In this exothermic reaction, 2.50 g of hydrogen peroxide decomposes and generates -7.22 kJ of heat, as indicated by the negative sign.
Explanation:The question asks for the amount of heat (q) generated when 2.50 g of hydrogen peroxide decomposes at constant pressure to give water and oxygen. Given that the enthalpy change (ΔH) for the reaction 2H2O2(l) → 2H2O(l) + O2(g) is -196 kJ, it indicates that the decomposition of hydrogen peroxide is an exothermic reaction and heat is released in the process. The negative sign of ΔH confirms this.
Firstly, we will have to find out how many moles of H2O2 are there in 2.5 g. Using the molar mass of H2O2 (34 g/mol), we get about 0.074 moles of H2O2. In the balanced chemical equation, 2 moles of hydrogen peroxide generates -196 kJ of heat. Therefore, for 0.074 moles, we calculate it by (-196 kJ * 0.074)/2 = -7.22 kJ.
The <-strong>value of q in this exothermic reaction when 2.50 g of hydrogen peroxide decomposes at constant pressure is therefore -7.22 kJ. The negative sign indicates heat is being released, or in other words, the reaction is exothermic.
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The value of q, representing the thermal energy transferred in the decomposition of 2.50g of hydrogen peroxide at constant pressure, is approximately -7.16 kJ.
Explanation:The question asks for the value of q, which represents the thermal energy transferred during a reaction, in an exothermic reaction where 2.50 g of hydrogen peroxide decomposes into water and oxygen gas. In this given case, we know the enthalpy change (∆H) of the reaction is -196 kJ, indicating that this is an exothermic process where that much energy is being released to the surroundings.
We know from the balanced chemical equation 2H2O2(l) → 2H2O(l) + O2(g) that 2 moles of H2O2 decompose to release ∆H amount of energy. Let's find out the number of moles in 2.50 g of H2O2. The molar mass of H2O2 is approximately 34.01 g/mol. So, the number of moles = mass/molar mass = 2.50 g / 34.01 g/mol = 0.0735 moles.
Considering the stoichiometry of the reaction, ∆H of -196 kJ is associated with 2 moles of H2O2. Therefore, the energy associated with 0.0735 mol will be:
q ( energy ) = ∆H x (moles of H2O2 /2) = -196 kJ x (0.0735 mol / 2 mol) = -7.1625 kJ
Therefore, the quantity of thermal energy (q) released in this decomposition reaction is approximately -7.16 kJ.
Learn more about Thermal Energy in Chemical Reactiona) Draw three structures of 3 sulfur containing acids. Minimize formal charges by expanding sulfur's octet . Include any nonbonded electrons on sulfur only.
b) Indicate the oxidation state of sulfur in each compound. Treat sulfur as more electronegative than carbon.
I. methyl sulfenic acid
II. methyl sulfinic acid
III. methyl sulfonic acid
You may find the structure of compounds as well the oxidation number of sulfur in each compound written with blue color. The nonbonded electrons are represented by black dots.
Explanation:
In the methyl sulfenic acid sulfur have a oxidation number of 0, because we have +1 from carbon and -1 from hydroxil group.
In the methyl sulfinic acid sulfur have a oxidation number of +2 , because we have +1 from carbon, -1 from hydroxil group and -2 from oxygen.
In the methyl sulfonic acid sulfur have a oxidation number of +4 , because we have +1 from carbon, -1 from hydroxil group, -2 from one oxygen and -2 form the other oxygen.
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Final answer:
The structures and oxidation states of the three sulfur containing acids are: I. Methyl sulfenic acid with sulfur's oxidation state of +2, II. Methyl sulfinic acid with sulfur's oxidation state of +4, and III. Methyl sulfonic acid with sulfur's oxidation state of +6.
Explanation:
I. Methyl sulfenic acid:
Structure: CH3-S(=O)(=O)H
Oxidation state of sulfur: +2
II. Methyl sulfinic acid:
Structure: CH3-S(=O)(=O)-H
Oxidation state of sulfur: +4
III. Methyl sulfonic acid:
Structure: CH3-S(=O)(=O)-OH
Oxidation state of sulfur: +6
A compound, C4H6O, exhibits IR absorption at 1705 cm-1. Its carbon NMR shifts are given below. The number of hydrogens at each carbon, determined by DEPT, is given in parentheses after the chemical shift. 13C NMR: δ 14.0 (3), 134.4 (2), 146.0 (0), 194.7 (1) Draw the structure of this compound
Answer:
Methacrylaldehyde
Explanation:
The first step is the calculation of the IHD (index hydrogen deficiency):
[tex]IHD=~\frac{2C+2+N-H-X}{2}[/tex]
[tex]IHD=~\frac{2(2)+2-6}{2}=2[/tex]
This value indicates that we have 2 double bonds. Now, if we check the IR info we can conclude that we have an oxo group (C=O) due to the signal in 1705 cm^-1 . So, the options that we can have are aldehyde or ketone.
If we analyze the NMR info we have a signal in 194.7 with only 1 hydrogen. This indicates that necessary we have an aldehyde due to the hydrogen. Also, for the signal in 14 we will have a [tex]CH_3[/tex], for the signal at 134.2 we will have a [tex]CH_2[/tex] and for the signal at 146.0 we will have a quaternary carbon (no hydrogens present).
So, we will have a [tex]CH_3[/tex], [tex]CH_2[/tex], C (without hydrogens), an aldehyde group and a double bond.
When we put all this together we will obtain the Methacrylaldehyde (see figure).
Determine whether each redox reaction occurs spontaneously in the forward direction.
a. Ni(s) + Zn2 + (aq) ¡ Ni2 + (aq) + Zn(s)
b. Ni(s) + Pb2 + (aq) ¡ Ni2 + (aq) + Pb(s)
c. Al(s) + 3 Ag+ (aq) ¡ Al3 + (aq) + 3 Ag(s)
d. Pb(s) + Mn2 + (aq) ¡ Pb2 + (aq) + Mn(s)
Answer:
a) non spontaneous
b) spontaneous
c) spontaneous
d) non spontaneous
Answer:
Explanation:
a. Ni(s) + Zn2 + (aq) ¡ Ni2 + (aq) + Zn(s) non spontaneous
b. Ni(s) + Pb2 + (aq) ¡ Ni2 + (aq) + Pb(s) spontaneous
c. Al(s) + 3 Ag+ (aq) ¡ Al3 + (aq) + 3 Ag(s) Reduction spontaneous
d. Pb(s) + Mn2 + (aq) ¡ Pb2 + (aq) + Mn(s) non spontaneous
You have 41.6 g of O2 gas in a container with twice the volume as one with CO2 gas in another container. The pressure and temperature of both containers are the same. Calculate the mass of carbon dioxide gas you have in the container.
