What is the equilibrium constant for the following reaction:HCO2H(aq) + CN–(aq) HCO2–(aq) + HCN(aq)Does the reaction favor the formation of reactants or products? The acid dissociation constant, Ka, for HCO2H is 1.8 x 10–4and the acid dissociation constant for HCN is 4.0 x 10–10.(A) K = 1.00. The reaction favors neither the formation of reactants nor products.(B) K = 2.2 x 10–6. The reaction favors the formation of products.(C) K = 2.2 x 10–6. The reaction favors the formation of reactants.(D) K = 4.5 x 105. The reaction favors the formation of products.(E) K = 4.5 x 105. The reaction favors the formation of reactants.

Answers

Answer 1

Answer:

(D) K = 4.5 x 10⁵. The reaction favors the formation of products

Explanation:

HCOOH + CN⁻ ⇆ HCOO⁻ + HCN

K = [HCOO⁻] [ HCN ] / [ HCOOH] [ CN⁻]

HCOOH ⇄ H ⁺ + COO⁻

K₁ = [ H⁺] [ COO⁻ ] / [HCOOH ]

HCN ⇆ H⁺ + CN⁻

K₂ = [ H⁺] [ CN⁻] / [ HCN ]

K₁ / K₂

= [ H⁺] [ COO⁻ ] / [HCOOH ]  X  [ HCN ] / [ H⁺] [ CN⁻]

= [ COO⁻ ][ HCN ] / [HCOOH ]  [ CN⁻]

= K

K = K₁ / K₂

= 1.8  x 10⁻⁴ / 4 x 10⁻¹⁰

= 4.5 x 10⁵

So equilibrium constant of the reaction

HCOOH + CN⁻ ⇆ HCOO⁻ + HCN

is very high . Hence reaction favours the formation of product.

option (D) is correct.


Related Questions

1. There are how many mol of oxygen in 3.5 mol of caffeine.

Answers

Answer:

7 mol

Explanation:

Caffeine molecular formula C8H10N4O2. It has 2 atoms of oxygen.

                          C8H10N4O2         - 2O

                           1 mol                        2 mol

                           3.5 mol                    x mol

x = 3.5*2/1 = 7 mol

An aqueous KNO3 solution is made using 72.5 g of KNO3 diluted to a total solution volume of 2.00 L. Calculate the molarity, molality, and mass percent of the solution. (Assume the density of 1.05 g/mL for the solution.)

Answers

Answer:

The molarity is 0.359[tex]\frac{moles}{L}[/tex]

The molality is 0.354 [tex]\frac{moles}{kg}[/tex]

The mass percent of the solution es 3.45%

Explanation:

Molarity is a unit of concentration that indicates the amount of moles of solute that appear dissolved in each liter of the mixture. It is determined by:

[tex]Molarity (M)=\frac{number of moles of solute}{volume}[/tex]

Being:

K: 39 g/moleN: 14 g/moleO: 16 g/mole

The molar mass of KNO₃ is:

KNO₃=  39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole

You can apply the following rule of three: if 101 grams of KNO₃ are present in 1 mole, 72.5 grams in how many moles are present?

[tex]moles of KNO_{3}=\frac{72.5 grams*1 mole}{101 grams}[/tex]

moles of KNO₃= 0.718

So you have:

moles of KNO₃= 0.718volume= 2 L

Applying this quantity in the definition of molarity:

[tex]molarity=\frac{0.718 moles}{2 L}[/tex]

Molarity= 0.359[tex]\frac{moles}{L}[/tex]

The molarity is 0.359[tex]\frac{moles}{L}[/tex]

Molality is a way of measuring the concentration of solute in solvent and indicates the amount of moles of solute in each kilogram of solvent.

Then the molality is calculated by:

[tex]Molality=\frac{moles of solute}{mass of solvent in kilograms}[/tex]

Density is defined as the property that matter, whether solid, liquid or gas, has to compress in a given space. That is, it is the amount of mass per unit volume. So, if the density of 1.05 g / mL for the solution indicates that in 1 mL of solution there are 1.05 grams of solution, in 2000 mL (where 2L = 2000 mL, because 1 L = 1000mL) how much mass is there?

