What is the edge length of a face-centered cubic unit cell that is made of of atoms, each with a radius of 154 pm

The correct answer is 3 liters

Explanation:

If a substance has a solubility of [tex]12 \frac{grams}{liter}[/tex], this means in 1 liter, the grams that can be dissolved are 12 grams. Now, considering Justin and Ellie need to dissolve 36 grams to calculate the number of liters just divide the total of grams into 12 as each liter dissolves only 12 grams. The process is shown below:

36 grams (the amount that will be dissolved) ÷ 12 (grames dissolved per liter) = 3 liters (liters to dissolved 36 grams)

Answer:

It would be 3 liters

Explanation:

Answer:

H₂CO₃   H₂O  ⇄  HCO₃⁻  +  H₃O⁺          Ka1

HCO₃⁻  +  H₂O  ⇄  CO₃⁻²  +  H₃O⁺        Ka2

CO₃⁻²  +  H₂O  ⇄  HCO₃⁻   +  OH⁻       Kb1

HCO₃⁻  +  H₂O  ⇄   H₂CO₃  +  OH⁻     Kb2

Explanation:

Formula for carbonic acid is: H₂CO₃

It is a dyprotic acid, because it can release two protons. We can also mention that is a weak one. The equilibrums are:

H₂CO₃   H₂O  ⇄  HCO₃⁻  +  H₃O⁺          Ka1

HCO₃⁻  +  H₂O  ⇄  CO₃⁻²  +  H₃O⁺        Ka2

When the conjugate strong bases, carbonate and bicarbonate take a proton from water, the reactions are:

CO₃⁻²  +  H₂O  ⇄  HCO₃⁻   +  OH⁻       Kb1

HCO₃⁻  +  H₂O  ⇄   H₂CO₃  +  OH⁻     Kb2

Notice, that bicarbonate anion can release or take a proton to/from water. This is called amphoteric,

Answer:

(1) -12 Kcal/mol

Explanation:

Our answer options for this question are:

(1) -12 Kcal/mol

(2) -13 Kcal/mol

(3) -15 Kcal/mol

(4) -16 Kcal/mol

With this in mind, we can start with the chemical reaction (Figure 1). In this reaction, two bonds are broken, a C-H and a Br-Br. Additionally, a C-Br and a H-Br are formed.

If we want to calculate the enthalpy value, we can use the equation:

ΔH=ΔHbonds broken-ΔHbonds formed

If we use the energy values reported, its possible to calculate the energy for each set of bonds:

ΔHbonds broken

C-H = 94.5 Kcal/mol

Br-Br = 51.5 Kcal/mol

Therefore:

105 Kcal/mol + 53.5 Kcal/mol = 146 Kcal/mol

ΔHbonds formed

C-Br = 70.5 Kcal/mol

H-Br = 87.5 Kcal/mol

Therefore:

70.5 Kcal/mol + 87.5 Kcal/mol = 158 Kcal/mol

ΔH of reaction

ΔH=ΔHbonds broken-ΔHbonds formed=(146-158) Kcal/mol = -12 Kcal/mol

I hope it helps!

Answer:

The salt breaks into positive chlorine ions and negative sodium icons

Explanation:

The question requested for the wrong option in the list. If we look at the option selected, we will notice that sodium ions are positively charged ions since sodium is a metal. Metals produce cations (positive ions) because they loose electrons. Therefore, a sodium ion can never be negatively charged.

Similarly, chlorine is a highly electronegative nonmetal. It gains electrons in an ionic bond. Hence chlorine ions can not be positive.

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.

Find more information about esters here:

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Answer:

Lithium chloride (LiCl), carbon dioxide, and water

Explanation:

Li₂CO₃ + HCl ⇒ LiCl + CO₂ + H₂O

When lithium carbonate reacts with hydrochloric acid, lithium chloride, water, and bubbles of carbon dioxide gas are given off. This is the result of a double displacement reaction followed by a decomposition reaction.

Hope that helps.

Answer:

Due to the resonance structures

Explanation:

In the question:

"Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus"

We have to take into account the structure of the amino acid arginine. In which, we have the amino and the carboxylic groups in the right and the guanidine group in the left.

In this group, we have a central carbon with three nitrogen atoms around and a double bond with the nitrogen on the top. This nitrogen on the top will accept the proton because the structure produced will have a positive charge on this nitrogen. Then, the double bond with the carbon can be delocalized into the nitrogen producing a positive charge in the carbon.