Answer:
The mass of carbon dioxide is 28.6 grams
Explanation:
Step 1: Data given
Mass of O2 = 41.6 grams
Volume of the O2 container = 2V
Volume of the CO2 container = V
Pressure and temperature are the same
Step 2: Ideal gas law
The ideal gas law = p*V=nRT
For O2: p*2V = n(O2)*R*T
For CO2: p*V = n(CO2)*R*T
n(O2)*R*T / P*2V = n(CO2)*R*T / P*V
Since P, R and T are contstant, we can simplify this to:
n(O2)/2 = n(CO2)
Step 3: Calculate moles of O2
Moles O2 = mass O2/ Molar mass O2
Moles O2 = 41.6 grams / 32 g/mol
Moles O2 = 1.3 mol O2
The number of moles CO2 = 0.65 mol
Mass of CO2 = Moles CO2 * Molar Mass CO2
Mass of CO2 = 0.65 mol * 44.01 g/mol
Mass of CO2 = 28.6 grams
The mass of carbon dioxide is 28.6 grams
The value of Kw at several different temperatures in given in the table below. What conclusion can be drawn on the basis of this information?
Temperature
0ºC
25ºC
45ºC
Kw
1.14 x 10-15
1.00 x 10-14
5.48 x 10-14
Pure water becomes more alkaline as the temperature is increased.
The ionization of water is an exothermic process.
Pure water becomes more acidic as the temperature is increased.
The pH of pure water decreases as the temperature is increased.
Answer:
The correct answer is:
The pH of pure water decreases as the temperature is increased.
Explanation:
Ionization constant of water :
[tex]2H_2O(l)\rightleftharpoons H_3O^+(aq) + OH^-(aq)[/tex]
Water will dissociate into equal amount of hydronium and hydroxide ions
[tex]K_w=[H_3O^+]\times [OH^-][/tex]
As we can see that from the given information that ionic product of water increase with increase in temperature which means concentration of hydronium ion concentration increases.
And according to definition of pH of the solution is a negative logarithm of hydronium ions concentration.
[tex]pH=-\log [H_3O^+][/tex]
The pH of the solution of solution is inversely proportional to the concentration of hydronium ions.
As the hydronium ion concentration of water increases with increase in temperature , the pH of the water will decrease.
At 1000 K, a sample of pure NO2 gas decomposes. 2 NO2(g) equilibrium reaction arrow 2 NO(g) + O2(g) The equilibrium constant KP is 158. Analysis shows that the partial pressure of O2 is 0.29 atm at equilibrium. Calculate the pressure of NO and NO2 in the mixture.
Answer: The pressure of NO and [tex]NO_2[/tex] in the mixture is 0.58 atm and 0.024 atm respectively.
Explanation:
We are given:
Equilibrium partial pressure of [tex]O_2[/tex] = 0.29 atm
For the given chemical equation:
[tex]2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)[/tex]
Initial: a
At eqllm: a-2x 2x x
Calculating for the value of 'x'
[tex]\Rightarrow x=0.29[/tex]
Equilibrium partial pressure of NO = 2x = 2(0.29) = 0.58 atm
Equilibrium partial pressure of [tex]NO_2[/tex] = a - 2x = a - 2(0.29) = a - 0.58
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{O_2}\times (p_{NO})^2}{(p_{NO_2})^2}[/tex]
We are given:
[tex]K_p=158[/tex]
Putting values in above expression, we get:
[tex]158=\frac{0.29\times (0.58)^2}{(a-0.58)^2}\\\\a=0.555,0.604[/tex]
Neglecting the value of a = 0.555 because it cannot be less than the equilibrium concentration.
So, [tex]a=0.604[/tex]
Equilibrium partial pressure of [tex]NO_2[/tex] = (a - 0.58) = (0.604 - 0.58) = 0.024 atm
Hence, the pressure of NO and [tex]NO_2[/tex] in the mixture is 0.58 atm and 0.024 atm respectively.
At a certain concentration of N2 and O3, the initial rate of reaction is 0.770 M / s. What would the initial rate of the reaction be if the concentration of N2 were halved? Be sure your answer has the correct number of significant digits.
The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
[tex]rate=k[N_2]^2[H_2]^2[/tex]
At a certain concentration of [tex]N_2[/tex] and [tex]H_2[/tex], the initial rate of reaction is 0.770 M/s. What would the initial rate of the reaction be if the concentration of [tex]N_2[/tex] were halved? Be sure your answer has the correct number of significant digits.
Answer : The initial rate of the reaction will be 0.192 M/s
Explanation :
Rate law expression for the reaction:
[tex]rate=k[N_2]^2[H_2]^2[/tex]
As we are given that:
Initial rate = 0.770 M/s
Expression for rate law for first observation:
[tex]0.770=k[N_2]^2[H_2]^2[/tex] ....(1)
Expression for rate law for second observation:
[tex]R=k(\frac{[N_2]}{2})^2[H_2]^2[/tex] ....(2)
Dividing 1 by 2, we get:
[tex]\frac{R}{0.770}=\frac{k(\frac{[N_2]}{2})^2[H_2]^2}{k[N_2]^2[H_2]^2}[/tex]
[tex]\frac{R}{0.770}=\frac{1}{4}[/tex]
[tex]R=0.192M/s[/tex]
Therefore, the initial rate of the reaction will be 0.192 M/s
If assuming first order, halving the concentration of N2 would halve the initial rate of the reaction, thus the reaction rate would be 0.385 M / s. But the exact rate would depend on the order of the reaction.
Explanation:Assuming that the reaction rate is directly proportional to the reactant concentration (a first order reaction), if you halve the concentration of N2, the initial rate of the reaction would also be halved. As such, the initial rate of the reaction would be 0.770 M / s divided by 2, which equals to 0.385 M / s.
However, if the reaction is second order with respect to N2 (meaning the rate of reaction is proportional to the square of the concentration of N2), halving the concentration would quarter the reaction rate, if it's zero order (rate independent of concentration), halving the concentration wouldn't change the reaction rate. Hence, additional information such as the order of the reaction with respect to N2 is needed for a more accurate answer.
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If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.6 C in an insulated beaker, and assuming that no heat is lost, what is the final temperature of the mixture?
I tried many times in trying to get the answer, but I keep getting it wrong. I appreciate who answers this writes it step by step. Thank you.
Applying the principle of conservation of energy and the heat equation, the final temperature of the ethanol-water mixture is calculated to be 22.7°C
The specific example used in the image is for a mixture of ethanol and water.
