The equilibrium constant (K p) for the interconversion of PCl 5 and PCl 3 is 0.0121:
PCl5 (g) → PCl3 (g) + Cl2 (g)
A vessel is charged with PCl 5 giving an initial pressure of 0.123 atm and yields PCl 3 and Cl 2. At equilibrium, the partial pressure of PCl 3 is ________ atm.
A) 0.0782.
B) 0.0455.
C) 0.0908.
D) 0.0330.
E) 0.123.

Answer:

A) 697 mL

Explanation:

First convert degrees Celsius to Kelvin.

That gives;

For T1; temp 1 (273 + 25.1)

= 298.1K

T2; temp 2 (273+60.7)

=333.7K

From the formula, (V1/T1) = (V2/T2)

To find V2 = (V1/T1) × T2

= (622.7/298.1) × 333.7.

= 697.06mL

~~ 697mL

Yes.

Each chemical substance has a specific chemical composition, so these chemical substances have their own chemical formula.

I hope this helps!

Answer: If matter goes through a chemical change then the physical properties are not likely to stay the same.

Explanation:

When chemical composition of a substance changes during a chemical reaction then it is called a chemical change.

Chemical change always leads to the formation of new substances. Properties like chemical reactivity, combustion, rusting etc are chemical changes.

For example, [tex]N_{2} + 3H_{2} \rightarrow 2NH_{3}[/tex]

Here, [tex]NH_{3}[/tex] will have different chemical as well as physical properties as compared to [tex]N_{2}[/tex] and [tex]H_{2}[/tex].

As physical properties are the properties that cause change in state of a substance.

Properties like boiling point, state of substance etc are physical properties.

Thus, we can conclude that if matter goes through a chemical change then the physical properties are not likely to stay the same.

Answer:

1) pure water

2) 0.75 m CaCl2

3) 1.0 m NaCl

4) 0.5 m KBr

5) 1.5 m glucose (C6H12O6)

Explanation:

Boiling point elevation is a colligative property. Coligative properties are properties that depend on the amount of solute present in the system. The boiling point of solvents increase due to the presence of solutes.

The boiling point elevation depends on the number of particles the solute forms in solution and the molality of the solute. The more the number of particles formed by the solute and the greater the molality of the solute, the greater the magnitude of boiling point elevation.

The order of decreasing hoping point elevation is;

1) 0.75 m CaCl2

2) 1.0 m NaCl

3) 0.5 m KBr

4) 1.5 m glucose (C6H12O6)

Answer:

a. [tex]Ag^0[/tex]

b. [tex]Ag^{+}[/tex]

c. [tex]Ca^{2+}F_2^-[/tex]

d. [tex]H_2^+S^{2-}[/tex]

e. [tex](C^{4+}O_3^{2-})^{-}[/tex]

f. [tex](Cr^{6+}O_4^{2-})^{2-}[/tex]

g. [tex]Cl_2^0[/tex]

h. [tex]Fe^0[/tex]

i. [tex]Cu^{2+}Cl_2^-[/tex]

j. [tex]C^{4-}H_4^+[/tex]

Explanation:

Hello there!

In this case, according to the concept of charge balance, which tell us that the overall charge is zero for any compound, except ions, it turns out possible to proceed as follows:

a. [tex]Ag^0[/tex]

b. [tex]Ag^{+}[/tex]

c. [tex]Ca^{2+}F_2^-[/tex]

d. [tex]H_2^+S^{2-}[/tex]

e. [tex](C^{4+}O_3^{2-})^{-}[/tex]

f. [tex](Cr^{6+}O_4^{2-})^{2-}[/tex]

g. [tex]Cl_2^0[/tex]

h. [tex]Fe^0[/tex]

i. [tex]Cu^{2+}Cl_2^-[/tex]

j. [tex]C^{4-}H_4^+[/tex]

Keep in mind lonely elements have 0 as their oxidation state.

Regards!

Answer:

Alkaline earth metals or akali metals

Answer:

highest boiling point.

