Final answer:
Heptane was used as the solvent instead of diethyl ether because it has a lower boiling point, does not react with the alkenes formed in the experiment, and does not evaporate as fast.
Explanation:
In this experiment, heptane was used as the solvent instead of diethyl ether for several reasons:
Heptane has a lower boiling point than that of diethyl ether. This means that it will evaporate at a slower rate, allowing for better separation and detection of the analytes in the gas chromatography analysis.Diethyl ether will react with the alkenes that are formed in the experiment. This can lead to the formation of unwanted by-products and inaccurate results.Heptane does not evaporate as fast as diethyl ether, which can be advantageous in GC analyses where longer retention times are desired.Which of these are paramagnetic in their ground state?
(a) Ga (b) Si (c) Be (d) Te
Answer:
(a) Ga (b) Si (d) Te
Explanation:
Paramagnetic are those which has unpaired electrons and diamagnetic are those in which all electrons are paired.
(a) Ga
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^1[/tex]
The electrons in 4p orbital = 1 (Unpaired)
Thus, the element is paramagnetic as the electrons are unpaired.
(b) Si
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^2[/tex]
The electrons in 3p orbital = 2 (Unpaired)
Thus, the element is paramagnetic as the electrons are unpaired.
(c) Be
The electronic configuration is -
[tex]1s^22s^2[/tex]
The electrons in 2s orbital = 2 (paired)
Thus, the element is diamagnetic as the electrons are paired.
(d) Te
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^4[/tex]
The electrons in 5p orbital = 4 (1 pair and 2 Unpaired)
Thus, the element is paramagnetic as the electrons are unpaired.
Which element would you expect to be more metallic?
(a) S or Cl (b) In or Al (c) As or Br
Explanation:
When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.
As sulfur (S) is a group 16 element and chlorine (Cl) is a group 17 element. Hence, sulfur (S) is more metallic in nature than chlorine.
This means that chlorine (S) is less metallic than chlorine (Cl).
Both indium (I) and aluminium (Al) are group 13 elements. And, when we move down a group then there occur an increase in non-metallic character of the elements. As indium belongs to group 13 and period 5 whereas aluminium belongs to group 13 and period 3.
Therefore, aluminium (Al) is more metallic than indium (In).
Arsenic (Ar) is a group 15 element and bromine (Br) is a group 17 element. Therefore, arsenic is more metallic than bromine.
Final answer:
The most metallic element can be determined by looking at their positions in the periodic table.
Explanation:
The most metallic element can be determined by looking at their positions in the periodic table. Metallic character generally increases going down a group and decreases going across a period. Using this information, we can predict that:
(a) S is more metallic than Cl because S is below Cl in the same group and is further down the periodic table.
(b) In is more metallic than Al because In is below Al in the same group and is further down the periodic table.
(c) Br is more metallic than As because Br is to the left of As in the same period and closer to the metals in the periodic table.
In this lab you will need to prepare solutions using dilutions. Starting with the stock 0.300 M NaOH solution, how would you prepare a 0.050 M NaOH solution (using 0.300 M NaCl as the diluent)? To prepare 24 mL of 0.050 M NaOH solution, you would add mL of 0.300 M NaOH stock solution and mL of 0.300 M NaCl solution.
Answer:
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.
Explanation:
Molarity of the NaOH solution = [tex]M_1=0.300 M[/tex]
Volume of the NaOH solution = [tex]V_1=?[/tex]
Molarity of the NaOH solution after dilution= [tex]M_2=0.050 M[/tex]
Volume of the NaOH solution after dilution= [tex]V_2=24 mmL[/tex]
[tex]M_1V_1=M_2V_2[/tex] (Dilution )
[tex]V_1=\frac{M_2V_2}{M_1}=\frac{0.050 M\times 24 mL}{0.300 M}=4 mL[/tex]
Volume of NaCl solution of 0.300 M = 24 mL - 4 mL = 20 mL
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution
Determination of the volume of the stock solution of NaOH needed
•Molarity of stock solution (M₁) = 0.3 M
•Molarity of diluted solution (M₂) = 0.05 M
•Volume of diluted solution (V₂) = 24 mL
•Volume of stock solution needed (V₁) =?
