Identify the compound in the following group that is most soluble in water. Match the words in the left column to the appropriate blanks in the sentences on the right. ResetHelp Of the three compounds butanoic acid, butane, and butanone, the one that is most soluble in water is . This is because its functional group can form the intermolecular forces with polar water.
butanoic acid strongest hydrocarbon butanone butane alcohol carboxylic acid What carboxylic acid is found in each of the following substances? Drag the appropriate descriptions to their respective bins. Reset rancid butter stinging red ants Methanoic acid Butanoic acid Propanoic acid Ethanoic acid Review Co Draw the structure of methyl butanoate, Draw the molecule on the canvas by choosing buttons from the Tools (for bonds and charges), Atoms toolbars. H: 129 uxo com H o с + N 1 0 S a CH Br р

Answer:

1. Participating in calcium homeostatis storage of calcium.

2. High capacity calcium (Ca) regulation and protection against herbivory

[tex]\large \boxed{\sf 2 \: functions \: of \: crystals \: are :- } [/tex]

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[tex] \sf \: \underline{ Calcium \: oxalate \: (CaOx) \: crystals} \: are \: distributed \: \\\sf among \: all \: taxonomic \: levels \\ \sf\: of \: photosynthetic \: organisms \: from \\ \sf \: small \: algae \: to \: angiosperms \: and \: giant \: gymnosperms .[/tex]

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[tex]\sf Bone \: is \: mostly \: made \: of \: \underline{mineral \: crystals} \: \\ \sf and \: the \: protein \: collagen. \: The \: mineral  \: crystals \: bone  \\ \sf\: provide \: strength \: and \: rigidity \: for \: the \: matrix \: upon \: \\ \sf \: and \: within \: which \: they \: are \: deposited.[/tex]

Answer:

0 J

Explanation:

Applying,

ΔE = q+w................ Equation 1

Where ΔE = change in internal energy of the system, q = Heat of the system, w = work of the system.

Note: q is positive, while w is negative

From the question,

Given: q = 425 J, w = -425 J

Substitute these values into equation 1

ΔE = 425-425

ΔE = 0 J

Hence the change in internal energy of the system is 0 J

Ethane consists of 6C−H bonds and 1C−C bond. Total number of bonds is 7. Each bond is made up of two electrons

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Answer:

The correct answer is - 2.557 KJ

Explanation:

In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.

we can calculate this energy by the following formula:

Q = met

where, m = mass,

e = specific heat

t = temperature

then,

Q = 78*0.14* (273-38.8)

here 0.14 = C(Hg)

= 2.557 Kj

Answer:

C. By super-cooling certain types of plasma.

Explanation:

Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.

Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.

When certain types of plasma are super cooled, Bose-Einstein condensate are formed.

Answer:

Option e.

Explanation:

Molarity is the concentration that indicates moles of solute in 1 L of solution.

We have another concentration, percent by mass.

Percent by mass indicates mass of solute in 100 g of solution.

Our solute is HNO₃, our solvent is water.

17.5 g of nitric acid is the mass of solute. We can convert them to moles:

17.5 g . 1mol / 63g = 0.278 moles

We do not have volume of solution. We assume the mass is 100 g because the percent by mass but we need density to state the volume.

Density = Mass / Volume

Mass / Density = Volume

Once we have the volume, we need to be sure the units is in L, to determine molarity

M = mol /L

Answer:

[tex]n=0.033mole[/tex]

Explanation:

From the question we are told that:

Initial volume [tex]V_1=0.40L[/tex]

Final Volume[tex]V_2=5.0L[/tex]

Work [tex]W=210J[/tex]

Temperature [tex]T=300k[/tex]

Generally the equation for Ideal gas is mathematically given by

[tex]W=nRTIn\frac{V_2}{V_1}[/tex]

[tex]n=\frac{W}{RTIn\frac{V_2}{V_1}}[/tex]

[tex]n=\frac{210}{8.32*300In\frac{5.0}{0.4}}[/tex]

[tex]n=0.033mole[/tex]

Answer:

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

Explanation:

An oxidation reaction reaction refers to a reaction in which electrons are lost. In this case, we are about to see the full balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution.

