How are radio telescopes

Interpretacao gerais das praticas sobre densidade de amostra solido e liquido

Savatey [412]

Answer:

O sólido tem densidade mais alta em comparação ao líquido.

Explicação:

A densidade da amostra sólida é maior do que a densidade do líquido porque há pouco espaço entre as partículas do sólido. A densidade tem relação inversa com o volume de uma substância, se uma substância ocupa mais espaço então sua densidade é menor, enquanto se a substância ocupa menos espaço então tem maior densidade. As substâncias sólidas ocupam menos espaço em comparação com as substâncias líquidas, então podemos dizer que a densidade do sólido é maior do que as substâncias líquidas.

6 0

3 years ago

One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 ℃, and the temperature af

gtnhenbr [62]

Answer:fH = - 3,255.7 kJ/mol

Explanation:Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0

3 years ago

12. Which compound can act as both a BrønstedLowry

oksian1 [2.3K]

1) The compound which can act as a Bronsted-Lowry acid and a Bronsted-Lowry base is definitely water - H2O. Remember that water is amphoteric which means it can either accept protons or donate them, so it is the most proper option among other represented. Here are examples of both base and acid with water : HCl+H20=H30+Cl ; NH3+H2O=NH4+OH

2) The acids in this equilibrium reaction CN– + H2O HCN + OH. Acid species always donate H+ to the species with which they react. In the second option you can see how H2O donates an H+ to CN-. If the reaction gets reversed we will obtain HCN that donates an H+ to OH that shows that it is an acid.

3) The products of self-ionization of water are OH⁻ and H₃O⁺. Self-ionization is an ionization reaction during which H2O deprotonates its hydrogen atoms to become a hydroxide ion -- OH−. After this process OH- protonates another water molecule forming H3O+.

4) The type of solution which is one with a pH of 8 is acidic. Here is a little table that can be a prompt for you if you ever come across such tasks - ph : 7 is neutral. pH lower than 7 are acidic, and pH higher than 7 basic ones.

5) The acid dissociation constant for an acid dissolved in water is equal to the equilibrium constant. I consider this option correct because we can obtan Kw only when dealing with Kb, and we can conclude that the hydrolysis constant of the conujugate base.

6) A 0.12 M solution of an acid that ionizes only slightly in solution would be termed dilute and weak. You can determine it depending on its concentration. Such value as 0.12M usually defined as a dilute solution of a weak acid due to the fact that acid represents its partial ionization which is a direct characteristic of a weak acid.

7) To solve this task we should appeal to Henry's law that says the solubility of a gaz is proportional to its partial pressure. And according to this we can understand that 202kPa is the half of 404kPa which means that the needed solubility must be divided by 2 7.5/2=3.75 g/L and that's all.

8) I think that the most important points which best show how the addition of a solute affects the boiling point, the freezing
point, and the vapor pressure of the solvent are : BOILING: additional attractive forces can only exist between solute and solvent and in order to boil they must be overcome for the solution;we should add KE to overcome the forces. FREEZING : to freeze we have to withdrawn KE as the solute particles are surrounded by solvent molecules. VAPOR : WHen solvent shells are being formed the solute particles reduces the number of solvent particles that have sufficient KE to vaporize.

9) $[H+][OH-]= Kw = 1.0 * 10^-14$
$[H+]= Kw/ [OH-]= 1.0x 10^-14 / 1 x 10^-11 =1 x 10^-3 mol/L 

pH = - log [H+]= - log 1 x 10^-3 = 3$
Since we got Ph of 3 in a result we can define solution as an acidic one, as I mentioned before.

10) Since the formula of the given acid is HA it undergoes like that : HA<=> H+ + A- .
ka = [H][A] / [HA].
Now we have only [H+] and to go further you need to write electroneutrality equation for the reaction :
[H+] = [OH-] + [A-] (since [H]>>>[OH]), then
[H+] = [A-]
Then mass balance equation :
Ct = 0.5M = [A-] + [HA]
[HA] = 0.5 - [A-] = 0.5 - [H+]
Finally here is what we have done and get :
$ka = [H]^2 / (0.5 - [H+]) 
$
$ka = 0.0001*0.0001/(0.5-0.0001) = 2.00x10^-8$

11) The main points that are common for acids : they form Hydrogen ions when dissloved in water, - Ex. Vinegar and Lemon, Ph >7, they have Increased hydrogen ions (H+). The facts about bases : they reduce the concentration of hydgoren ions in a solution which is opposite to asids,- Ex. Antiacid,and Ammonia ,Ph valuse above 7, they form hydrogen (OH-).
- The common points of both acids and bases : Hydrogen ions ,
both not neutral and water based.

