A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll
Carbon monoxide gas reacts with hydrogen gas to form methanol: CO (g_ + 2H2 (g) → CH3OH (g) A 1.50L reaction vessel, initially at 305 K, contains carbon monoxide gas at a partial pressure of 232 mmHg and hydrogen gas at a partial pressure of 397 mmHg. Identify the limiting reactant and determine the theoretical yield of methanol in grams.
:Answer : The limiting reactant is and the theoretical yield of methanol is, 0.96 grams.
Explanation :
First we have to calculate the moles of and .
where,
= pressure of CO gas = 232 mmHg = 0.305 atm (1 atm = 760 mmHg)
V = volume of gas = 1.65 L
T = temperature of gas = 305 K
= number of moles of CO gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
and,
where,
= pressure of gas = 374 mmHg = 0.492 atm (1 atm = 760 mmHg)
V = volume of gas = 1.65 L
T = temperature of gas = 305 K
= number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of
So, 0.0601 moles of react with moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 2 mole of react to give 1 mole of
So, 0.0601 moles of react with moles of
Now we have to calculate the mass of
Therefore, the theoretical yield of methanol is, 0.96 grams.
The theoretical yield of methanol is 0.496 g of methanol.
The reaction equation is CO (g) + 2H2 (g) → CH3OH (g).
From the partial pressures of each reactant, we can obtain the number of moles of reactants.
For CO;
P = 232 mmHg or 0.305 atm
V = 1.5 L
T = 305 K
n = ?
R = 0.082 atmL-1mol-1K-1
PV = nRT
n = PV/RT
n = 0.305 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K
n = 0.018 moles
For hydrogen;
P = 397 mmHg or 0.522 atm
V = 1.5 L
T = 305 K
n = ?
R = 0.082 atmL-1mol-1K-1
PV = nRT
n = PV/RT
n = 0.522 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K
n = 0.031 moles
From the reaction equation;
1 mole of CO reacted with 2 moles of H2
0.018 moles of CO will react with 0.018 moles × 2 moles/1 mole
= 0.036 moles of H2
We can see that there is not enough H2 to react with CO hence H2 is the limiting reactant.
2 moles of H2 yields 1 mole of methanol
0.031 moles of H2 yields 0.031 moles × 1 moles/2 mole
= 0.0155 moles of methanol
Mass of methanol produced = 0.0155 moles of methanol × 32 g/mol
= 0.496 g of methanol
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Calculate the frequency (Hz) and wavelength (nm)
of the emitted photon when an electron drops from
the n = 4 to the n=2 level in a hydrogen atom
Answer:
wavelength, λ = 486.6 nm
frequency, f = 6.16 * 10¹⁴ Hz
Explanation:
a. Wavelength
Using the wavelength equation; 1/λ = (1/hc) * 2.18 * 10⁻¹⁸ J * (1/nf² - 1/ni²)
Where nf is the final energy level; ni is the initial energy level; h is Planck's constant = 6.63 * 10⁻³⁴ J.s; c is velocity of light = 3 * 10⁸ m/s
1/λ = 1/(6.63 * 10⁻³⁴ J.s * 3 * 10⁸ m/s) * 2.18 * 10⁻¹⁸ J * (1/2² - 1/4²)
1/λ = 2.055 * 10⁶ m
λ = 4.866 * 10⁻⁷ m
wavelength, λ = 486.6 nm
b. Frequency
Using f = c/λ
f = (3 * 10⁸ m/s) / 4.866 * 10⁻⁷ m
frequency, f = 6.16 * 10¹⁴ Hz
What states can electrons exist in? A. Electron clouds or energy levels B. Positive and negative C. Up and down spin D. In phase and out of phase
Answer:
A. Electron clouds or energy levels
Explanation:
Electrons can exist in two states:
Stablized in electronic orbitalsFreely movingElectrons can exist in an electron cloud or energy level. Electron in an atoms have ability to change energy levels either by emitting or absorbing a photon that form the energy equal to the energy difference between the two levels.
Hence, the correct answer is A.
