Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.

Answer:

[tex]Molar \ solubility=3.12x10^{-5}M[/tex]

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

[tex]CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-[/tex]

The equilibrium expression is:

[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent [tex]x[/tex] is computed as follows:

[tex]3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M[/tex]

Thus, the molar solubility equals the reaction extent [tex]x[/tex], therefore:

[tex]Molar \ solubility=3.12x10^{-5}M[/tex]

Regards.

The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].

The dissociation of calcium fluoride has been given by:

[tex]\rm CaF_2\;\rightarrow\;Ca^2^+\;+\;2\;F^-[/tex]

The solubility constant, ksp has been given as:

[tex]ksp=\rm[Mg^2^+]\;[F^-]^2[/tex]

From the dissociation of Calcium nitrate, the concentration of Ca ion in the solution has been 0.01 M.

The dissociation of Calcium fluoride x M has been resulted in x M Ca and 2x M F ions.

The concentration of Ca in the solution has been resulted as x + 0.01 M.

The solubility product can be given as:

[tex]3.9\;\times\;10^-^1^1=[x+0.01]\;[2x]^2\\3.9\;\times\;10^-^1^1=[x+0.01]\;4x^2\\x=3.12\;\times\;10^-^5[/tex]

The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].

For more information about molar solubility, refer to the link:

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Answer:

C) 237.96682 amu

Explanation:

The symbol for neptunium-236 is given as;

²³⁶₉₃Np

This element has 93 protons and (236 - 93 = 143) neutrons.

Mass Number =Total mass of Protons + Total mass of neutrons

Total Mass pf protons = 93 * 1.007825 amu, = 93.727725 amu

Total mass of Neutrons = 143 * 1.008665 amu = 144.239095 amu

Mass = 144.239095 + 93.727725  = 237.96682 amu

Correct option is option C.

Answer:

Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1

Explanation:

The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:  

Vrms = [tex]\sqrt{3RT/M}[/tex]

where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol

For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R

Vrms = √(3 * R *4T)/0.04 = √300RT

For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R

Vrms = √(3 * R *T)/0.001 = √3000RT

Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316

Ratio of Vrms of argon to that of hydrogen = 0.316 : 1

Answer:

The correct answer is - option B -  lowest b.p.... isobutane < acetone < isopropanol ...highest b.p.

Explanation:

Isobutane has lowest boiling point due to no hydrogen bonding and no diole to dipole interaction found in them. Isobutane only shows weak dispersion force.

Acetone has dipole dipole interaction but due to lack of Hydrogen bonding they have low boiling point than isopropanol but higher than isobutanol.

Isopropanol is the compound that has ability to form hydrogen bonding with other molecule its boiling point is maximum among all three.

Thus, the correct answer is - option B -  lowest b.p.... isobutane < acetone < isopropanol ...highest b.p.

Answer:

3.11

Explanation:

Any buffer system can be described with the reaction:

[tex]HA~->~H^+~+~A^-[/tex]

Where is the acid and is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:

[tex]pH~=~pKa~+~Log(\frac{ [A^-]}{[HA]})[/tex]

With all this in mind, we can write the reaction for our buffer system:

-) Nitrous acid: [tex]HNO_2[/tex]

-) Sodium nitrate: [tex]NaNO_2[/tex]

[tex]HNO_2~->~H^+~+~NO_2^-[/tex]

In this case, the acid is [tex]HNO_2[/tex] with a concentration of 0.150 M and a volume of 45.0 mL (0.045 L). The base is [tex]NO_2^-[/tex] with a concentration of 0.175 M and a volume of 20.0 mL (0.020 L).

We can calculate the moles of each compound is we take into account the molarity equation ([tex]M=\frac{mol}{L}[/tex]). So:

-) moles of [tex]HNO_2[/tex]:

[tex]mol=0.150~M*0.045~L=0.00657[/tex]

-) moles of [tex]NO_2^-[/tex]:

[tex]mol=0.175~M*0.020~L=0.0035[/tex]

The total volume would be:

0.020 L + 0.045 L = 0.065 L

With this in mind, we can calculate the molarity of each compound:

-) Concentration of [tex]HNO_2[/tex]

[tex]M=\frac{0.00657~mol}{0.065~L}=0.101~M[/tex]

-) Concentration of [tex]NO_2^-[/tex]

[tex]M=\frac{0.0035~mol}{0.065~L}=0.0538~M[/tex]

The pKa reported is 3.39, therefore we can plug the values into the Henderson-Hasselbach equation:

[tex]pH~=~3.39~+~Log(\frac{[0.0538~M]}{[0.101~M]})~=~3.11[/tex]

The final pH value would be 3.11

I hope it helps!

