Answer: The standard cell potential of the cell is -0.71 V
Explanation:
The half reactions follows:
Oxidation half reaction: [tex]Mg\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V[/tex] ( × 3)
Reduction half reaction: [tex]Al^{3+}(aq.)+3e^-\rightarrow Al(s);E^o_{Al^{3+}/Al}=-1.66V[/tex] ( × 2)
The balanced cell reaction follows:
[tex]2Al^{3+}(aq.)+3Mg(s)\rightarrow 2Al(s)+3Mg^{2+}(aq.)[/tex]
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation, we get:
[tex]E^o_{cell}=-2.37-(-1.66)=-0.71V[/tex]
Hence, the standard cell potential of the cell is -0.71 V
Nitrogen became the most abundant gas in today's atmosphere because Choose one: nitrogen is highly reactive. nitrogen is relatively inert. nitrogen was the most abundant gas left over from the solar nebula. nitrogen is the only gas given off by volcanic outgassing.
Answer:
The correct answer is: 'Nitrogen is relatively inert'.
Explanation:
As, atomic number of nitrogen is 7 and its electronic distribution is 2, 5. Hence, there are 5 valence electrons present in a nitrogen atom. And, to attain stability it will gain three electrons from a donor atom. Hence, it will make a triple bond.
[tex][N]=1s^22s^22p^3[/tex]
Therefore, nitrogen has 5 valence electrons and makes 3 bonds in neutral compounds.Thus nitrogen will combine with another nitrogen atom to completes its octet to form [tex]N_2[/tex] gas molecule. Due to the presence of this triple bond, the gas molecule is almost inert as its bond dissociation energy is very high.
Hence, the correct option is:- nitrogen is relatively inert.
Consider the reaction 2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.74 moles of Na(s) react at standard conditions. S°surroundings = J/K g
The entropy change for the surroundings in the given reaction can be calculated using the formula ΔS° surroundings = -ΔH° reaction / T.
Explanation:The entropy change for the surroundings in the given reaction can be calculated using the formula: ΔS° surroundings = -ΔH° reaction / T. In this case, the enthalpy change of the reaction is -802 kJ mol⁻¹ and the temperature is 298 K. Plugging in these values, we can calculate the entropy change for the surroundings.
ΔS° surroundings = -(-802 kJ mol⁻¹) / 298 K
ΔS° surroundings = 2.69 kJ K⁻¹ mol⁻¹
Therefore, the entropy change for the surroundings is 2.69 kJ K⁻¹ mol⁻¹.
Use the reaction:
2AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2H2O(l)
What volume of 0.123 M AgNO3(aq) is needed to form 0.657 g of Ag2SO4(s)?
(A) 53.4 mL
(B) 10.7 mL
(C) 17.1 mL
(D) 34.2 L
(E) 34.2 mL
Answer:
It’s either 34.2 OR 34.3 either would be correct!
Explanation:
The correct answer is (E) [tex]34.17\text{ mL}[/tex].
To solve this problem, we will use stoichiometry to relate the moles of [tex]AgNO_3[/tex] needed to the moles of Ag2SO4 formed. Here is the step-by-step solution:
1. First, we need to write down the balanced chemical equation:
[tex]\[ 2\text{AgNO}_3(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Ag}_2\text{SO}_4(s) + 2\text{H}_2\text{O}(l) \][/tex]
2. From the equation, we can see that [tex]2[/tex] moles of [tex]AgNO_3[/tex] produce 1 mole of [tex]Ag_2SO_4[/tex] This gives us the mole ratio:
[tex]\[ \frac{2\text{ moles AgNO}_3}{1\text{ mole Ag}_2\text{SO}_4} \][/tex]
3. Calculate the molar mass of [tex]Ag_2SO_4[/tex] using the atomic masses of Ag [tex](107.87 g/mol)[/tex] and [tex]S (32.07 g/mol)[/tex]
[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 2 \times 107.87\text{ g/mol (Ag)} + 32.07\text{ g/mol (S)} + 4 \times 16.00\text{ g/mol (O)} \][/tex]
[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 2 \times 107.87 + 32.07 + 4 \times 16.00 \][/tex]
[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 215.74 + 32.07 + 64.00 \][/tex]
[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 311.81\text{ g/mol} \][/tex]
4. Use the given mass of [tex]Ag_2SO_4[/tex] to find the moles of [tex]Ag_2SO_4[/tex] produced:
[tex]\[ \text{Moles of Ag}_2\text{SO}_4 = \frac{\text{mass of Ag}_2\text{SO}_4}{\text{molar mass of Ag}_2\text{SO}_4} \][/tex]
[tex]\[ \text{Moles of Ag}_2\text{SO}_4 = \frac{0.657\text{ g}}{311.81\text{ g/mol}} \][/tex]
[tex]\[ \text{Moles of Ag}_2\text{SO}_4 \approx 0.002097\text{ moles} \][/tex]
5. Using the mole ratio from step 2, calculate the moles of AgNO3 needed:
[tex]\[ \text{Moles of AgNO}_3 = 2 \times \text{Moles of Ag}_2\text{SO}_4 \][/tex]
[tex]\[ \text{Moles of AgNO}_3 = 2 \times 0.002097 \][/tex]
[tex]\[ \text{Moles of AgNO}_3 = 0.004194\text{ moles} \][/tex]
6. Now, we can use the molarity of AgNO3 to find the volume needed:
[tex]\[ \text{Molarity (M)} = \frac{\text{moles of solute (mol)}}{\text{volume of solution (L)}} \][/tex]
[tex]\[ \text{Volume of AgNO}_3 = \frac{\text{moles of AgNO}_3}{\text{Molarity of AgNO}_3} \][/tex]
[tex]\[ \text{Volume of AgNO}_3 = \frac{0.004194\text{ moles}}{0.123\text{ M}} \][/tex]
[tex]\[ \text{Volume of AgNO}_3 \approx 0.03417\text{ L} \][/tex]
7. Convert the volume from liters to milliliters (since 1 L = 1000 mL):
[tex]\[ \text{Volume of AgNO}_3 = 0.03417\text{ L} \times 1000\text{ mL/L} \][/tex]
[tex]\[ \text{Volume of AgNO}_3 = 34.17\text{ mL} \][/tex]
A chemist prepares a solution of potassium dichromate K2Cr2O7 by measuring out 3.46μmol of potassium dichromate into a 50.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /mmolL of the chemist's potassium dichromate solution. Round your answer to 2 significant digits.
Answer: 0.069 mmol/L
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute = [tex]3.46\mu mol[/tex] = [tex]0.001\times 3.46=3.46\times 10^{-3}mmol[/tex]
1micro (µ) = 0.001 milli (m)
[tex]V_s[/tex] = volume of solution in ml= 50 ml
[tex]Molarity=\frac{3.46\times 10^{-3}\times 1000}{50ml}=0.069mmol/L[/tex]
Thus concentration in mmol/L of the chemist's potassium dichromate solution is 0.069.
Final answer:
To calculate the concentration of the potassium dichromate solution in mmol/L, convert 3.46μmol to mmol, then divide by the volume in liters, resulting in a concentration of 0.069 mmol/L.
Explanation:
The question asks to calculate the concentration of a potassium dichromate solution in millimoles per liter (mmol/L) given that 3.46μmol of potassium dichromate is dissolved in a 50 mL volumetric flask.
First, convert micromoles (μmol) to millimoles (mmol). Recall that 1μmol = 0.001 mmol:
3.46μmol = 3.46 x 0.001 = 0.00346 mmol
Next, we need to find the concentration in mmol/L. To do this, we use the formula for concentration:
Concentration (mmol/L) = (amount of solute in mmol) / (volume of solution in L)
Since the volume of the solution is 50 mL, we convert this to liters (L) because concentration is expressed in mmol/L. Recall that 1000 mL = 1 L.
