Explain why each molecule has a higher vapor pressure at a given temperature than the other in the pair. Match the words in the left column to the appropriate blanks in the sentences on the right. a. does not exhibit dispersion forces b. does not exhibit dipole-dipole forcescc. is polar d. has a smaller molar mass e. does not exhibit hydrogen bonding . 1. The main reasons why CH4 has a higher vapor pressure at a given temperature when compared to CH3Cl is that CH4____and_____. 2. The main reasons why H2CO has a higher vapor pressure at a given temperature when compared to CH3OH is that H2CO_____and_____.3. The main reason why CH3OH has a higher vapor pressure at a given temperature when compared to CH3CH2CH2OH is that CH3OH_____.

Yeah! I have the same question!!! I think it is heat.

Answer:

because they are both 2 deferent type of oxygen levels

Answer:

0.437

Explanation:

Divide the length value by 1000 to get the answer.

Answer:

the mass would not be the same because of the ashes in the jar

Explanation:

Answer: The total mass before and after burning is the same

Explanation: When the carbon and hydrogen of the hydrocarbon-containing substance (the paper) chemically combine with the oxygen, the remaining materials may appear as ash, the solid remains of a fire.

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Answer:

A

Explanation:

A because water turns into ice

Answer:

a

Explanation:

Answer:

2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ

Explanation:

Entropy increases with increase in the number of particles from left to right in a reaction. Hence, the reaction; 2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ is favoured by increase in entropy.

Similarly, the enthalpy change for the reaction is only +92.4 KJ. Hence for this reaction, both enthalpy and entropy changes drive the reaction in the same direction.

Answer:

0.0505

Explanation:

Step 1: Calculate the masses of solute and solvent

We have a 15.7% by mass nitric acid solution, that is, there are 15.7 g of nitric acid (solute) per 100 g of solution. The mass of water (solvent) is:

m(solution) = m(solute) + m(solvent)

m(solvent) = m(solution) - m(solute)

m(solvent) = 100 g - 15.7 g = 84.3 g

Step 2: Calculate the moles of nitric acid

The molar mass of nitric acid is 63.01 g/mol.

15.7 g × (1 mol/63.01 g) = 0.249 mol

Step 3: Calculate the moles of water

The molar mass of water is 18.02 g/mol.

84.3 g × (1 mol/18.02 g) = 4.68 mol

Step 4: Calculate the mole fraction of nitric acid

[tex]X(HNO_3) = \frac{nHNO_3}{nHNO_3+nH_2O} = \frac{0.249mol}{0.249mol+4.68mol} = 0.0505[/tex]

Answer:

In an NMR experiment, shielding refers to the reduced impact of the APPLIED MAGNETIC FIELD on a nucleus due to the presence of ELECTRON DENSITYaround the nucleus.

Shielding can be DECREASED by the presence of more electronegative groups around the nucleus.

A shielded nucleus appears further UPFIELD whereas a deshielded nucleus appears further downfield in an NMR spectrum.

Explanation:

1 There is reducece because alot of the electronegative atom binds more and thus reduce the electron density around the nucleus.

While a reduction in electron density bring about more nuclear magnetic field thus it resonates at higher frequency resulting downfield peak.

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Answer:

53.27 g.

Explanation:

From the question given above, the following data were obtained:

Density of gold = 19.3 g/mL

Volume of water = 9.27 mL

Volume of water + gold = 12.03 mL

Mass of gold =.?

Next, we shall determine the volume of the gold filling. This can be obtained as follow:

Volume of water = 9.27 mL

Volume of water + gold = 12.03 mL

Volume of gold =?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 12.03 – 9.27

Volume of gold = 2.76 mL

Finally, we shall determine the mass of the gold filling as follow:

Density of gold = 19.3 g/mL

Volume of gold = 2.76 mL

Mass of gold =.?

Density = mass/volume

19.3 = mass of gold /2.76

Cross multiply

Mass of gold = 19.3 × 2.76

Mass of gold = 53.27 g

The, the mass of the gold filling is 53.27 g.

Answer:

a heterogeneous mixture,-two substances present; two phases present

a homogeneous mixture- two substances present; one phase present

a pure substance- one substance present- one phase present

no correct response-one substance present; two phases present

Explanation:

When two substances are mixed, the mixture may be homogenous or heterogeneous. If a mixture is homogenous, then the two substances form a single phase, e.g ethanol and water.