The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.

We have a buffer made by combining 45.0mL of 0.150 M nitrous acid and 20.0mL of 0.175M sodium nitrate.

Nitrous acid is a weak acid and nitrate ion is its conjugate base.

It is a solution used to resist abrupt changes in pH when acids or bases are added.

We do so by multiplying the molar concentration by the volume in liters.

HNO₂: 0.150 mol/L × 0.0450 L = 6.75 × 10⁻³ mol

NaNO₂: 0.175 mol/L × 0.0200 L = 3.50 × 10⁻³ mol

The total volume will be the sum of the volumes of each solution.

V = 45.0 mL + 20.0 mL = 65.0 mL = 0.0650 L

HNO₂: 6.75 × 10⁻³ mol/0.0650 L = 0.104 M

NaNO₂: 3.50 × 10⁻³ mol/0.0650 L = 0.0538 M

We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.

pH = pKa + log [NaNO₂]/[HNO₂]

pH = 3.16 + log 0.0538/0.104 = 2.87

The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.

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Answer:

155 °C

Explanation:

Step 1: Given data

Step 2: Convert the initial temperature to Kelvin

We will use the following expression.

K = °C + 273.15 = 23.0°C + 273.15 = 296.2 K

Step 3: Calculate the final temperature

Assuming an ideal gas behavior, we can calculate the final temperature using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 4.20 L × 296.2 K/2.91 L

T₂ = 428 K

Step 4: Convert the final temperature to Celsius

We will use the following expression.

°C = K - 273.15 = 428 - 273.15 = 155 °C

Answer:

4.16x10⁻³m

Explanation:

Molality is defined as the ratio between moles of a solute, in this case glucose, and kg of solvent.

As there are 100g of solvent, the kg are 0.1. Thus, we only need to calculate from the mass of glucose its moles to solve the molality of the solution.

Moles glucose:

There are 75mg = 0.075g of glucose. To conver mass to moles it is necessary molar mass.

Molar mass glucose:

6C = 12.01g/mol*6 = 72.06g/mol

12H = 12*1.008g/mol = 12.10g/mol

6O = 6*16g/mol = 96g/mol

72.06 + 12.10 + 96 = 180.16g/mol

Moles of 0.075g of glucose:

0.075g * (1 mol / 180.16g) =

4.16x10⁻⁴ moles of glucose

Molality of the solution:

4.16x10⁻⁴ moles of glucose / 0.1kg of solvent =

The molarity of the solution is 4.16x10⁻³m

We know that the molarity refers to the ratio that arise between the moles of a solute.

Since there are 100 g of solvent so here the kg should be 0.1.

Likewise there is 75 mg so it should be 0.075g

Now the Molar mass glucose should be

6C = 12.01g/mol*6 = 72.06g/mol

12H = 12*1.008g/mol = 12.10g/mol

6O = 6*16g/mol = 96g/mol

So,

= 72.06 + 12.10 + 96

= 180.16g/mol

Now

Moles of 0.075g of glucose:

0.075g * (1 mol / 180.16g) =

4.16x10⁻⁴ moles of glucose

Now finally

Molality of the solution:

= 4.16x10⁻⁴ moles of glucose / 0.1kg of solvent

=4.16x10⁻³m

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electrostatic and ionic are definitely not the answer because they have high melting point

hydrogen bonds are too weak and not permanent.

so the answer is apolar as it is soluble in polar solvents (water)

Answer:

B. Nonpolar

Explanation:

The low melting point tells you the compound is not ionic, metallic, or a network solid.

It is almost certainly a molecular solid.

It does not conduct electricity, so it is not metallic (which we have already ruled out).

It is insoluble in polar solvents (water) and soluble in nonpolar solvents (gasoline).

Since like dissolves like, the molecule is nonpolar.

The type of links must be nonpolar.