Step 1: Understand the principle
The underlying principle is that heat lost by the hot object is equal to the heat gained by the cold object. This principle is known as the law of conservation of energy.
Step 2: Form an equation
We can express this principle mathematically using the following equation:
q₁ = -q₂
where:
q₁ is the heat lost by the hot object
q₂ is the heat gained by the cold object
We can further expand this equation using the following relationship between heat, mass, specific heat capacity, and temperature change:
q = mcΔT
where:
m is the mass of the object
c is the specific heat capacity of the object
ΔT is the change in temperature
Substituting this relationship into the first equation, we get:
m₁c₁(T₁ - T_f) = -m₂c₂(T_f - T₂)
where:
m₁ and c₁ are the mass and specific heat capacity of the hot object, respectively
m₂ and c₂ are the mass and specific heat capacity of the cold object, respectively
T₁ and T₂ are the initial temperatures of the hot and cold objects, respectively
T_f is the final temperature of the mixture
Step 3: Apply the formula to the specific case
In the example given, the hot object is ethanol and the cold object is water. The following values are provided:
m₁ = 45.0 g (mass of ethanol)
c₁ = 2.3 J/g°C (specific heat capacity of ethanol)
T₁ = 9.0°C (initial temperature of ethanol)
m₂ = 45.0 g (mass of water)
c₂ = 4.18 J/g°C (specific heat capacity of water)
T₂ = 28.6°C (initial temperature of water)
T_f = ? (final temperature of the mixture)
Step 4: Solve for the unknown
Substituting the known values into the equation and solving for T_f, we get:
(0.789 g/mL x 45.0 mL) x (2.3 J/g°C) x (T_f - 9.0°C) = -(1.0 g/mL x 45.0 mL) x (4.18 J/g°C) x (T_f - 28.6°C)
Solving this equation, we get:
T_f = 22.7°C
Therefore, the final temperature of the mixture is 22.7°C.
When 1.550 g of liquid hexane (C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87∘C to 38.13∘C. Find ΔErxn for the reaction in kJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘C. Express your answer in kilojoules per mole to three significant figures.2. The combustion of toluene has a ΔErxn of –3.91×103 kJ/mol. When 1.55 g of toluene (C7H8) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12∘C to 37.57∘C. Find the heat capacity of the bomb calorimeter. Express the heat capacity in kilojoules per degree Celsius to three significant
Explanation:
1). The given data is as follows.
[tex]T_{i} = 25.87^{o}C[/tex], [tex]T_{f} = 38.13^{o}C[/tex]
C = 5.73 [tex]kJ/^{o}C[/tex]
Hence, calculate the change in enthalpy of the reaction as follows.
[tex]\Delta E_{rxn} = -C \times \Delta T[/tex]
= [tex]-5.73 \times (38.13 - 25.87)^{o}C[/tex]
= -70.25 KJ
As, number of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{1.55}{(6 \times 12 + 14 \times 1)}[/tex]
= 0.018 mol
Therefore, enthalpy of reaction in kJ/mol hexane is as follows.
[tex]\Delta E_{rxn} = \frac{-70.25 KJ}{0.018 mol}[/tex]
= [tex]-3.90 \times 10^{3}[/tex] kJ/mol
Thus, we can conclude that [tex]\Delta E_{rxn}[/tex] for the reaction in kJ/mol hexane is [tex]-3.90 \times 10^{3}[/tex] kJ/mol .
2). As we know that,
Number of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{1.55}{(7 \times 12 + 8 \times 1)}[/tex]
= 0.017 mol
[tex]\Delta E_{rxn} = \E_{rxn} per mol \times \text{number of moles}[/tex]
= [tex]-3.91 \times 10^{3} \times 0.017[/tex]
= -65.875 kJ
As, [tex]\Delta E_{rxn} = C \times \Delta T [/tex]
-65.875 kJ = [tex]-C \times (37.57 - 23.12)^{o}C[/tex]
C = 4.56 [tex]kJ/^{o}C[/tex]
Hence, heat capacity of the bomb calorimeter is 4.56 [tex]kJ/^{o}C[/tex].
When 1.550 g of liquid hexane (C₆H₁₄) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 °C to 38.13 °C. The change in the internal energy of the reaction is -3.90 × 10³ kJ/mol.
When 1.55 g of toluene (C₇H₈) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 °C to 37.57 °C. The heat capacity of the bomb calorimeter is 4.55 kJ/°C.
1. When 1.550 g of liquid hexane (C₆H₁₄) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 °C to 38.13 °C. Find ΔErxn for the reaction in kJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘C. Express your answer in kilojoules per mole to three significant figures.First, we will calculate the moles corresponding to 1.550 g of hexane using its molar mass (86.18 g/mol).
[tex]n = 1.550 g \times \frac{1mol}{86.18 g} = 0.01799mol[/tex]
Then, we will calculate the heat (Q) absorbed by the bomb calorimeter using the following expression.
[tex]Q = C \times \Delta T = \frac{5.73kJ}{\° C} \times (38.13\° C - 25.87\° C) = 70.2 kJ[/tex]
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qb) and the heat released by the combustion (Qc) is zero.
[tex]Qb + Qc = 0\\\\Qc = -Qb = -70.2 kJ[/tex]
Finally, we will calculate the change in the internal energy of the reaction (ΔErxn) using the following expression.
[tex]\Delta Erxn = \frac{Qc}{n} = \frac{-70.2kJ}{0.01799mol} = -3.90 \times 10^{3} kJ/mol[/tex]
2. The combustion of toluene has a ΔErxn of –3.91 × 10³ kJ/mol. When 1.55 g of toluene (C₇H₈) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 °C to 37.57 °C. Find the heat capacity of the bomb calorimeter. Express the heat capacity in kilojoules per degree Celsius to three significant.First, we will calculate the moles corresponding to 1.55 g of toluene using its molar mass (92.14 g/mol).
[tex]1.55 g \times \frac{1mol}{92.14g} = 0.0168 mol[/tex]
The combustion of toluene has a ΔErxn of –3.91 × 10³ kJ/mol. The heat released by the combustion of 0.0168 moles of toluene is:
[tex]0.0168 mol \times \frac{-3.91\times 10^{3}kJ }{mol} = -65.7 kJ[/tex]
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qb) and the heat released by the combustion (Qc) is zero.
[tex]Qb + Qc = 0\\\\Qb = -Qc = 65.7 kJ[/tex]
We can calculate the heat capacity of the bomb calorimeter (C) using the following expression.