50 mL = 0.05 L
Now, we can calculate the concentration:
Concentration = 0.00346 mmol / 0.05 LConcentration = 0.0692 mmol/LAfter rounding to two significant digits, the concentration is 0.069 mmol/L.
For the reaction A + B − ⇀ ↽ − C + D , assume that the standard change in free energy has a positive value. Changing the conditions of the reaction can alter the value of the change in free energy ( Δ G ) . Classify the conditions as to whether each would decrease the value of Δ G , increase the value of Δ G , or not change the value of Δ G for the reaction. For each change, assume that the other variables are kept constant.
a. Adding a catalyst decrease the free energy value, increase the free energy value, or not change the free energyb. increasing [C] and [D] decrease the free energy value, increase the free energy value, or not change the free energyc. Coupling with ATP hydrolysis decrease the free energy value, increase the free energy value, or not change the free energyd.Increasing [A] and [B] decrease the free energy value, increase the free energy value, or not change the free energy
Final answer:
The conditions that can alter the value of ΔG for the reaction A + B ⇌ C + D are explained. Adding a catalyst does not change the value of ΔG. Increasing the concentrations of products decreases ΔG. Coupling with ATP hydrolysis decreases ΔG. Increasing the concentrations of reactants also decreases ΔG.
Explanation:
The reaction A + B ⇌ C + D has a positive standard change in free energy (ΔG°). Changing the conditions of the reaction can alter the value of ΔG. Let's classify the conditions:
a. Adding a catalyst: Adding a catalyst can decrease the activation energy and increase the rate of the reaction. However, it does not change the value of ΔG.
b. Increasing [C] and [D]: Increasing the concentrations of products (C and D) will shift the equilibrium towards the reactants and decrease the value of ΔG.
c. Coupling with ATP hydrolysis: Coupling the reaction with ATP hydrolysis, which is an exergonic reaction, can decrease the value of ΔG and make the overall reaction more spontaneous.
d. Increasing [A] and [B]: Increasing the concentrations of reactants (A and B) will shift the equilibrium towards the products and decrease the value of ΔG.
Which of the following conclusions can be drawn from JJ Thomson's cathode ray experiments?A.) Atoms contain electronsB.) Practically all the mass of an atom is contained in its nucleusC.) Atoms contain protons, neutrons, and electronsD.) Atoms have positively charged nuclues surrounded by electron cloud
Answer:
A.) Atoms contain electrons
Explanation:
J.J. Thomson conducted an experiment in which he took a gas at low pressure in a discharge tube and applied high voltage current. When he did so, he noticed that there are some particles emitting from cathode going towards the anode. Then he concluded that they are negatively charged particles and coined them electrons.
Hence, out of the options:- A.) Atoms contain electrons is correct.
According to Coulomb's Law, how does the force between a hydrogen nucleus and an electron in the n =2 level compare with that for one in the n = 1 level if the distance between the nucleus and the n = 2 level is twice as great as that between the nucleus and the n = 1 level?
one-quarter as strong
four times as strong
one-half as strong
twice as strong
Answer:
one-quarter as strong
Explanation:
Coulomb's law gives the mathematical expression to calculate the electrical force (F) between the charge of the nucleus (q₊) and the charge of an electron (q₋) separated by a distance (r).
[tex]F=\frac{k.q_{+}.q_{-}}{r^{2} }[/tex]
where,
k is the Coulomb's constant
The force between the nucleus and an electron in the level 1 is:
[tex]F_{1}=\frac{k.q_{+}.q_{-}}{r_{1}^{2} }[/tex]
Considering the distance to an electron from n = 2 is twice as great as the distance to an electron from n = 1, the force between the nucleus and an electron in the level 2 is:
[tex]F_{2}=\frac{k.q_{+}.q_{-}}{r_{2}^{2} }=\frac{k.q_{+}.q_{-}}{(2r_{1})^{2} }=\frac{1}{4} \frac{k.q_{+}.q_{-}}{(r_{1})^{2} }=\frac{1}{4}F_{1}[/tex]
530 mL of CO gas at a pressure of 1.2 atm and a temperature of 17ºC. What is the volume of the gas at 0ºC and 1 atm?
Answer:
598.74 mL
Explanation:
Using Ideal gas equation for same mole of gas as
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
Given ,
V₁ = 530 mL
V₂ = ?
P₁ = 1.2 atm
P₂ = 1 atm
T₁ = 17 ºC
T₂ = 0 ºC
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (17 + 273.15) K = 290.15 K
T₂ = (0 + 273.15) K = 273.15 K
Using above equation as:
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
[tex] \frac {{1.2}\times {530}}{290.15}=\frac {{1}\times {V_2}}{273.15}[/tex]
Solving for V₂ , we get:
V₂ = 598.74 mL
Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol − 1 −5.980 kJ·mol−1 . Calculate the equilibrium constant for the reaction at 25 °C. K ′ e q = Keq′= What is ΔG for the reaction at body temperature (37.0 °C) if the concentration of A is 1.8 M 1.8 M and the concentration of B is 0.55 M ?
The equilibrium constant for the reaction at 25 °C, [tex]K_{eq}[/tex] is 11.2.
ΔG for the reaction at body temperature (37.0 °C) is -9.04 kJ/mol.
Relation between equilibrium constant and standard Gibbs free energy:[tex]\triangle G^o= -RT*lnK_{eq}[/tex]
where,
[tex]\triangle G ^o[/tex] = standard Gibbs free energy = -5.980 kJ/mol = -5980 J/mol
R = gas constant = 8.314 J/K.mol
T = temperature = 298K
[tex]K_{eq}[/tex] = equilibrium constant = ?
→ Calculation for [tex]K_{eq}[/tex] :
Now, substituting the values in the above formula:
[tex]\triangle G^o= -RT*lnK_{eq}\\\\-5980J/mol=-(8.314J/K.mol)*(298K)* lnK_{eq}\\\\K_{eq}=11.2[/tex]
Thus, the value of [tex]K_{eq}[/tex] is 11.2.
→ Calculation for [tex]\triangle G_{rxn}[/tex]:
Chemical reaction:
A(aq) ⇌B(aq)
The formula used is:
[tex]\triangle G_{rxn}=\traingle G^o+RT lnQ\\\\\triangle G_{rxn}=\traingle G^o+RT ln\frac{[B]}{[A]}[/tex]
where,
[tex]\triangle G_{rxn}[/tex] = Gibbs free energy for the reaction = ?[tex]\triangle G^o[/tex] = standard Gibbs free energy = -30.5 kJ/molR = gas constant = 8.314 J/K.molT = temperature = 310 KQ = reaction quotient[A] = concentration of A = 1.8 M[B] = concentration of B = 0.55 MOn substituting the values in the above formula:
[tex]\triangle G_{rxn}=\traingle G^o+RT ln\frac{[B]}{[A]}\\\\\triangle G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)*(310K)*ln\frac{0.55}{1.8}] \\\\\triangle G_{rxn}=-9035.75J/mole=-9.04kJ/mol[/tex]
Therefore, the value of [tex]\triangle G_{rxn}[/tex] is -9.04 kJ/mol.
Find more information about Gibbs free energy here:
brainly.com/question/10012881
Balance the following equation: K2CrO4+Na2SO3+HCl→KCl+Na2SO4+CrCl3+H2O Generally coefficients of 1 are omitted from balanced chemical equations. When entering your answer, include coefficients of 1 as required for grading purposes. Enter the coefficients for each compound, separated by commas, in the order in which they appear in the equation (e.g., 1,2,3,4,5,6,7).
