A laboratory assistant needs to prepare 217 mL of 0.246 M solution. How many grams of calcium chloride will she need

Answer:

3

Explanation:

Resonance is a valence bond concept put forward by Linus Pauling to explain the fact that the observed properties of a molecule may be as a result of the fact that its actual structure lie somewhere between a given number of structural extremes called canonical structures or resonance structures.

There are three resonance structures for SO3 that obey the octet rule. All the S-O bonds in SO3 are equivalent in these resonance structures.

Seven equivalent resonance structures for the molecular of SO3 can be drawn without breaking the octet rule.

We can arrive at this answer because:

In an SO3 molecule, electrons are shared between atoms. This sharing can be done with seven resonance structures.

These structures are shown in the figure below.

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Answer:

2-methyl-2-pentyl-1,3-dioxolane

Explanation:

In this case, we have two reactions:

First reaction:

1-heptyne + mercuric acetate -------> Compound A

Second reaction:

Compound A + HOCH2CH2OH -------> Compound C

First reaction

In the first reaction, we have as a main functional group a triple bond. We have to remember that mercuric acetate in sulfuric acid will produce a ketone. The carbonyl group (C=O) would be placed in the most substituted carbon of the triplet bond (in this case, carbon 2). With this in mind, we will have as a product: heptan-2-one. (See figure 1).

Second reaction

In this reaction, we have as reagents:

-) Heptan-2-one

-) Ethylene-glycol [tex]HOCH_2CH_2OH[/tex]

-) Sulfuric acid [tex]H_2SO_4[/tex]

When we put ethylene-glycol with a ketone or an aldehyde we will form a cyclic acetal. In this case, this structure would be formed on carbon 2 forming 2-methyl-2-pentyl-1,3-dioxolane. (See figure 2).

I hope it helps!

Answer:

To prepare vegetables for finishing by grilling, sautéing, pan frying, deep frying, or stewing, you should parboil them to cook them to partial doneness.

Answer:

176.8 g of ammonia, NH3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6 g

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34 g.

From the balanced equation above,

6 g of H2 reacted to produce 34 g of NH3.

Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.

This can be obtained as follow:

From the balanced equation above,

6 g of H2 reacted to produce 34 g of NH3.

Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.

Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.

Answer:

[tex]128~ATP[/tex]

Explanation:

The metabolic pathway by which energy can be obtained from a fatty acid is called "beta-oxidation". In this route, acetyl-Coa is produced by removing 2 carbons from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the number of rounds that will take place for an 18-carbon fatty acid using the following equation:

[tex]Number~of~Rounds=\frac{n}{2}-1[/tex]

Where "n" is the number of carbons, in this case "18", so:

[tex]Number~of~Rounds=\frac{18}{2}-1~=~8[/tex]

We also have to calculate the amount of Acetyl-Coa produced:

[tex]Number~of~Acetyl-Coa=\frac{18}{2}~=~9[/tex]

Now, we have to keep in mind that in each round in the beta-oxidation we will have the production of 1 [tex]FADH_2[/tex] and 1 [tex]NADH[/tex]. So, if we have 8 rounds we will have 8 [tex]FADH_2[/tex] and 8 [tex]NADH[/tex].

Finally, for the total calculation of ATP. We have to remember the yield for each compound:

-) [tex]1~FADH_2~=~2~ATP[/tex]

-) [tex]1~NADH~=~3~ATP[/tex]

-) [tex]Acetyl~CoA~=~10~ATP[/tex]

Now we can do the total calculation:

[tex](8*2)~+~(8*3)~+~(9*10)=130~ATP[/tex]

We have to subtract  "2 ATP" molecules that correspond to the activation of the fatty acid, so:

[tex]130-2=128~ATP[/tex]

In total, we will have 128 ATP.

I hope it helps!

Answer:

0.057 M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴

Concentration of mercury (II) ion: 0.085 M

Step 2: Write the reaction for the solution of HgBr₂

HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻

Step 3: Calculate the bromide concentration needed for a precipitate to occur

The Ksp is:

Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²

[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M

Answer:

sodium hexachloroplatinate(IV)- Na2[PtCl6]

dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br

pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2

Explanation:

The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.

The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.