[tex]C = \frac{Qb}{\Delta T } = \frac{65.7 kJ}{37.57 \° C - 23.12 \° C } = 4.55 kJ/ \° C[/tex]
When 1.550 g of liquid hexane (C₆H₁₄) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 °C to 38.13 °C. The change in the internal energy of the reaction is -3.90 × 10³ kJ/mol.When 1.55 g of toluene (C₇H₈) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 °C to 37.57 °C. The heat capacity of the bomb calorimeter is 4.55 kJ/°C.Learn more: https://brainly.com/question/24245395
Given the DNA template strand 3' GCATTCAAG 5', write the amino acid sequence in the N‑terminal to C‑terminal direction. Note: Enter the amino acids using their three-letter designations. Put a hyphen between each amino acid. (for example, Glu‑Asp‑Val).
N-terminal Arg-Lys-Phe C-terminal
The universal codon code is attached for reference. The central dogma (DNA --> RNA--> Proteins) concept is used to translate the information in DNA into proteins.
Explanation:
We shall begin with the transcription of the DNA strand to form an mRNA. Remember that RNA has no Thiamine base but instead in its place is Uracil;
3' GCATTCAAG 5' DNA strand
5’ CGUAAGUUC 3’ RNA strand
We can then begin translating the mRNA into protein using the universal codon code (attached). Remember a codon is a 3 nucleotide sequence that codes for a specific amino acid;
5’ CGUAAGUUC 3’ RNA strand
N-terminal Arg-Lys-Phe C-terminal Peptide strand
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The DNA template strand 3' GCATTCAAG 5' can be transcribed into complementary DNA sequence 5' CGTAAGTTC 3'. This is then transcribed into mRNA sequence 5' CGUAAGUUC 3'. Using the genetic code, this is translated into the amino acid sequence: Arg‑Lys‑Phe.
Explanation:To determine the amino acid sequence, we first need to establish the complementary DNA sequence for the provided DNA template strand 3' GCATTCAAG 5'. DNA complementary sequence pairs adenine (A) with thymine (T) and cytosine (C) with guanine (G) so we get the complementary DNA sequence (5' to 3' direction) as 5' CGTAAGTTC 3'.
Next, this DNA sequence is transcribed into mRNA using the similar pairing rules but thymine is replaced by uracil (U) in RNA. Thus, the mRNA sequence is 5' CGUAAGUUC 3'. Then we translate this mRNA into amino acids. The genetic code translates each mRNA codon into an amino acid. Therefore, we get the following amino acid sequence: Arg‑Lys‑Phe. Arg stands for Arginine, Lys for Lysine and Phe for Phenylalanine. These abbreviations are internationally accepted standard to represent the corresponding amino acids.
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An atom of 135I has a mass of 134.910023 amu. Calculate the binding energy in MeV per atom. Enter your answer with 4 significant figures and no units. Use the masses:
mass of 1H atom = 1.007825 amu
mass of a neutron = 1.008665 amu
1 amu = 931.5 Me
Answer:
1132 MeV/atom
Explanation:
Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.
Also, atomic number of I = 5
3
Thus, the number of protons = 53
Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.
Mass number = Number of protons + Number of neutrons
135 = 53 + Number of neutrons
Number of neutrons = 82
Mass of neutron = 1.008665 amu
Mass of proton = 1.007825 amu
Calculated mass = Number of protons*Mass of proton + Number of neutrons*Mass of neutron
Thus,
Calculated mass = (53*1.007825 + 82*1.008665) amu = 136.125255 amu
Mass defect = Δm = |136.125255 - 134.910023| amu = 1.215232 amu
The conversion of amu to MeV is shown below as:-
1 amu = 931.5 MeV
So, Energy = 1.215232*931.5 MeV/atom = 1132 MeV/atom
Draw the structure of a soap surfactant molecule. Label the relevant features that enable it to be a good cleaning agent. Explain why surfactants dissolve in both water and water/oil mixtures. Use your knowledge of intermolecular forces in your explanation.
Explanation:
In the Figure can be seen the structure of a soap surfactant. It has two key parts: the hydrophobic tail and the hydrophilic head.
The first one is non-polar and formed only by H and C. The other, is polar due to the O atoms.Surfactant molecules bond with each other forming miscelles, expossing the part that is compatible with the solvent and protecting the other one.
This means thar the polar head enables the surfactant to dissolve in water and the non-polar tail enables it to dissolve in non polar solvents such as oil mixtures.
Nitrogen and hydrogen combine to form ammonia in the Haber process.
Calculate (in kJ) the standard enthalpy change image from custom entry toolH for the reaction written below, using the bond energies given.
N2(g) + 3H2(g) ---> 2NH3 (g)
Bond energy (kJ/mol): Nimage from custom entry toolN 945; H - ---- H 432; N ----- H 391
Answer:
-105 kJ
Explanation:
The enthalpy change of a reaction is the sum of the energy of the bonds of the reactants and the products. The bonds at the reactants are being broken, so it's an endothermic reaction, so the bond energy must be positive.
The bonds at the products are being formed, so the process is exothermic, and the bond energy must be negative. There are being broken 1 N≡N bond and 3 H-H bonds, and are being formed 6 N-H bonds:
Reactants: 945 + 3*432 = 2241 kJ
Products: 6*(-391) = -2346 kJ
ΔH = 2241 - 2346
ΔH = -105 kJ
A boy with pneumonia has lungs with a volume of 1.9 L that fill with 0.080 mol of air when he inhales. When he exhales, his lung volume decreases to 1.5 L. Enter the number of moles of gas that remain in his lungs after he exhales. Assume constant temperature and pressure. g
Answer: The number of moles of gas remaining in the lungs is 0.063 moles
Explanation:
The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.
The equation used to calculate number of moles is given by:
[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]
where,
[tex]V_1\text{ and }n_1[/tex] are the initial volume and number of moles
[tex]V_2\text{ and }n_2[/tex] are the final volume and number of moles
We are given:
[tex]V-1=1.9L\\n_1=0.080mol\\V_2=1.5L\\n_2=?[/tex]
Putting values in above equation, we get:
[tex]\frac{1.9}{0.080}=\frac{1.5}{n_2}\\\\n_2=\frac{1.5\times 0.080}{1.9}=0.063[/tex]
Hence, the number of moles of gas remaining in the lungs is 0.063 moles
A galvanic cell with E o cell = 0.30 V can be constructed using an iron electrode in a 1.0 M Fe(NO3)2 solution, and either a tin electrode in a 1.0 M Sn(NO3)2 solution, or a chromium electrode in a 1.0 M Cr(NO3)3 solution even though Sn2+/Sn and Cr3+/Cr have different reduction potentials. Give the overall balanced reaction for Fe-Sn cell. Do not include the states of matter.
Answer: The half reactions are written below.
