Answer:
66.67%
Explanation:
From the given information:
mass of cyclohexane = 2.9949 grams
density of cyclohexane = 0.779 g/mL
Recall that:
Density = mass/volume
∴
Volume = mass/density
So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL
= 3.8445 mL
Also,
mass of propylbenzene = 1.6575 grams
density of propylbenzene = 0.862 g/mL
Volume of propylbenzene = 1.6575 g/ 0.862 g/mL
= 1.9229 mL
The volume % composition of cyclohexane from the mixture is:
[tex]= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100[/tex]
[tex]= (\dfrac{3.8445}{3.8445+1.9229})\times 100[/tex]
[tex]= (\dfrac{3.8445}{5.7674})\times 100[/tex]
= 66.67%
When 3-methyl-1-pentene is treated with in dichloromethane, the major product is 1-bromo-3-methyl-2-pentene.
a. True
b. False
Answer:
True
Explanation:
When Methyl Pentene is introduced in a chemical reaction with dichloromethane then the major product will be bromomethylpentene. There can be small amount of bromo methyl pentene than the amount of methyl pentene introduced for reaction.
write the balanced equation for
[B]⁴[C][D]/[A]²
công thức của định lý pytago
The sum of the squares of two sides of a right angle is equal to the square of the hypotenuse
Given the equation: 2C6H10(l) 17 O2(g) ---> 12 CO2(g) 10 H2O(g) MM( g/mol): 82 32 44 18 If 115 g of C6H10 reacts with 199 g of O2 and 49 g of H2O are formed, what is the percent yield of the reaction
Answer:
74%
Explanation:
Step 1: Write the balanced equation
2 C₆H₁₀(l) + 17 O₂(g) ⇒ 12 CO₂(g) + 10 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio (TMR) of C₆H₁₀ to O₂ is 164:544 = 0.301:1.
The experimental mass ratio (EMR) of C₆H₁₀ to O₂ is 115:199 = 0.578:1.
Since EMR > TMR, the limiting reactant is O₂.
Step 3: Calculate the theoretical yield of H₂O
The theoretical mass ratio of O₂ to H₂O 544:180.
199 g O₂ × 180 g H₂O/544 g O₂ = 65.8 g H₂O
Step 4: Calculate the percent yield of H₂O
%yield = (experimental yield/theoretical yield) × 100%
%yield = (49 g/65.8 g) × 100% = 74%
Answer:
Percentage yield of H₂O = 74.24%
Explanation:
The balanced equation for the reaction is given below:
2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O
Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:
Molar mass of C₆H₁₀ = 82 g/mol
Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g
Molar mass of O₂ = 32 g/mol
Mass of O₂ from the balanced equation = 17 × 32 = 544 g
Molar mass of H₂O = 18 g/mol
Mass of H₂O from the balanced equation = 10 × 18 = 180 g
SUMMARY:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂.
Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.
Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.
Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:
From the balanced equation above,
544 g of O₂ reacted to produce 180 g of H₂O.
Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.
Thus, the theoretical yield of H₂O is 66 g.
Finally, we shall determine the percentage yield. This can be obtained as follow:
Actual yield of H₂O = 49 g
Theoretical yield of H₂O = 66 g
Percentage yield of H₂O =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield of H₂O = 49/66 × 100
Percentage yield of H₂O = 74.24%
GIVING BRAINLIEST
Which equations are used to calculate the velocity of a wave?
O velocity = distance ~ time
velocity = wavelength x frequency
velocity = distance/time
velocity = wavelength/frequency
velocity = distance/time
velocity = wavelength x frequency
velocity = distance ~ time
velocity = wavelength/frequency
Answer:
velocity = distance/time
velocity = wavelength × frequency
Both of these are commonly known equations to calculate velocity with different variables.
Which of the following have only a -C-O-C- functional group?
Answer:
B) ethers
Explanation:
The functional group of an organic compound defines its specificity. The functional group is responsible for the chemical behavior of an organic compound. For example, alkenes are known to have a carbon-carbon double bond (C=C) functional group.