[tex]mass=\frac{2000 mL*1.05 grams}{1 mL}[/tex]

mass= 2100 grams

Since mass solution = mass water + mass KNO₃

then mass water = mass solution - mass KNO₃

Being mass solution 2100 grams and mass KNO₃ 72.5 grams, and replacing you get: mass water= 2100 grams - 72.5 grams

mass water= 2,027.5 grams

Then, being:

moles of KNO₃= 0.718mass of solvent in kilograms= 2.0275 kg (being 2,027.5 grams= 2.0275 kilograms because 1,000 grams= 1 kilogram)

Replacing in the definition of molality:

[tex]molality=\frac{0.718 moles}{2.0275 kg}[/tex]

molality= 0.354 [tex]\frac{moles}{kg}[/tex]

The molality is 0.354 [tex]\frac{moles}{kg}[/tex]

The mass percent of a solution is the number of grams of solute per 100 grams of solution. Then the mass percent is the mass of the element or solute divided by the mass of the compound or solute and the result of which is multiplied by 100 to give a percentage.

[tex]mass percent=\frac{mass of solute}{mass of solution} *100[/tex]

So, in this case:

[tex]mass percent=\frac{72.5 grams}{2100 grams} *100[/tex]

mass percent= 3.45 % KNO₃ by mass

The mass percent of the solution es 3.45%

The molarity of the solution is 0.36 mol/L. The molality of the solution is 0.34 m. The mass percent of the solution is   3.33%.

Number of moles of  KNO3 = mass/molar mass =  72.5 g/101 g/mol = 0.72 moles

Molarity = Number of moles / volume =  0.72 moles/ 2.00 L = 0.36 mol/L

The molality = Number of moles of solute/Mass of solution in kilograms

mass of solution =  1.05 g/mL × 2000 mL = 21000 g or 2.1Kg

Molality of solution =  0.72 moles/2.1 Kg = 0.34 m

Mass percent of solution = mass of solute/mass of solution × 100/1

Mass percent of solution =  72.5 g/ (72.5 g + 2100 g) × 100/1

= 3.33%

Lear more:https://brainly.com/question/11897796

A scientists compares two samples of white powder

Answers

Answer:

answer is in exaplation

Explanation:

Answer. Chemical reaction had occurred and both the powders are different substances.

Explanation:

As density is an intensive property of the substance.Which means that different substance have different densities.

Density = \frac{mass}{volume}

volume

mass

Density of powder 1, d_1=\frac{0.5g}{45cm^3}=0.11g/cm^3d

1

=

45cm

3

0.5g

=0.11g/cm

3

Density of powder 2, d_2=\frac{1.3g}{65cm^3}=0.02g/cm^3d

2

=

65cm

3

1.3g

=0.02g/cm

3

On comparing both the densities of the powders we can say that both the substances are different. So we can conclude that the chemical reaction had occurred.

Calculate the pH of this solution 0.0043 M of H2SO4=

Answers

Answer:

pH = - log [concentration]

pH = - log (0.0043M)

pH = 2.37

Based on VSEPR theory and your observations from the Molecular Geometry lab consider the following questions What is the predicted hybridization at an atom which is surrounded by a double bond and two single bonds?
a) Sp
b) sp^2
c) sp^3

Answers

Answer:

b) sp^2

Explanation:

Hybridization refers to the concept that atomic orbitals fuse to form newly hybridized orbitals, which in turn, influences molecular geometry and bonding properties. In chemistry, orbital hybridisation (or hybridization) is the implies the mixing of atomic orbitals to form hybrid orbitals (with different energies, shapes, etc., different from that of the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds according to the principles of the valence bond theory.

In 1931 Linus Pauling proposed the idea of “mixing” the orbitals or “hybridizing” them to account for certain observed bonding patterns. Pauling proposed a sort of a combination of the orbitals giving you an orbital that has partial characters.

Hybridization is merely a mathematical construct. It is never an actual “process” that occurs within orbitals . Hybridization is a mathematical model that describes how the atomic orbitals would’ve looked like based on the observable molecular orbitals.

sp2 hybridization leads to the formation of a double bond. sigma bonds may also be formed depending on the valency of the central atom. In alkenes, an sp2 hybridized carbon atom forms a double bond in addition to two sigma bonds to other atoms.

The predicted hybridization is:

b) [tex]sp^2[/tex]

What does Hybridization tell us?

It is the integration of atomic orbitals to shape new orbitals with exclusive energies and shapes than the unique orbitals.

Given: An atom that's surrounded with the aid of using a double bond and unmarried bonds.