The full equation is;

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

So, two electrons were lost in the process.

Answer:

pH= 1.9 then [tex]H_{3} PO_{4}[/tex]

pH = 5.0 , [tex]CH_{3} COOH[/tex]

pH = 3.9 , HCOOH

As we know range left [tex]pH= pKa+/- 1[/tex]

endothermic absorbs heat

exothermic gives heat

a. endothermic

b. exothermic

c. endothermic

d. exothermic

a. Ice melting - endothermic

b. Water condensing on the surface - exothermic

c. Baking a cake - endothermic

d. The chemical reaction inside an instant cold pack - endothermic

e. A car using gasoline - exothermic

An exothermic reaction can be described as a thermodynamic chemical reaction that emits energy from the system to its surroundings usually in the form of light, heat, or sound.

While an endothermic reaction can be described as an opposite of an exothermic reaction where the energy gains in the form of heat. In exothermic chemical reactions, the bond energy is transformed into thermal energy.

In exothermic reactions, the reaction happens the form of the kinetic energy of molecules when the energy is released. The release of energy is due to the electronic transition of electrons from one energy level to another.

The burning of gasoline, and water condensation is also an exothermic reaction in which energy is released while ice melting and baking cake is an endothermic reaction.

Learn more about the exothermic process, here:

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Answer:

2.772 seconds

Explanation:

Given that;

t1/2 = 0.693/k

Where;

t1/2 = half life of the reaction

k= rate constant

Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant

t1/2 = 0.693/2.5 x 10-1 s-1

t1/2= 2.772 seconds

Answer:

41 g

Explanation:

The equation of the reaction is;

Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)

Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles

1 mole of sodium phosphate reacts with 1 mole of chromium nitrate

x moles of sodium phosphate react as with 0.25 moles of chromium nitrate

x= 1 × 0.25/1

x= 0.25 moles

Mass of sodium phosphate = 0.25 moles × 163.94 g/mol

Mass of sodium phosphate = 41 g

Answer:

10.71%

Explanation:

The dissociation of acetic acid can be well expressed as follow:

CH₃COOH ⇄   CH₃COO⁻  + H⁺

Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:

Then:

The I.C.E Table is expressed as follows:

                     CH₃COOH       ⇄   CH₃COO⁻        +           H⁺  

Initial              0.0014                       0                                0

Change            - x                           +x                               +x

Equilibrium   (0.0014 - x)                 x                                 x

Recall that:

Ka for acetic acid CH₃COOH  = 1.8×10⁻⁵

∴

[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]

By rearrangement:

[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]

Multiplying through  by (-) and solving the quadratic equation:

[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]

[tex](-0.00015 + x) (0.000168 + x) =0[/tex]

x = 0.00015 or x = -0.000168

We will only consider the positive value;

so x=[CH₃COO⁻] = [H⁺] = 0.00015

CH₃COOH = (0.0014 - 0.00015) = 0.00125

However, the percentage fraction of the dissociated acetic acid is:

[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]

= 10.71%

Answer:

See explanation

Explanation:

a) The balanced reaction equation is;

Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH + PbCO3

b) When we include states of matter;

Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH(aq) + PbCO3 (s)

c) Complete ionic equation;

2Na^+(aq) + CO3^2-(aq) + Pb^2+(aq) + 2OH^-(aq) ----> 2Na^+(aq) + 2OH^-(aq) + PbCO3(s)

Net Ionic equation;

Pb^2+(aq) + CO3^2-(aq) ----> PbCO3(s)

Answer:

3.00 L

Explanation:

P₁V₁ = P₂V₂

V₁ = 1.00 L

P₁ = (x) atm

P₂ = [tex]\frac{1}{3}[/tex] · (P₁) = [tex]\frac{x}{3}[/tex]

V₂ = unknown

(x atm)(1.00 L) = ( [tex]\frac{x}{3}[/tex] atm)(V₂)

divide both sides by ( [tex]\frac{x}{3}[/tex] atm)