Likewise, organic compounds known as ETHERS are known to possess an ethoxy functional group i.e. oxygen atom bonded to two alkyl groups (R- OR; where R is an alkyl group). Members of ether functional group includes dimethyl ether (CH3-O-CH3), diethyl ether (C2H5-O-C2H5).
explain in details how triacylglycerol have an advantage over carbohydrates as stored fuel
Answer:
As stored fuels, triacylglycerols have two significant advantages over polysaccharides such as glycogen and starch. The carbon atoms of fatty acids are more reduced than those of sugars, and oxidation of triacylglycerols yields more than twice as much energy, gram for gram, as that of carbohydrates.
Explanation:
Manganese-55 has _____neutrons.
55 Mn
25
A. 55
B. 30
C. 25
QUESTION:- Manganese-55 has _____neutrons.
OPTIONS :-
A. 55
B. 30
C. 25
ANSWER:- NUMBER OF NEUTRONS IS EQUAL TO THE DIFFERENCE BETWEEN THE MASS IF THE ATOM AND ATOMIC NUMBER
SO DIFFERENCE IS EQUAL TO :- 55-25 = 30 NEUTRONS.
SO THERE IS 30 NEUTRONS IN SINGLE ATOM OF THE MANGANESE-55 ATOM.
Answer:
the mass of an atom is the sum of proton and neutron which are both concentrated in nocleus of an atom. from the question the mass is given as 55 and the proton is 25.
Consider a galvanic (voltaic) cell that has the generic metals X and Y as electrodes. If X is more reactive than Y (that is, X more readily reacts to form a cation than Y does), classify the following descriptions by whether they apply to the X or Y electrode.
i. anode
ii. cathode
iii. electrons in the wire flow toward
iv. electrons in the wire flow away
v. cations from salt bridge flow toward
vi. anions from salt bridge flow toward
vii. gains mass
viii. loses mass
Answer:
X
anode
electrons in the wire flow away
anions from salt bridge flow toward
loses mass
Y
cathode
electrons in the wire flow toward
cations from salt bridge flow toward
gains mass
Explanation:
In a galvanic cell, oxidation occurs at the anode while reduction occurs at the cathode. The metal that is more reactive functions as the anode while the less reactive metal functions as the cathode.
Electrons leave the anode and travel via a wire to the cathode. At the anode cations give up electrons and enter into the solution.
At the cathode, cations pick up electrons and are deposited on the cathode leading to a gain in mass at the cathode.
Positive ions from the salt bridge flow towards the cathode while negative ions from the salt bridge flow towards the anode.
A tank contains isoflurane, an inhaled anesthetic, at a pressure of 0.30 atm and 17.9°C. What is the pressure, in atmospheres, if the gas is warmed to a
temperature of 27.4°C, if n and V do not change?
Explanation:
here's the answer to your question
The pressure of the isoflurane gas at the temperature of 27.4 °C is 0.31 atm.
What is Gay Lussac's law?Gay-Lussac's law states that the pressure of a gas is directly proportional to the absolute temperature when the volume of the gas is kept constant.
Mathematically, Gay Lussca's law can be written as follows:
P/T = k
It is also expressed as the pressure of the gas being directly proportional to temperature.
P ∝ T (where V is constant)
or
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex] ................(1)
Given, the initial pressure and initial temperature of the gas:
P₁ = 0.30 atm
and T₁ = 17.9 °C. = 17.9 + 273 = 290.9 K
The final temperature of gas T₂ = 27.4°C = 27.4 + 273 = 300.4 K
Now, from the equation (1): [tex]P_2=\frac{P_1\times T_2}{T_1}[/tex]
P₂ = (0.30) × (300.4)/290.0
P₂ = 0.31 atm
Therefore, the pressure of the isoflurane at 27.4°C is equal to 0.31 atm.
Learn more about Gay Lussac's law, here:
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who much the velocity of a body when it travels 600m in 5 min
Answer:
2 m/s
Explanation:
Applying the formulae of velocity,
V = d/t............. Equation 1
Where V = Velocity of the body, d = distance, t = time
From the question,
Given: d = 600 m, t = 5 minutes = (5×60) = 300 seconds.
Substitute these values into equation 1
V = 600/300
V = 2 m/s.
Hence the velocity of the body when it travels is 2 m/s
Un sistema formado por una única sustancia, ¿será siempre homogéneo? ¿Porqué? Piensa a partir de las definiciones y trata de corroborar o negar usando ejemplos concretos.
Una sustancia homogénea es una sustancia que se compone de una sola fase.