In this structure (the carbocation), we can have several resonance structures. In the blue option, we can produce a double bond with the nitrogen on the right. In the purple option, we can produce a double bond with the nitrogen on the left.

In conclusion, if the nitrogen in the top on the guanidine group accepts an hydrogen atom and we will have several resonance structures that can stabilize the molecule. Due to this, the nitrogen in the top its the best option to accept hydrogens.

See figure 1

I hope it helps!

Answer:

Boiling

Explanation:

When a liquid is heated, the vapor pressure rises steadily. When water attains a temperature of 100°C or 212°F its vapor pressure is now equal to the atmospheric pressure at sea level, this is what we mean by boiling.

When this occurs, water continues to evaporate untill the vapor pressure inside the bubbles becomes high enough to stop water bubbles from collapsing again from the pressure of the water around it so the bubbles rise and break the surface.

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate [tex]SO_4~^-^2[/tex] and the chloride [tex]Cl^-[/tex]:

Sulfate

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.([tex]Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2[/tex]), which are NOT soluble.

Chloride

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ([tex]Pb^+^2~,Ag^+~,Hg_2~^+^2[/tex]), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I),  lead" and our possibilities are:

"Calcium, strontium, or barium".

I hope it helps!

Answer:

[tex]M=0.684M[/tex]

Explanation:

Hello,

In this case, considering that the solution is formed by NaCl as the solute and water as the solvent, we can compute the molarity as shown below:

[tex]M=\frac{mol_{solute}}{V_{solution}}[/tex]

Whereas the volume of the solution must be in liters. In such a way, since the addition of sodium chloride does not significantly changes the volume of the solution we can say it remains in 100 mL (0.100 L) and the moles of sodium chloride are computed by using its molar mass (58.45 g/mol):

[tex]mol_{solute}=4g*\frac{1mol}{58.45g} =0.0684mol[/tex]

Therefore, the molarity is:

[tex]M=\frac{0.0684mol}{0.100L} \\\\M=0.684\frac{mol}{L}=0.684M[/tex]

Regards.

Answer:

The empirical formulae for the compound is CH3.

Answer:

The temperature at which the vapor pressure would be 0.350 atm is 201.37°C

Explanation:

The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.

Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.

The equation for Clausius- Clapeyron Equation can be expressed as:

[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]

where ;

[tex]P_1[/tex] is the vapor pressure at temperature 1

[tex]P_ 2[/tex] is the vapor pressure  at temperature 2

∆Hvap = enthalpy of vaporization

R = universal gas constant

Given that:

[tex]P_1[/tex] = 1 atm

[tex]P_ 2[/tex]  = 0.350 atm

∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol

[tex]T_1[/tex] = 282 °C  = (282 + 273) K = 555 K

R = 8.314 J/mol/k

Substituting the above values  into the Clausius - Clapeyron equation, we have:

[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]

[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }[/tex]

[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }[/tex]

[tex]\dfrac{1}{T_2} = 0.002108048708[/tex]

[tex]T_2 = \dfrac{1}{0.002108048708}[/tex]

[tex]\mathbf{T_2 }[/tex] =  474.37 K

To °C ; we have [tex]\mathbf{T_2 }[/tex] =   (474.37 - 273)°C

[tex]\mathbf{T_2 }[/tex] =  201.37 °C

Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C

The temperature of the liquid at the given vapor pressure is 201.5 ⁰C.

The given parameters;

The temperature of the liquid will be determined by applying Clausius- Clapeyron Equation;

[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )[/tex]

where;

[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )\\\\ln(\frac{0.35}{1} ) = \frac{28.5 \times 10^3}{8.314} (\frac{T_2 - 555}{555T_2} )\\\\-1.049 = 6.176- \frac{3427.95}{T_2} \\\\\frac{3427.95}{T_2} = 6.176 + 1.049\\\\\frac{3427.95}{T_2} = 7.225\\\\T_2 = \frac{3427.95}{7.225} \\\\T_2 = 474.5 \ K\\\\T_2 = 474.5 - 273 = 201.5 \ ^0C[/tex]

Thus, the temperature of the liquid at the given vapor pressure is 201.5 ⁰C.

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Answer:

The correct option is C.

Note the full question and structure of the moleculesis found in the attachment below.