( 1.00x )( [tex]\frac{3}{x}[/tex] ) = V₂

x cancels out

(1.00)(3) = V₂

V₂ = 3.00 L

Answer:

the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Explanation:

Given the data in the question;

Co = 53 or [ 53 wt% B-47 wt% A ]

W∝ = 0.5 = Wβ

Cβ = 92 or [ 92 wt% B-8 wt% A ]

Now, lets set up the Lever rule for W∝ as follows;

W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]

so we substitute our given values into the expression;

0.5 = [ 92 - 53 ] / [ 92 - C∝ ]

0.5 = 39 /  [ 92 - C∝ ]

0.5[ 92 - C∝ ] = 39

46 - 0.5C∝  = 39

0.5C∝ = 46 - 39

0.5C∝ = 7

C∝ = 7 / 0.5

C∝ = 14  or [ 14 wt% B-86 wt% A ]

Therefore, the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Answer:

See explanation

Explanation:

The molecule IF5 possesses five I-F polar bonds. However, the presence of polar bonds does not automatically imply that the molecule will be polar.

The geometry of the molecule is very important in determining the polarity of a compound. Since IF5 has a lone pair of electrons, the molecule is bent and as such there is a permanent dipole moment created in the molecule thereby making IF5 polar in nature.

Answer:

0.17 g

Explanation:

Since the volume of gas collected is 132 mL, we need to find the number of moles of gas present in 132 mL.

So, number of moles, n = volume of gas, v/molar volume, V

n = v/V where v = 132 mL = 0.132 L and V = 22.4 L

So, substituting the values of the variables into the equation, we have

n = v/V

n = 0.132 L/22.4 L

n = 0.005893 mol

We then need to calculate the molar mass of CO, M = atomic mass of carbon + atomic mass of oxygen = 12 g/mol + 16 g/mol = 28 g/mol

Also, number of moles of gas, n = m/M where m = mass of CO and M = molar mass of CO

m = nM

m = 0.005893 mol × 28 g/mol

m = 0.165004 g

m ≅ 0.17 g to 2 significant digits

Answer:

density of second liquid = 650 kg/m³

Explanation:

Given that:

The volume of the plastic block submerged inside the water  = 0.5 V

The force on the plastic block  = [tex]\rho_1V_1g[/tex]

[tex]= 0.5p_1 V_g[/tex]

when the block is floating, the weight supporting the force (buoyancy force) is:

W [tex]= 0.5p_1 V_g[/tex]

[tex]\rho Vg = 0.5p_1 V_g[/tex]

[tex]\rho = 0.5 \rho _1[/tex]

where;

water density [tex]\rho _1[/tex] = 1000

[tex]\rho = 0.5 (1000)[/tex]

[tex]\rho = 500 kg/m^3[/tex]

In the second liquid, the volume of plastic block in the water = (100-23)%

= 77% = 0.7 V

The force on the plastic block is:

[tex]= 0.77p_2 V_g[/tex]

when the block is floating, the weight supporting the force (buoyancy force) is:

[tex]W = 0.77p_2 V_g[/tex]

[tex]\rho Vg = 0.77 \rho_2 V_g \\ \\ \rho = 0.77 \rho_2 \\ \\ 500 = 0.77 \rho_2 \\ \\ \rho_2 = 500/0.77[/tex]

[tex]\mathbf{ \rho_2 \simeq 650 \ kg/m^3}[/tex]

Answer:

[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]

Explanation:

A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).

Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.

Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.

Answer: Chemicals like acids and bases are harmful and must be neutralized before draining.

Explanation:

A strong acid or strong base is required to be diluted or neutralized before it is discarded in the drain as if is discarded without diluting and neutralization it can spill and splash from sink or drain and can harm people in chemistry lab, moreover the fumes of the discarded chemical on spilling can cause respiratory tract burning and can even cause fire hazard so it must be converted into less harmful form and then must be drained.  

Answer:

(A) Slipping and breaking

(B) Clean and dry

(C) Cracks

Explanation:

This describes the process of unpacking a glassware for use.