Explanation:

Hello there!

In this case, according to the given information, it turns out necessary for us to recall the definition of colligative properties, more specifically that of boiling point elevation, defined in terms of the temperature change, van't Hoff's factor, molality of the solution and boiling point elevation constant:

[tex]\Delta T_b=i*m*K_b[/tex]

In such a way, since all the given solutions have the same molality and solvent (water), they all have the same m and Kb, for which we focus on the van't Hoff's factor which is 1 for CH2O (nonionizing), 2 for NaCl (Na and Cl ions are released), 2 for HF (H and F ions are released) and 3 for AlCl₃ (one Al and three Cl ions are released).

Therefore, since D. 0.50 mol AICI, in 1.0 kg water  has the greatest van't Hoff's factor, we infer it has the highest boiling point.

Regards!

Answer:

Write a chemical equation for KOH(aq) showing how it is an acid or a base according to the Arrhenius definition.

Express your answer as a chemical equation. Identify all of the phases in your answer.

Write a chemical equation for NH4+(aq) showing how it is an acid or a base according to the Arrhenius definition.

Express your answer as a chemical equation. Identify all of the phases in your answer.

Explanation:

According to this theory, an acid is a substance that is a proton donor when the substance is dissolved in water.

The base is the one that gives OH- ions when dissolved in water.

KOH(aq) is a base.

Because it releases OH- ions when dissolved in water.

[tex]KOH(aq)->K^+(aq)+OH^-(aq)[/tex]

The ammonium ion is an acid.

It can donate a proton when dissolved in water.

[tex]NH_4^+(aq)->NH_3(aq)+H^+(aq)[/tex]

Answer:

hexane

I hope it's helps you

Explanation:

So, first you will want to write the balanced chemical equation for this reaction.

Butane = [tex]C_4H_{10}[/tex]

[tex]2C_4H_{10}+13O_2=>10H_2O+8CO_2[/tex]

^ This ends up being your balanced chemical equation. Now, you can do the math!

[tex]1.57gC_4H_{10}*\frac{1molC_4H_{10}}{58.12gC_4H_{10}}*\frac{10molH_2O}{2molC_4H_{10}}*\frac{18gH_2O}{1molH_2O}[/tex]

After plugging this into a calculator, your final mass of water should be:

2.43gH2O

Answer:

a little less than 109.5°

Explanation:

SCl2 has four regions of electron density around the central atom of the molecule. This implies that it has a tetrahedral electron domain geometry with an expected bond angle of 109.5° according to valence shell electron pair repulsion theory.

However, there are two lone pair of electrons on the central atom of the molecule which decreases the bond angle a little less than 109.5° owing to repulsion between electron pairs.

Answer: The property of all bases is that they are substances which neutralizes an acid to form a salt and water only.

Explanation:

A base is also s substance that can accept or combine with a proton; a proton acceptor.

Bases include the oxides, hydroxides and carbonates of metals. These include MgO,Na2O, NaOH,Ca(OH)2. Most metals burn in oxygen to form metallic oxides which are basic. Examples are the MgO and Na2O.

Other properties or characteristics of bases includes:

--> They have a bitter taste.

--> They turn red litmus paper blue.

--> They are soapy to touch.

--> Aqueous solutions of bases are also electrolytes. Bases can be either strong or weak, just as acids can.

Bases (sodium hydroxide) can be used in the preparation of soaps, glass, paper and rayon. While some bases (magnesium hydroxide) can also be used to manufacture toothpaste and laxatives.

Answer:

- Stronger reducing agent than Chlorine

- Oxidized to it's elemental form

- Change the colour of the aqueous layer.

Explanation:

Halides are electronegative elements in group seven of the Periodic table which have gained electrons to complete their electronic configuration.

They include F-, CL-, Br- and I-.

As you descend the group electro negativity decreases as the number of outermost shells increases. Hence F- is the most electronegative while I- is the least electronegative.