Answer:
Up and DOWN spin
Explanation:
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. Assume that the volume of the solutions are additive . What would be the Ka for NH4
Answer:
Following are the answer to this question:
Explanation:
The value of pH solution is =5.17 So, the p^{OH}:
[tex]p^{OH}[/tex]=14-56.17
=8.823
The volume of the [tex]NH_{3}[/tex] = 40.00 ml
convert into the liter= 0.040L
The value of the concentrated [tex]NH_{3}[/tex] =0.10 M
The volume of the [tex]NH_{4}Cl[/tex]= 50.00 ml
convert into the liter= 0.050L
The value of concentrated [tex]NH_{4}Cl[/tex]= 0.10 M
The volume of the [tex]H_{2}So_{4}[/tex]= 30 ml
convert into the liter= 0.030L
The value of concentrated [tex]H_2So_4[/tex]=0.05 M
Calculating total volume=(0.40+0.050+0.030)
=0.120 L
calculating the new concentrated value of [tex]NH_3[/tex] = [tex]\frac{0.10\times 0.040}{0.120}= 0.33 \ M[/tex]
calculating the new concentrated value of [tex]NH_4Cl[/tex]= [tex]\frac{0.050\times 0.10}{0.120}= 0.04166 \ M[/tex]calculating the new concentrated value of [tex]H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M[/tex] when 1 mol [tex]H_2So_4[/tex] produced 2 mols [tex]H^{+}[/tex] so, 0.0125 in [tex]H_2So_4[/tex]produced:
[tex]=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}[/tex]
create the ICE table:
[tex]NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}[/tex]
I (m) 0.033(m) 0.025 0.04166
C -0.025 -0.025 + 0.025
E 8.3\times 10^{-3} 0 0.0667
now calculating pH:
when ph= 8.83:
[tex]P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769[/tex]
How does a balanced chemical equation show the conservation of mass?
A. It shows that the number of each type of atom stays the same.
B. It shows that the mass of the products is greater than the mass of
the reactants when a reaction increases the moles of substances.
C. It shows that the total number of moles of substances stays the
same.
D. It shows that the mass of the reactants is greater than the mass
Answer:
A. It shows that the number of each type of atom stays the same.
Explanation:
Though you may see a change in the way they are arranged, the same number of atoms are present before and after. Balanced chemical equations show equal numbers of atoms of each element on each side of the equation.
Determine the number of moles of the anhydrous salt present after heating, assuming that the contents of the aluminum cup after heating are pure anhydrous KAl(SO 4 ) 2 .
Answer:
0.2 moles, assuming weight of dried salt
Explanation:
In order to determine the number of moles, we need to be aware of the mass of the substance in question.
Assuming the mass of the dehydrated [tex]KAl(SO_{4} )_{2}.H_{2} O[/tex] is 50g.
No. of moles = mass of substance/ molar mass of the substance.
= [tex]\frac{50g}{39+27+32*2+16*4*2\\)g/mol}[/tex]
= 0.2 moles moles.
which of the following compounds exhibits dipole -dipole forces as its strongest attraction between molecules? o2,ch3Br,CCl4,He,BrCH2CH2OH
Answer:
B. CH3Br
Explanation:
Dipole -Dipole interactions take place in polar molecules.
CH3Br exhibits dipole -dipole forces as its strongest attraction between molecules because it is a polar molecule due to the slightly negative dipole present on the Br molecule.
While O2 is a nonpolar molecule due to its linear structure, CCl4 has zero resultant dipole moment, Helium is non-polar and BrCH2CH2OH is a non polar compound having net dipole moment is zero.
Hence, the correct option is B. CH3Br.
The compound that exhibits dipole -dipole forces is CH3Br
Dipole -Dipole interactions:It should be taken place in polar molecules. Also, CH3Br should be the strongest attraction that lies between the molecules since it is treated as the polar molecule because of the slightly negative. While on the other hand, O2 should be non-polar molecule because of the linear structure.
Therefore, The compound that exhibits dipole -dipole forces is CH3Br
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Which molecule or ion has a trigonal planar shape?
Answer:B
Explanation: A P E X
The volume of a sample of oxygen is 300mL when the pressure is 1 atm and the temperature is 27 C . At what temperature is the volume 1.00 L and the pressure.500 atm?
Answer:
T2 = 500K
Explanation:
Given data:
P1 = 1atm
V1 = 300ml
T1= 27 + 273 = 300K
T2 = ?
V2 = 1.00ml
P2 = 500atm
Apply combined law:
P1xV1//T1 = P2xV2/T2 ...eq1
Substituting values into eq1:
1 x 300/300 = 500 x 1/T2
Solve for T2:
300T2 = 500 x 300
300T2 = 150000
Divide both sides by the coefficient of T2:
300T2/300 = 150000/300
T2 = 500K
assume that amonia can be prepared by the folowing reaction in the gas phase at STP. If the reaction conditions are maintainted at STP, how many liters of NH3 can be produced by the reaction of 12.0 L of H2 and the exact required volumen of N2
Answer:
8.00L of ammonia can be produced
Explanation:
The reaction is:
N₂(g) + 3H₂(g) → 2NH₃(g)
Where 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
Avogadro's law states that, under constant pressure and temperature, equal volumes of gases contains equal number of moles.