Explanation:

Triglycerides or triacylglycerols are non-polar, hydrophobic lipid molecules composed of three fatty acids linked by ester bonds to a molecule of glycerol.

The fatty acids linked to the glycerol molecule are denoted by R and may be of the same kind or different. when the R group is the same, the R is attached in all the three positions for ester bonding in the glycerol molecule but when they are different are denoted by R, R' and R'' respectively.

During the hydrolysis of triglycerides, the three fatty acids molecules are obtained as well as a glycerol molecule.

From the question, when 1 mole of the triglyceride is hydrolysed, 2 moles of RCOOH, 1 mole of R'COOH and 1 mole of glycerol is obtained. The triglyceride must then be composed of two fatty acids which are the same denoted by R, and a different fatty acid molecule denoted by R'.

The correct option therefore, is C

Answer:

B. 58°C

Explanation:

Hello,

In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:

[tex]Q=mCp\Delta T=mCp(T_2-T_1)[/tex]

In such a way, solving for the final temperature [tex]T_2[/tex] we obtain:

[tex]T_2=T_1+\frac{Q}{mCp}[/tex]

Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:

[tex]T_2=85\°C+\frac{-15kJ*\frac{1000J}{1kJ} }{135g*4.184\frac{J}{g\°C} }\\\\T_2=58\°C[/tex]

Thereby, answer is B. 58°C .

Regards.

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:

[tex][H^+]=0.000123M[/tex]

[tex]pH=3.91[/tex]

Explanation:

Hello,

In this case, dissociation reaction for acetic acid is:

[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]

For which the equilibrium expression is:

[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]

Which in terms of the reaction extent [tex]x[/tex] could be written as:

[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]

Thus, solving by using a solver or quadratic equation we obtain:

[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]

And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:

[tex][H^+]=0.000123M[/tex]

Now, the pH is computed as follows:

[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]

Best regards.

Answer:

Bronsted lowry base = Proton acceptor = H2O

Bronsted lowry acid = Proton donor = HF-

Explanation:

The equation is given as;

HF-(aq) + H2O(l) --> F-(aq) + H3O(aq)

A bronsted lowry base is any specie that can accept hydrogen ion (proton) from another molecule.

Basically a bronsted lowry base is  a proton acceptor while a bronsted lowry acid is a proton donor.

In the reaction above, upon comparing both the reactants and products;

Bronsted lowry base = Proton acceptor = H2O

Bronsted lowry acid = Proton donor = HF-

Answer:

b. carbon diamond turning to carbon graphite at normal pressure.

Explanation:

Graphite is the stable form of the carbon element under normal ambient pressure.  The process of conversion of carbon diamond to carbon graphite is a spontaneous process under typical ambient conditions, since diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions, and the process of conversion can take place without a continuous input of energy from an external source.

Answer:

Explanation:

MnO₂(s) + 4 HCl(aq)  = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g                                                                     22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

=  175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require    87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .

Answer:

27°C or 300K

Explanation

We were told that the pressureof the system decreased by 10 times implies that P2= P1/10

Where P2=final pressure

P1= initial pressure

Wew were also told that the volume of the system increased by 5 times this implies that V2= 5×V1

Where T2= final temperature =-123C= 273+(-123C)=150K

T1= initial temperature

But from gas law

PV=nRT

As n and R are constant

P1V1/T1 = P2V2/T2

T1= P1V1T2/P2V2

T1=2×T2

T1=2×150

T1=300K

=300-273

=27°C

the initial temperature (°C) of a system is 27°C

Answer: So other scientist can replicate the experiment and see if they get the same results in other words, test reliability.

Explanation:

Answer:

The rms speed of its molecules becomes. (T) has become four times. Therefore, v_(rms) will become two times,...

Answer:

310 mL

Explanation:

Step 1: Given data

Step 2: Calculate the initial volume

We have a concentrated HBr solution and we want to prepare a diluted one. We can do so using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.400 M × 5.50 L / 7.10 M

V₁ = 0.310 L = 310 mL

Explanation:

Entropy refers to the degree of disorderliness of a system.

a. A snowman melts on a spring day.

Entropy is increasing because there is a change in state of matter from solid to liquid. Liquid particles have more freedom f movement compared to solids.

b. A document goes through a paper shredder.

Entropy increases because random, disorganized bits of paper are left.

c. A water bottle cools down in a refrigerator.