Answer:

pH = 3.49

Explanation:

We have a buffer system formed by a weak acid (HNO₂) and its conjugate base (NO₂⁻ coming from KNO₂). We can calculate the pH  of a buffer ssytem using the Henderson-Hasselbach equation.

pH = pKa + log [base] / [acid]

pH = -log Ka + log [NO₂⁻] / [HNO₂]

pH = -log 4.50 × 10⁻⁴ + log 0.290 M / 0.210 M

pH = 3.49

The pH of the solution containing 0.210 M nitrous acid (HNO₂) and 0.290 M potassium nitrite (KNO₂) is 3.49

We'll begin by calculating the the pKa of acid. This can be obtained as follow:

Acid dissociation constant (Ka) = 4.50×10¯⁴

pKa = –Log Ka

pKa = –Log 4.50×10¯⁴

pKa = 3.35

Concentration of HNO₂, [HNO₂] = 0.210 M

Concentration of KNO₂, [KNO₂] = 0.290 M

The pH of the solution can obtain by using the Henderson-Hasselbach equation as illustrated below:

pH = pKa+ log [NO₂⁻] / [HNO₂]

pH = 3.35 + log (0.290 / 0.210)

Thus, the pH of the solution is 3.49

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Answer:

Kindly check the explanation section.

Explanation:

PS: kindly check the attachment below for the required diagram that is the diagram showing solid sodium chloride looks like at the atomic level.

The chemical compound known as sodium chloride, NaCl has Molar mass: 58.44 g/mol, Melting point: 801 °C and

Boiling point: 1,465 °C. The structure of the solid sodium chloride is FACE CENTRED CUBIC STRUCTURE. Also, solid sodium chloride has a coordination number of 6: 6.

In the diagram below, the positive sign shows the sodium ion while the thick full stop sign represent the chlorine ion.

The NaCl has been the ionic structure with an equal number of sodium and chlorine ions bonded.

In the structure, there has been each Na ion bonded with the Cl ions. There has been the transfer of electrons between the structure in order to attain a stable configuration.

The expected structure of the NaCl would be the image attached below.

The image has been the cubic structure of NaCl. With the presence of Na ions at the vertex of the structure, there has been the presence of the Cl ion with every Na ion for the electron transfer.

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Answer:

1. pH = 2.98

2. pH = 4.02

3. pH = 8.12

Explanation:

1. Initial molarity of benzoic acid (Molar mass: 122.12g/mol; Ka = 6.14x10⁻⁵) is:

0.235 ₓ (1mol / 122.12g) = 1.92x10⁻³ moles / 0.100L = 0.01924M

The equilibrium of benzoic acid with water is:

C6H5CO2H(aq) + H2O(l) → C6H5O-(aq) + H3O+(aq)

And Ka is defined as the ratio between equilibrium concentrations of products over reactants, thus:

Ka = 6.14x10⁻⁵ = [C6H5O⁻] [H3O⁺] / [C6H5CO2H]

The benzoic acid will react with water until reach equilibrium. And equilibrium concentrations will be:

[C6H5CO2H] = 0.01924 - X

[C6H5O⁻] = X

[H3O⁺] = X

Replacing in Ka:

6.14x10⁻⁵ = [X] [X] / [0.01924 - X]

1.1815x10⁻⁶ - 6.14x10⁻⁵X = X²

1.1815x10⁻⁶ - 6.14x10⁻⁵X - X² = 0

Solving for X:

X = -0.0010→ False solution. There is no negative concentrations

X = 0.0010567M → Right solution.

pH = - log [H3O⁺] and as [H3O⁺] = X:

pH = - log [0.0010567M]

2.

pH of a buffer is determined using H-H equation (For benzoic acid:

pH = pka + log [C6H5O⁻] / [C6H5OH]

pKa = -log Ka = 4.21 and [] could be understood as moles of each chemical

The benzoic acid reacts with NaOH as follows:

C6H5OH + NaOH → C6H5O⁻ + Na⁺ + H₂O

That means NaOH added = Moles C6H5O⁻ And C6H5OH = Initial moles (1.92x10⁻³ moles - Moles NaOH added)

7.00mL of NaOH 0.108M are:

7x10⁻³L ₓ (0.108 mol / L) = 7.56x10⁻⁴ moles NaOH = Moles C₆H₅O⁻

And moles C6H5OH = 1.92x10⁻³ moles - 7.56x10⁻⁴ moles = 1.164x10⁻³ moles C₆H₅OH

Replacing in H-H equation:

pH = 4.21 + log [7.56x10⁻⁴ moles] / [ 1.164x10⁻³ moles]

3. At equivalence point, all C6H5OH reacts producing C6H5O⁻. The moles are 1.164x10⁻³ moles

Volume of NaOH to reach equivalence point:

1.164x10⁻³ moles ₓ (1L / 0.108mol) = 0.011L. As initial volume was 0.100L, In equivalence point volume is 0.111L and concentration of C₆H₅O⁻ is:

1.164x10⁻³ moles / 0.111L = 0.01049M

Equilibrium of  C₆H₅O⁻ with water is:

C₆H₅O⁻(aq) + H₂O(l) ⇄  C₆H₅OH(aq) + OH⁻(aq)

Kb = [C₆H₅OH] [OH⁻]/ [C₆H₅O⁻]

Kb = kw / Ka = 1x10⁻¹⁴ / 6.14x10⁻⁵ = 1.63x10⁻¹⁰

Equilibrium concentrations of the species are:

C₆H₅O⁻ = 0.01049M - X

C₆H₅OH = X

OH⁻ = X

Replacing in Kb expression:

1.63x10⁻¹⁰ = X² / 0.01049- X

1.71x10⁻¹² - 1.63x10⁻¹⁰X - X² = 0

Solving for X:

X = -1.3x10⁻⁶ → False solution

X = 1.3076x10⁻⁶ → Right solution

[OH⁻] =  1.3076x10⁻⁶

as pOH = -log [OH⁻]

pOH = 5.88

And pH = 14 - pOH

Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

Answer:

I believe the answer The case study was influenced by bias, and led to incorrect conclusions being drawn. plz correct me if I am wrong

Explanation:

Answer: options B

Explanation:

Answer:

Triacontyl palmitate

Explanation:

In this case, we have a reaction between an acid and an alcohol. When we put together these kind of compounds an ester is produced. This reaction is called "esterification".

In our case, the alcohol is a structure with 30 carbon in which the "OH" group is bonded on carbon 1. The name of this compound is "n-triacontanol". The acid is a structure in which we have 16 carbon in which the "COOH" group is placed on carbon 1. The name of this compound is "palmitic acid". The ester produced by the acid and the alcohol is "Triacontyl palmitate".

See figure 1.

I hope it helps!

Answer:

d. Oxidation and reduction

Explanation:

For this question we have to remember the definition of each type of reaction:

-) Hydrogenation

In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.

-) Alkylation

In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.

-) Hydrolysis

In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.

-) Halogenation

In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.

-) Ammoniation

In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.

-) Oxidation and reduction

In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:

[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction

[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation

Answer:

PbCl₂ will precipitate from solution.

Explanation:

Statements are:

Insufficient information is given.

Both NH4NO3 and PbCl2 precipitate from solution.

No precipitate forms.

PbCl2 will precipitate from solution.

NH4NO3 will precipitate from solution.

The reaction of ammonium chloride (NH₄Cl) with lead nitrate (Pb(NO₃)₂) is:

Pb(NO₃)₂ + 2NH₄Cl → PbCl₂ + 2 NH₄NO₃

Talking of rules of solubility, all nitrates are soluble in water, that means NH₄NO₃ is soluble and no precipitate is formed.

In the same way, all chlorides are soluble except silver chloride and lead chloride. That means:

Explanation:

a. Esters

b. Carboxylic acids

c. Esters (ethyl hexanoate smells like pineapple)

d. Carboxylic acids (produces a carboxylic salt)

For the given phrases the following description is better.

a. Esters

b. Carboxylic acids

c. Esters (ethyl hexanoate smells like pineapple)

d. Carboxylic acids (produces a carboxylic salt)

An ester is a synthetic compound got from a corrosive in which somewhere around one - OH hydroxyl bunch is supplanted by an - O-alkyl (alkoxy) bunch, as in the replacement response of a carboxylic acid and a liquor.

Carboxylic acid is any of a class of natural mixtures in which a carbon (C) particle is clung to an oxygen (O) molecule by a twofold bond and to a hydroxyl bunch (―OH) by a solitary bond.

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Answer:

See explanation

Explanation:

The reaction of chlorine with the pictured compound will occur via free radical mechanism. The stability of the free radical formed will depend on its structure.

The order of stability of free radicals is methyl < primary < secondary < tertiary. Hence a tertiary carbon free radical is the most stable.

Looking at the compound, the radical will form at the position shown in the image attached since it will lead to a secondary free radical which is more stable.

The structure that should be drawn is shown below.

It should be within the pictured compound that will arise via a free radical mechanism. The stability should be based on the structure. The stability of the order of free radicals should be methyl < primary < secondary < tertiary. Thus, a tertiary carbon free radical should be most stable.

Here look at the compound, the radical should form at the position that should be shown in the image that resulted in the secondary free radical i.e. more stable.