In terms of oxidising and reducing abilities amongst the halogens, since an oxidizing agent readily accepts electrons and is thereby reduced, oxidizing power decreases down the group.

For example, Fluorine being the strongest oxidising agent in the group readily accepts electrons from other members of the group and is reduced to the fluoride ion

F + e = F -

Therefore in terms of oxidizing abilities,

F > Cl > Br > I

Conversely, , as the oxidising power decreases down the group, the reducing powers increases

Therefore, in terms of reducing powers,

I > Br > Cl > F

In the test for halide ions using aqueous chlorine, since chlorine is a stronger oxidizing agent/weaker reducing agent than Bromine or iodine, it readily accepts their electrons forming the chloride ion.

Cl2 + 2 Br- = 2 Cl- + Br2

The bromide ion (assuming the unknown halide is bromide) being a stronger reducing agent/weaker oxidizing agent than Chlorine would readily lose it's electrons and get oxidized to it's elemental form changing the colour of the aqueous layer to brown.

That is : Br2- = Br2 + 2e

The fill in the blanks could be filled with stronger, elemental form and mineral oil.

In the case when the halide is not known so it should be stronger. The halide should be oxidized with respect to the elemental form and it should change the color of mineral oil. Due to this, halide should be oxidized for elemental halogen i.e. more soluble for mineral oil.

Learn more about the water here: https://brainly.com/question/21281574

Answer:

This would make the reaction more energetically favourable

Explanation:

Fatty acid synthesis starts with the carboxylation of acetyl CoA to malonyl CoA. This irreversible reaction is the committed step in fatty acid synthesis. The synthesis of malonyl CoA is catalyzed by acetyl CoA carboxylase, which contains a biotin prosthetic group.

Answer:

[HI] = 0.704mol

[H2] = 0.048mol

[I2] = 0.148mol

Explanation:

Based on the equilibrium:

H2(g)+I2(g) --> 2HI(g)

The equilibrium constant, K, is defined as:

K = 70 = [HI]² / [H2] [I2]

Where [] could be taken as the moles in equilibrium of each reactant

To know the direction of the equilibrium we need to find Q with the initial moles of each species:

Q = [0.200mol]² / [0.300mol] [0.400mol]

Q = 0.333

As Q < K, the reaction will shift to the right producing more HI. The equilibrium moles are:

[HI] = 0.200mol + 2X

[H2] = 0.300mol - X

[I2] = 0.400mol - X

Replacing in K:

70 = [0.200 + 2X ]² / [0.300 - X] [0.400 - X]

70 = 0.04 + 0.8 X + 4 X² / 0.12 - 0.7 X + X²

8.4 - 49 X + 70 X² = 0.04 + 0.8 X + 4 X²

8.36 - 49.8X + 66X² = 0

Solving for X:

X = 0.252 moles. Right solution

X = 0.502 moles. False solution. Produce negative moles.

Replacing:

[HI] = 0.200mol + 2*0.252 mol

[H2] = 0.300mol - 0.252 mol

[I2] = 0.400mol - 0.252 mol

[HI] = 0.704mol

[H2] = 0.048mol

[I2] = 0.148mol

Answer:

The student completed the experiment correctly and there were no errors in the experiment.

Explanation:

When a pea size amount of K2CO3 is dropped into a solution of HCl, the following reaction occurs;

K2CO3(s) + 2HCl(aq) ----> 2KCl(aq) + CO2(g) + H2O(l)

The gas CO2 does not support burning hence, when the student performed a splint test and observed that the splint was extinguished when he placed the splint into the test tube of the gas.

Hence, the experiment was properly conducted and the student completed the experiment correctly and there were no errors in the experiment.

Answer:

I don't know what to say . just for points

Answer:

Please find the complete question and its solution file in the attachment.

Explanation:

Answer:

d. van der waals force

Explanation:

Van der Waals force :

the weakest intermolecular forceand consist of dipole-dipole force and dispersion force.

Answer: There are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.