[tex]sp^2[/tex]  hybridization ends in the formation of a double bond. sigma bonds can also be shaped relying at the valency alkenes, an   [tex]sp^2[/tex]sigma bonds to different atoms.

Thus, correct option is b.

Find more information about Hybridization here:

brainly.com/question/26079947

Arrange the following compounds in order of increasing solubility in

water and explain your sequence.

C7H15OH C6H13OH C6H6 C2H5OH​

Answers

Answer:

C6H6<C7H15OH<C6H13OH<C2H5OH​

Explanation:

Organic substances are ordinarily nonpolar. This means that they do not dissolve in water. However, certain homologous series of organic compounds actually dissolve in water because they possess certain functional groups that effectively interact with water via hydrogen bonding.

A typical example of this is alcohol family. All members of this homologous series contain the -OH functional group. This group can effectively interact with water via hydrogen bonding, leading to the dissolution of low molecular weight alcohols in water.

Low molecular weight alcohols are miscible with water in all proportions. This implies that they are highly soluble in water. However, as the size of the alkyl moiety in the alcohol increases, the solubility of the alcohol in water decreases due to less effective interaction of the -OH group with water via hydrogen bonding. This explains the fact that C2H5OH​ is the most soluble alcohol in the list.

C6H6 is insoluble in water since it is purely a hydrocarbon with no -OH group capable of interaction with water via hydrogen bonding.

The boiling of water is a:_______.
a. chemical change because a gas (steam) is given off.
b. chemical change because heat is needed for the process to occur.
c. physical change because the water merely disappears chemical and physical damage.
d. physical change because the gaseous water is chemically the same as the liquid.

Answers

Answer:

D

Explanation:

trust me its correct i think

Fishing trawlers in a certain bay catch a large variety of marketable fish along with a species of eel that is toxic. They normally kill the eels and throw them back into the sea. What term is used to refer to the eel? The eel species is called a(n) ______ of the fishing operation.

Answers

Answer:

1. Non-target

2. Bycatch

Explanation:

In the fishing industry, the main aim of the industry is to capture fishes that can be used or eaten and sell. A variety of fishes are captured for this purpose and since they are used therefore are known as Target catch.

But there are some species which has to be discarded because they are toxic and not useful. These non-useful species like eel which gets captured in the net while capturing other fishes are known as Non-target fish.

The eel fish which gets captured is known as bycatch fishes in the fishing operation.

Thus, Non-target and Bycatch are the correct answer.

Answer:

Its just bycatch

Explanation:

The eel species is called a(n) bycatch of the fishing operation.

molar mass of A1C1 3

Answers

Answer:

Gold(III) chloride

Give two examples (i.e. list 2 elements that are examples) of: a. an atom with a half-filled subshell b. an atom with a completely filled outer shell c. an atom with its outer electrons occupying a half-filled subshell and a filled subshell

Answers

Answer:

an atom with a half-filled subshell - hydrogen

an atom with a completely filled outer shell - argon

an atom with its outer electrons occupying a half-filled subshell and a filled subshell- copper

Explanation:

The outermost shell or the valence shell of the atom is the last shell in the atom. Chemical reactions occur at this outer most shell. The number of electrons on the outermost shell of an atom determines the group to which it belongs in the periodic table as well as its chemical properties.

Hydrogen has a half filled 1s sublevel. Only one electron is present in this sublevel.

Let us consider argon

1s2 2s2 2p6 3s2 3p6

The outermost ns and np levels are completely filled. Thus the outermost shell is completely filled.

In the last case; let us look at the electronic configuration of nitrogen;

1s2 2s2 2p3

The outermost 2p subshell is exactly half filled while the 2s sublevel is fully filled. The outermost shell of nitrogen is made up of 2s2 and 2p3 sublevels.

Describe, in detail, to a freshman undergraduate how to make 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media. [Include things like how many grams of each component that you use, how much antibiotic (in ml) to add (stock solutions – 100 mg/ml ampicillin, 25 mg/ml kanamycin), and in what type of container you perform the sterilization step.] Show your calculations.

Answers

Answer:

Explanation:

The objective here is to prepare  1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.