As in the reaction conditions are mantained at STP (Pressure and temperature are constant) you can say of the reaction that:
1 liter of nitrogen reacts with 3 liters of hydrogen to produce 2 liters of ammonia
Thus, if 12.0L of hydrogen reacts and 3L of hydrogen produce 2L of ammonia, liters of ammonia produced are:
12L H₂(g) ₓ (2L NH₃(g) / 3L H₂(g)) =
8.00L of ammonia can be producedWhat's the electron configuration of a Ca+2 ion?
A. [Kr]
B. [Ar]
C. [Ne]
D. [He]
Answer:
B
Explanation:
The Ca+2 ion means that 2 electrons have been given away. So when you try and find the answer, you have to count backwards from Calcium. When you do, you get K+ first and then Argon which is either column 8 orc column 18 depending on your periodic table.
The element you hit is Argon.
The answer is B
Answer:
B ar
Explanation:
pen foster answer
Identify the Lewis acid and Lewis base from among the reactants in each of the following equations. Match the words in the left column to the appropriate blanks in the sentences on the right.
1. Fe3+ (aq)+6CN (aq) Fe(CN) (aq)______is the Lewis acid and_____is the Lewis base. is the Lewis
2. CI- (aq) + AlCl3 (aq) AlCl4-____is the Lewis acid and______is the Lewis base.
3. AlBr3 + NH3 H3NAlBr3______is the Lewis acid and______is the Lewis base.
A. AlCl3
B. CN-
C. AlBr3
D. Cl-
E. NH3
F. Fe3+
Answer:
1. Lewis acid: F. Fe₃⁺, Lewis base: B. CN⁻
2. Lewis acid: A. AlCl₃, Lewis base: D. Cl⁻
3. Lewis acid: C. AlBr₃, Lewis base: E. NH₃
Hope this helps.
The Lewis acid is chemical substance which possesses an empty orbital and accepts an electron pair from a Lewis base ( donor ), in order to create a Lewis adduct ( molecule created from the bonding of Lewis base and acid ).
The Lewis acid from reaction 1 is Fe₃⁺ while the Lewis base is CN⁻ also the Lewis acid from reaction 2 is AICI₃ while the Lewis base is CI⁻
Hence we can conclude that the Lewis acids and Lewis bases of the reactions in the question are as listed above.
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For a single substance at atmospheric pressure, classify the following as describing a spontaneous process, a nonspontaneous process, or an equilibrium system.
A) Solid melting below its melting point
B) Gas condensing below its condensation point
C) Liquid vaporizing above its boiling point
D) Liquid freezing below its freezing point
E) Liquid freezing above its freezing point
F) Solid melting above its melting point
G) Liquid and gas together at boiling point with no net condensation or vaporization
H) Gas condensing above its condensation point
I) Solid and liquid together at the melting point with no net freezing or melting
Answer:
Spontaneous process- This is the process that occurs on its own without the application of any external energy or other factor. They include
B) Gas condensing below its condensation point
C) Liquid vaporizing above its boiling point
D) Liquid freezing below its freezing point
F) Solid melting above its melting point
Non spontaneous - This is the process that doesn’t occurs on its own and requires the application of any external energy or factor. They include
A) Solid melting below its melting point
E) Liquid freezing above its freezing point
H) Gas condensing above its condensation point
Equilibrium system
G) Liquid and gas together at boiling point with no net condensation or vaporization
I) Solid and liquid together at the melting point with no net freezing or melting
A) Solid melting below its melting point - nonspontaneous process
B) Gas condensing below its condensation point - spontaneous process
C) Liquid vaporizing above its boiling point - spontaneous process
D) Liquid freezing below its freezing point - spontaneous process
E) Liquid freezing above its freezing point - nonspontaneous process
F) Solid melting above its melting point - spontaneous process
G) Liquid and gas together at boiling point with no net condensation or vaporization - Equilibrium system
H) Gas condensing above its condensation point - nonspontaneous process