Explanation:
Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
We are given a chemical cell which is Fe-Sn cell. The half reaction follows:
Oxidation half reaction: [tex]Fe\rightarrow Fe^{2+}+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V[/tex]
Reduction half reaction: [tex]Sn^{2+}+2e^-\rightarrow Sn;E^o_{Sn^{2+}/Sn}=-0.14V[/tex]
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced.
Total cell reaction: [tex]Fe+Sn^{2+}\rightarrow Fe^{2+}+Sn[/tex]
Hence, the half reactions are written above.
The overall balanced reaction for a Fe-Sn galvanic cell is: Fe + Sn^2+ -> Fe^2+ + Sn. This reaction is obtained by combining the respective half-reactions for the oxidation of iron and the reduction of tin.
Explanation:To determine the overall balanced reaction for a Fe-Sn galvanic cell, we need to consider the half-reactions for the oxidation of iron (Fe) and the reduction of tin (Sn).
In this case, the half-reaction showing the oxidation of iron is: Fe -> Fe^2+ + 2e^-
And the half-reaction showing the reduction of tin is: Sn^2+ + 2e^- -> Sn
When these two reactions are combined to give the overall reaction for the galvanic cell, we get: Fe + Sn^2+ -> Fe^2+ + Sn
This is the overall balanced reaction for the Fe-Sn galvanic cell. Please note that the reduction potentials of Sn^2+/Sn and Cr^3+/Cr are different which explains why you would expect different cell potentials if iron was coupled with either tin or chromium.
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At a certain temperature the rate of this reaction is first order in HI with a rate constant of :7.21s?12HI(g)??H2(g) + I2(g)
Suppose a vessel contains HI at a concentration of 0.440M
Calculate the concentration of HI in the vessel 0.210 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
Answer:
[HI] = 0.097 M
Explanation:
Let's consider the following reaction.
2 HI(g) ⇄ H₂(g) + I₂(g)
The order of reaction for HI is 1. Thus, we can calculate the concentration of HI ([HI]) at certain time using the following expression:
ln [HI] = ln [HI]₀ - k. t
where,
[HI]₀ is the initial concentration of HI
k is the rate constant
t is the time elapsed
When [HI]₀ = 0.440 M and t = 0.210 s, the concentration of HI is
ln [HI] = ln (0.440) - 7.21 s⁻¹ × 0.210 s
ln [HI] = -2.33
[HI] = 0.097 M
Which of the following does not correctly describe Sn2 reactions of alkyl halides? A) The mechanism consists of a single step with no intermediates. B) Tertiary halides react faster than secondary halides. C) The transition state species has a pentavalent carbon atom. D) Rate of reaction depends on the concentrations of both the alkyl halide and the nucleophile.
Answer:
The correct answer is B the tertiary halides reacts faster than primary halides.
Explanation:
During SN2 reaction the nucleophile attack the alkyl halide from the opposite side resulting in the formation of transition state in which a bond is not completely broken or a new bond is not completely formed.
After a certain period of time the nucleophile attach with the substrate by substituting the existing nuclophile.
An increase in the bulkiness in the alkyl halide the SN2 reaction rate of that alkyl halide decreases.This phenomenon is called steric hindrance.
So from that point of view the that statement tertiary halides reacts faster that secondary halide is not correct.
The correct answer is C) The transition state species has a pentavalent carbon atom.
Explanation:The correct answer is C) The transition state species has a pentavalent carbon atom.
In an Sn2 reaction, the mechanism consists of a single step with no intermediates, which eliminates option A as a correct description. Option B is also incorrect because tertiary halides react slower than secondary halides due to steric hindrance. Option D is true, as the rate of the Sn2 reaction depends on the concentrations of both the alkyl halide and the nucleophile.
Thus, option C does not correctly describe Sn2 reactions of alkyl halides.
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Given the standard enthalpy changes for the following two reactions:
(1) N2(g) + 2O2(g)2NO2(g)...... ΔH° = 66.4 kJ
(2) 2N2O(g)2N2(g) + O2(g)......ΔH° = -164.2 kJ
what is the standard enthalpy change for the reaction:
(3) 2N2O(g) + 3O2(g)4NO2(g)......ΔH° = _________?
The standard enthalpy of formation, which has been established for a huge variety of compounds, is enthalpy change. The reactants undergo chemical changes and combine to produce products in any general chemical reaction. Here the enthalpy change is -297 kJ.
The reaction enthalpy is the change in enthalpy, denoted by the symbol ΔrH. By deducting the total enthalpies of all the reactants from the total enthalpies of the products, the reaction enthalpy is determined.
Here we will multiply the equation (1) with "2" and then will subtract it from equation "2".
N₂(g) + 2O₂ (g)→ 2NO₂(g) ...... ΔH° = 66.4 kJ (1) × (2)
2N₂O(g) → 2N₂(g) + O₂(g) .......ΔH° = -164.2 kJ (2)
2N₂(g) + 4O₂ (g)→ 4NO₂(g) ....... ΔH° = 132.8 kJ
2N₂O(g) + 3O₂(g) → 4NO2(g)
ΔH° = ΔH°(eq 2) - 2ΔH°(eq 1) = -164.2 -(2X66.4) = -297 kJ
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Final answer:
Using Hess's Law and manipulating the given reactions to achieve the target reaction, the standard enthalpy change for the reaction 2N2O(g) + 3O2(g) -> 4NO2(g) is found to be -31.4 kJ.
Explanation:
To find the standard enthalpy change ΔH° for the reaction 2N2O(g) + 3O2(g) → 4NO2(g), we need to apply Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of individual steps that lead to the overall reaction.
The reactions given are:
(1) N2(g) + 2O2(g) → 2NO2(g)...... ΔH° = 66.4 kJ
(2) 2N2O(g) → 2N2(g) + O2(g)......ΔH° = -164.2 kJ
To use these reactions to find the enthalpy change for (3), we need to manipulate them so that when added, they result in the target reaction. We notice that reaction (1) must be doubled to provide the 4NO2(g) as in reaction (3). This also doubles its enthalpy change. Then, we can simply add the first and second reactions together to get the desired reaction.
So, we have:
(1) x 2: 2N2(g) + 4O2(g) → 4NO2(g)...... ΔH° = 2 x 66.4 kJ = 132.8 kJ
(2): 2N2O(g) → 2N2(g) + O2(g)......ΔH° = -164.2 kJ
To find the ΔH° for (3), we add the ΔH° from the doubled reaction (1) to the ΔH° from reaction (2):
132.8 kJ + (-164.2 kJ) = -31.4 kJ
Hence, the standard enthalpy change for the reaction 2N2O(g) + 3O2(g) → 4NO2(g) is -31.4 kJ.