On the other hand, a heterogeneous mixture is divided into two distinct phases, e.g kerosene and water mixture.

A single substance can only give one phase, no substance can simultaneously give two phases of the same substance in the same physical state.

Similarly, a pure substance always consists of a single phase.

Answer:

18.1 mi/h

Explanation:

Step 1: Given data

Distance traveled (d): 100 yd

Time elapsed (t): 11.3 s

Step 2: Convert "d" to miles

We will use the relationship 1 mi = 1,760 yd.

100 yd × (1 mi/1,760 yd) = 0.0568 mi

Step 3: Convert "t" to hours

We will use the relationship 1 h = 3,600 s.

11.3 s × (1 h/3,600 s) = 3.14 × 10⁻³ h

Step 4: Calculate the average speed (s)

We will use the following expression.

s = d / t

s = 0.0568 mi / 3.14 × 10⁻³ h

s = 18.1 mi/h

Answer:

For one atom of lithium -6 I will need_16.85____ atom of lithium -7 for an average atomic mass of 6.944 .

Explanation:

1 ) Let required atom be m .

[tex]\frac{6\times 1 + 7\times m}{1 + m} = 6.944[/tex]

6 + 7m = 6.944 + 6.944m

7 m - 6.944 m = .944

.056 m = .944

m = 16.85

Here, we are required to find how many atoms of lithium -7 is to combine with 1 atom of lithium -6 for an average atomic mass of 6.944.

For one atom of lithium -6 I will need 16.86 atoms of lithium -7 for an average atomic mass of 6.944.

Let the total number of lithium -6 and lithium -7 required be; x.

Therefore, the number of lithium -7 required is;

(x-1)

Therefore;

{(1/x) × 6 + ((x-1)/x)× 7} = 6.944

Therefore, (6 + 7x - 7)/x = 6.944

And, 7x -1 = 6.944x

0.056x = 1

And, x = 1÷0.056 = 17.86 atoms.

Therefore, the number of lithium -7 atoms needed is;

x - 1 = 17.86 - 1 = 16.85 atoms of lithium -7.

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Answer:

Impurities will be trapped in the crystals of the benzoic acid crystallized in this manner.

Explanation:

After benzoic acid is dissolved in hot water, it should have been allowed to cool gradually before it is transferred into an ice bath.

This gradual cooling will aid the separation of impurities so that when the vessel is now submerged in an ice bath, only pure benzoic acid is recrystalized.

If the vessel is immediately submerged into an ice bath, impurities will be trapped in the crystals of the benzoic acid.

Atomic m. Element Atomic no. Symbol

1.0079u Hydrogen 1 H

4.0026u Helium 2 He

6.941u Lithium 3 Li

★ Atomic number : The number of protons in one atom of an element is known as atomic number.

The symbol for every chemical element that has an atomic number less than 5 and atomic mass greater than 8.3 u are hydrogen H, helium He, lithium Li and atomic masses 1,4 and 6.94 respectively in the periodic table.

A periodic table is an arrangement of the elements in which they are classified on the basis of the number of electrons, neutrons, and protons present in their nucleus and shell of them.

Hydrogen has one electron in it and atomic mass is also one, helium comes in the second number with 4 amu and lithium comes in the third number containing 6.94 mass with it.

Therefore, hydrogen H, helium He, lithium Li and atomic masses 1,4 and 6.94 respectively in the periodic table contain symbols for every chemical element that has an atomic number less than 5 and atomic mass greater than 8.3 u are hydrogen H.

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Answer:

(a) [tex]Cu^{2+}S^{2-}[/tex]: copper (II) sulfide or cupric sulfide.

(b)  [tex]Ag^+_2(SO_4)^{2-}[/tex]: silver sulfate.

(c) [tex]Al^{3+}(ClO_3)_3^-[/tex]: aluminum chlorate.

(d) [tex]Co^{2+}(OH)^-_2[/tex]: cobalt (II) hydroxide.

(e) [tex]Pb^{2+}(CO_3)^{2-}[/tex]: lead (II) carbonate.

Explanation:

Hello,

In this case, we proceed by exchanging the subscripts between the cation and anion for each case as follows:

(a) CuS: cation is cupper (II) and anion is sulfide, thus the name is copper (II) sulfide or cupric sulfide.