I) Solid and liquid together at the melting point with no net freezing or melting - Equilibrium system
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The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism:
Answer:
The two-step mechanism is a slow mechanism and a fast mechanism. When we combine them, the result is
2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)
Explanation:
We know that the decomposition of hydrogen peroxide is catalyzed by iodide ion, which means that the iodide ion will react with the hydrogen peroxide. There is a slow mechanism and a fast one:
H₂O₂(aq) + I₋(aq) ⇒ H₂O(l) + IO₋(aq) this is the slow reaction
IO₋(aq) + H₂O₂(aq)⇒ H₂O(l) + O₂(g) + I₋ (aq) this is the fast reaction
If we cancel the same type of molecules and ions, the final result is:
2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)
The two-step mechanism represents the slow mechanism and a fast mechanism. At the time of combining them, the result is 2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)
Two-step mechanism:
The decomposition of hydrogen peroxide should be catalyzed by iodide ion, which represents the iodide ion will react with the hydrogen peroxide. There is a slow mechanism and a fast one
Now
H₂O₂(aq) + I₋(aq) ⇒ H₂O(l) + IO₋(aq) this is the slow reaction
IO₋(aq) + H₂O₂(aq)⇒ H₂O(l) + O₂(g) + I₋ (aq) this is the fast reaction
Now in case of cancelling, the same type of molecules and ions, the final result is 2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)
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Methanol is produced industrially by catalytic hydrogenation of carbon monoxide according to the following equation: CO(g) + 2 H2(g) → CH3OH(l) If the yield of the reaction is 40%, what volume of CO (measured at STP) would be needed to produce 1.0 × 106 kg CH3OH?
Answer:
1.7 × 10⁹ L
Explanation:
Step 1: Write the balanced equation
CO(g) + 2 H₂(g) → CH₃OH(l)
Step 2: Calculate the moles corresponding to 1.0 × 10⁶ kg CH₃OH
The molar mass of CH₃OH is 32.04 g/mol.
[tex]1.0 \times 10^{6} kg \times \frac{10^{3}g }{1kg} \times \frac{1mol}{32.04g} = 3.1 \times 10^{7} mol[/tex]
Step 3: Calculate the theoretical yield of CH₃OH
The real yield of CH₃OH is 3.1 × 10⁷ mol and the percent yield is 40%. The theoretical yield is:
[tex]3.1 \times 10^{7} mol (R) \times \frac{100mol(T)}{40mol(R)} = 7.8 \times 10^{7}mol(T)[/tex]
Step 4: Calculate the moles of CO required to produce 7.8 × 10⁷ mol of CH₃OH
The molar ratio of CO to CH₃OH is 1:1. The moles of CO required are 1/1 × 7.8 × 10⁷ mol = 7.8 × 10⁷ mol
Step 5: Calculate the volume of 7.8 × 10⁷ mol of CO at STP
The volume of 1 mole of CO at STP is 22.4 L.
[tex]7.8 \times 10^{7}mol \times \frac{22.4L}{mol} = 1.7 \times 10^{9}L[/tex]
Jane is doing an experiment with plants. She makes a good scientific guess that one will grow taller than the other. What is this guess called? A. Prediction B. Procedure C. Observation D. Data
Answer:
A
Explanation:
which material is a conductor of electric current?
Answer:Conduction materials include metals, electrolytes, superconductors, semiconductors, plasmas and some nonmetallic conductors such as graphite and Conductive polymers. Copper has a high conductivity.
Explanation:
Interpret the following equation for a chemical reaction using the coefficients given:
Cl2(g) + F2(g) 2ClF(g)
On the particulate level:
________ of Cl2(g) reacts with ______ of F2(g) to form______ of ClF(g).
On the molar level:
______ of Cl2(g) reacts with______ of F2(g) to form______ of ClF(g).
Answer and Explanation:
Given the following chemical equation:
Cl₂(g) + F₂(g) ⇒ 2ClF(g)
The coefficients are: 1 for Cl₂, 1 for F₂ and 2 for ClF. The coefficients indicate the number of units of each ompound that participates in the reaction. It gives the proportion of reactants and products in the reaction. These units can be molecules or moles. In this reaction, we can say:
On the particulate level: 1 molecule of Cl₂(g) reacts with 1 molecule of F₂(g) to form 2 molecules of ClF(g).
On the molar level: 1 mol of Cl₂(g) reacts with 1 mol of F₂(g) to form 2 mol of ClF(g).