Entropy  decreases because temperature is directly proportional to entropy.

d. Silver tarnishes

Entropy increases because random bits of the sliver particles are formed.

e. Dissolved sugar precipitates out of water to form rock candy.

Entropy decreases because the random dissolved sugar precipitates are ordered into a rock candy.

Answer:

a) NH3/NH4Cl (pKa = 9.25)

b) [tex]\frac{[Base]}{[Acid]} =0.708[/tex]

c)

Explanation:

Hello,

a) In this case, for a buffering capacity, if we want to select the best buffer, we should ensure that the buffer's pKa approaches the desired pH, therefore, since the buffer NH3/NH4Cl has a pKa of 9.25 that is very close to the desired pH of 9.10, we can pick it as the best choice.

b) In this case, we use the Henderson-Hasselbach equation in order to compute the molar ratio:

[tex]pH=pKa+log(\frac{[Base]}{[Acid]} )\\\\log(\frac{[Base]}{[Acid]} )=9.10-9.25=-0.15\\\\\frac{[Base]}{[Acid]} =10^{-0.15}\\\\\frac{[Base]}{[Acid]} =0.708[/tex]

c) Finally, for the ratio of masses, we use the molar mass of both ammonia as the base (17 g/mol) and ammonium chloride as the acid (53.45 g/mol) to compute it, assuming 1.00 L as the volume of the solution:

[tex]\frac{m_{Base}}{m_{Acid}} =0.708\frac{molBase}{molAcid}*\frac{17gBase}{1molBase} *\frac{1molAcid}{53.45gAcid}\\ \\\frac{m_{Base}}{m_{Acid}} =0.225[/tex]

Regards.

The best choice to create a buffer with pH 9.10 is NH₃/NH₄Cl (pKa=9.25), ratio of molarities and masses for NH₃/NH₄Cl are 0.708 & 0.225 respectively.

pH of buffer solution will be calculated by using the Henderson Hasselbalch equation as:

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

        log([NH₃]/[NH₄Cl]) = 9.10 - 9.25 =

        log([NH₃]/[NH₄Cl]) = -0.15

       [NH₃]/[NH₄Cl] = [tex]10^{-0.15}[/tex] = 0.708

        M = n/V, where

        M = molarity

        V = volume = 1L

        n = moles = W(mass) / M(molar mass)

        Mass(NH₃)/Mass(NH₄Cl) = 0.708 {(mol of NH₃×17g of NH₃NH₃) /          

                                                              (mol of NH₄Cl×53.45g of NH₄Cl)

       Mass(NH₃)/Mass(NH₄Cl) = 0.225

Hence required values are calculated above.

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Answer:

Water

Explanation:

When a base react to and hydrogen ion, we can produce water.

According to these equation

H⁺  +  OH⁻   ⇄   H₂O       Kw: 1×10¹⁴

Remember that OH⁻ is determined by a strong base.

This reaction is called neutralization. You can also produce water with a weak base, because OH⁻ are released. For example, let's mention ammonia which is a weak base, it takes protons from water (H⁺)

NH₃  +  H₂O  ⇄  OH⁻  +  NH₄⁺          Kb

When the ammonium ion (acid), reacts to a base, you produce water.

NH₄⁺   +  NaOH →   NH₃ +  H₂O  +  Na⁺

Answer : HBr polar

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Answer:

a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]

b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]

c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]

Explanation:

Hello,

In this case, since those rate relationships have the stoichiometric coefficient at the denominators of the fractions preceding each ratio and the negative terms account for reactants and positive for products, we have:

a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]

b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]

c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]

Best regards.

Answer:

After 47.9 days, will remain 14.5mg of the isotope

Explanation:

The radioactive decay follows always first-order kinetics where its general law is:  

Ln[A] = -Kt + ln[A]₀

Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.

We can find rate constant from half-life as follows:

[tex]t_{1/2} = \frac{ln2}{K}[/tex]

K = ln 2 / 27.7 days

K = 0.025 days⁻¹

Replacing, initial amount of isotope is 48.0mg = [A]₀ , K is 0.025 days⁻¹ and t = 47.9 days:

Ln[A] = -Kt + ln[A]₀

Ln[A] = -0.025 days⁻¹*47.9 days + ln (48.0mg)

ln [A] = 2.6726

[A] = e^ (2.6726)

[A] = 14.5mg