5 0

3 years ago

Draw the structure of the compound C9H10O2 that might exhibit the 13C-NMR spectrum below. Impurity peaks are omitted from the pe

zhenek [66]

Complete question

Draw the structure of the compound $C_{9}H_{10}O_{2}$ that exhibits the $^{13}C-NMR$ spectrum shown on the first uploaded image(on the second and third uploaded image is closer look at the $^{13}C-NMR$ spectrum ) . Impurity peaks are omitted from the peak list. The triplet at 77 ppm is $CDC_{l3}$ .

Answer:

The structure that might exhibit the $^{13}C-NMR$ spectrum is shown on the fifth uploaded image

Explanation:

In order to get a good understanding of the answer above we need to know that

• Proton NMR spectrum: proton NMR spectroscopy is one of the techniques, which is useful to predict the structure of the compound.

• In $^{\rm{1}}{\rm{H NMR}}$ spectroscopy, peaks are observed at the point where the wavelength of proton nuclei matched to substance nuclei wavelength.

• In same manner there are other spectroscopies are present like $^{{\rm{13}}}{\rm{C NMR}}$

, IR and mass spectroscopy.

• Infrared spectroscopy is used to determine the functional groups present in a compound.

• Infrared bands observed when there is change in dipole moment occurs between the atoms. Infrared bands describe about the bond stretches, which causes due to the dipole moment present in the molecule.

Fundamentals

Double bond equivalence: number of double bonds or number of rings in the structure can be calculated by using double bond equivalence formula.

$DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})$

Where,

$N_{c} =$ number of carbon atoms

$N_{H}=$ number of hydrogen atoms

$N_{Cl} =$ number of chlorine atoms

$N_{N}=$ number of nitrogen atoms

The table for the $^{{\rm{13}}}{\rm{C NMR}}$ is shown on the fourth uploaded image

Molecular formula of the compound is ${{\rm{C}}_9}{{\rm{H}}_{{\rm{10}}}}{{\rm{O}}_{\rm{2}}}$

Double bond equivalence of the compound is calculated below.

$DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})$

Where,

$N_{c} =$ 9

$N_{H}=$ 10

$N_{Cl} =$ 0

$N_{N}=$ 0

$DBE = N_{c} + 1 - (\frac{(10+0) -0}{2}})$

$DBE =5$

Therefore, the compound has five double bonds, which indicating that there is chance of getting aromatic rings too.

Note:

Double bond equivalence is calculated as 5 which indicates that there are 5 double bond (may rings) in the structure of the compound.

Double bond equivalence is calculated by using this formula.

$DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})$

13C NMR data of the compound is explained below.

1.A peak at 166.5 ppm, which indicates the presence of ester group

2.Peaks at 132.7, 130.5, 129.5, 128.2 ppm (aromatic carbons) are indicating a mono substituted aromatic ring

3.A peak at 60.9 ppm means methylene group attached to oxygen atom

4.A peak at 14.3 ppm, which indicates the presence of methyl group

According to this data and the using the double bond equivalence, structure of the compound shown on the fifth uploaded image .

Note:

According to given spectral data, structure of the compound has been predicted. It is clear that; -ester functional group is present in the structure because there is a peak at 166.5ppm. According to given proton $^{13}C NMR$ data, above structure has been drawn. Therefore, the compound is ethyl benzoate.

7 0

4 years ago

What is dissociation constant?

iragen [17]

In Chemistry, the dissociation constant describes a dissociation reaction in which a compound is broken up. It measures how likely the reaction will occur. For example for reaction AB -> A + B, the dissociation constant is equal to concentration of A x concentration of B / concentration of AB at equilibrium.

6 0

4 years ago

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