same as:

K.E = 3/2(1.38 × 10⁻²³J/K)(303.15K) = 6.2 × 10⁻²⁵J/mol

Average kinetic energy per molecule of He will be calculated as:

K.E of He molecule = ½ × 0.004 × 1.37×10³

K.E = 2.74J

Hence, calculations for the given points are described above.

To know more about root mean square velocity, visit the below link:

https://brainly.com/question/2020688

Need help asap!!! Can someone please help me?

Answers

Answer:

The answer is option C.

voltage (V) = Current( I ) × Resistance (R)

V = IR

I = V/ R

V = 10V R = 20Ω

I = 10/20 = 1/2

I = 0.50A

Hope this helps

ionic compound (NH4)2SO4. b. Explain why it cannot be molecular by referring to the Lewis diagrams you drew above as well as the octet rule.
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