A soda manufacturing company is experimenting with changing the taste of its product as the concentration of carbon dioxide changes. To track their results, they must determine how concentration changes with pressure. The concentration of CO2 under a partial pressure of 0.719 atm is 429.7 ppm. At what pressure (in atm) would the CO2 need to be so that the concentration of CO2 is 235.3 ppm at the same temperature
Answer:
0.394 atm
Explanation:
Mathematically, when we increase the pressure of a gas, we are increasing its concentration and when we decrease the pressure, we are decreasing its concentration.l at same temperature
What this means is that pressure and concentration are directly proportional.
Representing concentration by c and pressure by p, we have;
P1/C1 = P2/C2
From the question;
P1 = 0.719 atm
P2 = ?
C1 = 429.7 ppm
C2 = 235.3 ppm
Now, we can rewrite the equation to be;
P1C2/C1 = P2
Substituting the values we have;
0.719 * 235.3/429.7 = 0.394 atm
Ethane burns according to the following reaction: 2C2H6 + 7O2 → 4CO2 + 6H2O + 2.86 x 103 kJ How much heat can be generated when 185 grams of oxygen gas (MW = 32 g/mol) are consumed?
Answer:
question is not clear please send clear question
Calculate the pH of a buffer solution obtained by dissolving 25.025.0 g of KH2PO4(s)KH2PO4(s) and 38.038.0 g of Na2HPO4(s)Na2HPO4(s) in water and then diluting to 1.00 L.
Answer:
pH = 7.37
Explanation:
The buffer H₂PO₄⁻/HPO₄²⁻ has as pKa 7.21. To find the pH of a buffer you can use H-H equation:
pH = pKa + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]
Where [HPO₄²⁻] and [H₂PO₄⁻] are molar concentrations of each species. As volume is 1.00L, [HPO₄²⁻] and [H₂PO₄⁻] are MOLES.
Moles of 25.0g of KH₂PO₄ (Molar mass: 136.086g/mol):
25.0g KH₂PO₄ ₓ (1mol / 136.086g) = 0.1837 moles KH₂PO₄ = moles H₂PO₄⁻
Moles of 38.0g of Na₂HPO₄ (Molar mass: 141.96g/mol):
38.0g KH₂PO₄ ₓ (1mol / 141.96g) = 0.2677 moles Na₂HPO₄ = moles HPO₄²⁻
Replacing in H-H equation:
pH = pKa + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]
pH = 7.21 + log₁₀ [0.2677] / [0.1837]
pH = 7.37
Nitrogen has different oxidation states in the following compounds: nitrite ion, nitrous oxide, nitrate ion, ammonia, and nitrogen gas. Arrange these species in order of increasing nitrogen oxidation state. Select the correct answer below: A. ammonia, nitrogen gas, nitrite, nitrous oxide, nitrate B. nitrogen gas, ammonia, nitrous oxide, nitrite, nitrate C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate D. ammonia, nitrogen gas, nitrate, nitrite, nitrous oxide
Answer:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Explanation:
To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.
Nitrite ion (NO₂⁻)
1 × N + 2 × O = -1
1 × N + 2 × (-2) = -1
N = +3
Nitrous oxide (NO)
1 × N + 1 × O = 0
1 × N + 1 × (-2) = 0
N = +2
Nitrate ion (NO₃⁻)
1 × N + 3 × O = -1
1 × N + 3 × (-2) = -1
N = +5
Ammonia (NH₃)
1 × N + 3 × H = 0
1 × N + 3 × (+1) = 0
N = -3
Nitrogen gas (N₂)
2 × N = 0
N = 0
The order of increasing nitrogen oxidation state is:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Why is phosphorus unusual compared to other group 15 elements? Select the correct answer below: A. There are compounds that contain phosphorus-phosphorus bonds with uncommon oxidation states. B. Phosphorus is relatively unreactive. C. Phosphorus only forms compounds where the oxidation number of phosphorus is 5+. D. Phosphorus is the most electronegative of the group 15 elements.
Answer:
There are compounds that contain phosphorus-phosphorus bonds with uncommon oxidation states.
Explanation:
Phosphorus is a member of group 15 in the periodic table. Its common oxidation States are -3 and +5. Phosphorus is believed to firm some of its compounds by the participation of hybridized d-orbitals in bonding although this is also disputed by some scientists owing to the high energy of d - orbitals.
Phosphorus form compounds having phosphorus-phosphorus bonds in unusual oxidation states such as diphosphorus tetrahydride, H2P-PH2, and tetraphosphorus trisulfide, P4S hence the answer.
Reduction occurs at which electrode?