Using the dilution formula, the volume of the stock solution needed can be obtained as follow:
M₁V₁ = M₂V₂
0.3 × V₁ = 0.05 × 24
0.3 × V₁ = 1.2
Divide both side by 0.3
V₁ = 1.2 / 0.3
V₁ = 4 mL
Determination of the volume of NaCl needed•Volume of NaOH needed = 4 mL
•Volume of diluted solution of NaOH = 24 mL
•Volume of NaCl needed =?
Volume of NaCl needed = (Volume of diluted solution of Na) – (Volume of NaOH needed)
Volume of NaCl needed = 24 – 4
Volume of NaCl needed = 20 mL
Therefore, we can conclude as follow:
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.
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Draw an orbital diagram showing valence electrons, and write the condensed ground-state electron configuration for each:
(a) Mn (b) P (c) Fe
Answer:
The complete answer is in the diagram.
Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.27 m NH4I A. Highest boiling point 2. 0.11 m FeCl3 B. Second highest boiling point 3. 0.16 m CaI2 C. Third highest boiling point 4. 0.50 m Sucrose(nonelectrolyte) D. Lowest boiling point Submit AnswerRetry Entire Group9 more group attempts remaining
Answer:
1. 0.27 m [tex]NH_4I[/tex] : Highest boiling point
2. 0.11 m
: Lowest boiling point
3. 0.16 m
: Third highest boiling point
4. 0.5 m sucrose : Second highest boiling point
Explanation:
[tex]\Delta T_b=i\times k_b\times m[/tex]
[tex]\Delta T_b[/tex] = elevation in boiling point
i = Van'T Hoff factor
[tex]k_b[/tex] = boiling point constant
m = molality
1. For 0.27 m [tex]NH_4I[/tex]
[tex]NH_4I\rightarrow NH_4^{+}+I^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]2\times 0.27=0.54m[/tex]
2. For 0.11 m [tex]FeCl_3[/tex]
[tex]FeCl_3\rightarrow Fe^{3+}+3Cl^{-}[/tex]
i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]4\times 0.11=0.44m[/tex]
3. For 0.16 m [tex]CaCl_2[/tex]
[tex]CaCl_2\rightarrow Ca^{2+}+2Cl^{-}[/tex]
, i= 3 as it is a electrolyte and dissociate to give 3 ions, concentration of ions will be [tex]3\times 0.16=0.48m[/tex]
4. For 0.5 m sucrose
, i= 1 as it is a non electrolyte and does not dissociate to give ions, concentration will be 0.5 m
Thus as concentration of solute follows the order : [tex]NH_4I[/tex] > sucrose > [tex]CaCl_2[/tex] > [tex]FeCl_3[/tex], the boiling point will also follow the same order.
State Hund’s rule in your own words, and show its application in the orbital diagram of the nitrogen atom.
Answer:
.
Explanation:
Answer:
Hund's rule: states that electrons always enter an empty orbital before they pair up.
In this exercise, we notice that orbital p has three suborbitals, then,
we must start filling each suborbitals with one electrons and after that we start to pairing them up.
The result must be,
1st suborbital with 2 electrons
2nd suborbital with 1 electron
3 rd suborbital with 1 electron
PJ’s unknown solid 1) dissolves in hot ethanol, 2) is essentially insoluble in hexane, and 3) is insoluble in cold water, but sparingly soluble in warm water. Outline the recrystallization procedure you would suggest she use here.
Answer:
1. The solid is dissolved in the hot solvent (ethanol or water);
2. If the impurities are not dissolved, they are separated by filtration;
3. The solution is then cold, and the crystals of the solid are formed;
4. The solution is filtrated and the pure solid is obtained.
Explanation:
The recrystallization is a process to separate impurities from a solid. The solid with the impurities is dissolved in a hot solvent, and, is insoluble in the cold one. But the impurities must be soluble in the cold solvent, or insoluble in the hot one.
If the impurities are soluble in the cold solvent, then, the crystals of the analyte will be removed, and if they are insoluble in the hot solvent, then its crystal is removed first.
So let's assume that the solid is insoluble in hot ethanol, and soluble in hot water. Because its insoluble in hexane, the recrystallization is not possible with it. So the procedure would be:
1. The solid is dissolved in the hot solvent (ethanol or water);
2. If the impurities are not dissolved, they are separated by filtration;
3. The solution is then cold, and the crystals of the solid are formed;
4. The solution is filtrated and the pure solid is obtained.
Draw the structure of the aromatic compound para-aminochlorobenzene (para-chloroaniline). Draw the molecule on the canvas by choosing buttons from the Tools
Answer:
Explanation:
In the picture you have the answer.