[tex]Cu^{2+}S^{2-}[/tex]

(b) Ag₂SO₄: cation is silver ion and anion is sulfate, thus the name is silver sulfate.

[tex]Ag^+_2(SO_4)^{2-}[/tex]

(c) Al(ClO₃)₃: cation is aluminum ion and anion is chlorate, thus the name is aluminum chlorate.

[tex]Al^{3+}(ClO_3)_3^-[/tex]

(d) Co(OH)₂: cation is cobalt (II) and anion is hydroxide, thus the name is cobalt (II) hydroxide.

[tex]Co^{2+}(OH)^-_2[/tex]

(e) PbCO₃: cation is lead (II) and anion is carbonate, thus the name is lead (II) carbonate.

[tex]Pb^{2+}(CO_3)^{2-}[/tex]

Regards.

Answer:

Options

Explanation:

Answer:

(c) CH₂F₂

Explanation:

Hydrogen bonds are weak intermolecular forces. They are the strongest kind of intermolecular forces, although they are weaker than the covalent bonds.

Hydrogen bonds arise from molecules which contain a hydrogen atom which is bonded to one of the most electronegative elements such as N, O or F.

(a) HF,  → has H-F bond

(b) CH₃NH₂,   → has N-H bond

(c) CH₂F₂,  → has no H-F bond ( F- C- F)

(d) HOCH₂CH₂OH, → has O-H bond

Therefore, only CH₂F₂ does not exhibit hydrogen bonding.

Answer:

Protons and Neutrons are found in the nucleus

Answer:

The chemical characteristics of carbon affect the characteristics of organic molecules due to its tetravalent nature. It has four valence electrons in which it shares with other elements in order to form an octet configuration.

Carbon atoms are also capable of forming double and triple bonds with other atoms. These properties help determine the functional group present and gives us a knowledge of the chemical features such as polarity, melting and boiling present in the compound.

Answer:

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

Explanation:

The first mixture

0.160 M C₆H₅OHCOOH and 0.160 M C₆H₅OHCOONa,

Ka = 1.05x 10⁻³ .

The second mixture

0.0892 M H₂NCH₂COOH and 0.0892 M H₂NCH₂COOK,

Ka = 4.5 x 10⁻³ .

The third mixture

.0725 M H₂NCH₂COOH and 0.0725 M H₂NCH₂COONa,

Ka = 4.5 x 10⁻³

fourth mixture

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

In all the mixtures the ratio of acid and its salt are same and equal to one so this ratio will not determine their relative buffering capacity .

Now we know that weak acid has more buffering capacity so mixture having acid of less Ka will have more buffering capacity .

Ka is less if

Ka = 1.05 x 10⁻³ .

Dilute acids have greater buffering capacity

So ultimate answer is

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the I.D.H (hydrogen deficiency index):

[tex]I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}[/tex]

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

[tex]I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1[/tex]

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (1-bromoprop-1-ene). In the second structure, we can move the Br atom to the second carbon (2-bromoprop-1-ene), in the third structure we can move the Br to carbon 3 (3-bromoprop-1-ene). Finally, we can have a cyclic structure with a Br atom (bromocyclopropane).

See figure 1

I hope it helps!

Answer:

7.18 × 10⁵ kg

Explanation:

Step 1: Given data

Volume of the iceberg (V): 783 m³

Density of ice (ρ): 0.917 g/cm³

Step 2: Convert "V" to cm³

We will use the relationship 1 m³ = 10⁶ cm³.

783 m³ × (10⁶ cm³/1 m³) = 7.83 × 10⁸ cm³

Step 3: Calculate the mass (m) of the iceberg

We will use the following expression.

ρ = m/V

m = ρ × V

m = 0.917 g/cm³ × 7.83 × 10⁸ cm³

m = 7.18 × 10⁸ g

Step 4: Convert "m" to kg

We will use the relationship 1 kg = 10³ g.

