Answer:
I hope it works
Explanation:
As we know that
w=m*g
given m=0.145 , g=9.8
hence we get
w= (9.8)*(0.145)
w=1.421 m/sec 2
if its help-full thank hit the stars and brain-list it thank you
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. The reactant is H2NCNHCH2CH2CH2CHCO minus, with an NH group, with a lone pair at the N atom, double-bonded to the first (from left to right) carbon, an NH2 group attached to the fifth carbon, an O atom double-bonded to the sixth carbon and a lone pair of electrons at the first and the second N atoms of the chain. The product has the same structure as the reactant, except that not an NH group with a lone pair, but an NH2 plus group is double-bonded to the first carbon. In addition, an NH3 plus group is attached to the fifth carbon instead of the NH2 group.
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!
When scientists are ready to publish the result of their experiments why is it important for them to include a description of the procedure they used
Answer: So other scientist can replicate the experiment and see if they get the same results in other words, test reliability.
Explanation:
Janet observes that bubbles rise inside water when water is heated. Which of the following best names and explains the change that causes bubbles to rise?
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.
What is the initial temperature (°C) of a system that has the pressure decreased by 10 times while the volume increased by 5 times with a final temperature of -123°C?
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
Which of the following is the balanced reaction, given the rate relationships below.
a. rate = − 13 Δ[CH4] Δt = − 12 Δ[H2O] Δt = − Δ[CO2] Δt = 14 Δ[CH3OH] Δt
b. rate = − 12 Δ[N2O5] Δt = 12 Δ[N2] Δt = 15 Δ[O2] Δt
c. rate = − 12 Δ[H2] Δt = − 12 Δ[CO2] Δt = − Δ[O2] Δt = 12 Δ[H2CO3] Δt
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.
What is the compound formed from the combination of the base and a hydrogen ion
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⁺
11.
What is the value of AH in k cal/mol for the following reaction?
CH3CH2CH3 + Br2
CH3 -CH-CH3 + HBr
Br
(1)
-12
(2)
-13
(3)
-15
(
4)
-16
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!
It would require ? Liters of water to dissolve 36 grams of the substance.
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:
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
described by the chemical equation
MnO,(s) + 4 HCl(aq)
MnCl(aq) + 2 H2O(l) + Cl (8)
How much MnO(s) should be added to excess HCl(aq) to obtain 175 mL C12(g) at 25 °C and 715 Torr?
mass of MnO2
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 .
For each of the processes, determine whether the entropy of the system is increasing or decreasing.
a. A snowman melts on a spring day.
b. A document goes through a paper shredder.
c. A water bottle cools down in a refrigerator.
d. Silver tarnishes
e. Dissolved sugar precipitates out of water to form rock candy.
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.
Determine whether each phrase describes carboxylic acids or esters.a. Do not form hydrogen bonds amongst themselves and have higher vapor pressureb. Form hydrogen bonds amongst themselves and have lower vapor pressurec. Notable for their pleasant fragrancesd. Their reactions with base are kn. own as saponificationse. Usually have a sour odorf. Their reactions with base are known as neutralizations
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)
Esters and carboxylic acids: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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is a polyprotic acid. Write balanced chemical equations for the sequence of reactions that carbonic acid can undergo when it's dissolved in water.
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,
when the temperature of an ideal gas is increased from 27C to 927C then kinetic energy increases by
Answer:
The rms speed of its molecules becomes. (T) has become four times. Therefore, v_(rms) will become two times,...
A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but addition of (SO4)2- ion resulted in a precipitate. Which cation is present
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!
Which of the following processes is spontaneous? Select the correct answer below:
a. carbon graphite turning to carbon diamond at normal pressure
b. carbon diamond turning to carbon graphite at normal pressure
c. carbon diamond turning to carbon graphite at high pressure
d. none of the above
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.
A sample of a hydrocarbon is found to contain 7.99g carbon and 2.01g hydrogen. What is the empirical formula for this compound
Answer:
The empirical formulae for the compound is CH3.
In the following net ionic equation, identify each reactant as either a Bronsted-Lowry acid or a Bronsted-Lowry base.
HF-(aq) + H2O(l) rightarrow F-(aq) + H3O(aq)
B-L_____ B-L_____
The formula of the reactant that acts as a proton donor is_____.