Answer:
The coefficients found are: 2,3,10,4,3,2,5
The sum of the coefficients is: 2+3+10 + 4 + 3 + 2 + 5 = 29
Explanation:
K2CrO4+Na2SO3+HCl→KCl+Na2SO4+CrCl3+H2O
On the left side we 2 times K , on the right side we have 1 time K
So we have to multiply KCl by 2
K2CrO4+Na2SO3+HCl→2KCl+Na2SO4+CrCl3+H2O
On the left side wehave 1 Cl, on the right side we have 5 Cl
So the left side we have to multiply by 5
K2CrO4+Na2SO3+5 HCl→2KCl+Na2SO4+CrCl3+H2O
On the left side we have 5 times H, on the right side, we have 2 times H
So on the left side the HCl we should multiply by 10 (instead of 5)
On the right side, we should multiply H2O by 5
K2CrO4+Na2SO3+ 10HCl→2KCl+Na2SO4+CrCl3+ 5H2O
So on the left side we have 10 times Cl, on the right side KCl and CrCL3 should be doubled
K2CrO4+Na2SO3+ 10HCl→4 KCl+Na2SO4+ 2CrCl3+ 5H2O
On the right side we have 4 times K, and 2 times Cr. This mean on the left side K2CrO4 should be multiplied by 2
2K2CrO4+Na2SO3+ 10HCl→4 KCl+Na2SO4+ 2CrCl3+ 5H2O
On the left side we have 8 times O plus 3x times O
On the right side we have 5 times O plus 4x times O
To equal this we should multiply on the left side Na2SO3 by 3
and Na2SO4 on the right side also by 3
Now the entire equation is balanced
2 K2CrO4 + 3 Na2SO3 + 10 HCl → 4 KCl + 3 Na2SO4 + 2 CrCl3 + 5 H2O
The coefficients found are: 2,3,10,4,3,2,5
The sum of the coefficients is: 2+3+10 + 4 + 3 + 2 + 5 = 29
Final answer:
To balance the chemical equation, coefficients are placed in front of the compounds to ensure the same number and types of atoms on both sides. The balanced equation is: 3K2CrO4 + 3Na2SO3 + 14HCl → 3KCl + 3Na2SO4 + 2CrCl3 + 7H2O and the coefficients are 3,3,14,3,3,2,7.
Explanation:
To balance the given chemical equation, we must ensure that the number of atoms of each element is the same on both sides of the equation. Let's add coefficients to balance:
Start with Chrome atoms: There is 1 Cr on the left side and 3 on the right side, so place a coefficient 3 in front of K2CrO4.
Next, look at sulfur atoms: There are 3 S on the left and 3 S on the right, so we are now balanced for sulfur.
For potassium atoms, put a coefficient 3 in front of KCl on the product side.
Balance sodium by having a coefficient 3 in front of Na2SO4 since there are 6 Na on the reactant side.
Hydrogen atoms are balanced by placing a coefficient 6 in front of HCl to get 6 H atoms on both sides.
Now, balance the chloride ions: We have 6 Cl from HCl, but we need 9 Cl on the products side (3 from KCl and 6 from CrCl3). So the final coefficient for HCl is 14.
Finally, place a coefficient 7 in front of H2O to balance out the oxygen atoms.
The balanced equation is therefore: 3K2CrO4 + 3Na2SO3 + 14HCl → 3KCl + 3Na2SO4 + 2CrCl3 + 7H2O. The coefficients in the order that they appear in the equation are: 3,3,14,3,3,2,7.
When 1 mole of H2CO(g) reacts with O2(g) to form CO2(g) and H2O(l) according to the following equation, 563 kJ of energy are evolved. H2CO(g) + O2(g)CO2(g) + H2O(l) Is this reaction endothermic or exothermic? What is the value of q? Write a balanced thermochemical equation
Final answer:
The reaction where 1 mole of H2CO(g) reacts with O2(g) to form CO2(g) and H2O(l) is exothermic, evolving -563 kJ of energy. This is represented in the balanced thermochemical equation with a negative enthalpy change (ΔH = -563 kJ).
Explanation:
When 1 mole of H2CO(g) reacts with O2(g) to form CO2(g) and H2O(l), the reaction is exothermic because energy is released. The value of q in this context represents the amount of energy evolved during the reaction, which is -563 kJ (negative sign indicates energy is released by the system). Therefore, the balanced thermochemical equation incorporating the heat of reaction would be as follows:
H2CO(g) + O2(g) → CO2(g) + H2O(l) ΔH = -563 kJ
An exothermic reaction is characterized by the release of energy to the surroundings, with a negative enthalpy change (ΔH), indicating that the products are at a lower energy level than the reactants. Conversely, an endothermic reaction absorbs energy, with a positive enthalpy change, indicating that the products are at a higher energy level than the reactants.
Final answer:
The reaction in question is exothermic, with the value of q is -563 kJ of energy being evolved. The balanced thermochemical equation reflects this by including a negative ΔH value.
Explanation:
The reaction of [tex]H_2CO[/tex] (g) with [tex]O_2[/tex] (g) to form [tex]CO_2[/tex] (g) and [tex]H_2O[/tex] (l) is an exothermic reaction because energy is released in the process.
The value of q for this reaction is -563 kJ, indicating that 563 kJ of energy are evolved, where the negative sign signifies the direction of energy flow out of the system. The balanced thermochemical equation for the reaction would be:
[tex]H_2CO[/tex] (g) + [tex]O_2[/tex] (g) → [tex]CO_2[/tex] (g) + [tex]H_2O[/tex] (l) ΔH = -563 kJ/mol
Here, the ΔH value represents the enthalpy change for the reaction, and since it is negative, the reaction is considered exothermic.
A fossil was analyzed and determined to have a carbon-14 level that is 70 % that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil? Express your answer with the appropriate units. View Available Hint(s) t t t = nothing nothing
Answer: 2948
Explanation:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}[/tex]
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]1.21\times 10^{-4}years^{-1}[/tex]
t = age of sample = ?
a = let initial amount of the reactant = 100
a - x = amount left after decay process = [tex]\frac{70}{100}\times 100=70[/tex]
[tex]t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}[/tex]
[tex]t=2948years[/tex]
Thus the fossil is 2948 years old.
The age of the fossil with 70% of the Carbon-14 level of living organism is 2948 years old
To determine the age of the fossil, we need to consider the decay of Carbon-14.
Given that the Carbon-14 level in the fossil is 70% that of living organisms, we can calculate the age using the concept of half-life.
Half life is the amount of time taken by a radioactive material to decay to half of its original value.t{(1)/(2)}=(0.693)/(k)k= {0.69}{t{(1)/(2)}}=(0.693)/(5730)=1.21* 10⁻⁴years⁻¹Expression for rate law for first order kinetics is given by:
t=(2.303)/(k)\log(a)/(a-x)where,
k = rate constant = 1.21* 10⁻⁴years⁻¹t = age of sample = ?a = let initial amount of the reactant = 100a - x = amount left after decay process = (70)/(100)* 100=70t=(2.303)/(1.21* 10⁻⁴)\log(100)/(70)t=2948yearsThus the fossil is 2948 years old.
A 30.0 mL sample of hydrogen gas (H2) is collected over water at 20.00∘C and has a total pressure of 700.0 torr. The partial pressure of water vapor at 20.00∘C is 17.5 torr. Calculate the mole fraction of H2 gas in the sample.