Recordemos que definimos una fase en química como "cantidad química y físicamente uniforme u homogénea de materia que se puede separar mecánicamente de una mezcla no homogénea y que puede consistir en una sola sustancia o una mezcla de sustancias" según Ecyclopedia Britiannica.
El hecho de que un sistema esté compuesto por una sola sustancia no lo hace es autóctono. A veces, un sistema puede estar compuesto por partículas sólidas de una sustancia en equilibrio con su líquido. El sistema contiene solo una sustancia pero en diferentes fases, por lo tanto, el sistema contiene una sustancia pero no es homogéneo.
Por tanto, el hecho de que un sistema contenga una sola sustancia no significa necesariamente que sea homogéneo.
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A 0.204 g sample of a CO3 2- antacid is dissolved with 25.0ml of 0.0981 M HCL. The hydrochloric acid that is not neutralized by the antacid is titrated to a bromophenol blue endpoint with 5.83 ml of 0.104 M NaOH. Assuming the active ingredient in the antsacid sample is CaCO3, calculate the mass of CaCO3 in the sample.
Answer:
0.0922 g
Explanation:
Number of moles of acid present = 25/1000 × 0.0981
= 0.00245 moles
Number of moles of base = 5.83/1000 × 0.104
= 0.000606 moles
Since the reaction of HCl and NaOH is 1:1
Number of moles of HCl that reacted with antacid = 0.00245 moles - 0.000606 moles
= 0.001844 moles
From the reaction;
CaCO3 + 2HCl ----> CaCl2 + H2O + CO2
1 mole of CaCO3 reacts with 2 moles of HCl
x moles of CaCO3 reacts with 0.001844 moles ofHCl
x = 1 × 0.001844/2
= 0.000922 moles
Mass of CaCO3 = 0.000922 moles × 100 g/mol
= 0.0922 g
A reaction vessel is charged with phosphorus pentachloride, which partially decomposes to phosphorus trichloride and molecular chlorine according to the following reaction:
PCl5(g)â PCl3(g)+Cl2(g)
When the system comes to equilibrium at 250.0°C, the equilibrium partial pressures are: PPCl5 = 0.688 atm and PPCl3 = PCl2 = 0.870 atm.
Required:
What is the value of Kp at this temperature?
Answer:
Kp = 1.10.
Explanation:
Hello there!
In this case, according to the given information about the chemical reaction at equilibrium, it turns out possible for us to find the partial pressures-based equilibrium expression for the decomposition of phosphorous pentachloride by applying the law of mass action whereas the pressure of products is divided by that of the reactants as shown below:
[tex]Kp=\frac{p_{PCl_3}p_{Cl_2}}{p_{PCl_5}}[/tex]
Now, we plug in the given pressures to obtain:
[tex]Kp=\frac{0.870}{0.688} \\\\Kp=1.10[/tex]
Regards!
A reactant. Q. decomposes at a second order. The slope of the graph 1/[Q] (1/M) vs time (s) is -0.04556. If the initial
concentration of Q for the reaction is 0.50 M, what is the concentration in M. of Q after 10.0 minutes?
Answer:
0.034 M
Explanation:
1/[A] = kt + 1/[A]o
[A] = ?
k= 0.04556
t= 10.0 minutes or 600 seconds
[A]o = 0.50 M
1/[A] = (0.04556 × 600) + 1/0.50
[A] = 0.034 M
A monatomic ion with a charge of 2 has an electronic configuration of 1s22s22p6. This ion is a(n) _______ . What is the chemical symbol of the noble gas this ion is isoelectronic with
Answer:
A. Cation
B. Ne
Explanation:
The ion is positively charged by 2, making it a cation.
The electron configuration of the nearest noble gas Neon is 1s22s22p6
1. A monatomic ion with a charge of 2 has an electronic configuration of 1s22s22p6 is Neon.
2. chemical symbol of the noble gas is Kr (krypton).
Isoelectronic atom or ion has the same number of valence electrons. Krypton has 36 electrons and 36 protons (atomic number 36).