Answer:

A    ⇒ 20

B    ⇒ 40

C    ⇒ Ca

D    ⇒ 10

E     ⇒ 9

F     ⇒ F

Explanation:

edge 2021

Answer:

the person above is correct

Explanation:

Answer:

The products are KNO3 + PbCl2.....

Espero que te sirva.

Answer:

The correct answer is - 1.02 V

Explanation:

From the reduction-oxidation reaction:

Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)

Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:

Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s)                    Eº= ‑0.25 V

Oxidation (anode) :  2 x (Fe²⁺ → Fe³⁺ + e-)(aq)                Eº= -0.77 V

                                -------------------------------------

                     Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)

In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):

Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V

Answer:

B) hydroxide concentration

Explanation:

Hello,

In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.

After that, due to the fact that the pH is related with the pOH as shown below:

pH=14-pOH

We can directly compute the pH.

Best regards.

Answer:

Following are the solution to this equation:

Explanation:

In the given-question, an attachment file of the choices was missing, which can be attached in the question and its solution can be defined as follows:

In the given question "Option (iii)" is correct, which is defined in the attachment file.

When 2,2,3-trimethylcyclohexanone reacts with hydroxylamine it will produce the 2,2,3-trimethylcyclohexanoxime.

Answer:

The answer is "Tertiary carbon".

Explanation:

Accent to the results, the carbon-bromine bond is weak, whenever, the bromine is connected to tertiary carbon so, bonding energy is separation for methyl-carbon, which is connected to the bromine = 70 kcal/mol and for the primary energy to the secondary energy is=  68 kcal/mol, and for tertiary CO2 = 65 kcal/mol.

The stronger the energy dissociating connection and the weaker, its power dissociation connection and its weaker bond becomes connecting with a tertiary carbon, that's why "Tertiary carbon" is the correct answer.

Answer:

whats the ph  ofpoh=9.78

Explanation:

Answer:

[tex]m=0.127g[/tex]

Explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}[/tex]

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

[tex]m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g[/tex]

Best regards.

Answer:

An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction

Answer:

226Ra88→222Rn86+4He2

Explanation:

An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.

During α-decay(alpha decay), an atomic nucleus emits an alpha particle.

Answer: C) [tex]CO^{2-}_{3}[/tex]

Explanation: The Valence Shell Electron Pair Repulsion Model (VSEPR Model) shows bonding and nonbonding electron pairs present in the valence, outermost, shell of an atom connecting to other atoms. It also gives the molecular geometric shape of a molecule.

To determine molecular geometry:

1) Draw Lewis Structure, i.e., a simplified representation of the valence shell electrons;

2) Count the number of electron pairs (count multiple bonds as 1 pair);

3) Arrange electron pairs to minimise repulsion;

4) Position the atoms to minimise the lone pair;

5) Name the molecular geometry from the atom position;

Trigonal planar molecular geometry is a model which molecule's shape is triangular and in one plane. Such molecule has three regions of electron density extending out from the central atom and the repulsion will be at minimum when angle between any two is 120°.

The Lewis structure of each molecule is shown in the attachment.

Analysing each one, it can be concluded that molecule with trigonal planar geometry is [tex]CO^{2-}_{3}[/tex]

Answer:

The three major types of bond are ionic, polar covalent, and covalent bonds. Ionic occurs majorly between metals and non-metals, which allows sharing of electrons to form an ionic compound. Whereas covalent bonding calls for complete transfer of electrons between atoms. Polar covalent bonds have unequaly shared electron-pair between two atoms.

Explanation:

a. Cu (Copper)- ionic bonding

b. KCl (Potassium Chloride) - ionic bonding

c. Si (Silicon) - covalent bonding

d. CdTe (Cadmium Telluride) - polar covalent bonding

e. ZnTe (Zinc Telluride)- polar covalent bonding

Answer:

1. Exothermic 2.  Exothermic 3. Endothermic 4. Endothermic 5. Exothermic.

Explanation:

1. An A-A and a C-C bond results in 2 A-C bonds which are lower than the A-A and C-C bonds so this reaction is exothermic.

2. A B-B bond and a C-C bond results in 2 B-C bonds which are lower than the first 2 bonds so this reaction is also exothermic.

3. There is no bond for single A, a single B-C bond results in a A-B bond and a C molecule. A-B bond is stronger than the B-C bond so the reaction absorbed energy along the way. This shows that it is endothermic.