Given that :

the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml since 100 mg/ml = 100000  μg/ml

However, using formula C₁V₁=C₂V₂ (Ampicilin),

where:

C₁ = 100000 μg/ml,

V₁=?,

C₂= 50  μg/ml,

V₂=1000 ml

100000 μg/ml × V₁ = 50  μg/ml × 1000 ml

V₁ =  50  μg/ml × 1000 ml/100000 μg/ml

V₁ =   0.5 ml

Given that:

the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml , 25 mg/ml = 25000 μg/ml

Now by using formula C₁V₁=C₂V₂ (Kanamycin),

C₁ = 25000 μg/ml,

V₁=?,

C₂= 100  μg/ml,

V₂=1000 ml

25000 μg/ml × V₁ = 100  μg/ml × 1000 ml

V₁ =  100  μg/ml × 1000 ml/25000 μg/ml

V₁ =   4 ml

Thus; in 1 lite of Lb+ Kan+Amp preparation;

0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.

Finally;

Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.

Acetonitrile, CH3CN, is a polar organic solvent that dissolves many solutes, including many salts. The density of a 1.80 M acetonitrile solution of LiBr is 0.826 g/mL. Calculate the concentration of the solution in units of (a) molality; (b) mole fraction of LiBr; (c) mass percentage of CH3CN.

Answers

Answer:

(a) [tex]m=2.69m[/tex]

(b) [tex]x_{LiBr}=0.099[/tex]

(c) [tex]\% LiBr=18.9\%[/tex]

Explanation:

Hello,

In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:

(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:

[tex]m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g[/tex]

Next, we compute the mass of the solution:

[tex]m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g[/tex]

Then, the mass of the solvent (acetonitrile) in kg:

[tex]m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg[/tex]

Finally, the molality:

[tex]m=\frac{1.80mol}{0.670kg} \\\\m=2.69m[/tex]

(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):

[tex]n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol[/tex]

Then, the mole fraction of lithium bromide:

[tex]x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099[/tex]

(c) Finally, the mass percentage with the previously computed masses:

[tex]\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%[/tex]

Regards.

A 75 gram solid cube of mercury (II) oxide has a density of 2.4 x 103 kg/m3 .
What is the length of one side of the cube in cm?

The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?

The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?

Answers

Answer:

0.031 m

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

Chemical change

Element

Explanation:

A 75 gram solid cube of mercury (II) oxide has a density of 2.4 × 10³ kg/m³.  What is the length of one side of the cube in cm?

Step 1: Convert the mass to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]75g \times \frac{1kg}{1,000g} = 0.075kg[/tex]

Step 2: Calculate the volume (V) of the cube

[tex]0.075kg \times \frac{1m^{3} }{2.4 \times 10^{3} kg} = 3.1 \times 10^{-5} m^{3}[/tex]

Step 3: Calculate the length (l) of one side of the cube

We will use the following expression.

[tex]V = l^{3} \\l = \sqrt[3]{V} = \sqrt[3]{3.1 \times 10^{-5} m^{3} }=0.031m[/tex]

The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?

The balanced chemical equation is:

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

This is a chemical change because new substances are formed.

The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?

The liquid gray substance is Hg(l), which is an element because it is formed by just one kind of atoms.

The mass of an object with 500 J of kinetic energy moving with a velocity of 5 m/s is kg.

Answers

Answer:

[tex]m=20kg[/tex]

Explanation:

Hello,

In this case, we define the kinnetic energy as:

[tex]K=\frac{1}{2} m*v^2[/tex]

Thus, for finding the mass we simply solve for it on the previous equation given the kinetic energy and the velocity:

[tex]m=\frac{2*K}{v^2}=\frac{500kg*\frac{m^2}{s^2} }{(5\frac{m}{s})^2} =\frac{500kg*\frac{m^2}{s^2} }{25\frac{m^2}{s^2}}\\\\m=20kg[/tex]

Best regards.

Answer:

The answer is 40 kg

Explanation:

You will this formula below:

m=[tex]\frac{2*\\KE}{v^{2} }[/tex]

Now we know our formula, now we plug in the given numbers:

m=[tex]\frac{2(500J)}{(5m/s)^2}[/tex]

Simplify and we get:

m=40 kg

I hope this was helpful.

4 molecules of glucose has how many carbon hydrogen and oxygen

Answers

Answer:

24 carbons

48 hydrogens

24 oxygen

Explanation:

reasons for good care on computer​

Answers

answer

1)maximise your software efficiency

2)Prevention against viruses and malware

3)Early detection of problematic issues

4)prevent data loss

5)Speed up your computer

Write the limiting forms (or Canonical forms) of the following ions:

i. H3O+

, ii. CO3

2-

, iii. NO3-​

Answers

Answer:

Canonical structures of a chemical specie explain its observed properties from a valence bond theory perspective.