Now, let's analize the structure, so you can know why the structure in the picture is the correct structure.
The aniline is the name that receives the benzene with a NH2 group as one of it's substituent. Now, This group is a really strong activating group and in the nomenclature priority, it has more order priority than any halide.
Now, it says that the chloro it's on the para position. The "para" position in a aromatic ring, in this case, the benzene, refers to the position of this substituent to the first substitued position. In this case, the NH2 it's on the position 1 or carbon 1, the para position, means that it's on position 4 of the ring. The ortho position is carbon 2, and meta position is carbon 3 of the benzene. So, according to this, the p-chloroaniline it's on picture attached.
You have a freshly prepared 1 M (molar) solution of glucose in water. You carefully pour out a 100 mL sample of that solution. How many glucose molecules are included in that 100 mL sample?
Answer:
[tex]6.022\times 10^{22} [/tex]glucose molecules are included in that 100 mL sample.
Explanation:
Concentration of freshly prepared glucose solution = 1 M = 1 mol/L
1 L = 1000 ml
This means that 1 mole of glucose is present in 1000 mL of water.
If we have 100 mL of solution. then number of moles of glucose will be L;
[tex]\frac{1}{1000}\times 100 mL=0.1 mole[/tex]
1 mole = [tex]N_A=6.022\times 10^{23} [/tex] molecules/atoms
Number of molecules of glucose in 0.1 mole :
= [tex]0.1 mol\times 6.022\times 10^{23} molecules=6.022\times 10^{22} moleules[/tex]
The exponents in a rate law can be found:
Select the correct answer below:
O from the stoichiometric coefficients of the reaction
O from the molar masses of the compounds
O by experiment only
O none of the above
Answer:
by experiment only
Explanation:
According to the law of mass action:-
The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.
Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.
The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.
Thus, the exponents in the rate law, the order is determined by experiment only.
Exponents in a rate law can only be determined by experiment, not from the stoichiometric coefficients of the reaction or the molar masses of the compounds.
Explanation:The exponents in a rate law cannot be determined from the stoichiometric coefficients of the reaction or the molar masses of the compounds. The correct answer is that they can be determined by experiment only. This is because the rate law and its exponents are indicative of the reaction mechanism, or the step-by-step sequence of elementary reactions by which overall chemical change occurs. These can't usually be predicted just from the overall reaction equation (which provides the stoichiometric coefficients) and require experimental data for determination.
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If the pH of a phosphoric acid (H3PO4) solution is adjusted to 6.50, what is the most abundant species and which is the second most abundant species? For H3PO4 Ka1 = 7.11 x 10-3, Ka2 = 6.34 x 10-8, and Ka3 = 4.22 x 10-13.
Explanation:
For the given values of [tex]K_{a}[/tex] we will have the values of [tex]pK[/tex] as follows.
As, [tex]pK_{a} = -log K_{a}[/tex]
Therefore,
[tex]pK_{a1}[/tex] = 2.15, [tex]pK_{a2}[/tex] = 7.20
[tex]pK_{a3}[/tex] = 12.38
Now, at pH 6.50
[tex]H_{3}PO_{4} \rightarrow H_{2}PO^{-}_{4} + H^{+}[/tex]; [tex]K_{a1}[/tex]
At pH = 2.15; [tex]H_{3}PO_{4} = H_{2}PO^{-}_{4}[/tex]
[tex]H_{2}PO^{-}_{4} \rightarrow HPO^{2-}_{4} + H^{+}[/tex]; [tex]K_{a2}[/tex]
At pH 7.20; [tex]H_{2}PO^{-}_{4} = HPO^{2-}_{4}[/tex]
[tex]HPO^{2-}_{4} \rightarrow PO^{3-}_{4} + H^{+}[/tex]; [tex]K_{a3}[/tex]
Hence, we can conclude that most abundant species is [tex]H_{2}PO^{-}_{4}[/tex] and the second most abundant species is [tex]HPO^{2-}_{4}[/tex].
The rate constant for a certain reaction is k = 4.50×10−3 s−1 . If the initial reactant concentration was 0.400 M, what will the concentration be after 11.0 minutes?