4. An A-A bond and a B-B bond results in 2 A-B bonds which are stronger than the first two bonds so this reaction is also endothermic.

5. An A-B bond and a C molecule result in an A-C bond and a B molecule. A-C bond is weaker than the A-B bond so there is energy released. This reaction is exothermic.

I hope this answer helps.

Answer:

The correct answer is : option D. The test was inconclusive because he forgot to add heat.

Explanation:

Benedict's test is a test that is used to confirm the presence of the simple carbohydrates (mono saccharides and some disaccharides). It is a reagent made by mixture of solution of CuSO4 with sodium citrate and Na2CO3.

Benedict's reagent is added to the substance to test and then heated if it turns yellow to orange or red the presence of simple sugar is confirmed.

Thus, the correct answer is : option D. The test was inconclusive because he forgot to add heat.

Answer:

The test was inconclusive because the student forgot to add heat.

Explanation:

If the test revealed it was not glucose, then the student could run these tests. The student, however, does not need these substances to run the glucose test properly.

Answer:

ITS NOT D. ITS B. 4.52x10^-9 M

Explanation:

Answer:

4.52 ×10–9 M

Explanation:

Answer:

2HClO(aq) + Ba(OH)₂(aq) →  Ba(ClO)₂(aq) + 2H₂O(l)

Explanation:

The reaction corresponds to a neutralization reaction between an acid and a base, as follows:

2HClO(aq) + Ba(OH)₂(aq)  →  Ba(ClO)₂(aq) + 2H₂O(l)            

From the equation above we have that the acid HClO reacts with the base Ba(OH)₂ to obtain a salt Ba(ClO)₂ and water.

In the balanced reaction, we have that 2 moles of HClO react with 1 mol of Ba(OH)₂ to produce 1 mol of Ba(ClO)₂ and 2 moles of water.

I hope it helps you!    

Answer:

d) NH4F

Explanation:

Hello,

In this case, the base resulting from mixing a weak acid and a weak base is d) NH4F since ammonium hydroxide is a wear base and hydrofluoric acid is a weak acid.

Ammonium hydroxide is a weak base since it is not completely ionized in ammonium and hydroxyl ions:

[tex]NH_4OH\rightarrow NH_4^++OH^-[/tex]

Moreover, hydrofluoric acid is a weak acid since it is not completely ionized in hydrogen and fluoride ions:

[tex]HF\rightleftharpoons H^++F^-[/tex]

For the both of the substances, the limit is established by the basic and the acid dissociation constant respectively.

Regards.

Answer:

Animals whose certain organs closely match those of humans or have similar genetic makeup are used in vaccine tests because the results can closely resemble those same results on humans.

Explanation:

Answer:

they use them to test the effectiveness of the vaccine.

Explanation:

Answer:

Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature

Explanation:

The basis of spontaneity in a chemical reaction is that ∆G must be negative. ¡∆G is known as the change in free energy of a system. If ∆G is negative, then the reaction will occur without any external help (the reaction is spontaneous at room temperature).

∆G is given by;

∆G= ∆H -T∆S

Where;

∆H= change in enthalpy of the system

T= absolute temperature of the system

∆S= change in entropy

Hence; when ∆H -T∆S gives a negative result, the reaction proceeds without any external help.

Answer:

B. 2,5-dimethyl-3-heptene

Explanation:

Answer:

B. 2,5-dimethyl-3-heptene

Explanation:

Answer:

[tex]Zn(s) + 2H+(aq)\Rightarrow Zn^{2+}(aq) + H_{2}(g)[/tex]

Explanation:

Given that,

Anode : Zn electrode in a solution of 0.050 M Zn(NO₃)₂

Cathode : Pt electrode with 0.500 atm H₂(g) in 0.010 M HNO₃

Anode :

[tex]Zn(s)\Rightarrow Zn^{2+}(aq) + 2e^{-}[/tex]

Cathode :

[tex]2H+(aq)+2e^{-}\Rightarrow H_{2}(g)[/tex]

We need to write the overall balanced cell reaction

Using anode and cathode

[tex]Zn(s) + 2H+(aq)\Rightarrow Zn^{2+}(aq) + H_{2}(g)[/tex]

Hence, This is required answer.