Explanation:

Resonance is a valence bond concept introduced by Linus Pauling to explain the observed properties of certain chemical species such as bond lengths, bond angles, bond order , etc.

There are certain chemical species for which a single chemical structure does not suffice in explaining its observed properties. For instance, the bond order in CO3^2- is about 1.33. Its bond length, shows that the C-O bond present in CO3^2- is neither a pure C-O single bond nor a pure C-O double bond. Hence the structure of CO3^2- is 'somewhere in between' three contributing canonical structures as shown in the image attached to this answer. The resonance structures of NO3^- are also shown.

.Draw the born-Haber lattice energy cycle for sodium chloride. Explain the concept of resonance using the nitrate ion structure.

Answers

Answer:

Born-Haber cycle is consist on four to five steps. 1: ionization energy 2: electron affinity 3: dissociation energy 4: sublimation energy and last is Hess law.Nitrate ion have 3 localized sigma bonds and 1 delocalized pie bond according to the resonance structure.

Explanation:

Step 1: NaCl(s) → Na(s) + 1/2 Cl2(g)  ΔHf   (ionization energy) in this step energy is required to change the phase of the compound

Step 2: Na(s) + 1/2 Cl2(g) → Na(g) + 1/2 Cl2(g)  ΔHa  (elements needed to be in gaseous state for born-haber cycle so metal changes from solid to gas state by changing the enthalpy.

Step 3: Na(g) + 1/2 Cl2(g) → Na(g) + Cl (g)  1/2ΔHd  

Step 4: Na(g) + Cl(g) → Na⁺(g) + Cl⁻(g)  IE+EA  ( in this step both ionization energy and electron affinity was involved because in metal (Na) electron is added which needs the energy and this energy draw from the step 3 and Chlorine require releasing electron to be in ionic state so when electron leaves the orbit energy releases.

Step 5: final step is Hess Law which is the combination of all the steps which step 4 again go back to step 5 and this cycle continues by repeating same steps    Na⁺(g) + Cl⁻(g)→NaCl(s)

at this step heat of formation is calculated

Heat of formation= atomization energy+ dissociation energy+ sum of ionization energies + sum of electron affinity + lattice energy.

2:  if we look at the electron configuration of the nitrogen it has 5 electrons in its outermost shell which indicates it can make 5 bonds 4 bonds and 1 lone pair usually and Oxygen has 6 electrons in its outermost shell. So nitrate ion have the total number of 24 electrons including the 1 electron which shows on the compound.

So when they make nitrate ion NO₃⁻¹ it shows that nitrate has 3 resonance structures. Nitrogen's three sigma bonds are attached to oxygen and fourth one make 1 pie bond which can rotate, delocalized and change its position anytime from one Oxygen atom to other oxygen atom.

Classify an element having the following ground state electron configuration as an alkali metal, alkaline earth metal, nonmetal, halogen, transition metal, or noble gas.

a. [Ne]3s1
b. [Ne]3s23p3
c. [Ar]4s23d104p5
d. [Kr]5s24d1
e. [Kr]5s24d105p6

Answers

Explanation:

Alkali metal refers to group1 elements.

Alkali earth metal refers to group 2 elements.

Non metals refers to elements in grouos 4 to group 7.

Halogen refers to group 17 elements

Transition Metal refers to group 3 to group 12 elements

Noble gases refer to elements in group 18.

To obtain the group number from the electronic configuration, we calculate the total number of electrons in the principal quantum number (coefficient of the letters).

a. [Ne]3s1

Principal quantum number = 3

Number of electrons present = 1

This element belongs to group 1. It is an Alkali Metal.

b. [Ne]3s23p3

Principal quantum number = 3

Number of electrons present = 2 + 3 = 5

This element belongs to group 15 (5A). It is a Non metal

c. [Ar]4s23d104p5

Principal quantum number = 4

Number of electrons present = 2 + 5 = 7

This element belongs to group 17 (7A). It is an Halogen.

d. [Kr]5s24d1

This configuration belongs to the element yttrium and has an incomplete d shell. Hence it is a transition metal.

e. [Kr]5s24d105p6

Principal quantum number = 5

Number of electrons present = 2 + 6 = 8

This element belongs to group 18 (8A). It is a Noble gas.