Answer: The concentration of reactant after the given time is 0.0205 M
Explanation:
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]4.50\times 10^{-3}s^{-1}[/tex]
t = time taken for decay process = 11.0 min = 660 s (Conversion factor: 1 min = 60 s)
[tex][A_o][/tex] = initial amount of the reactant = 0.400 M
[A] = amount left after decay process = ?
Putting values in above equation, we get:
[tex]4.50\times 10^{-3}s^{-1}=\frac{2.303}{660s}\log\frac{0.400}{[A]}[/tex]
[tex][A]=0.0205M[/tex]
Hence, the concentration of reactant after the given time is 0.0205 M
A 34.57 mL sample of an unknown phosphoric acid solution is titrated with a 0.127 M sodium hydroxide solution. The equivalence point is reached when 28.2 mL of sodium hydroxide solution is added. What is the concentration of the unknown phosphoric acid solution?
The concentration of the unknown phosphoric acid solution can be determined using the concept of stoichiometry in titrations. In this case, the concentration of the sodium hydroxide solution and the volume used at the equivalence point are known. By using the given information, the concentration of the unknown phosphoric acid solution can be calculated to be approximately 0.150 M.
Explanation:The concentration of the unknown phosphoric acid solution can be determined using the concept of stoichiometry in titrations. In this case, the concentration of the sodium hydroxide solution and the volume used at the equivalence point are known. By using the equation:
NaOH (aq) + H3PO4 (aq) → NaH2PO4 (aq) + H2O (l)
the moles of sodium hydroxide can be determined. Then, by considering the volume and concentration of the phosphoric acid solution, the concentration of the unknown phosphoric acid solution can be calculated. In this case, the concentration of the unknown phosphoric acid solution is approximately 0.150 M.
How many electrons in an atom can have each of the following quantum number or sublevel designations?
(a) n = 2, l = 1, ml = 0
(b) 5p
(c) n = 4, l = 3
Answer: (a) 2 (b) 6 (c) 14
Explanation:
In the Azimuthal quantum number(l) electrons in a particular subshell (such as s, p, d, or f) are defined by values of l (0, 1, 2, or 3).
s is l=0, p is l=1, d is l=2, f is l=3.
The magnetic quantum number (ml) The value of ml can range from -l to +l, including zero. Thus the s, p, d, and f subshells contain 1, 3, 5, and 7 orbitals each, with values of m within the ranges 0, ±1, ±2, ±3 respectively. Each shell can have 2 x l + 1 sublevels, and each of these sublevel can accommodate up to two electrons.
(a) n=2, l=1, ml=0. If l=1 then 2 x 1+ 1=3 sublevels, 3*2=6 electrons. When l=1, ml =-1,0,+1, ml=0 accommodate two(2)electrons
(b) 5p. p is l=1 If l=1 then 2 x 1+ 1=3 sublevels, 3*2= electrons. This means in the 5 shell, the p orbital has 3 subshell and accommodate 6 electrons.
(c) n = 4, l = 3 if l=3 then 2 x 3+ 1=7 sublevel 7*2=14 electrons. This means the in the 4 shell, the f orbital has 7 subshell and accomdate 14 elections.
(a) The maximum number of electrons an atom can have with the sublevel designation is 2 electrons.
(b) The maximum number of electrons for p-orbital is 6 electrons.
(c) The maximum number of electrons for f-orbital is 14 electrons.
The energy level of each of an atom with the given sublevel designations determines the maximum number of electrons the atom can occupy.
(a) For n = 2, I = 1, ml = 0,
The energy level is calculated as;
Energy level = 2(1) + 1 = 3 sub-orbtials
= -1 0 1 : ( 2 electrons each)
Thus, the maximum number of electrons an atom can have with the sublevel designation is 2 electrons.
(b) 5p
p-orbital has 3 sub-orbitals and maximum of 6 electrons
(c) n= 4, I = 3
energy level = 2(3) + 1 = 7 sub-orbitals
7 sub-orbitals corresponds f-orbital and it has maximum 14 electrons.
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What is penetration? How is it related to shielding? Use the penetration effect to explain the difference in relative orbital energies of a 3p and a 3d electron in the same atom.
Answer: penetration is the ability of an electron in a given orbital to approach the nucleus closely. Shielding refers to the fact that core electrons reduce the degree of nuclear attraction felt by the orbital electrons. Shielding is the opposite of penetration. The most penetrating orbital is the least screening orbital. The order of increasing shielding effect/decreasing penetration is s<p<d<f.