41Ca decays by electron capture. The product of this reaction undergoes alpha decay. What is the product of this second decay reaction? 41Ca decays by electron capture. The product of this reaction undergoes alpha decay. What is the product of this second decay reaction? Ti Cl Sc Ar Ca
41Ca (Calcium 41) undergoes electron capture to become 41K (Potassium 41). This 41K then undergoes alpha decay to become 37Cl (Chlorine 37).
Explanation:The starting nuclide is 41Ca (Calcium 41), which undergoes electron capture (EC). This process involves the nucleus of an atom drawing in a nearby electron and combining it with a proton to form a neutron. The result is a nuclide of an element with atomic number one less than the original. Therefore, 41Ca becomes 41K (Potassium 41).
Next, 41K undergoes alpha decay, a type of radioactive decay where an atomic nucleus emits an alpha particle (consisting of two protons and two neutrons - essentially a helium nucleus). This results in a decrease of both atomic and mass number (by 2 and 4 respectively). Applying this to 41K, we get 37Cl (Chlorine 37) as the product of the alpha decay.
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Which of the following statements concerning galvanic cells is/are true?
A.) The two half-cells are connected by a salt bridge
B.) Electrons flow from the anode to the cathode
C.) Reduction occurs at the cathode
D.) All of the above are true
Answer: Correct answer is D.) Every option is true and correct for a galvanic cell.
All the given statements are true: a galvanic cell consists of two half-cells connected by a salt bridge, electrons flow from the anode to the cathode, and reduction occurs at the cathode.
Explanation:The statement 'D.) All of the above are true' is the correct selection regarding galvanic cells. A.) In a galvanic cell, the two half-cells are indeed connected by a salt bridge. The salt bridge serves to balance the charge, allowing ions to migrate and maintain a neutral charge in the cell. B.) Electrons flow from the anode to the cathode. This is the basic concept of electricity – the flow of electrons. The anode is the electrode where oxidation (loss of electrons) happens, and the electrons released from the oxidation reaction at the anode flow through a wire to the cathode. C.) Reduction occurs at the cathode, meaning that it gains electrons. The cathode is the electrode where reduction (gain of electrons) happens in redox (oxidation-reduction) reactions.
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Silane is a gas that is used to prepare extremely pure silicon for use in semiconductors, such as computer chips. A sample of silane has a mass of 15.0 g and occupies a volume of 4.81 L at a pressure of 2.35 atm and a temperature of 22 ºC. What is the molar mass of silane?
Answer:
Molar mass of silane = 32.15 g/mol
Explanation:
Given:
Pressure = 2.35 atm
Temperature = 22 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (22 + 273.15) K = 295.15 K
Volume = 42.9 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
2.35 atm × 4.81 L = n × 0.0821 L.atm/K.mol × 295.15 K
⇒n = 0.4665 moles
Given, Mass = 15.0 g
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]0.4665\ mole= \frac{15.0\ g}{Molar\ mass}[/tex]
Molar mass of silane = 32.15 g/mol
Methanol, CH 3 OH CH3OH , is formed from methane and water in a two‑step process. Step 1: CH4 ( g ) + H2O ( g ) ⟶ CO ( g ) + 3H2 ( g ) Δ S ∘ = 214.7 J / K Step 2: CO ( g ) + 2 H 2 ( g ) ⟶ CH 3 OH ( l ) Δ S ∘ = − 332.3 J / K Calculate Δ H∘ and ΔG∘ at 298 K for step 1.
Answer:
ΔH° = 206.1 kJ
ΔG° = 142.1 kJ
Explanation:
Let's consider the first step in the synthesis of methanol.
Step 1: CH₄(g) + H₂O(g) ⟶ CO(g) + 3 H₂(g) ΔS° = 214.7 J / K
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
ni are the moles of reactants and products
ΔH°f(p) are the standard enthalpies of formation of reactants and products
ΔH° = [1 mol × ΔH°f(CO(g)) + 3 mol × ΔH°f(H₂(g))] - [1 mol × ΔH°f(CH₄(g)) + 1 mol × ΔH°f(H₂O(g))]
ΔH° = [1 mol × (-110.5 kJ/mol) + 3 mol × (0 kJ/mol)] - [1 mol × (-74.81 kJ/mol) + 1 mol × (-241.8 kJ/mol)]
ΔH° = 206.1 kJ
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
ΔG° = 206.1 kJ - 298 K × (214.7 × 10⁻³ kJ/K)
ΔG° = 142.1 kJ
The values of ΔH∘ and ΔG∘ for the reaction can be calculated using their respective formulas and knowing the standard enthalpy and free energy of formation of all the reactants and products. The data is usually found in Chemistry textbooks, specifically in Appendix G.
Explanation:To calculate ΔH∘ and ΔG∘ at 298K for step 1, we use the Gibb's Helmholtz equation, ΔG = ΔH - TΔS. However, the problem doesn't provide enough information to calculate these directly. Let's presume, however, that we know the values of ΔH∘ for methane and water. Using Appendix G data (usually found in Chemistry textbooks), we could apply the formula ΔH = [ΔH products] - [ΔH reactants] to get ΔH∘.
Similarly, for ΔG∘, if the standard free energy of formation of the compounds is known, we could apply the formula ΔG = [ΔG products] - [ΔG reactants].
Since these values are usually tabulated, this could give us the values of ΔH∘ and ΔG∘ for step 1 of the reaction.
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Draw the Lewis structure (including resonance structures) for the acetate ion (CH3COO−). For each resonance structure, use square brackets to denote the overall charge. Draw the Lewis dot structures for the acetate ion. Include all hydrogen atoms and nonbonding electrons.
Answer:
Lewis structure is shown in the image below.
Explanation:
Acetate ion (CH₃COO⁻)
Valence electrons of carbon = 4
Valence electrons of oxygen = 6
Valence electrons of hydrogen = 1
Charge = 1 (Negative which means that the electrons are being added)
The total number of the valence electrons = 2(4) + 2(6) + 3(1) + 1 = 24
The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,
The Lewis structure is shown in the image below.
The Lewis structure of the acetate ion consists of a carbon atom bonded to two oxygen atoms. There are also resonance structures with the double bond and a negative charge on different atoms. The Lewis dot structure includes lone pairs of electrons for each atom.
Explanation:The Lewis structure of the acetate ion (CH3COO-) can be represented by drawing the individual Lewis structures for each resonance form. The main structure consists of a carbon atom single-bonded to two oxygen atoms, with one oxygen atom also bonded to a hydrogen atom. The remaining carbon atom is double-bonded to one of the oxygen atoms and also bonded to three hydrogen atoms. Each O atom has three lone pairs of electrons.
