The Ka of HX is mathematically given as
Ka = 9.11 *10^-8
What is the Ka of HX?
Generally, the equation for the molarity of HX is mathematically given as
M HX = moles HX / volume solution
Therefore
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
Therefore, ICE-chart
[H+] = [H3O+]
10^-3.70 = 10^-3.70 = 1.995 *10^-4
The concentration at the equilibrium is
[HX] = (0.437 - x)M
[H3O+] = 1.995*10^-4 M
x=1.995*10^-4
In conclusion
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
Read more about Solubility
https://brainly.com/question/23946616
Final answer:
The Ka of HX, given its concentration and the pH of solution, can be determined through a series of calculations leading to a Ka value of 9.17×10-8.
Explanation:
To find the Ka of HX given a 0.365-mol sample dissolved in 835.0 mL of solution with a pH of 3.70, we start by calculating the concentration of HX, then use the pH to find the concentration of H+ ions, which will help determine the acid dissociation constant (Ka).
First, convert volume from mL to L: 835.0 mL = 0.835 L. The concentration of HX (Molarity, M) is moles of solute (HX) divided by volume of solution in liters, which is 0.365 mol / 0.835 L = 0.4371 M.
Given pH = 3.70, we calculate the concentration of H+ ions ([H+]) as 10-pH = 10-3.70 = 2.00×10-4 M.
Assuming all H+ comes from the dissociation of HX and ignoring autoionization of water because HX is a weak acid, the Ka expression for HX is Ka = [H+][X-] / [HX]. Assuming the concentration of [X-] is equal to [H+] because each molecule of HX donates one H+, Ka becomes 2.00×10-4 M * 2.00×10-4 M / 0.4371 M = 9.17×10-8, after considering the initial concentration of HX and the change in concentration due to dissociation.
Which of the following statement(s) is/are correct? i) The mass defect is the difference in mass between that of a nucleus and the sum of the masses of its component nucleons. ii) The splitting of a heavier nucleus into two nuclei with smaller mass numbers is known as nuclear fission. iii) The first example of nuclear fission involved bombarding with nuclei. A. i) only B. ii) only C. iii) only D. i) and ii) only E. i) and iii) only F. ii) and iii) only G. i), ii) and iii) H. None of the choices are correct.
Answer:
Correct option -D
Explanation:
Mass defect:
The difference between the sum of the masses of individual nucleons that form an atomic nucleus and the mass of the nucleus.
Hence, i- is correct
Nuclear fission:
It is the process of splitting a nucleus into two nuclei with smaler masses.
Hence, ii- is correct
Nuclear fission involved bombarding with nuclei [tex]^{235}_{92}U[/tex] with [tex]^{4}_{2}He[/tex]
Nuclear fission of [tex]^{235}_{92}U[/tex] with neutrons forms two smaller mass nuclei.
[tex]^{1}_{0}n+^{235}_{92}U \rightarrow ^{91}_{36}Kr+^{142}_{56}B+3^{1}_{0}n+Energy[/tex]
Hence, iii- is incorrect
Therefore, option -D is correct
Statements i) and ii) are correct. Mass defect refers to the difference in mass between that of a nucleus and the sum of the masses of its component nucleons. Nuclear fission refers to the splitting of a heavier nucleus into two nuclei with smaller mass numbers.
Explanation:The correct statement(s) from the ones provided are: i) Mass defect is the difference in mass between that of a nucleus and the sum of the masses of its component nucleons. This is because the mass of a nucleus is not just the sum of the individual masses of the protons and neutrons. Some mass is converted into energy to provide the binding force that holds the nucleus together, so the mass of the nucleus has a slight deficit (or defect) compared to the combined masses of its components.
ii) Nuclear fission involves the splitting of a heavier nucleus into two nuclei with smaller mass numbers. This process usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when uranium-235 atoms were bombarded with slow-moving neutrons that split the uranium nuclei into smaller fragments.
Thus, the correct option is D. i) and ii) only.
Learn more about Nuclear Physics here:https://brainly.com/question/29134126
#SPJ11
the freezing point depression constants of the solvents cyclohexane and naphthalene are 20.1°C/m and 6.94°C/m respectively. Which would give a more accurate determination by freezing point depression of the molar mass of a substance that is soluble in either solvent? Why?
Explanation:
It'd be better to use cyclohexane. The possible explanation is that the freezing temperature will change by 20.1 degrees for each mole of substance added to 1 kg of cyclohexane, although the same amount added to naphthalene will change its freezing point just by 6.94 degrees.
It is so much easier to identify a larger change more adequately than a smaller one. You would actually not have a 1 molal solution in operation, so the variations in freezing points would be even smaller than the ones already described.
Final answer:
Cyclohexane, with its higher freezing point depression constant, would give a more accurate determination of molar mass than naphthalene because it results in more significant temperature changes, making precise measurements easier.