Explanation:
The order of penetrating power is 1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f....
Since the 3p orbital is more penetrating than the 3d orbital, it will lie nearer to the nucleus and thus possess lower energy.
Menthol (FW = 156.3 g/mol), the strong-smelling substance in many cough drops, is a compound of carbon, hydrogen, and oxygen. When 0.1595 g of menthol was burned in a combustion apparatus, 0.449 g of CO2 and 0.184 g of H2O formed. What is menthol's molecular formula? enter as C#H#O#
Answer: the molecular formula is C10H20O
Explanation:Please see attachment for explanation
An experiment calls for you to use 100 mL of 0.25 M HNO3 solution. All you have available is a bottle of 3.4 M HNO3. How many milliliters of the 3.4 M HNO3 solution do you need to prepare the desired solution?
Answer: 7.35mL
Explanation:
C1 = 3.4M
V1 =?
C2 = 0.25M
V2 = 100mL
C1V1 = C2V2
3.4 x V1 = 0.25 x 100
V1 = (0.25 x 100) /3.4
V1 = 7.35mL
Final answer:
To prepare the desired 100 mL of 0.25 M HNO3 solution, approximately 7.35 mL of the 3.4 M HNO3 stock solution is required, with the remaining volume filled with water for dilution.
Explanation:
To prepare 100 mL of a 0.25 M HNO3 solution from a 3.4 M HNO3 stock solution, we use the dilution formula M1V1 = M2V2, where M1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, M2 is the desired concentration, and V2 is the final volume of the solution.
Here, M1 = 3.4 M, M2 = 0.25 M, and V2 = 100 mL. We need to find V1. Plugging the values into the equation:
V1 = (M2 x V2) / M1
V1 = (0.25 M x 100 mL) / 3.4 M
After calculating, we find that V1 ≈ 7.35 mL. Therefore, you need approximately 7.35 mL of the 3.4 M HNO3 stock solution to prepare the desired 100 mL of 0.25 M HNO3 solution. The rest of the volume up to 100 mL should be filled with distilled water to achieve the correct dilution.
The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K. PCl3(g) Cl2(g) PCl5(g) Calculate the equilibrium concentrations of reactant and products when 0.400 moles of PCl3 and 0.400 moles of Cl2 are introduced into a 1.00 L vessel at 500 K.
Answer:
[PCl₅] = 0.336 M
[Cl₂] = [PCl₃] = 0.064 M
Explanation:
We are given the equilibrium constant, kc , and the moles of reactants and products. Thus, our strategy here sholud be to express the quantities at equilibrium in terms of its constant by setting up our ICE table helper.
First lets start by writing the expression for the equilibrium constant:
PCl₃ (g) + Cl₂ (g) ⇄ PCl₅(g)
Kc = [PCl₅] / [Cl₂] / [PCl₃] =83.3
and setup the ICE table
We are given the moles introduced in the 1.00 L vessel, so we can calculate the molarities as M = mol/L
[Cl₂] [PCl₃] [PCl₅]
i 0.400 M 0.400 M 0
c -x -x +x
e 0.400 - x 0.400 - x x
x / (0.400 - x )² = 83.3
(0.16 - 0.8x + x²) x 83.3 = x
13.33 -66.66x + 83.3x² = x
13.33 - 67.66 x + 83.3 x² = 0
Solving the quadratic equation we have x₁ = 0.476 and x₂ = 0.336
The first solution is phisically impossible, since it will give us a negative quantity at equilibrim for the reactants.
With the second solution x = 0.336, the equilibrium concentrations are:
[PCl₅] = 0.336 M
[Cl₂] = [PCl₃] = 0.400 - 0.336 = 0.064 M
Police often monitor traffic with "K-band" radar guns, which operate in the microwave region at 22.235 GHz (1 GHz = 10⁹ Hz). Find the wavelength (in nm and Å) of this radiation.