Answer: The mole fraction of hydrogen gas at 20°C is 0.975
Explanation:
We are given:
Vapor pressure of water vapor at 20°C = 17.5 torr
Total pressure at 20°C = 700.0 torr
Vapor pressure of hydrogen gas at 20°C = (700.0 - 17.5) torr = 682.5 torr
To calculate the mole fraction of hydrogen gas at 20°C, we use the equation given by Raoult's law, which is:
[tex]p_{H_2}=p_T\times \chi_{H_2}[/tex]
where,
[tex]p_{H_2}[/tex] = partial pressure of hydrogen gas = 682.5 torr
[tex]p_T[/tex] = total pressure = 700.0 torr
[tex]\chi_{H_2}[/tex] = mole fraction of hydrogen gas = ?
Putting values in above equation, we get:
[tex]682.5torr=700.0torr\times \chi_{H_2}\\\\\chi_{H_2}=\frac{682.5}{700.0}=0.975[/tex]
Hence, the mole fraction of hydrogen gas at 20°C is 0.975
Answer: 0.975
Explanation:
First, calculate the partial pressure of hydrogen gas. The partial pressure of hydrogen gas, PH2, is the difference between the total pressure and the partial pressure of water vapor.
PH2=700.0 torr−17.5torr=682.5torr
Now, use the following equation to calculate the mole fraction of hydrogen gas.
PH2 = XH2 × Ptotal
XH2 = PH2 / Ptotal
= 682.5 torr / 700.0 torr = 0.975
What is the wavelength (in nm) of the photon absorbed for a transition of an electron from ninitial = 2 that results in the least energetic spectral line in the visible series of the H atom?
Answer: 656.6 nm.
Explanation:
Using Rydberg's Equation for hydrogen atom:
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
Where,
[tex]\lambda[/tex] = Wavelength of radiation
[tex]R_H[/tex] = Rydberg's Constant
[tex]n_f[/tex] = Higher energy level = 3 (least energetic for visible series)
[tex]n_i[/tex]= Lower energy level = 2
We have:
[tex]n_f=3, n_i=2[/tex]
[tex]R_H=1.097\times 10^7 m^{-1}[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \frac{5}{36}[/tex]
[tex]\frac{1}{\lambda}=0.1523\times 10^{7} m[/tex]
[tex]\lambda=6.566\times 10^{-7}m=656.6nm[/tex]
([tex]1 m= 10^9nm[/tex])
The wavelength of the photon emitted when the hydrogen atom undergoes a transition from n = 2 to n = 3 is 656.6 nm.
At 20°C, the ion-product constant of water, Kw, is 6.88 x 10-15 . What is the pH of pure water at 20°C?
A) 7.000
B) 6.501
C) 7.181
D) 7.081
E) none of these
Answer:
The correct answer is D) 7.081
Explanation:
The water equilibrium is the following:
H₂O ⇄ H⁺ + OH⁻ Kw
Where Kw= [H⁺] x [OH⁻]
In pure water: [H⁺] = [OH⁻]= C
If we introduce C in the expression for Kw:
Kw= [H⁺] x [OH⁻]= C x C= C²
⇒ C= [tex]\sqrt{Kw}[/tex]= [tex]\sqrt{6.88 x 10^{-15} }[/tex]= 8.29 x 10⁻⁸
pH= -log [H⁺] = -log C = -log (8.29 x 10⁻⁸) = 7.081
Final answer:
At 20°C, the pH of pure water is 7.000.
Explanation:
The ion-product constant of water (Kw) is the product of the concentrations of hydrogen ions (H3O+) and hydroxide ions (OH-) in water. At 20°C, the value of Kw is 6.88 x 10^-15. Since pure water is neutral, the concentration of hydrogen ions and hydroxide ions in pure water are equal. Therefore, at 20°C, the concentration of hydrogen ions and hydroxide ions in pure water is 6.88 x 10^-15 M.
The pH of a solution is a measure of the concentration of hydrogen ions. It is calculated as the negative logarithm (base 10) of the hydrogen ion concentration. Therefore, at 20°C, the pH of pure water is 7.000 (approximately) since the hydrogen ion concentration and the hydroxide ion concentration in pure water are both 6.88 x 10^-15 M.
What is the [H3O+] and the pH of a benzoic acid-benzoate buffer that consists of 0.17 M C6H5COOH and 0.42 M C6H5COONa? (Ka of benzoic acid = 6.3 × 10^−5). Be sure to report your answer to the correct number of significant figures.
Answer:
[tex][H_{3}O^{+}]=x M = 2.5\times 10^{-5}M[/tex] and pH = 4.6
Explanation:
Construct an ICE table to calculate changes in concentration at equilibrium.
[tex]C_{6}H_{5}COOH+H_{2}O\rightleftharpoons C_{6}H_{5}COO^{-}+H_{3}O^{+}[/tex]
I(M): 0.17 0.42 0
C(M): -x +x +x
E(M): 0.17-x 0.42+x x
So, [tex]\frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}=K_{a}(C_{6}H_{5}COOH)[/tex]
or, [tex]\frac{(0.42+x)x}{(0.17-x)}=6.3\times 10^{-5}[/tex]
or, [tex]x^{2}+0.4201x-(1.071\times 10^{-5})=0[/tex]
So, [tex]x=\frac{-0.4201+\sqrt{(0.4201)^{2}+(4\times 1\times 1.071\times 10^{-5})}}{(2\times 1)}M[/tex]
([tex]ax^{2}+bx+c=0\Rightarrow x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a};x< 0.17M[/tex])
So, [tex]x=2.5\times 10^{-5}[/tex]M
Hence [tex][H_{3}O^{+}]=x M = 2.5\times 10^{-5}M[/tex]
[tex]pH=-log[H_{3}O^{+}]=-logx=-log(2.5\times 10^{-5})=4.6[/tex]
The pH is 4.44, and the [H3O+] is 3.63 × 10^-5 M.
The question asks about the calculation of the hydronium ion concentration ([H3O+]) and the pH of a buffer solution consisting of benzoic acid (C6H5COOH) and sodium benzoate (C6H5COONa).
Write the equilibrium expression for benzoic acid dissociation:
C6H5COOH ↔ C6H5COO- + H3O+.
Use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA]),
where [A-] = concentration of benzoate ion and [HA] = concentration of benzoic acid.
Insert the values into the Henderson-Hasselbalch equation:
pH = 4.20 + log (0.42/0.17).
Calculate the pH.
Determine [H3O+] from pH:
[H3O+] = 10-pH.
Using the provided concentrations and the pKa of benzoic acid (4.20), the pH and [H3O+] can be calculated with high precision. For this specific buffer:
The pH is calculated to be 4.44.
The [H3O+] can then be found as 3.63 × 10-5 M.
UsingLewissymbolsandLewisstructures,diagramthefor- mation of SiCl4 from Si and Cl atoms, showing valence-shell electrons. (a) How many valence electrons does Si have ini- tially? (b) How many valence electrons does each Cl have initially? (c) How many valence electrons surround the Si in the SiCl4 molecule? (d) How many valence electrons sur- round each Cl in the SiCl4 molecule? (e) How many bonding pairs of electrons are in the SiCl4 molecule?
Answer:
(a) 4
(b) 7
(c) 8
(d) 8
(e) 4
Explanation:
In order to draw the Lewis structure, we need to take into account the octet rule: atoms will share electrons to have 8 electrons in their valence shell.
(a) How many valence electrons does Si have initially?
Si has 4 valence electrons. It has to share 4 pairs of electrons to reach the octet.
(b) How many valence electrons does each Cl have initially?
Cl has 7 valence electrons. Each Cl has to share 1 pair of electrons to reach the octet.
(c) How many valence electrons surround the Si in the SiCl₄ molecule?
Si is surrounded by 8 electrons in SiCl₄.
(d) How many valence electrons surround each Cl in the SiCl₄ molecule?