What is Neon?Neon is a chemical element with the symbol Ne and atomic number 10. It is a noble gas. Neon is a colorless, odorless, inert monatomic gas under standard conditions, with about two-thirds the density of air. It was discovered (along with krypton and xenon) in 1898 as one of the three residual rare inert elements remaining in dry air after nitrogen, oxygen, argon, and carbon dioxide were removed. Neon was the second of these three rare gases to be discovered and was immediately recognized as a new element of its bright red emission spectrum. The name neon is derived from the Greek word, the neuter singular form of (neos), meaning 'new'. Neon is chemically inert, and no uncharged neon compounds are known. The compounds of neon currently known include ionic molecules, molecules held together by van der Waals forces, and clathrates.
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g 32.53 g of a solid is heated to 100.oC and added to 50.0 g of water in a coffee cup calorimeter and the contents are allowed to sit until they finally have the same temperature. The water temperature changes from 25.36 oZ to 34.4 oC. What is the specific heat capacity (in J/goC) of the solid
Answer:
0.886 J/g.°C
Explanation:
Step 1: Calculate the heat absorbed by the water
We will use the following expression
Q = c × m × ΔT
where,
Q: heatc: specific heat capacitym: massΔT: change in the temperatureQ(water) = c(water) × m(water) × ΔT(water)
Q(water) = 4.184 J/g.°C × 50.0 g × (34.4 °C - 25.36 °C) = 1.89 × 10³ J
According to the law of conservation of energy, the sum of the energy lost by the solid and the energy absorbed by the water is zero.
Q(water) + Q(solid) = 0
Q(solid) = -Q(water) = -1.89 × 10³ J
Step 2: Calculate the specific heat capacity of the solid
We will use the following expression.
Q(solid) = c(solid) × m(solid) × ΔT(solid)
c(solid) = Q(solid) / m(solid) × ΔT(solid)
c(solid) = (-1.89 × 10³ J) / 32.53 g × (34.4 °C - 100. °C) = 0.886 J/g.°C
The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 °C to 37 °C? a. 5.8 × 10-4 J b. 0.13 J c. 29 J d. 2.0 J e. -0.13 J
Answer:
c. 29 J
Explanation:
Step 1: Given data
Specific heat capacity of Pb (c): 0.13 J/g.K (= 0.13 J/g.°C)Mass of Pb (m): 15 gInitial temperature: 22 °CFinal temperature: 37 °CStep 2: Calculate the temperature change
ΔT = 37 °C - 22 °C = 15 °C
Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece
We will use the following expression.
Q = c × m × ΔT
Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J
How many grams of sodium chloride are contained in 250.0 g of a 15% NaCl solution?
Please explain and show work.
Given concentration of NaCl=15%
Means ,
In every 100g of Solution 15g of NaCl is present .
Now
Given mass=250gSo ,
[tex]\\ \Large\sf\longmapsto 250\times 15\%[/tex]
[tex]\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}[/tex]
[tex]\\ \Large\sf\longmapsto 37.5g[/tex]
37.5g of NaCl present in 250g of solution.
Answer:
Given concentration of NaCl=15%
Means ,
In every 100g of Solution 15g of NaCl is present .
Now
Given mass=250g
So ,
➡250 × 15%
➡250×15/100
➡37.5g
37.5g of NaCl present in 250g of solution.
Na Sa Bant HCL -> 50g Hao pt Soy
North America and south africa
A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen. The decomposition of a sample of diphosphorus pentoxide forms 0.775 g phosphorus to every 1.00 g oxygen.
Required:
How many grams of P205 are formed when 5.89 g of P react with excess oxgen?
Answer:
There is 13.48 grams of P2O5 formed
Explanation:
Step 1: Data given
A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen.
Mass of P = 5.89 grams
Molar mass of O2 = 32.0 g/mol
atomic mass of P = 30.97 g/mol
molar mass of P2O5 = 141.94 g/mol
Step 2: The balanced equation
4P(s)+5O2(g)⇔ 2P2O5(s)
Step 3: Calculate moles of P
Moles P = Mass P / atomic mass P
Moles P = 5.89 grams / 30.97 g/mol
Moles P = 0.190 moles
Step 4: Calculate moles of P2O5
For 4 moles P we need 5 moles O2 to produce 2 moles P2O5
For 0.190 moles of P we'll have 0.190/2 = 0.095 moles P2O5
Step 5: Calculate mass of P2O5
Mass P2O5 = moles P2O5 * molar mass P2O5
Mass P2O5 = 0.095 moles * 141.94 g/mol
Mass P2O5 = 13.48 grams
There is 13.48 grams of P2O5 formed
To draw a Lewis structure for a polyatomic ion, begin by calculating A, the available electrons, and N, the needed electrons. What is N for CIO3-, the chlorate ion?