Answer:
Wavelength = 13492242 nm
Wavelength = 134922420 Å
Explanation:
The relation between frequency and wavelength is shown below as:
[tex]c=frequency\times Wavelength [/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
Given, Frequency = [tex]22.235\ GHz=22.235\times 10^{9}\ Hz[/tex]
Thus, Wavelength is:
[tex]Wavelength=\frac{c}{Frequency}[/tex]
[tex]Wavelength=\frac{3\times 10^8}{22.235\times 10^{9}}\ m[/tex]
[tex]Wavelength=0.013492242\ m=13492242\times 10^{-9}\ m[/tex]
Also, 1 m = [tex]10^{-9}[/tex] nm
So,
Wavelength = 13492242 nm
Also, 1 m = [tex]10^{-10}[/tex] Å
Wavelength = 134922420 Å
Consider the following molecule. In common nomenclature, what Greek letter would be assigned to the carbon indicated by an asterisk?
alpha
beta
gamma
epsilon
Answer: beta
Explanation: the beta carbon is the second after the alpha carbon( carbon bonding to the functional group)
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25 ∘C. k1k2 =
Final answer:
The rate law and the Arrhenius equation help calculate the ratio of rate constants for reactions with similar frequency factors. In this case, with identical frequency factors, the ratio of the rate constants is 1.
Explanation:
The rate law for a reaction:
rate = k[A] [B]
The Arrhenius equation:
k = Ae(-Ea/RT)
For the given question:
Given k1 and k2 are the same, k1/k2 = 1.
The ratio of rate constants depends on the exponential term in the Arrhenius equation, specifically the difference in activation energies. If the activation energies are very close, the ratio [tex]\( \frac{k_1}{k_2} \)[/tex] will be close to 1, indicating similar rate constants for the two reactions at 25 °C.
The ratio of reaction rate constants [tex](\(k_1/k_2\))[/tex] for two reactions with similar frequency factors can be expressed using the Arrhenius equation, which relates the rate constant (k) to temperature. The Arrhenius equation is given by:
[tex]\[ k = A \cdot e^{-\frac{E_a}{RT}} \][/tex]
Where:
- ( k ) is the rate constant,
- ( A) is the frequency factor (pre-exponential factor),
- ( E_a ) is the activation energy,
- ( R ) is the gas constant (8.314 J/(mol·K)),
- ( T ) is the temperature in Kelvin.
Assuming the frequency factors[tex](\( A_1 \) and \( A_2 \))[/tex] are the same, the ratio of rate constants [tex](\( \frac{k_1}{k_2} \))[/tex] can be simplified to the ratio of the exponential term:
[tex]\[ \frac{k_1}{k_2} = e^{-\frac{E_{a1} - E_{a2}}{RT}} \][/tex]
Given that the reactions are at 25 °C (298 K), the ratio [tex]\( \frac{k_1}{k_2} \)[/tex]will depend on the difference in activation energies [tex](\( E_{a1} - E_{a2} \)).[/tex]
Which of the following does not affect the rate of a reaction? Question 1 options: A) volume B) catalyst C) temperature D) nature of reactant g
Answer: option A. volume
Explanation:
What is the solubility of ethylene (in units of grams per liter) in water at 25 °C, when the C2H4 gas over the solution has a partial pressure of 0.684 atm? kH for C2H4 at 25 °C is 4.78×10-3 mol/L·atm.
Answer: The solubility of ethylene gas in water is [tex]9.16\times 10^{-2}g/L[/tex]
Explanation:
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{C_2H_4}=K_H\times p_{C_2H_4}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]4.78\times 10^{-3}mol/L.atm[/tex]
[tex]C_{C_2H_4}[/tex] = molar solubility of ethylene gas = ?
[tex]p_{C_2H_4}[/tex] = partial pressure of ethylene gas = 0.684 atm
Putting values in above equation, we get:
[tex]C_{C_2H_4}=4.78\times 10^{-3}mol/L.atm\times 0.684atm\\\\C_{C_2H_4}=3.27\times 10^{-3}mol/L[/tex]
Converting this into grams per liter, by multiplying with the molar mass of ethylene:
Molar mass of ethylene gas = 28 g/mol
So, [tex]C_{C_2H_6}=3.27\times 10^{-3}mol/L\times 28g/mol=9.16\times 10^{-2}g/L[/tex]
Hence, the solubility of ethylene gas in water is [tex]9.16\times 10^{-2}g/L[/tex]
The solubility of ethylene in water at 25 °C with a partial pressure of 0.684 atm is approximately 0.0919 grams per liter.