The resonance structures can be depicted by moving the double bond between the carbon atoms and by moving the lone pairs of electrons around. One of the resonance structures places a negative charge on the single-bonded oxygen, while the other resonance structure places the negative charge on the double-bonded oxygen.
The Lewis dot structure for the acetate ion includes lone pairs of electrons for all the atoms, as well as the bonding pairs of electrons representing the chemical bonds between the atoms. The oxygen atoms each have three lone pairs of electrons, the carbon atoms each have one lone pair of electrons, and the hydrogen atoms each have no lone pairs of electrons.
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The standard enthalpy change for the following reaction is 873 kJ at 298 K.
2 KCl(s) 2 K(s) + Cl2(g) ΔH° = 873 kJ
What is the standard enthalpy change for this reaction at 298 K?
K(s) + 1/2 Cl2(g) KCl(s)_________KJ
Answer: - 436.5 kJ.
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.
The given chemical reaction is,
[tex]2KCl(s)\rightarrow 2K(s)+Cl_2(g)[/tex] [tex]\Delta H_1=873kJ[/tex]
Now we have to determine the value of [tex]\Delta H[/tex] for the following reaction i.e,
[tex]K(s)+\frac{1}{2}Cl_2(g)\rightarrow KCl(s)[/tex] [tex]\Delta H_2=?[/tex]
According to the Hess’s law, if we divide the reaction by half then the [tex]\Delta H[/tex] will also get halved and on reversing the reaction , the sign of enthlapy changes.
So, the value [tex]\Delta H_2[/tex] for the reaction will be:
[tex]\Delta H_2=\frac{1}{2}\times (-873kJ)[/tex]
[tex]\Delta H_2=-436.5kJ[/tex]
Hence, the value of [tex]\Delta H_2[/tex] for the reaction is -436.5 kJ.
The standard enthalpy change for the formation of KCl from K(s) and Cl2(g) is -436.5 kJ, calculated by reversing and halving the given reaction's enthalpy change of 873 kJ for 2 moles of KCl.
Explanation:The student asks about the standard enthalpy change for the reaction where potassium (K) reacts with chlorine gas (Cl2) to form potassium chloride (KCl).
To find the standard enthalpy change for the formation of KCl from K(s) and Cl2(g), we use the given reaction 2 KCl(s) → 2 K(s) + Cl2(g) with a ΔH° of 873 kJ. Since this is the reverse reaction of the formation of KCl, and because enthalpy is a state function, the enthalpy change of the forward reaction is the negative of the reverse reaction. Furthermore, the given reaction involves 2 moles of KCl, so we need to divide the enthalpy by 2 to find the enthalpy change for the formation of 1 mole of KCl.
Therefore, the standard enthalpy change for the reaction K(s) + 0.5 Cl2(g) → KCl(s) is -873 kJ / 2 = -436.5 kJ.
Which is not true of the standard conditions for electrochemical cell measurements?
Gases are at 1 bar pressure.
Liquids are in the pure state.
Solids are in the pure state.
Solutes are at 1.0 M concentration.
The cell voltage is always positive.
Answer:
Gases are at 1 bar pressure.
Explanation:
Option a is incorrect . Standard Pressure in chemistry is always taken in atm and not in bar. The standard condition for electro chemical cell measurements are
Gases must be at 1 atm pressure.
Liquids must be in pure state
Solids must also be in pure state
Solutes are at 1.0 M concentration
And the cell voltages are positive.
In the context of electrochemical cells, the statement 'The cell voltage is always positive' is incorrect as cell voltage can be negative in an electrolytic cell.
Explanation:The statement that is not true of the standard conditions for electrochemical cell measurements is that 'The cell voltage is always positive.' While the cell voltage can be positive in a galvanic/voltaic cell as it produces an electrical current, it can be negative in an electrolytic cell, where electrical energy is being consumed to bring about a chemical change.
The other statements about gases at 1 bar pressure, liquids and solids being in a pure state, and solutes at a 1.0 M concentration are indeed part of the standard conditions for measuring the properties of an electrochemical cell.
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Write the electron configuration for the element titanium, Ti.
Express your answer in order of increasing orbital energy as a string without blank space between orbitals. For example, the electron configuration of Li could be entered as 1s^22s^1 or [He]2s^1.
Answer:
1s^22s^22p^63s^23p^63d^24s^2
Explanation:
Titanium with the chemical symbol Ti, is a transition element, and in fact the second transition element after scandium. It has an atomic number of 22. It is the second element in the first transition series.
To write the electronic configuration, we consider the atomic number of titanium. The atomic number of titanium is 22. Hence, since an atom is electrically neutral, the number of electrons in the titanium metal is 22.
It should be noted that while filling,the maximum number of electrons the s subshell is 2. The maximum number of electrons in the p subshell is 6 while the number in the d shell is 10
Determine the electron-group arrangement, molecular shape, and ideal bond angle for the following molecule: PH3 Electron-group arrangement: tetrahedral trigonal pyramidal V-shaped trigonal planar Molecular shape: tetrahedral trigonal pyramidal T-shaped bent Ideal bond angle: degrees.
Answer:
The molecular shape and ideal bond angle of the [tex]PH_{3}[/tex] is trigonalbipyramidal and [tex]109.5^{o}[/tex] respectively.
Explanation:
The structure of [tex]PH_{3}[/tex] is as follows.(in attachment)
From the structure,
Phosphor atom has one lone pair and three hydrogens are bonded by six electrons.
Therefore, total electrons invovled in the formation [tex]PH_{3}[/tex] is eight.
Hence, four electron groups which indicate the tetrahedral shape. But one pair is lone pair i.e, present on the phosphor atom.
Therefore, ideal geometry of the [tex]PH_{3}[/tex] molecule is Trigonalbipyramidal.
The ideal angle of trigonalbipyrmidal is [tex]109.5^{o}[/tex].
All three bonds of P-H has [tex]109.5^{o}[/tex].
Therefore, ideal bond angle is [tex]109.5^{o}[/tex].
The electron-group arrangement of PH3 is tetrahedral, the molecular shape is trigonal pyramidal, and the ideal bond angle is 104.5°.
Explanation:In the case of PH3, the electron-group arrangement is tetrahedral and the molecular shape is trigonal pyramidal. The ideal bond angle for PH3 is 104.5°.
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Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl.
Calculate the after the following volumes of titrant have been added.
a)0 mL
b)20.0mL
c)59.2mL
d)60.0mL
e)71.3mL
f)73.3mL
Answer:
60.0mL
Explanation:
For the process of titration , the formula we use is -
M₁V₁ = M₂V₂
Where ,
M₁ = initial concentration
V₁ = initial volume
M₂ = final concentration
V₂ = final volume .