Each Cl is surrounded by 8 electrons in SiCl₄.
(e) How many bonding pairs of electrons are in the SiCl₄ molecule?
There are 4 bonding pairs between Si and Cl.
Silicon (Si) forms SiCl₄ by bonding with four chlorine (Cl) atoms. Si initially has 4 valence electrons, while each Cl has 7. In SiCl₄, Si has 8 valence electrons, and each Cl has 8. There are four bonding pairs of electrons in the molecule, formed by single bonds between Si and Cl.
Lewis Structure for the Formation of SiCl₄:
(a) **Valence Electrons in Si (Silicon):** Silicon is in Group 14 of the periodic table, so it has 4 valence electrons.
(b) **Valence Electrons in Each Cl (Chlorine) Atom:** Chlorine is in Group 17, so each Cl atom has 7 valence electrons.
(c) **Valence Electrons Surrounding Si in SiCl₄ Molecule:** In SiCl₄, Silicon forms four single bonds with four Cl atoms. Each bond involves two electrons. Therefore, around Silicon, there are 4 * 2 = 8 valence electrons.
(d) **Valence Electrons Surrounding Each Cl in SiCl₄ Molecule:** Each Cl atom is involved in one bond with Si, contributing one pair of electrons. Therefore, each Cl has a total of 8 valence electrons around it (7 from its own valence shell plus 1 from the shared bond).
(e) **Bonding Pairs of Electrons in SiCl₄ Molecule:** Since Si forms four single bonds with four Cl atoms, there are four bonding pairs of electrons in the SiCl₄ molecule.
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All of the following statements are consistent with the kinetic molecular theory of gases EXCEPT:
A. The gas molecules collide with each other and with the surfaces around them.
B. Strong attractive forces hold the gas molecules together.
C. The average kinetic energy of the molecules of a gas is proportional to the temperature of the gas in kelvins.
D. The size of the gas molecules is negligible compared to the total volume of the gas.
E. none of the above.
Answer:
B. Strong attractive forces hold the gas molecules together.
Explanation:
Kinetic molecular theory postulates:-
The gas is composed of small molecules are they are in continuous random motion and having elastic collisions with one another and also with the walls of the container. The molecules of the gas does not exert any kind of repulsive or attractive forces on each other and they their size is negligible as compared to the difference between them. Pressure exerted by the molecules of the gas results from the collisions which is happening between the molecules of the gas and the walls of the container. Average kinetic energy of molecules of the gas is directly proportional to absolute temperature.Hence, The correct option which is not a postulate of kinetic molecular theory is:- B. Strong attractive forces hold the gas molecules together.
Identify which redox reactions occur spontaneously in the forward direction. Check all that apply.
a.Fe(s)+Mn2+(aq)→Fe2+(aq)+Mn(s)
b. 2Ag+(aq)+Fe(s)→2Ag(s)+Fe2+(aq)
c. Mg2+(aq)+Zn(s)→Mg(s)+Zn2+(aq)
d. 2Al(s)+3Pb2+(aq)→2Al3+(aq)+3Pb(s)
Answer:
For a: The reaction is not spontaneous.
For b: The reaction is spontaneous.
For c: The reaction is not spontaneous.
For d: The reaction is spontaneous.
Explanation:
For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.
Relationship between standard Gibbs free energy and standard electrode potential follows:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
For a reaction to be spontaneous, the standard electrode potential must be positive.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex] .......(1)
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
For a:The chemical reaction follows:
[tex]Fe(s)+Mn^{2+}(aq.)\rightarrow Fe^{2+}(aq.)+Mn(s)[/tex]
We know that:
[tex]E^o_{Fe^{2+}/Fe}=-0.44V\\E^o_{Mn^{2+}/Mn}=-1.18V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-1.18-(-0.44)=-0.74V[/tex]
As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.
For b:The chemical reaction follows:
[tex]Fe(s)+2Ag^{+}(aq.)\rightarrow Fe^{2+}(aq.)+2Ag(s)[/tex]
We know that:
[tex]E^o_{Fe^{2+}/Fe}=-0.44V\\E^o_{Ag^{+}/Ag}=0.80V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=0.80-(-0.44)=1.24V[/tex]
As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.
For c:The chemical reaction follows:
[tex]Zn(s)+Mg^{2+}(aq.)\rightarrow Zn^{2+}(aq.)+Mg(s)[/tex]
We know that:
[tex]E^o_{Zn^{2+}/Zn}=-0.76V\\E^o_{Mg^{2+}/Mg}=-2.37V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-2.37-(-0.76)=-1.61V[/tex]
As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.
For d:The chemical reaction follows:
[tex]2Al(s)+3Pb^{2+}(aq.)\rightarrow 2Al^{3+}(aq.)+3Pb(s)[/tex]
We know that:
[tex]E^o_{Al^{3+}/Al}=-1.66V\\E^o_{Pb^{2+}/Pb}=-0.13V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-0.13-(-1.66)=1.53V[/tex]
As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.
The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning the formation and decomposition of ozone in the stratosphere. Rowland and Molina hypothesized that chlorofluorocarbons (CFCs) in the stratosphere break down upon exposure to UV radiation, producing chlorine atoms. Chlorine was previously identified as a catalyst in the breakdown of ozone into oxygen gas. Using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine with ozone. ( 1 ) ClO ( g ) + O 3 ( g ) ⟶ Cl ( g ) + 2 O 2 ( g ) Δ H ∘ rxn = − 122.8 kJ ( 2 ) 2 O 3 ( g ) ⟶ 3 O 2 ( g ) Δ H ∘ rxn = − 285.3 kJ ( 3 ) O 3 ( g ) + Cl ( g ) ⟶ ClO ( g ) + O 2 ( g ) Δ H ∘ rxn = ?
Answer:
The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.
Explanation:
[tex]ClO ( g ) + O_3 ( g )\rightarrow Cl ( g ) + 2 O_2 ( g ),\Delta H^o_{1,rxn} =-122.8 kJ [/tex]..[1]
[tex]2 O_3 ( g )\rightarrow 3O_2 ( g ),\Delta H^o_{2,rxn} = -285.3 kJ [/tex]..[2]
[tex]O_3(g) + Cl(g)\rightarrow ClO (g)+O_2(g),\Delta H^o_{3,rxn}=?[/tex]..[3]
The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:
[2] - [1] = [3]
[tex]\Delta H^o_{3,rxn}=\Delta H^o_{2,rxn}-\Delta H^o_{1,rxn}[/tex]
[tex]=-285.3 kJ-(-122.8 kJ)=162.5 kJ[/tex]
The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.
In an aqueous mixture of aluminum, lead, and iron salts, which of these will be reduced first upon application of an electric current?
(A) Al3+ + 3e− → Al (s) Eo = −1.66V
(B) Pb2+ + 2e− → Pb (s) Eo = −0.13V
(C) Fe2+ + 2e− → Fe (s) Eo = −0.45V
(D) Fe3+ + 3e− → Fe (s) Eo = −0.036V
Answer:
D) Fe3+ + 3e− → Fe (s) Eo = −0.036V
Explanation:
In electrolysis we have to add electrical energy for redox reactions non-spontaneous because ΔºG is positive.
These 4 reduction reactions are all non-spontaneous because their reduction potentials are negative (ΔGº = - nFεº , ΔGº will be positive). So the ion most easily reduced is the least negative, in this case Fe3+ + 3e− → Fe (s) Eo = −0.036V.