Explanation:To calculate the solubility of ethylene in water, we can use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The equation is given by:
S = kH * P
Where:
First, we can calculate the solubility of ethylene in units of moles per liter:
S = (4.78×10-3 mol/L·atm) * (0.684 atm) = 0.00327 mol/L
Then, we can convert the solubility to grams per liter using the molar mass of ethylene:
Molar mass of C2H4 = 2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol
Grams/Liter = (0.00327 mol/L) * (28.05 g/mol) = 0.0919 g/L
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How is the molar heat of sublimation related to the molar heats of vaporization and fusion? On what law are these relationships based?
Answer: Hsub=Hfus+Hvap
Explanation:
The molar heat of vaporization measured in kilojoules per mole, or kJ/mol is the energy needed to make vapor one mole of a liquid. .
The molar heat of sublimation measured in kilojoules per mole, or kJ/mol is the energy needed to sublime one mole of a solid,
the molar heat of fusion measured in kilojoules per mole, or kJ/mol is the energy needed to melt one mole of a solid.
Hess law helps to explain the relationship in physical chemistry stating that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps.
In this context Hess’s law helps to see the several steps involved as the heat of sublimation energy is equal to the sum of vaporization energy and fusion energy.
The molar heat of sublimation is the sum of the molar heats of vaporization and fusion, based on Hess's Law. This law enables understanding that sublimation is similar to a sequential process of melting followed by vaporization, reflecting the energy required to overcome intermolecular forces.
The molar heat of sublimation is directly related to the molar heats of vaporization and fusion. The relationship between these heats is guided by the Hess's Law, which asserts that the total enthalpy change during a chemical reaction is the same no matter how many steps the reaction is carried out in.
Thus, for a given substance, the molar heat of sublimation can be approximated by the sum of its molar heat of fusion (solid to liquid transition) and molar heat of vaporization (liquid to gas transition). This is because when fusion is followed by vaporization at the temperature and pressure of the triple point, the net change corresponds to sublimation.
The molar enthalpies related to phase changes are all positive for endothermic processes (where heat is absorbed). Conversely, negative enthalpy changes are associated with the reverse, exothermic processes. The differences in these heats reflect the extent to which intermolecular forces must be overcome going from one phase to another.
Draw a Lewis structure for SO 2 in which all atoms have a formal charge of zero. Do not consider ringed structures.
Answer : The Lewis-dot structure of [tex]SO_2[/tex] is shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, [tex]SO_2[/tex]
As we know that sulfur and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]SO_2[/tex] = 6 + 2(6) = 18
According to Lewis-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.
Now we have to determine the formal charge for each atom.
Formula for formal charge :
[tex]\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}[/tex]
[tex]\text{Formal charge on S}=6-2-\frac{8}{2}=0[/tex]
[tex]\text{Formal charge on }O_1=6-4-\frac{4}{2}=0[/tex]
[tex]\text{Formal charge on }O_2=6-4-\frac{4}{2}=0[/tex]
15.2 g of NO2(g) is placed in a sealed 10.0 L flask at room temperature. The total pressure of the system is found to be 0.50 atm. What are the partial pressures of NO2and N2O4are present?
Answer:
Partial pressure of NO2 = 0.37 atm
Partial pressure of N2O4 = 0.13 atm
Explanation:
Number of moles of NO2 = mass/MW = 15.2g/46g/mole = 0.33 mole
Volume of flask = 10L
1 mole of the mixture of gas contains 22.4L of the gas
10L of the gas contains 10/22.4 mole = 0.45 mole
Total mole of gas mixture in the tank = 0.45 mole
Total pressure of the system = 0.5atm
Partial pressure of NO2 = 0.33mole/0.45mole × 0.5atm = 0.37 atm
Partial pressure of N2O4 = 0.5atm - 0.37atm = 0.13atm
If the exact outer limit of an isolated atom cannot be measured, what criterion can we use to determine atomic radii? What is the difference between a covalent radius and a metallic radius?
Answer:
Calculate the atomic radii of two touching or overlapping atoms.
Explanation:
No doubt, we can't find the atomic boundary of a single atom, but when atoms are in the form of pairs it becomes very easy to measure the atomic radii of two and then dividing it by 2 to get an estimate of atomic radius of a single atom.
It is also called as covalent radius which is half of the total inter-nuclear distance between two same bonded atoms (Homo-nuclear).