A = 26
N = ?
Answer:
16
Explanation:
Because the sum of all electron in that compound should be 41 and as it has one electron extra ,total no. of electrons are 42 .
So if we add 26 +16 we get 42
Hence it's correct answer
Ammonia reacts with oxygen to produce nitrogen monoxide and water:
4 NH3(g) + 5 O2(g) ---> 4 NO(g) + 6 H2O(g)
Which of the following are stoichiometric amounts of the two reactants?
a) 1.0 g, 1.25 g
b) 0.75 mol, 0.9375 mol
Answer:
b) 0.75 mol, 0.9375 mol
Explanation:
According to this question, ammonia reacts with oxygen to produce nitrogen monoxide and water. The chemical equation is as follows:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Based on this balanced equation, 4 moles of ammonia (NH3) reacts with 5 moles of oxygen (O2).
A stoichiometric amount of the two reactants (NH3 and O2) must represent the ratio 4:5.
Given the provided options;
0.75 mol of ammonia (NH3) will react with (0.75 × 5/4) = 0.935 mol of O2 for them to be in stoichiometry.
N.B: 1 mol of NH3 will react with 1.25mol of O2 and not 1g, 1.25g.
At 445oC, Kc for the following reaction is 0.020. 2 HI(g) <--> H2 (g) + I2 (g) A mixture of H2, I2, and HI in a vessel at 445oC has the following concentrations: [HI] = 1.5 M, [H2] = 2.50 M and [I2] = 0.05 M. Which one of the following statements concerning the reaction quotient, Qc, is TRUE for the above system?
a. Qc = Kc; the system is at equilibrium.
b. Qc is less than Kc; more H2 and I2 will be produced.
c. Qc is less than Kc; more HI will be produced.
d. Qc is greater than Kc; more HI will be produced.
Explanation:
The given balanced chemical equation is:
[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]
The value of Kc at 445oC is 0.020.
[HI]=1.5M
[H2]=2.50M
[I2]=0.05M
The value of Qc(reaction quotient ) is calculated as shown below:
Qc has the same expression as the equilibrium constant.
[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]
Qc>Kc,
Hence, the backward reaction is favored and the formation of Hi is favored.
Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.
What is the concentration of Htions at a pH = 11?
mol/L
What is the concentration of Htions at a pH = 6?
mol/L
How many fewer Htions are there in a solution at a
pH = 11 than in a solution at a pH = 6?
Determine the boiling point of a solution that contains 150.0 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 950 mL of benzene (d = 0.877 g/mL). Pure benzene has a boiling point of 80.1°C and a boiling point elevation constant of 2.53°C/m.
Answer:
Boiling T° of solution → 83.6°C
Explanation:
To solve this, we apply Elevation of boiling point, property
ΔT = Kb . m . i
As we talk about organic solute, i = 1. No ions are formed.
m = molality (moles of solute in 1kg of solvent)
We determine mass of solvent by density
D = m /V so D . V = m
950 mL . 0.877 g/mL = 833.15 g
We convert to kg → 833.15 g . 1 kg/ 1000g = 0.833 kg
Moles of solute (naphtalene): 150 g . 1 mol/ 128.16g = 1.17 mol
m = 1.17mol / 0.833 kg = 1.41 mol/kg
We replace data:
Boiling T° of solution - 80.1°C = 2.53°C/m . 1.41 m . 1
Boiling T° of solution = 2.53°C/m . 1.41 m . 1 + 80.1°C → 83.6°C
Answer:
The answer is c or 17.1 g
One crystalline form of silica (SiO2) has a cubic unit cell, and from X-ray diffraction data it is known that the cell edge length is 0.700 nm. If the measured density is 2.32 g/cm3, how many (a) Si4 and (b) O2- ions are there per unit cell
Answer:
There are 8Si atoms and 16 O atoms per unit cell
Explanation:
From the question we are told that:
Edge length [tex]l=0.700nm=>0.7*10^9nm[/tex]
Density [tex]\rho=2.32g/cm^3[/tex]
Generally the equation for Volume is mathematically given by
[tex]V=l^3[/tex]
[tex]V=(0.7*10^9)m^3[/tex]
[tex]V=3,43*10^-{22}cm[/tex]
Where
Molar mass of (SiO2) for one formula unit
[tex]M=28+32[/tex]
[tex]M=60g/mol[/tex]
Therefore
Density of Si per unit length is
[tex]\rho_{si}=\frac{9.96*10^{23}}{3.43*10^22}[/tex]
[tex]\rho=0.29[/tex]
Molar mass of (SiO2) for one formula unit
[tex]M=28+32[/tex]
[tex]M=60g/mol[/tex]
Therefore
There are 8Si atoms and 16 O atoms per unit cell
Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and products in your balanced equation. You do not need to include the states with the identities of the oxidizing and reducing agents.