Use Appendix E in the textbook. Which of the following substances can oxidize RuO42−(aq) to RuO4−(aq) under standard conditions? Use Appendix E in the textbook. Which of the following substances can oxidize to under standard conditions? Ni2+(aq) I2(s) Cr2O72−(aq) ClO3−(aq) Pb2+(aq)
Answer:
Cr₂O₇⁻²(aq) and ClO₃⁻(aq)
Explanation:
At a redox reaction, one substance must be reduced (gain electrons) and others must be oxidized (lose electrons). To evaluate the potential of the substance to be reduced, it's placed a reaction, in standards conditions, with H₂.
The potential reduction is quantified by E°, and as higher is the value of E°, as easy is to the compound to be reduced. So, at a redox reaction, the compound with the greatest E° will be reduced, and the other will be oxidized, in a spontaneous reaction. The values of E° are:
RuO₄⁻(aq) to RuO₄²⁻(aq) E° = + 0.59 V (the reduction reaction is the opposite of the oxidation reaction).
Ni⁺²(aq) E° = -0.257 V
I₂(s) E° = +0.535 V
Cr₂O₇⁻²(aq) E° = +1.33 V
ClO₃⁻(aq) E° = +0.890 V
Pb²⁺(aq) E° = -0.125 V
So, the substances that have E° higher than the E° of the RuO₄⁻²(aq) are Cr₂O₇⁻²(aq) and ClO₃⁻(aq), which are the substances that can oxidize RuO₄⁻(aq) to RuO₄²⁻(aq).
The correct substances that can oxidize RuO42(aq) to RuO4(aq) under standard conditions are Cr2O72(aq) and ClO3(aq).
To determine which substances can act as oxidizing agents for RuO42−(aq), we need to look at the standard reduction potentials (E°) for the given half-reactions. An oxidizing agent is a substance that accepts electrons, and for a redox reaction to occur spontaneously, the substance with the higher reduction potential will oxidize the substance with the lower reduction potential.
The half-reaction for the reduction of RuO4’(aq) to RuO4(aq) is as follows:
[tex]\[ \text{RuO}_4(aq) + e^- \rightarrow \text{RuO}_4^{2-}(aq) \][/tex]
We do not have the standard reduction potential for this half-reaction, but we can compare the potentials of the given substances to determine if they can act as oxidizing agents.
From Appendix E, we can find the standard reduction potentials for the given substances:
1.[tex]\( \text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \) with \( E^\circ = -0.25 \) V[/tex]
2.[tex]\( \text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq) \) with \( E^\circ = +0.54 \) V[/tex]
3.[tex]\( \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \) with \( E^\circ = +1.33 \) V[/tex]
4. [tex]\( \text{ClO}_3^-(aq) + 6\text{H}^+(aq) + 6e^- \rightarrow \text{Cl}^-(aq) + 3\text{H}_2\text{O}(l) \) with \( E^\circ = +1.45 \) V[/tex]
5.[tex]\( \text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \) with \( E^\circ = -0.13 \) V[/tex]
Comparing these potentials with the half-reaction for RuO42(aq), we can see that Ni2+(aq) and Pb2+(aq) have negative reduction potentials, which means they cannot oxidize RuO42(aq) because they are stronger reducing agents than RuO42(aq).
I2(s) has a positive reduction potential, but it is lower than the reduction potentials of Cr2O72(aq) and ClO3(aq). Therefore, I2(s) can oxidize RuO42’(aq), but there are stronger oxidizing agents available.
Cr2O72(aq) and ClO3(aq) have the highest reduction potentials in the list, which means they are the strongest oxidizing agents and can oxidize RuO4(aq) to RuO4(aq) under standard conditions.
Thus, the correct substances that can oxidize RuO42(aq) to RuO4(aq) under standard conditions are Cr2O72(aq) and ClO3(aq)
The last step of glycolysis converts phosphoenolpyruvate to pyruvate. phosphoenolpyruvate + ADP pyruvate kinase −−−−−−−−−→ pyruvate + ATP Δ G = − 33 kJ / mol ( − 7.5 kcal / mol ) The Δ G ° ' of the reverse reaction is + 31 kJ / mol ( + 7.5 kcal / mol ) . Instead of reversing the pyruvate kinase reaction, the step is bypassed in gluconeogenesis. Several steps for the conversion of pyruvate to phosphoenolpyruvate via gluconeogenesis are given. Place the steps in the correct order. You will not place all of the steps.
Answer:
Steps
i) In thee mitochondrion, pyruvate carboxylase converts pyruvate to oxloacetate.
ii) Malate dehydrogenase in the mitochondrion reduces oxaloacete to to malate.
iii) Malate dehydrogenase in the cytoplasm oxidizes malate to oxaloacetate.
iv) Phosphoenolpyruvate carboxykinase decarboxylates and phosphorylates oxaloacetate.forming phosphoenolpyruvate.
Explanation:
Gluconeogenesis is the synthesis of glucose from non - carbohydrate compounds. The substrates for gluconeogenesis are lactate, pyruvate, amino acids, propionate and glycerol.
Gluconeogenesis occurs only in cytosol but the precursor is produced in mitochondria. In the conversion of pyruvte to phosphoenolpyruvate occur in mitochondria and cytosol.
Step -1:
Pyruvate carboxylase is a biotin dependent enzyme located in mitochondria. It converts pyruvate to oxlaoacetate and carbondioxide in the presence of ATP.Oxlaocetate synthesized in mitochondrial matrix has to be transported to cytosol for gluconeogenesis. Oxaloacetate is impermeble, cannot be sent out of mitochondria. So it has to be converted to malate.
Step -2:
Malate dehydrogenase in mitochondria converts oxaloacetate synthesized in mitochondrial matrix to malate. And then it is transported to cytosol.
Step 3:
Malate dehydrogenase responsible for reversible reaction in cytosol converts malate to oxaloacetate.
Step -4
Phosphoenolpyruvate carboxy-kinase in cytosol converts oxaloacetate to PEP. The enzyme transfer high energy phosphate bond from GTP to oxaloacetate to from PEP and liberated carbondioxide.
Therefore, the steps of glycolysis converts phosphoenolpyruvate to pyruvate are as follows.
i) In thee mitochondrion, pyruvate carboxylase converts pyruvate to oxloacetate.
ii) Malate dehydrogenase in the mitochondrion reduces oxaloacete to to malate.
iii) Malate dehydrogenase in the cytoplasm oxidizes malate to oxaloacetate.
iv) Phosphoenolpyruvate carboxykinase decarboxylates and phosphorylates oxaloacetate.forming phosphoenolpyruvate.
Gluconeogenesis uses pyruvate as a starting point to generate glucose, going through several steps that largely mirror glycolysis in reverse order. However, the first step in gluconeogenesis varies, converting pyruvate into oxaloacetate before continuing with the conversion into phosphoenolpyruvate. It's a complex process that allows the cell to regulate the production and use of glucose efficiently.
Explanation:The process of converting pyruvate to phosphoenolpyruvate via gluconeogenesis is not a simple reversal of the glycolysis steps. It starts with the conversion of pyruvate to oxaloacetate. Oxaloacetate serves as a substrate for the enzyme phosphoenolpyruvate carboxykinase (PEPCK), which transforms oxaloacetate into phosphoenolpyruvate (PEP). From this step, gluconeogenesis nearly mimics glycolysis, but in reverse. PEP is converted back into 2-phosphoglycerate, then into 3-phosphoglycerate. Then, 3-phosphoglycerate is converted into 1,3 bisphosphoglycerate and then into glyceraldehyde-3-phosphate. Next, two molecules of glyceraldehyde-3-phosphate combine to form fructose-1-6-bisphosphate, which is then converted into fructose 6-phosphate and then into glucose-6-phosphate. Finally, a series of reactions generates glucose. This sequence enables glycolysis and gluconeogenesis to be regulated independently of each other.