If two adjacent mettalic ions are joined by such pairing then the same half of the distance between the nucleus is termed as metallic radii.
n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form of the acid?
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,
[tex]pK_a=8.0[/tex]
pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}[/tex]
Now put all the given values in this expression, we get:
[tex]6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}[/tex]
[tex]\frac{[Deprotonated]}{[Protonated]}=0.01[/tex]
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
[tex]\frac{[Protonated]}{[Deprotonated]}=100[/tex]
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
The ratio of the protonated to the deprotonated form of the acid in the solution is 100:1.
The ratio of the protonated to the deprotonated form of the acid is given by the equation:
[tex]\[ \text{Ratio} = \frac{[\text{A}^-]}{[\text{HA}]} = 10^{(\text{pH} - \text{pKa})} \][/tex]
Given that the pKa of the acid is 8.0 and the pH of the solution is 6.0, we can plug these values into the equation:
[tex]\[ \text{Ratio} = \frac{[\text{A}^-]}{[\text{HA}]} = 10^{(6.0 - 8.0)} \] \[ \text{Ratio} = \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-2} \] \[ \text{Ratio} = \frac{[\text{A}^-]}{[\text{HA}]} = \frac{1}{100} \][/tex]
Therefore, the ratio of the deprotonated form to the protonated form of the acid is 1:100. To find the ratio of the protonated to the deprotonated form, we take the reciprocal of this value:
[tex]\[ \text{Ratio of protonated to deprotonated} = \frac{[\text{HA}]}{[\text{A}^-]} = 100:1 \][/tex]
The answer is: 100:1.
The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is testosterone. Calculate the vapor pressure of the solution at 25 °C when 7.752 grams of testosterone, C19H28O2 (288.4 g/mol), are dissolved in 208.0 grams of diethyl ether. diethyl ether = CH3CH2OCH2CH3 = 74.12 g/mol.
Answer: The vapor pressure of solution is 459.17 mmHg
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For testosterone:Given mass of testosterone = 7.752 g
Molar mass of testosterone = 288.4 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol[/tex]
For diethyl ether:Given mass of diethyl ether = 208.0 g
Molar mass of diethyl ether = 74.12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol[/tex]
Mole fraction of a substance is calculated by using the equation:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]
[tex]\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}[/tex]
[tex]\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095[/tex]
The formula for relative lowering of vapor pressure will be:
[tex]\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}[/tex]
where,
[tex]p^o[/tex] = vapor pressure of solvent (diethyl ether) = 463.57 mmHg
[tex]p^s[/tex] = vapor pressure of the solution = ?
i = Van't Hoff factor = 1 (for non electrolytes)
[tex]\chi_{\text{solute}}[/tex] = mole fraction of solute (testosterone) = 0.0095
Putting values in above equation, we get:
[tex]\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg[/tex]
Hence, the vapor pressure of solution is 459.17 mmHg
At 39.5 o C, the vapor pressure of pure acetone (MM = 58.08 g/mol) is 400.0 torr. If 15.0 grams of an unknown molecule is dissolved in 485.0 g acetone, the vapor pressure decreases to 361.8 torr. What is the molar mass of the solute?
Answer: The molar mass of unknown molecule is 157.07 g/mol
Explanation:
The equation used to calculate relative lowering of vapor pressure follows:
[tex]\frac{p^o-p_s}{p^o}=i\times \chi_{solute}[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex] = relative lowering in vapor pressure
i = Van't Hoff factor = 1 (for non electrolytes)
[tex]\chi_{solute}[/tex] = mole fraction of solute = ?
[tex]p^o[/tex] = vapor pressure of pure acetone = 400 torr
[tex]p_s[/tex] = vapor pressure of solution = 361.8 torr
Putting values in above equation, we get:
[tex]\frac{400-361.8}{400}=1\times\chi_{A}\\\\\chi_{A}=0.0955[/tex]
This means that 0.0955 moles of unknown molecule is present in the solution
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of unknown molecule = 0.0955 moles
Mass of unknown molecule = 15.0 grams
Putting values in above equation, we get:
[tex]0.0955mol=\frac{15.0g}{\text{Molar mass of unknown molecule}}\\\\\text{Molar mass of unknown molecule}=\frac{15.0g}{0.0955mol}=157.07g/mol[/tex]
Hence, the molar mass of unknown molecule is 157.07 g/mol