Answer:
negative charge electrode
Explanation:
In cathode positive ions are picked up to perform reduction.At the same time negative ions are picked up at anode to get oxidized from electrolyte.
Answer:
The electrode that removes ions from the solution :) a p e x
The free energy obtained from the oxidation (reaction with oxygen) of glucose (C6H12O6) to form carbon dioxide and water can be used to re-form ATP by driving the above reaction in reverse. Calculate the standard free energy change for the oxidation of glucose.
Answer:
The correct answer is -2878 kJ/mol.
Explanation:
The reaction that takes place at the time of the oxidation of glucose is,
C₆H₁₂O₆ (s) + 6O₂ (g) ⇒ 6CO₂ (g) + 6H₂O (l)
The standard free energy change for the oxidation of glucose can be determined by using the formula,
ΔG°rxn = ∑nΔG°f (products) - ∑nΔG°f (reactants)
The ΔG°f for glucose is -910.56 kJ/mol, for oxygen is 0 kJ/mol, for H2O -237.14 kJ/mol and for CO2 is -394.39 kJ/mol.
Therefore, ΔG°rxn = 6 (-237.14) + 6 (-394.39) - (-910.56)
ΔG°rxn = -2878 kJ/mol
Question 7 options: The cell potential of an electrochemical cell made of an Fe, Fe2 half-cell and a Pb, Pb2 half-cell is _____ V. Enter your answer to the hundredths place and do not leave off the leading zero, if needed.
Answer: Thus the cell potential of an electrochemical cell is +0.28 V
Explanation:
The calculation of cell potential is done by :
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Fe^{2+}/Fe]}= -0.41V[/tex]
[tex]E^0_{[Pb^{2+}/Pb]}=-0.13V[/tex]
As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.
[tex]E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Fe^{2+}/Fe]}[/tex]
[tex]E^0=-0.13- (-0.41V)=0.28V[/tex]
Thus the cell potential of an electrochemical cell is +0.28 V
Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was reacted with excess oxygen and 6.8g of product was obtained. What was the percent yield of the reaction?
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
4 moles of Al produce 2 moles of Al₂O₃
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
8.7g of Al₂O₃ can be produced (Theoretical yield)Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
78.2%Which of the following metals is paramagnetic?
A. Magnesium
B. Sodium
C. Beryllium
D. Calcium
Answer:
sodium
Explanation:
(Na) atom is paramagnetic and sodium is a na atom.
Q1. Calculate the amount of copper produced in 1.0 hour when aqueous CuBr2 solution was electrolyzed by using a current of 4.50 A. Q2. In another electroplating experiment, if electric current was passed for 3 hours and 2.00 g of silver was deposited from a AgNO3 solution, what was the current used in amperes
Answer:
[tex]\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}[/tex]
Explanation:
Q1. Mass of Cu
(a) Write the equation for the half-reaction.
Cu²⁺ + 2e⁻ ⟶ Cu
The number of electrons transferred (z) is 2 mol per mole of Cu.
(b) Calculate the number of coulombs
q = It
[tex]\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}[/tex]
(c) Mass of Cu
We can summarize Faraday's laws of electrolysis as
[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}[/tex]
Note: The answer can have only two significant figures because that is all you gave for the time.
Q2. Current used
(a) Write the equation for the half-reaction.
Ag⁺ + e⁻ ⟶ Ag
The number of electrons transferred (z) is 1 mol per mole of Ag.
(a) Calculate q
[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}[/tex]
(b) Calculate the current
t = 3 h = 3 × 3600 s = 10 800 s
[tex]\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}[/tex]
Note: The answer can have only one significant figure because that is all you gave for the time.
An analytical laboratory balance typically measures mass to the nearest 0.1 mg. You may want to reference (Page) Section 21.6 while completing this problem. Part A What energy change would accompany the loss of 0.1 mg in mass
Answer:
The energy change is [tex]E = 9.0 *10^{9}\ J[/tex]
Explanation:
From the question we are told that
Mass loss is [tex]m_l = 0.1 \ mg = 0.1 *10^{-3} mkg = 0.1 *10^{-6} \ kg[/tex]
Generally the energy change that would accompany this loss is mathematically represented as
[tex]E = m * c^2[/tex]
Where c is the speed of light with values [tex]c = 3.0*10^{8} \ m/s[/tex]
[tex]E = 0.1 *10^{-6} * [3.0 *10^{8}]^2[/tex]
[tex]E = 9.0 *10^{9}\ J[/tex]