NO_2(g) rightarrow NO_3^-(aq) +NO_2^- (aq) [basic]
The oxidizing agent is:______.
The reducing agent is:_______.
Answer:
a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. i. NO₂⁻ is the oxidizing agent
ii. NO₃⁻ is the reducing agent.
Explanation:
a. Balance the following skeleton reaction
The reaction is
NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)
The half reactions are
NO₂ (g) → NO₃⁻ (aq) (1) and
NO₂ (g) → NO₂⁻ (aq) (2)
We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)
We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms
NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)
The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.
So, NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
We now add equation (4) and (5)
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
+ NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
2NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq)
We now add two hydroxide ions to both sides of the equation.
So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + 2OH⁻ (aq)
The hydrogen ion and the hydroxide ion become a water molecule
2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H₂O (l)
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
So, the required reaction is
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. Identify the oxidizing agent and reducing agent
Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = 0
x - 4 = 0
x = 4
Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = -1
x - 4 = -1
x = 4 - 1
x = 3
Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = -1
x + 3(-2) = -1
x - 6 = -1
x = 6 - 1
x = 5
i. The oxidizing agent
The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus NO₂⁻ is the oxidizing agent
ii. The reducing agent
The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and NO₃⁻ is the reducing agent.
Three peptides were obtained from a trypsin digestion of two different polypeptides. Indicate the possible sequences from the given data.
a. Val-Gly-Arg
b. Ala-Val-Lys
c. Ala-Gly-Phe
Answer:
A) Val-Gly-Arg-Ala-Val-Lys-Ala-Gly-Phe
B) Ala-Val-Lys-Val-Gly-Arg-Ala-Gly-Phe
Explanation:
The possible sequences that could be obtained from the available dat provided are :
A) Val-Gly-Arg-Ala-Val-Lys-Ala-Gly-Phe
B) Ala-Val-Lys-Val-Gly-Arg-Ala-Gly-Phe
Trypsin is generally a catalyst for the breakdown of proteins into smaller peptides ( during the hydrolysis of peptide bonds )
Determine the mmol of both starting materials (factoring in that formic acid is not pure, but rather 88% weight/volume, or 88g/100 ml), showing your work. Determine the limiting reagent in this synthesis. Lastly, calculate the theoretical yield of benzimidazole that you could expect to form.
Solution :
Molecular Molar Mass Volume Density Mass Moles nmoles
formula (g/mol) (mL) (g/mL) (g)
[tex]$C_6H_8N_2$[/tex] 108.14 0.108 0.001 1
HCOOH 46.02 0.064 1.22 0.07808 0.0017 1.7
mmoles of o-phenylenediamine = 1 mmoles
mmoles of formic acid = 1.7 [tex]\approx[/tex] 2 mmoles
From the reaction of o-phenylenediamine and formic acid, we see,
1 mmole of o-phenylenediamine reacts with 1 mmole of formic acid.
But here, 2 mmoles of the formic acid , this means that the formic acid is an excess reagent and the o-phenylenediamine is the limiting reagent here.
The amount of product depends on the limiting reagent that is o-phenylenediamine. So, 1mmole of o-phenylenediamine will give 1mmole of product.
molar mass of Benzimidazole = [tex]118.14[/tex] g/mol
mmoles of Benzimidazole formed = [tex]1[/tex] mmol
Mass of benzimidazole formed = molar mass x [tex]\frac{nmoles}{1000}[/tex]
[tex]$=\frac{118.14 \times 1}{1000}$[/tex]
= 0.11814 g
So the theoretical yield of Benzimidazole is = 0.118 g = 118mg