7.18 × 10⁸ g × (1 kg/10³ g) = 7.18 × 10⁵ kg

The questions asks us to calculate the wavelength of light produced if an electron moves from n = 6 to n = 5 state in a hydrogen atom. The answer would be:

λ = [tex]7.46 * 10^{-6}[/tex]  m

m = [tex]7.46 * 10^{-4}[/tex] cm

Explanation:

Please see my graphic below:

The wavelength of light produced if an electron moves from n=6 state to n=5 state of an electron in a hydrogen atom is 7.59 × [tex]10^-6[/tex] m

Using the Rydberg formula;

1/λ = 1.09737 × [tex]10^-6[/tex] (1/[tex]nf^2[/tex] - 1/[tex]ni^2[/tex])

Where;

λ = wavelength

nf = final state of the electron

ni = initial state of the electron

Substituting values;

1/λ = 1.09737 × [tex]10^7[/tex] ([tex]1/5^2[/tex] - [tex]1/6^2[/tex])

1/λ = 1.09737 × [tex]10^7[/tex] (1/25 - 1/36)

1/λ = 1.09737 × [tex]10^7[/tex] (0.04 - 0.028)

1/λ = 1.09737 × [tex]10^7[/tex] (0.012)

λ = 7.59 × [tex]10^-6[/tex] m

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Answer:

O C. The atom has 7 protons and 7 electrons.

If the proton amd electron of atoms are equal it is said to be electrically nuetral

Answer:

Explanation:

Plant cell because its shape is rectangular

Answer:

E) None of these

Explanation:

For this question, we must remember the definition of a chiral carbon. In the chiral carbons, we have four different groups around the carbon.

So, an achiral molecule would be a molecule that does not have chiral carbons. That is, in an achiral molecule in all carbons we have at least one repeating group.

In all the molecules, which we have in the question, we have absolute configuration. That is, we have chiral carbons. In molecule A we have two chiral carbons (R and R), in molecule b we have two chiral carbons (R and S), in molecule c we have two chiral carbons (S and S) and in molecule d we have two chiral carbons (R and S). Therefore in all molecules, we have chiral carbons.

Now, if we analyze molecules a and c, we have a plane of symmetry. If we have a plane of symmetry, even though there are chiral carbons, there will be no optical activity. That is, these molecules do not have the ability to deflect polarized light despite having chiral carbons. Therefore molecules a and c are meso compounds.

But optical activity is not related to chirality.

See figure 1 to further explanations

I hope it helps!

Answer:

The answer to your question is given below

Explanation:

2,4,4,5-tetramethyl-2-hexene.

To draw the structure of the above compound, we must bear the following in mind:

1. The compound is an alkene i.e it contains carbon to carbon double bond (C=C)

2. The parent name of the compound is hexene i.e it contain 6 carbon atoms.

3. The double is located at carbon 2.

4. The substituent group attached to the compound is methly.

5. There are four methly group attached to the compound of which one is located at carbon 2, two at carbon 4 and one at 5.

With the above information, we can easily draw the structure of the compound.

Please see attached photo for the structure of the compound.

Answer:

Explanation:

Simple organic molecules having only carbon and hydrogen are alkanes, alkenes and alkynes.

Ways of making these molecules are referred to as synthesis

Synthesis of

(1) alkanes can be achieved by halogenation of an alkene or alkyne as shown in the reaction below

C₂H₄ + H₂ ⇒ C₂H₆

The reaction above shows an alkene (ethene) undergoing halogenation to form an alkane (ethane). This reaction occurs in the presence of a platinum or nickel catalyst

C₂H₂ + 2H₂  ⇒ C₂H₆

The reaction above shows an alkyne (ethyne) undergoing halogenation to form an alkane (ethane). This reaction also occurs in the presence of a nickel or platinum catalyst

(2) Alkanes can be achieved by Wurtz reaction; treating alkyl halides with sodium metal as shown below

CH₃CH₂Br + 2Na + BrCH₂CH₃ ⇒ CH₃CH₂CH₂CH₃ + 2NaBr

This reaction proceed in the presence of a dry ether.

(3) alkanes from Kolbe's electrolytic method; this involves the synthesis of alkanes through the electrolysis of sodium or potassium salt of a carboxylic acid

2CH₃COO⁻Na⁺ + 2H₂O  ⇒ CH₃CH₃ + 2CO₂ + H₂ + 2NaOH

(4) alkenes from dehydration of alcohol, as shown in the reaction below

C₂H₅OH ⇒ C₂H₄    (-H₂O)

The reaction above proceeds by heating the alcohol (ethanol) at high temperature in the presence of strong acid such as sulphuric acid.