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Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the equation: CO2(s)→CO2(g). When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 L of water heated to 90 ∘C. Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 28 ?
Final answer:
To calculate the mass of dry ice needed to completely sublime in water, we can use the concept of stoichiometry and the ideal gas law equation. The moles of CO2 can be determined using the ideal gas law equation, and then converted to moles of water using the stoichiometry of the balanced equation. Finally, the mass of dry ice can be calculated using the moles of CO2 and the molar mass of CO2.
Explanation:
In order to calculate the mass of dry ice that should be added to the water, we need to use the concept of stoichiometry. The equation for the sublimation of carbon dioxide is CO2(s) -> CO2(g). We can use the molar ratio between CO2(s) and CO2(g) to convert the volume of water to moles of CO2. From there, we can use the molar mass of CO2 to calculate the mass of dry ice needed.
First, we need to calculate the moles of CO2 that can be produced using the ideal gas law equation:
n = PV / RT
where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature. Since temperature and pressure are not given, we cannot directly solve for moles of CO2. However, we can assume that the pressure and temperature remain constant throughout the process, so the ideal gas equation can still be used to find the moles of CO2.
The next step is to use stoichiometry to convert moles of CO2 to moles of water. The balanced equation is:
CO2(s) + H2O(l) -> H2CO3(aq)
From the balanced equation, we can see that the mole ratio between CO2 and water is 1:1. Therefore, if we know the moles of CO2, we also know the moles of water. The molar mass of CO2 is 44.01 g/mol, so we can calculate the mass of dry ice needed using the equation:
mass = moles * molar mass
Since we know the volume of water (15.0 L) and the final temperature (28°C), we can plug those values into the ideal gas equation and calculate the moles of CO2. Then, we can use the moles of CO2 to calculate the mass of dry ice needed. The final mass will depend on the pressure and temperature at which the dry ice is added, as well as the final temperature of the water.
Final answer:
The student's question involves calculating the mass of dry ice needed to produce fog in water until the temperature drops to 28°C. To perform this calculation, details such as the heat capacity of water, the heat of sublimation of CO2, and the complete temperature value are critical, but these are not provided in the question.
Explanation:
Dry ice, known chemically as solid carbon dioxide (CO2), sublimes at room temperature and pressure, meaning it converts directly from solid to gas. In the context of the question, when dry ice is added to warm water, the heat transfer from the water to the dry ice accelerates this sublimation process. The interaction cools the water down while providing a fog effect until either all the dry ice sublimes or the water temperature drops too much to sustain rapid sublimation.
To calculate the mass of dry ice required to create these special effects until the water cools to 28°C, one would need to determine the amount of heat the water can provide before reaching this temperature, and then use the heat of sublimation for CO2 to find the corresponding mass of dry ice. However, since the question doesn't provide specific heat capacity, heat of sublimation values, or the completion temperature value (presumably 28°C), an exact calculation cannot be performed without this information.
Handling dry ice requires caution due to its extremely cold temperature, and it is essential to be in a well-ventilated area since the resulting CO2 gas is heavier than air and can displace oxygen.
All of the following statements correctly describe a contemporary approach to create the Work Breakdown Structure (WBS) for a project EXCEPT:
a. Project teams may use a template or a WBS from a previous project as a starting point.
b. Project teams may jump start the WBS construction with a brainstorming session.
c. Project teams may adopt a top-down approach to construct the WBS.
d. Project teams may rely upon the project manager to create all levels of the WBS.
Answer: option d
Explanation:
Project teams must be proactive. Each member of the team should empowered and each member can share ideas and experience in order to make the WBS more precise.
below regarding an electrochemical cell in an automotive lead-acid battery. The cell's anode is made of lead and the cathode is made of lead(IV) oxide. Both are submerged in 4.30 M sulfuric acid. The half-reactions are: PbO_2(s) + 3H^+ (aq) + HSO_4^- (aq) + 2e^- rightarrow PbSO_4(s) + 2H_2O(l) E degree = 1.685 V PbSO_4(s) + H^+(aq) + 2e^- rightarrow Pb(s) + HSO_4^- (aq) E degree = -0.356 V (a) Calculate the value of E degree. (b) Determine the initial value of E_cell. Assume that the first ionization of H_2SO_4 is complete and that [H^+] almostequalto [HSO_4^-]. (c) Find E_cell when the H^+ concentration has dropped by 76.00%. Again, assume [H^+] almostequalto [HSO_4^-].
Answer:
a. 2.041 V
b. 2.116 V
c. 2.043 V
Explanation:
a.
In an electrochemical cell;
reduction occurs at the cathode & oxidation at the anode.
Thus, from the question; reduction at cathode=
[tex]PbO_{2(s)} + 3H_{(aq)}^{+} +HSO_{4}_{(aq)} +2e^{-}[/tex] → [tex]PbSO_{4(s)} + 2H_{2} O_{(l)} E^{0}_{(cathode)} = 1.685V[/tex]
Oxidation at anode;
[tex]Pb_{s} + HSO^{-}_{4}[/tex] → [tex]PbSO_{4(s)}+H^{+}+2e^{-} E^{0}_{anode} =0.356V[/tex]
The net process in the Cell is obtained by the algebraic sum of the two half-cell reactions:
Overall reaction:
[tex]PbO_{2(s)}+Pb_{s}+2HSO^{-}_{4(aq)}+ 2 H^{+}_{(aq)}[/tex] → [tex]PbSO_{4(s)} + 2H_{2} O_{(l)}[/tex]
[tex]E^{0} _{cell} = E^{0} _{cathode}- E^{0} _{anode}[/tex]
= 1.685V - (- 0.356V)
= 2.041V
b.
To determine the initial value of E-cell :
In the electrochemical cell; both the anode and the cathode were submerged into 4.30M of Sulfuric acid; therefore:
[tex]H_{2} SO_{4}[/tex] → [tex]H^{+} + HSO_{4}^{-}[/tex]
Assuming:
[tex]H^{+} = HSO_{4}^{-}[/tex] such that (;) [tex]H_{2} SO_{4}[/tex] = 4.30M
let : a= [tex]H^{+}[/tex] & b= [tex]HSO_{4}^{-}[/tex]
[tex]E^{i} _{cell}[/tex] =
[tex]E^{0} _{cell}[/tex] - [tex]\frac{0.0591}{2}[/tex] × log [tex]\frac{1}{(a)^{2} (b)^{2} }[/tex]
=2.041V - [tex]\frac{0.0591}{2}[/tex] × log [tex]\frac{1}{(4.30)^{2} (4.30)^{2} }[/tex]
= 2.041V - 0.02955 × log [tex]\frac{1}{341.8801}[/tex]
= 2.041V - 0.02955 × log [tex]{0.0029250026}[/tex]
= 2.041V - 0.02955 × (-2.5339)
= 2.041V + 0.075V
=2.116V
c.
If concentration of [tex]H^{+}[/tex] is dropped by 76.00%
[tex]H^{+} = HSO_{4}^{-}[/tex] = 4.30M
= 4.30M - 4.30M [tex]\frac{76}{100}[/tex]
= 1.032M
[tex]E_{cell}[/tex] = 2.041V - [tex]\frac{0.0591}{2}[/tex] × log [tex]\frac{1}{(1.032)^{2} (1.032)^{2} }[/tex]
=2.041V - 0.02955 × log [tex]\frac{1}{1.13427612058}[/tex]
=2.041V - 0.02955 × log (0.88161954735)
=2.041V - 0.02955 (- 0.0547)
=2.041V + (0.0020)
= 2.043V
The standard electrode potential (E°) of the cell in the lead-acid battery is 2.041 V. The initial value of E_cell in the same condition is also 2.041 V. As [H+] falls by 76%, E_cell can be calculated with the Nernst Equation.