Hence , from the question ,
The given data is ,
M₁ = 0.050 M
V₁ = 30.0mL
M₂ = 0.025 M
V₂ = ?
Now, to determine the unknown quantity , the formula can be applied ,
Hence ,
M₁V₁ = M₂V₂
Putting the respective values ,
0.050 M * 30.0mL = 0.025 M * ?
solving the above equation ,
V₂ = ? = 60.0mL
This titration reaction occurs between a weak base (NH3) and a strong acid (HCl). The distinct pH stages refer to the volume of HCl added. Initially, the pH is determined by the NH3. However, as more HCl is added and reacts with NH3, the pH is determined by the balance of NH3 and NH4+ (from NH3 + HCl -> NH4Cl). If there’s an excess of HCl, the pH drops becoming more acidic.
Explanation:This question relates to the titration process in chemistry. Let's start by identifying the reactants: NH3 (a weak base) and HCl (a strong acid). The chemical reaction for this titration can be written as NH3 + HCl -> NH4Cl.
To calculate after specific volumes of titrant have been added, we must first understand the principle of titration. In titration, the reaction proceeds until one of the reactants is completely used up. The point at which this happens is known as the equivalence point.
a) At 0 mL HCl, no reaction has occurred so the pH is determined by the NH3 present.
b) At 20.0 mL of HCl, a partial reaction has occurred. The moles of NH3 originally present were 30.0 mL * 0.050 M = 1.5 mmol. After 20 mL of HCl added (0.5 mmol), we have 1.0 mmol NH3 left.
c) At 59.2 mL of HCl added, all the NH3 has reacted because the moles of HCl added are equal to the moles of NH3 initially present.
d) At 60.0 mL of HCl, an excess of HCl has been added and we have a solution of NH4Cl plus excess HCl.
e) At 71.3 mL and f) at 73.3 mL, we still have an excess of HCl, leading to a more acidic solution.
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Consider the reaction BF3 + NH3 → F3B―NH3
What changes in hybridization (if any) of the B atom are a result of this reaction?
A. Before the reaction boron is sp2 hybridized, and after the reaction it is sp3d hybridized.
B. Before the reaction boron is sp3 hybridized, and after the reaction it is sp3 hybridized.
C. Before the reaction boron is sp2 hybridized, and after the reaction it is sp3 hybridized.
What changes in hybridization (if any) of the N atom are a result of this reaction?
A. Before the reaction nitrogen is sp3 hybridized, and after the reaction it is sp2 hybridized.
B. Before the reaction nitrogen is sp3 hybridized, and after the reaction it is sp3 hybridized.
C. Before the reaction nitrogen is sp3d hybridized, and after the reaction it is sp3 hybridized.
Answer:
The correct answer is given below
Explanation:
B before the reaction boron is sp2 hybridized and after the reaction boron is it is sp3 hybridized.
Before the reaction boron is joined to 3 fluorine atoms by 3 covalent bonds between them one is s and rest of the 2 are p.That"s why before the reaction boron is sp2 hybridized.
Whereas after the reaction boron is joined to 3 fluorine atoms by 3 covalent bonds and one nitrogen atom by one co ordinate bond.That"s why after reaction boron is sp3 hybridized.
B before the reaction nitrogen is sp3 hybridized and after the reaction it is sp3 hybridized.
Before the reaction Nitrogen is joined to 3 hydrogen atoms by 3 covalent bonds and contain one lone pair of electron.That"s why before the reaction Nitrogen atom is sp3 hybridized.
whereas after the reaction Nitrogen is joined to 3 hydrogen atoms by 3 covalent bonds and with one boron atom by a co ordinate bond.That"s why after reaction Nitrogen atom is sp3 hybridized.
Boron atom's hybridization changes from sp2 to sp3 as a result of the reaction while Nitrogen's hybridization remains sp3 both before and after the reaction.
Explanation:In the given reaction, BF3 + NH3 → F3B―NH3, the hybridization of both Boron (B) and Nitrogen (N) atoms change. For the Boron atom, before the reaction, it is sp2 hybridized. After the reaction, when it bonds with NH3, it becomes sp3 hybridized. Hence, the change in Boron's hybridization as a result of this reaction is from sp2 to sp3.
For the Nitrogen atom, before the reaction, it is sp3 hybridized. After the reaction, it remains sp3 hybridized because it still maintains four regions of electron density. So, there is no change in Nitrogen's hybridization as a result of the reaction. It remains sp3 both before and after the reaction.
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A mixture of gaseous CO and H2, called synthesis gas, is used commercially to prepare methanol (CH3OH), a compound considered an alternative fuel to gasoline. Under equilibrium conditions at 550.3 K, [H2] = 0.07710 mol/L, [CO] = 0.02722 mol/L, and [CH3OH] = 0.0401 mol/L. What is the value of Kc for this reaction at 550.3 K?
Answer: The value of [tex]K_c[/tex] for the reaction at 550.3 K is 247.83
Explanation:
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{c}[/tex]
For a general chemical reaction:
[tex]aA+bB\rightarrow cC+dD[/tex]
The expression for [tex]K_{c}[/tex] is written as:
[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]
The chemical equation for the production of methanol follows:
[tex]CO+2H_2\rightleftharpoons CH_3OH[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[CH_3OH]}{[CO][H_2]^2}[/tex]
We are given:
[tex][CH_3OH]=0.0401mol/L[/tex]
[tex][CO]=0.02722mol/L[/tex]
[tex][H_2]=0.07710mol/L[/tex]
Putting values in above equation, we get:
[tex]K_c=\frac{0.0401}{0.02722\times (0.07710)^2}\\\\K_c=247.83[/tex]
Hence, the value of [tex]K_c[/tex] for the reaction at 550.3 K is 247.83
Final answer:
The equilibrium constant (Kc) for the synthesis of methanol from CO and H₂ at 550.3 K is 30.36, calculated using the given equilibrium concentrations.
Explanation:
The question concerns calculating the equilibrium constant (Kc) at 550.3 K for the synthesis of methanol from CO and H₂. The balanced chemical equation for this process is CO(g) + 2H₂(g) => CH₃OH(g). Given equilibrium concentrations: [H₂] = 0.07710 mol/L, [CO] = 0.02722 mol/L, and [CH₃OH] = 0.0401 mol/L, the equilibrium constant (Kc) can be calculated using the expression Kc = [CH₃OH]/([CO][H₂]²). Plugging in the given values yields Kc = (0.0401)/((0.02722)(0.07710)²) = 30.36.