You should set out support like a cork ring or clamp (these are simple machines that'll hold the glassware in place) before removing the glassware from a glassware kit; to place the glassware in and to stop it from slipping and breaking.

Thoroughly check that the glassware is clean and dry and that it does not have any cracks, before using it.

Answer: Organic compounds ending with the name (-ene) indicate that the compounds contain double bonds in their molecules.

Explanation:

Organic compounds are those molecules that contains carbon atoms (as their main element), hydrogen and oxygen which are usually present. The presence of numerous organic compounds is due to the following properties of carbon:

--> the exceptional ability of carbon atoms to catenate, that is, to combine with one another to form straight chains, branched chains or ring compounds containing many carbon atoms.

--> The ease with which carbon combines with hydrogen, oxygen, Nitrogen and halogens

--> The ability of carbon atoms to form single, DOUBLE or triple bonds.

The organic compound that has the name ending with -ene are known as the alkenes. The members of the alkene series are formed from the alkanes by the removal of two hydrogen atoms and the introduction of a DOUBLE BOND in the carbon chain. They are named after the corresponding alkanes by changing the -ane ending to -ene.

Note: the systematic name of a compound is formed from the root hydrocarbon by adding a suffix and prefixes to denote the substitution of the hydrogen atoms.

Answer:

1.) Propanone (ketone)

2.) Ethanal( aldehyde)

3.) 3-phenyl-2-propenal (aldehyde)

4.) Butanone (ketone)

5.) Ethanol ( alcohol)

6.) 2-propanol (alcohol)

Explanation:

In organic chemistry, ALCOHOL ( also known as alkanol) are compounds in which hydroxyl groups are linked to alkyl groups. They can be considered as being derived from the corresponding alkanes by replacing the hydrogen atoms with hydroxyl groups. The hydroxyl group is the functional group of the alcohol as it is responsible for their characteristic chemical properties. A typical example of alcohol is ethanol and 2-propanol.

Alkanals or ALDEHYDES have the general formula RCHO while alkanones or KETONES have the general formula RR'CO where R and R' may be alkyl or aryl groups. The main similarity between these two classes of compounds is the presence of the carbonyl group. In aldehydes, there is a hydrogen atom attached to the carbon In the carbonyl group while there is none on the ketones.

Some common examples of ketones are Propanone, Butanone while examples of aldehydes are Ethanal and 3-phenyl-2-propenal

he says he doesnt know sorry

Answer:

a. 2..86 b. 4.86 c. 10.7 d. 8.7

Explanation:

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 2.86

b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.

pH = pKa + log0.99x/0.01x

pH = pKa + log0.99/0.01

pH = 2.86 + log99

pH = 2.86 + 1.996

pH = 4.856

pH ≅ 4.86

c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 10.7

d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.

pH = pKa + log0.01x/0.99x

pH = pKa + log1/99

pH = 10.7 - log99

pH = 10.7 - 1.996

pH = 8.704

pH ≅ 8.7

Answer:

Explanation:

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Answer :  The work done for path A and path B is -685.3 J and -478.1 J  respectively.

Explanation :

To calculate the work done for path A :

First we have to calculate the moles of the gas.

where,

= initial pressure of gas  = 2.57 atm

= initial volume of gas  = 3.42 L

n = moles of gas  = ?

R = gas constant = 0.0821 atm.L/mol.K

T = temperature of gas  = 298 K

Now put all the given values in the above formula, we get:

According to the question, this is the case of isothermal reversible expansion of gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

The expression used for work done will be,

where,

w = work done on the system = ?

n = number of moles of gas  = 0.359 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 298 K

= initial volume of gas  = 3.42 L

= final volume of gas  = 7.39 L

Now put all the given values in the above formula, we get :

Thus, the work done of path A is, -685.3 J

To calculate the work done for path B :

The formula used for isothermally irreversible expansion is :

where,

w = work done

= external pressure = 1.19 atm

= initial volume of gas = 3.42 L

= final volume of gas = 7.39 L

Now put all the given values in the above formula, we get :

Thus, the work done of path B is, -478.1 J