Explanation:The electrochemical cell in a lead-acid battery consists of an anode made of lead and a cathode made of lead(IV) oxide, both submerged in sulfuric acid. The half reactions are as follows:
PbO_2(s) + 3H+ (aq) + HSO_4- (aq) + 2e- → PbSO_4(s) + 2H_2O(l) with E° = 1.685 VPbSO_4(s) + H+(aq) + 2e- → Pb(s) + HSO_4- (aq) with E° = -0.356 V(a) You can calculate the cell's standard electrode potential (E°) by subtracting the anode potential from the cathode potential, as per the formula E°cell = E°cathode - E°anode. Therefore, E° = 1.685 V - (-0.356 V) = 2.041 V.
(b) The initial value of E_cell is the same as the standard electrode potential, since initially [H+] = [HSO_4-] = 1M as per the standard conditions, which means E_cell = E° = 2.041 V.
(c) If the H+ concentration drops by 76%, it becomes 0.24M. For calculating the new E_cell we have to utilize the Nernst Equation, which is E=E°-(0.059/n)log(Q), where n is the number of electrons transferred (in this case 2), and Q is the reaction quotient ([Pb][HSO4-])/([PbSO4][H+]^2). Plugging in the values, we can determine the new E_cell.
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At 25.0 ⁰C the henry's law constant for hydrogen sulfide(H2S) gas in water is 0.087 M/atm. Caculate the mass in grams of H2S gas that can be dissolved in 400.0 ml of water at 25.00 C and a H2S partial pressure of 2.42atm.
Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{H_2S}=K_H\times p_{liquid}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]0.087M/atm[/tex]
[tex]p_{H_2S}[/tex] = partial pressure of hydrogen sulfide gas = 2.42 atm
Putting values in above equation, we get:
[tex]C_{H_2S}=0.087M/atm\times 2.42atm\\\\C_{H_2S}=0.2105M[/tex]
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of solution = 0.2105 M
Molar mass of hydrogen sulfide = 34 g/mol
Volume of solution = 400.0 mL
Putting values in above equation, we get:
[tex]0.2105M=\frac{\text{Mass of hydrogen sulfide}\times 1000}{34g/mol\times 400.0mL}\\\\\text{Mass of }H_2S=\frac{0.2105\times 34\times 400}{1000}=2.86g[/tex]
Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.
Select the correct value for the indicated bond angle in each of the following compounds: O-S-O angle of SO2 F-B-F angle of BF3 Cl-S-Cl angle of SCI2 O-C-O angle of CO2 F-P-F angle of PF3 H-C-H angle of CH4
The bond angles are determined by their molecular structures - SO2 is 120°, BF3 is 180°, SCl2 is 120°, CO2 is 180°, PF3 is slightly less than 109.5°, and CH4 is 109.5°.
Explanation:The bond angles in various compounds are determined by the molecule's electron-pair geometry and molecular structure. For the SO2 compound, O-S-O angle corresponds to a bent molecular structure with a bond angle of 120° (electron-pair geometry: trigonal planar). The F-B-F angle in BF3 has a linear molecular structure that leads to a bond angle of 180° (electron-pair geometry: linear). In SCl2, the Cl-S-Cl angle is 120° due to its bent structure (electron-pair geometry: trigonal planar). For CO2, the O-C-O angle is 180° because of its linear structure (electron-pair geometry: linear). In PF3, the F-P-F angle is slightly less than 109.5° because of its trigonal pyramidal structure (electron-pair geometry: tetrahedral). Finally, in CH4, the H-C-H is 109.5° as it has a tetrahedral structure (electron-pair geometry: tetrahedral).
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The equilibrium constant for the gas phase reaction: N2O5(g) ---> 2 NO2(g) + ½ O2(g) is 95 at 25ºC. What is the value of the equilibrium constant for the following reaction at 25ºC?
O2(g) + 4 NO2(g) ---> 2 N2O5(g)
1/95
(95)^2
1/(95)^2
(95)^½
Answer: The value of equilibrium constant for reverse reaction is [tex](\frac{1}{95})^2[/tex]
Explanation:
The given chemical equation follows:
[tex]N_2O_5(g)\rightarrow 2NO_2(g)+\frac{1}{2}O_2(g)[/tex]
The equilibrium constant for the above equation is 95.
We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:
[tex]O_2(g)+4NO_2(g)\rightarrow 2N_2O_5(g)[/tex]
The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.
If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant of initial reaction.
The value of equilibrium constant for reverse reaction is:
[tex]K_{eq}'=(\frac{1}{95})^2[/tex]
Hence, the value of equilibrium constant for reverse reaction is [tex](\frac{1}{95})^2[/tex]
The equilibrium constant for the reaction O2(g) + 4 NO2(g) --> 2 N2O5(g) at 25ºC is (1/95)^2. This is due to Le Chatelier's principle stating that K for the reverse reaction is the reciprocal of the original one and it's squared when the reaction is doubled.
Explanation:The second reaction in your question is the reverse reaction of the first one, but also multiplied by two. According to Le Chatelier's principle, the equilibrium constant (K) of the reverse reaction is the reciprocal of the original one. So the K of the reverse of 1: N2O5(g) --> 2 NO2(g) + ½ O2(g) would be 1/95. However, the actual reaction to the question is twice that.
When you double a reaction, you square its equilibrium constant. Thus, the K for the reaction: O2(g) + 4 NO2(g) --> 2 N2O5(g) would be (1/95)^2.
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What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?
A) 0.57 %
B) 0.32 %
C) 2.2 %
D) 0.18 %
E) 0.24 %
Answer:
B) 0.32 %
Explanation:
Given that:
[tex]K_{a}=1.8\times 10^{-5}[/tex]
Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-
[tex]\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}[/tex]
The expression for dissociation constant of acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}[/tex]
[tex]1.8\times 10^{-5}=\frac{x^2}{1.8-x}[/tex]
[tex]1.8\left(1.8-x\right)=100000x^2[/tex]
Solving for x, we get:
x = 0.00568 M
Percentage ionization = [tex]\frac{0.00568}{1.8}\times 100=0.32 \%[/tex]
Option B is correct.
Final answer:
After setting up the equilibrium expression for acetic acid dissociation and solving the equation, the percent ionization of a 1.8 M HC2H3O2 solution is determined to be approximately 0.1%. The closest answer option given is D) 0.18%.
Explanation:
To calculate the percent ionization for a 1.8 M HC2H3O2 solution, we need to set up an equilibrium expression based on the acid dissociation constant (Ka) and initial concentration of acetic acid. We start with the dissociation reaction:
HC2H3O2 \<=> H+ + C2H3O2-
Let's assume that the concentration of H+ (x) is equal to the concentration of C2H3O2- (x) at equilibrium due to stoichiometry:
Ka = [H+][C2H3O2-]/[HC2H3O2]
If we let x represent the molarity of H+ and C2H3O2- ions at equilibrium and assuming that x is small enough that (1.8 M - x) ≈ 1.8 M, we can simplify the expression to:
Ka = x^2/1.8 M
The equation becomes:
1.8 \ 10^-5 = x^2/1.8
Solving for x gives us:
x ≈ \ 10^-5 \ 1.8)}\> ≈ 1.8 \ 10^-3 M
The percent ionization is then calculated by:
Percent ionization = (x / [HC2H3O2] initial) \ 100%
Percent ionization = (1.8 \ 10^-3 M / 1.8 M) \ 100% = 0.1%
Since none of the options given matches the calculated value strictly, and provided that we've made an approximation, we'll have to choose the closest answer, which is:
D) 0.18%