The formula of the reactant that acts as a proton acceptor is_______.
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-
What is the molarity of 4 g of NaCl dissolved in 100mL of water?
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.
How many milliliters of 7.10 M hydrobromic acid solution should be used to prepare 5.50 L of 0.400 M HBr
Answer:
310 mL
Explanation:
Step 1: Given data
Initial concentration (C₁): 7.10 MInitial volume (V₁): ?Final concentration (C₂): 0.400 MFinal volume (V₂): 5.50 LStep 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
which of the following compounds are polar: CH2Cl2, HBr?
Answer : HBr polar
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o prepare vegetables for finishing by grilling, sautéing, pan frying, deep frying, or stewing, you should _______ them to cook them to partial doneness
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.
The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.350 atm? (∆Hvap = 28.5 kJ/mol)
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;
boiling point temperature, = 282 ⁰Cvapor pressure, P₂ = 0.35 atmenthalpy of vaporization, ∆Hvap = 28.5 kJ/molThe 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;
R is ideal gas constant = 8.314 J/mol.kT₁ is the initial temperature in Kelvin = 282 + 273 = 555 KP₁ is the initial pressure = 1 atm[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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What is the result in the double displacement reaction of hydrochloric acid and
lithium carbonate?
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.
Which of the following buffer systems would be the best choice to create a buffer with pH 9.10?
a) HF/KF (pKa = 3.14)
HNO2/KNO2 (pKa = 3.39)
NH3/NH4Cl (pKa = 9.25)
HClO/KClO (pKa = 7.46)
b) for the best buffer system, calculate the ratio of the molarities of the buffer components required to make the buffer
c) for the best buffer system, calculate the ratio of the masses of the buffer components required to make 1.00 L of the buffer
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.
How do we calculate the pH of buffer solution?pH of buffer solution will be calculated by using the Henderson Hasselbalch equation as:
pH = pKa + log([base]/[acid])
From the above reaction it is clear that valu of pH is directly proportional to the value of pKa. So, the pKa value for NH₃/NH₄Cl is comparatively high which will close to the 9.10 pH.Ratio of the molarities for the NH₃/NH₄Cl buffer solution will be calculated by using the Henderson Hasselbalch equation as:log([NH₃]/[NH₄Cl]) = 9.10 - 9.25 =
log([NH₃]/[NH₄Cl]) = -0.15
[NH₃]/[NH₄Cl] = [tex]10^{-0.15}[/tex] = 0.708
Ratio of masses for the NH₃/NH₄Cl buffer solution will be calculated by using the below equation as: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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If 1 mol of a pure triglyceride is hydrolyzed to give 2 mol of RCOOH, 1 mol of R'COOH, and 1 mol of glycerol, which of the following compounds might be the triglyceride?
CHOC(O)R
A. CHOC(O)R
CHOC(O)R
CH,OC(O)R
B. CHOC(O)R
CH2OC(O)R
CHOC(O)R
C. CHOC(O)R
CHOC(O)R
CHOC(O)R
D. CHOC(O)R
CHOC(O)R
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
When salt is added to water, all of the following happens except? A. The salt breaks into positive chlorine ions and negative sodium icons B. the positive part of the water molecule is attracted to the negative ions C. The negative part of the water molecule is attracted to the positive ions D. The water molecules surround the dissociated ions
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.
How did Earth come to have an oxygen atmosphere?A.Precambrian rocks released oxygen into the atmosphere.B.Volcanoes released oxygen into the atmosphere.C.Early organisms created oxygen from other gases in the atmosphere.D.Oxygen was the primary gas originally in Earth's atmosphere.
Calculate the [H+] and pH of a 0.0010 M acetic acid solution. The Ka of acetic acid is 1.76×10−5. Use the method of successive approximations in your calculations.
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.
A 135 g sample of H20 at 85°C is cooled. The water loses a total of 15 kJ of energy in the cooling
process. What is the final temperature of the water? The specific heat of water is 4.184 J/g.°C.
A. 112°C
B. 58°C
C. 70°C
D. 84°C
E. 27°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.
Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 48.0 mg of this isotope, what mass remains after 47.9 days have passed?
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
After 47.9 days, will remain 14.5mg of the isotope