A gaseous hydride of Nitrogen
contains its own volume of Nitrogen
and twice its volume of Hydrogen
and has vapour density 16. The
formula of the hydride is.
Select one:
a. NH2
b. NH3
c. N3H
• d. N2H4

Answers

Answer:

N2H4

Explanation:

A hydride is a binary compound of hydrogen and another element. Binary compounds contain only two atoms. We have to x-ray the hydrides of nitrogen given in the question in order to make our choice. Remember that we were told that that the hydride contains its own volume of nitrogen and twice its volume of hydrogen.

Now consider the hydride N2H4.

N2H4(g) -----> N2(g) + 2H2(g)

The volume ration of nitrogen gas to hydrogen gas in N2H4 is 1:2.

The molecular mass of the compound is;

N2H4= 2(14) + 4(1)= 28+4= 32

Since

molecular mass= 2 vapour density

Vapour density= molecular mass/2

Vapour density= 32/2

Vapour density = 16

Therefore the hydride of nitrogen referred to in the question is N2H4

What is a good title for this chart?

Answers

Answer:

pH of the acid

Explanation:

Methods in electrochemistry that can be used for the separation of proteins and enzymes?

Answers

Answer:

Some of the methods in Methods in electrochemistry that can be used for the separation of proteins and enzymes are as follows:

Redox transformations: In this method enzymes or proteins would be adsorbed on the electrode surface and facilitates direct electron transfer that causes denaturation and loss of their electrochemical activities and bioactivities. It is widely used in biosensors and biofuel cells.

Protein electrophoresis: In this process proteins are seperated by placing them in a gel matrix in the presence of an electrical field. In this method a negative charge is applied so that proteins move towards a positive charge.

The molar enthalpy of
mole of a liquid.
is the heat required to vaporize one

thermochemical equation

combustion

released

vaporization

fusion

absorbed

heat

Answers

Answer:

vaporization

Explanation:

The molar enthapy of _vaporization______ is the heat required to vaporize one mole of a liquid”

ilicon has three naturally occurring isotopes (Si-28, Si-29, and Si-30). Masses and natural abundances for two isotopes are listed here. Isotope Mass (amu) Abundance (%) Si-28 27.9769 92.2 Si-29 28.9765 4.67 Si-30 ? ? Part A Find the natural abundance of Si-30 . Express your answer using two significant figures.

Answers

Answer:

Part A

The natural abundance of Si-30 = 3.1

Part B

The mass of Si-30 = 29.9551 amu

Explanation:

Part A

The sum of the natural abundances = 100

Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%.

Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67%

Si-30's mass and natural abundance are unknown.

Natural abundance for Si-30 = 100% - 92.2% - 4.67% = 3.13% = 3.1% to 2 s.f.

Part B

The total atomic mass of an element is an addition combination of the mass and natural abundances of all the isotopes of that element.

Molar mass of Silicon normally = 28.0855 amu

Let the mass of Si-30 be m

28.0855 = (27.9769×0.922) + (28.9765×0.0467) + (m×0.0313)

28.0855 = 27.14790435 + 0.0313m

0.0313m = 28.0855 - 27.14790435 = 0.93759565

m = (0.93759565/0.0313) = 29.9551325879 amu = 29.9551 amu

Hope this Helps!!

Part A:  The natural abundance of Si-30 = 3.1

  Lets solve the question:

The sum of the natural abundances = 100 Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%. Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67% Si-30's mass and natural abundance are unknown.

Natural abundance (NA) refers to the abundance of isotopes of a chemical element as naturally found on a planet.

Natural abundance for Si-30 = [tex]100\% - 92.2\% - 4.67\% = 3.13\% = 3.1\%[/tex] in significant figures.

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Calculate the pH of a 0.020 M H2CO3 solution. At 25 °C, Ka1 = 4.3 × 10-7. H2CO3(aq) + H2O(l) ↔ H3O+(aq) + HCO3-(aq)

Answers

Answer:

Explanation:

H₂CO₃(aq) + H₂O(l) ↔ H₃O⁺(aq) + HCO₃⁻(aq)

Let d be the degree of dissociation

.02( 1-d )                              .02d          .02d

Dissociation constant Ka₁ is given

4.3 x 10⁻⁷  = .02d x .02d / .02( 1-d )

= .004 d² / .02  ( neglecting d in denominator )

= .02 d²

d² = 215 x 10⁻⁷

d = 4.636 x 10⁻³

= .004636

concentration of H₃O⁺

= d x .02

= .004636 x .02

= 9.272 x 10⁻⁵

pH = - log [ H₃O⁺ ]

= - log ( 9.272 x 10⁻⁵ )

5 - log 9.272

= 5 - .967

= 4.033 .

