Answer:
26.87g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2Fe + O2 —> 2FeO
Next, we shall determine the masses of Fe and O2 that reacted and the mass of FeO produced from the balanced equation.
This is illustrated below:
Molar mass of Fe =56 g/mol
Mass of Fe from the balanced equation = 2 x 56 = 112 g
Molar mass of O2 = 16x2 = 32 g/mol
Mass of O2 from the balanced equation = 1 x 32 = 32 g
Molar mass of FeO = 56 + 16 =72 g/mol
Mass of FeO from the balanced equation = 2 x 72 = 144 g
From the balanced equation above,
112 g of Fe reacted with 32 g of O2 to produce 144 g of FeO.
Next, we shall determine the limiting reactant.
This is illustrated below:
From the balanced equation above,
112 g of Fe reacted with 32 g of O2.
Therefore, 20.9 g of Fe will react with = (20.9 x 32)/112 = 5.97 g of O2.
From the calculations made above, we can see that only 5.97 g out of 9.19 g of O2 given were required to react completely with 20.9 g of Fe.
Therefore, Fe is the limiting reactant and O2 is the excess reactant.
Finally, we shall determine the mass of FeO produced from the reaction.
In this case, the limiting reactant will be used, as it will give the maximum yield of the reaction since all of it is used up in the reaction.
The limiting reactant is Fe and the maximum mass of FeO produced can be obtained as follow:
From the balanced equation above,
112 g of Fe reacted to produce 144 g of FeO.
Therefore, 20.9 g of Fe will react to produce = (20.9 x 144)/112 = 26.87g of FeO.
Therefore, the maximum mass of iron(II) oxide, FeO produced is 26.87g.
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:
Which molecule or ion has a trigonal planar shape?
Answer:B
Explanation: A P E X
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:
Consider the following system, which is at equilibrium, 3C(s) + 3H2(g) <--> CH4(g) + C2H2(g) The result of removing some C(s) from the system will be:
Answer:
The equilibrium position will shift to the left, thus, favouring the backward reaction.
Explanation:
The equation for the reaction is given below:
3C(s) + 3H2(g) <==> CH4(g) + C2H2(g)
According to Le Chatelier's principle, if an external constrain such as change in concentration, temperature or pressure is imposed on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.
From Le Chatelier's principle, removing some C(s) implies removing some of the concentration of the reactants.
This will shift the equilibrium position to the left, thus, favouring the backward reaction (i.e forming more reactants) because removing C(s) implies that the products are now more than the reactants and as such, they will react to form more reactants.
The equilibrium position will shift to the left, thus, favouring the backward reaction.
The following information should be considered:
The equation for the reaction is given below:[tex]3C(s) + 3H2(g) <==> CH4(g) + C2H2(g)[/tex]
In the case when an external constrain like as change in concentration is applied on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.This will shift the equilibrium position to the left, thus, favouring the backward reaction.Learn more: https://brainly.com/question/10283285?referrer=searchResults
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
A saturated solution of lead(II) iodide, PbI2 has an iodide concentration of 3.0 x 10^-3 mol/L.
a) What is the molar solubility of PbI2?
b) Determine the solubility constant, Ksp, for lead(II) iodide.
c) Does the molar solubility of lead (II) iodide increase, decrease, or remain unchanged with the addition of potassium iodide to the solution? EXPLAIN.
Answer:
a) 1.5 x 10^-3 mol/L
b) 1.35×10^-8
c) decrease
Explanation:
The solubility of lead II iodide is given by the equation;
PbI2(s) -----> Pb^2+(aq) + 2I^-
By looking at the ICE table, I^-=2x= 3.0 x 10^-3 mol/L/2 = 1.5×10^-3 mol/L
Hence molar solubility of PbI2 = 1.5 x 10^-3 mol/L
Ksp= [Pb^2+] [2I^-]^2 =
Let the molar solubility of each ion be x, therefore;
Ksp= 4x^3
Ksp= 4(1.5 x 10^-3 mol/L)^3= 1.35×10^-8
Addition of kI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI2 in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le Chateliers principle.
a) The molar solubility of PbI₂ is [tex]1.5 * 10^{-3} mol/L[/tex]
b) The solubility constant is [tex]1.35*10^{-8}[/tex]
c) The molar solubility of lead (II) will decrease.
Molar Solubility:The solubility of lead II iodide is given by the equation;
[tex]PbI_2(s) ----- > Pb^{2+}(aq) + 2I^-[/tex]
By looking at the ICE table,
[tex]I^-=2x= 3.0 * 10^{-3} mol/L/2 =[/tex] [tex]1.5 * 10^{-3} mol/L[/tex]
Hence, molar solubility of PbI2 = [tex]1.5 * 10^{-3} mol/L[/tex]
[tex]Ksp= [Pb^{2+}] [2I^-]^2[/tex]
Let the molar solubility of each ion be x, therefore;
[tex]Ksp= 4x^3\\\\Ksp= 4(1.5 * 10^{-3} mol/L)^3\\\\Ksp= 1.35*10^{-8}[/tex]
The addition of KI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI₂ in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le- Ch-ateliers principle.
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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]
What is the balanced oxidation half-reaction for the following reaction? Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq) Question 8 options: There is no reaction Cu2+(aq) + 2e– → Cu(s) Fe2+(aq) + 2e– → Fe(s) Cu(s) → Cu2+(aq) + 2e– Fe(s) → Fe2+(aq) + 2e–
Answer:
Fe(s) → Fe2+(aq) + 2e-
Explanation:
Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq)
Oxidation is the loss of electrons. When the oxidation number of an element increases, that means there is a loss of electrons and that element is being oxidized.