Answer:

0.16 M

Explanation:

Data provided as per the question is below:-

Volume of [tex]HNO_3[/tex] = 0.22 L

The Volume of NaOH = 0.18 L

Morality of NaOH = 0.2

According to the given situation, the calculation of the concentration of the [tex]HNO_3[/tex] is shown below:-

For equivalence,

Number of the equivalent of [tex]HNO_3[/tex] = Number of equivalents of NaOH

[tex]= \frac{0.18\times0.2}{0.22}[/tex]

[tex]= \frac{0.036}{0.22}[/tex]

= 0.16363 M

or

= 0.16 M

Answer:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases.

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Explanation:

Hello,

In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.

With the aforementioned, we can conclude that the chemical reaction:

[tex]N_2O_4(g) \rightleftharpoons 2 NO_2(g)[/tex]

Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:

[tex]G=H-TS[/tex]

As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:

[tex]K=exp(-\frac{\Delta G}{RT} )[/tex]

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Regards.

Answer:

[tex]\Delta G=-359\frac{kJ}{mol}[/tex]

Explanation:

Hello,

In this case, we must remember that the Gibbs free energy is defined in terms of the enthalpy, temperature and entropy as shown below:

[tex]\Delta G=\Delta H -T\Delta S\\[/tex]

In such a way, for the given data, we obtain it, considering the conversion from J to kJ for the entropy in order to conserve the proper units:

[tex]\Delta G=-292\frac{kJ}{mol} -(298)(224\frac{J}{mol}*\frac{1kJ}{1000J} )\\\\\Delta G=-359\frac{kJ}{mol}[/tex]

Best regards.

Answer:

B- 358 kj

Explanation: I took the test

Answer:

In 5 US liquid gallons, there are 80 cups.

Explanation:

To get from gallons to cups, just multiply the amount of gallons you have by 16.

Answer:

The frequency of the photon is 7.41*10¹⁶ Hz

Explanation:

Planck states that light is made up of photons, whose energy is directly proportional to the frequency of radiation, according to a constant of proportionality, h, which is called Planck's constant. This is expressed by:

E = h*v

where E is the energy, h the Planck constant (whose value is 6.63*10⁻³⁴ J.s) and v the frequency (Hz or s⁻¹).

So the frequency will be:

[tex]v=\frac{E}{h}[/tex]

Being E= 4.91*10⁻¹⁷ J and replacing:

[tex]v=\frac{4.91*10^{-17} J}{6.63*10^{-34} J.s}[/tex]

You can get:

v= 7.41*10¹⁶ [tex]\frac{1}{s}[/tex]= 7.41*10¹⁶ Hz

The frequency of the photon is 7.41*10¹⁶ Hz

Answer:Liquid

Explanation:

Answer:

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

Answer:

ELECTROLYTIC CELL

An electrochemical cell in which a non-spontaneous chemical reaction takes place when electric current is passed through the solution,is called an electrolytic cell.

EXAMPLE

Nelsons cell and Downs cell

VOLTAIC CELL OR GALVANIC CELL

The electrochemical cell in which a spontaneous chemical reaction takes place and generates electric current is called galvanic and voltaic cell.

EXAMPLE

Daniel cell

Explanation:

Electrolytic cell by the non-spontaneous reactions covert the chemical energy into electical and volataic cells are those in which spontaneous redox reaction takes place.

Cell is a device which converts chemical energy into electrical energy.

Electrolytic Cell: An electrolytic cell is an electrochemical cell in which a non-spontaneous redox chemical reaction occurs when an electric current is conducted through the solution.

Example- Electrolysis of sodium chloride, by which formation of sodium metal and chlorine gas takes place.

Voltaic cell: Voltaic cells are electrochemical cells in which a spontaneous redox chemical reaction occurs and creates electric current. These cells are also known by the name of Galvanic cells.

Example- It is used in the form of batteries which can be portable easily.

Hence, in electrolytic cells non - spontaneous reaction occur and in voltaic cell spontaneous reaction is occured.

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Answer:

The empirical formulae for the compound is CH3.

Answer:

1 L

Explanation:

ppm means parts per million. Generally the relationship between mass and litre is given as;

1 ppm = 1 mg/L

This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.

Explanation:

The spontaneity of a system is deduced by the sign of the gibbs free energy value. If it is negative, it means the process / reaction is spontaneous however a positive value indicates the such process is not spontaneous.

Gibbs free energy, enthalpy and entropy are related by the following equation;

ΔG = ΔH - TΔS

A positive value of enthalpy, H and entropy, S means that G would always be a negative value at all temperatures.

Answer:

n = 0.207 mole

Explanation:

We have,

P = 1 atm

V = 5 liter

R = 0.0821 L.atm/mol.K

T = 293 K

We need to find the value of n. The relation is as follows :

PV = nRT

Solving for n,

[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}[/tex]

So, the value of n is 0.207 mol.