What kind of solid is crystalline boron (B)?
A. lonic solid
B. Metallic solid
C. Molecular solid
D. Network solid

Answers

Answer:

D

Explanation:

gr. 2.3 at 25°C; valence +3. Boron is a nonmetallic element existing as a dark brown to black amorphous powder or as an extremely hard, usually jet-black to silver-gray, brittle, lustrous, metallike crystalline solid

it is a network solid, a lattice of many covalent bonds (like diamond, except that it is black rather than transparent).

Network solid kind of solid is crystalline boron (B). Hence, option D is correct.

What is Network solid?

A network solid is a solid where all the atoms are covalently bonded in a continuous network.

Boron is a nonmetallic element existing as a dark brown to black amorphous powder or as an extremely hard, usually jet-black to silver-grey, brittle, lustrous, metallike crystalline solid

It is a network solid, a lattice of many covalent bonds (like a diamond, except that it is black rather than transparent).

Hence, option D is correct.

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Choose the INCORRECT statement. A. Temperatures of two bodies are equal when the average kinetic energies of the two bodies become the same. B. The heat capacity is the quantity of heat required to change the temperature of the system by one degree. C. The specific heat is the heat capacity for one mole of substance. D. Most metals have low specific heats, as metals can be heated quickly. E. The law of conservation of energy can be written: qsystem qsurroundings

Answers

Answer:

Option C

The specific heat is the heat capacity for one mole of a substance.

Explanation:

The incorrect statement is  The specific heat is the heat capacity for one mole of a substance.

This is because the specific heat capacity is the amount of heat required to raise the temperature of 1 gramme of a substance by 1 degree Celcius.

Note that the unit in question here for the specific heat capacity of the substance is in grammes.

The definition given in the options is actually for the molar heat capacity of the substance, not the specific heat capacity.

A 1.00 liter solution contains 0.42 moles nitrous acid and 0.32 moles sodium nitrite .
If 0.16 moles of nitric acid are added to this system, indicate whether the following statements are true or false.
(Assume that the volume does not change upon the addition of nitric acid.)
A. The number of moles of HNO2 will decrease.
B. The number of moles of NO2- will remain the same.
C. The equilibrium concentration of H3O+ will increase.
D. The pH will decrease.
E. The ratio of [HNO2] / [NO2-] will increase

Answers

Answer:

E. The ratio of [HNO2] / [NO2-] will increase

D. The pH will decrease.

Explanation:

Nitrous acid ( HNO₂ ) is a weak acid and NaNO₂ is its salt . The mixture makes a buffer solution .

pH = pka + log [ salt] / [ Acid ]

= 3.4 + log .32 / .42

= 3.4 - .118

= 3.282 .

Now .16 moles of nitric acid is added which will react with salt to form acid

HNO₃ + NaNO₂ = HNO₂ + NaNO₃

concentration of nitrous acid will be increased and concentration of sodium nitrite ( salt will decrease )

concentration of nitrous acid = .42 + .16 = .58 M

concentration of salt = .32 - .16 = .16 M

ratio of [HNO₂ ] / NO₂⁻]

= .42 / .32 = 1.3125

ratio of [HNO₂ ] / NO₂⁻] after reaction

= .42 + .16 / .32 - .16

= 58 / 16

= 3.625 .

ratio will increase.

Option E is the answer .

pH after reaction

= 3.4 + log .16 / .58

= 2.84

pH will decrease.

Round to 3 significant figures.
1.4593

Answers

Answer : The correct answer is 1.46

Explanation :

The following rules are used to round off a number to the required number of significant figures:

(1) If the rightmost digit to be removed is more than 5, the preceding number is increased by one.

(2) If the rightmost digit to be removed is less than 5, the preceding number is not changed.

(3) If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but it is increased by one if it is an odd number.

(4) The same procedure is follow for decimal values.

As we are given, 1.4593

In the given answer, there are 5 significant figures. Now we have to convert it into 3 significant figures.

According to the rules, round off the given measurement in three significant figures as 1.46

Therefore, the correct answer is 1.46

Unscramble the following words to form a complete
sentence about the cycles of nature:

limited is through environment Matter recycled the on Earth is and

Answers

Answer:

recycled is limited through enviroment and matter on earth

Explanation:

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