Explanation:

Given: [tex]M_{1}[/tex] = 1.0 M,    [tex]V_{1}[/tex] = ?

[tex]M_{2}[/tex] = 0.1 M,    [tex]V_{2}[/tex] = 200 mL

Formula used is as follows.

[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]

Substitute values into the above formula as follows.

[tex]M_{1}V_{1} = M_{2}V_{2}\\1.0 M \times V_{1} = 0.1 M \times 200 mL\\V_{1} = \frac{0.1 M \times 200 mL}{1.0 M}\\= 20 mL[/tex]

Thus, we can conclude that there are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.

Answer:

Explanation:

From the given information:

Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.

Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.

If the molecular weight of camphor = 152.24 g/mol

and it mass = 200 mg

The its no of moles = 200 mg/ 152.24 g/mol

= 1.3137 mmol

Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol

= 6.831 mmol

since the molar mass of NaBH4 = 37.83 g/mol

Then, using the same formula:

No of moles = mass/molar mass

mass = No of moles × molar mass

mass = 6.831 mmol × 37.83 g/mol

mass of NaBH4 used = 258.42 mg  

Answer:

An unknown compound's 2,4-DNP product melting range is between 144-146'C. It does not give a silver mirror on the Tollens test and is slow to react to the chromic acid test. What would this compound be

Explanation:

From the given data it is clear that the unknown compound gives positive test with 2,4-DNP reagent.

That means it has a carbonyl group.Either aldehyde or ketone.

It does not give a silver mirror on the Tollens test and is slow to react to the chromic acid test.

That means aldehyde is absent.

So, the other carbonyl group that is ketone is present in the given unknown compound.

Answer:

88.9%

Explanation:

Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:

Después calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:

Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:

Finalmente calculamos el rendimiento porcentual:

Answer: The molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Explanation:

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol[/tex]

Now, molarity of ethanol solution is as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M[/tex]

(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol[/tex]

Now, molarity of sucrose solution is as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L} (1 mL = 0.001 L)\\= 0.94 M[/tex]

(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol[/tex]

Now, molarity of sodium chloride solution is as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M[/tex]

Thus, we can conclude that the molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Answer:

e. The lipid bilayer is a permeability barrier for integral membrane proteins.

Explanation:

Integral membrane proteins enter the lipid bilayer by translocon channel which gives nascent polypeptide segment a chance to partition itself into lipid bilayer hydrophobic core.

Complete Question

Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=4 and the atom emits a photon of light with a wavelength of 486 nm. Group of answer choices

Answer:

[tex]n=2[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda=486nm=>486*10^{-9}[/tex]

Generally the equation for Atom Transition is mathematically given by

[tex]\frac{1}{\lambda}=R_{\infty }(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]

Where

Rydberg constant [tex]R_{\infty}=1.097*10^7[/tex]

Therefore

[tex]\frac{1}{486*10^{-9}}=1.097*10^7*(\frac{1}{n_1^2}-\frac{1}{4^2})[/tex]

[tex](\frac{1}{n_1^2}-\frac{1}{4^2})=\frac{1}{486*10^{-9}*1.097*10^7}[/tex]

[tex]n_1^2=3.98[/tex]

[tex]n=1.99[/tex]

[tex]n=2[/tex]

Using the Rydberg formula, the final state of the electron is n=2.

Using the Rydberg formula;

1/λ = R(1/nf^2 - 1/ni^2)

Where;

λ = wavelength

nf = final state

ni = initial state

R = Rydberg constant

When λ =  486 × 10^-9 m and ni = 4, R = 1.097  × 10^7 m-1

1/486 × 10^-9 =  1.097  × 10^7(1/nf^2 - 1/4^2)

0.188 = 1/nf^2 - 0.0625

1/nf^2 = 0.188 +  0.0625

nf = 2

Missing parts;

Determine the end (final) value of n in the hydrogen atom transition, if electron starts in n-4 and the atom emits a photon of light with a wavelength of 486.

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