The oxidation half equation in this reaction is;
Fe(s) → Fe2+(aq)
The loss of electron is represented in the product side and is given by;
Fe(s) → Fe2+(aq) + 2e-
Re-order each list of elements in the table below, if necessary, so that the elements are listed in order of decreasing electronegativity.
Answer:
O, S, Te
Cl, Br, Se
Explanation:
Main group elements have an electronegativity that increases across a period (from left to right) and decreases down a group.
Each atom has its own value which you can find on the electronegativity chart.
Metals have low values
Nonmetals have high values
I'm your case:
O = 3.5
S = 2.5
Te = 2.1
Cl = 3.0
Br = 2.8
Se = 2.4
O, S, Te, and Cl, Br, Se are the correct order for the elements listed in order of decreasing electronegativity.
What is electronegativity?
Electronegativity. is the tendency or the efficiency of an atom to attract the lone pair of electrons towards itself to become stable and complete their octave is known as electronegativity.
In the periodic table oxygen, fluorine and nitrogen are the three top, most electronegative elements, and from moving up to down in periodic table electronegativity. decreases and left to right increases.
Therefore, the correct order for the elements listed in order of decreasing electronegativity is O, S, Te, and Cl, Br, Se.
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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.
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
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.
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
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
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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When mixed, solutions of silver nitrate, AgNO3, and sodium sulfate, Na2SO4, form a precipitate of silver sulfate, Ag2SO4. The balanced equation is:
Answer:
2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3
Explanation:
The general schemefor a reaction is given as;
Reactants --> Products
In this question, the reactants are AgNO3 and Na2SO4. The product is Ag2SO4.
The equation is given as;
AgNO3 + Na2SO4 --> Ag2SO4
The other poduct formed in this reaction is NaNO3.
The full reaction is given as;
AgNO3 + Na2SO4 --> Ag2SO4 + NaNO3
The above reaction is not balanced because there are unequal number of atoms of the elements on both sides of the reaction.
The balanced equation is given as;
2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3
In this equation, there are equal number of moles of the atoms on both sides.
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 0.100g sample of Mg when combined with O2 yields 0.166g of Mgo, a
second Mg sample with a mass of 0.144g is also combined with O2. What
mass of MgO is produced from the second sample?
Answer:
[tex]m_{MgO}=0.239gMgO[/tex]
Explanation:
Hello,
In this case, the chemical reaction between magnesium and oxygen to yield magnesium oxide is:
[tex]2Mg+O_2\rightarrow 2MgO[/tex]
In such a way, for 0.144 g of magnesium reacting with sufficient oxygen, the mass of magnesium oxide, whose molar mass is 40.3 g/mol (2:2 mole ratio) turns out:
[tex]m_{MgO}=0.144gMg*\frac{1molMg}{24.3gMg} *\frac{2molMgO}{2molMg}* \frac{40.3gMgO}{1molMgO} \\\\m_{MgO}=0.239gMgO[/tex]
Best regards.
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
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.
3. There are many different primary standards that could be used in a standardization titration. What are the criteria for a primary standard
Answer:
High purity.
Stability (low reactivity)
Low hygroscopicity (to minimize weight changes due to humidity)
Explanation:
There are different primary standards that could be used in a standardization titration in order to achieve the best and accurate result possible. These standards include high purity,stability and low hygoscropicity .
A high purity means the reactants lack impurities which could affect the result. Stability also ensures that there is non reactivity with elements/compounds in the atmosphere while low hygroscopicity ensures weight changes are minimized due to humidity.
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
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.
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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One kilogram of butane (C4H10) is burned with 25 kg of air that is at 308C and 90 kPa. Assuming that the combustion is complete and the pressure of the products is 90 kPa, determine (a) the percentage of theoretical air used and (b) the dew-point temperature of the products.
Explanation:
Combustion of butane gives following reaction
C4H10 + 13/2 O2 → 5H20 + 4CO2
Therefore, One Kg of Butane consists of 1000/58 moles . That is 17.24 moles.
17.24 moles of butane reacts with (6.5×17.24) moles of O2 = 112.06 moles = 3.586 Kg O2.
Weight % of O2 in Air = 20 %
Mass of theoretical air used is 3.586×5 = 17.93 Kg = 71.72 %
The dew point temperature of H2O is :
D = (237.3×B)/(1-B)
B = ln(E/6.108)/17.27
E is vapor pressure at given temp.
At T = 30° C , E = 31.8 mm of Hg = 41.84 Milli Bars
B = 0.1114
D = 29.74°C
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
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 correctly lists the three land uses that the Bureau of Land Management was originally created to manage? mining, recreation, wildlife refuges recreation, developing oil and gas, battlefields grazing, mining, developing oil and gas developing oil and gas, battlefields, wildlife refuges
Answer: C
Explanation:
Right on edge 2020
The Bureau of Land Management was originally created to manage land for grazing, mining, developing oil and gas.
What is land management?Land management refers to the activities which are done in order to protect and preserve the land as well the resources found on land.
The Bureau of Land Management was created to manage land in the US.
The Bureau of Land Management was originally created to manage land for grazing, mining, developing oil and gas.
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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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