Explanation:
The determination of the molar mass of a substance using freezing point depression will be more accurate with a solvent that has a higher freezing point depression constant (Kƒ). In this case, cyclohexane, with a Kƒ of 20.1°C/m, would give a more accurate determination of molar mass compared with naphthalene, which has a Kƒ of 6.94°C/m, assuming that the substance is equally soluble in both. This is because a higher Kƒ value indicates a greater change in freezing point per mole of solute, making the temperature change easier to measure precisely.
Glucose (C6H12O6) is a key nutrient for generating chemical potential energy in biological systems. We were provided 16.55 g of glucose. Please calculate:
a) The mass percent of carbon in glucose.
b) The mass of CO2 produced by the combustion of 16.55 g glucose with sufficient oxygen gas.
c) How many oxygen molecules needed for the completely combustion of 16.55 g glucose?
Answer:
a) 40 %
b) [tex]4.04~g~CO_2[/tex]
c) [tex]5.53x10^2^3~molecules~of~O_2[/tex]
Explanation:
For a) we will have to calculate the molar mass of [tex]C_6H_1_2O_6[/tex], so the first step is to find the atomic mass of each atom and multiply by the amount of atoms in the molecule.
C => 12*(6) = 72
H => 1*(12) = 12
O => 6*(16) = 96
Molar mass = 180 g/mol
Then we can calculate the percentage by mass:
[tex]Percentage~=~\frac{72}{180}*100=40[/tex]
For b) we have to start with the reaction of glucose:
[tex]C_6H_1_2O_6~+~6O_2~->~6CO_2~+~6H_2O[/tex]
Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the following conversion ratios:
-) 180 g of glucose = 1 mol glucose
-) 1 mol glucose = 6 mol carbon dioxide
-) 1 mol carbon dioxide = 44 g carbon dioxide
[tex]16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~CO_2}{1~mol~C_6H_1_2O_6}\frac{44~g~CO_2}{1~mol~CO_2}=4.04~g~CO_2[/tex]
For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the following conversion ratios:
-) 180 g of glucose = 1 mol glucose
-) 1 mol glucose = 6 mol oxygen
-) 1 mol oxygen = 6.023x10^23 molecules of O2
[tex]16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~O_2}{1~mol~C_6H_1_2O_6}\frac{6.023x10^2^3~molecules~O_2}{1~mol~O_2}=~5.53x10^2^3~molecules~of~O_2[/tex]
This detailed answer explains the mass percent of carbon in glucose, the mass of CO2 produced by its combustion, and the number of oxygen molecules needed for complete combustion.
a) Mass Percent of Carbon in Glucose: Glucose has a molar mass of 180.16 g/mol. The molar mass of the carbon in one mole of glucose is 72.06 g (6 carbons in a molecule of glucose). Therefore, the mass percent of carbon in glucose is 40%.
b) Mass of CO2 Produced: The combustion of 16.55 g of glucose will produce 44.01 g of CO2.
c) Oxygen Molecules Needed: For the complete combustion of 16.55 g of glucose, 23 molecules of O2 are required.
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.
Gaseous cyclobutene undergoes a first-order reaction to form gaseous butadiene. At a particular temperature, the partial pressure of cyclobutene in the reaction vessel drops to one-eighth its original value in 124 seconds. What is the half-life for this reaction at this temperature?
Answer:
41.3 minutes
Explanation:
Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.
[tex]t_{1/2}= \frac{0.693}{K}[/tex]
So, fraction of original pressure = [tex]\frac{1}{2}^2[/tex]
n here is number of half life
therefore, [tex]\frac{1}{8}= \frac{1}{2}^3[/tex]
⇒ n= 3
it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.
Answer : The half-life of this reaction at this temperature is, 41.5 seconds.
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = ?
t = time passed by the sample = 124 s
a = let initial amount of the reactant = X
a - x = amount left after decay process = [tex]\frac{1}{8}\times (X)=\frac{X}{8}[/tex]
Now put all the given values in above equation, we get
[tex]k=\frac{2.303}{124s}\log\frac{X}{(\frac{X}{8})}}[/tex]
[tex]k=0.0167s^{-1}[/tex]
Now we have to calculate the half-life.
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=\frac{0.693}{0.0167s^{-1}}[/tex]
[tex]t_{1/2}=41.5s[/tex]
Therefore, the half-life of this reaction at this temperature is, 41.5 seconds.
Write complete ionic and net ionic equations for the reaction between sulfuric acid (H2SO4) and calcium carbonate (CaCO3).
H2SO4(aq) + CaCO3(s) —> H2O(I) + CO2(g) + CaSO4(aq)
Answer:
Complete ionic equation:
2H²⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
Net ionic equation:
2H²⁺(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq)
Explanation:
Chemical equation:
H₂SO₄(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + CaSO₄(aq)
Balanced chemical equation:
H₂SO₄(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + CaSO₄(aq)
Complete ionic equation:
2H²⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
Net ionic equation:
2H²⁺(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq)
Answer:
Complete ionic equation: 2 H⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
Net ionic equation: 2 H⁺(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq)
Explanation:
The molecular equation includes all the species in the molecular form.
H₂SO₄(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + CaSO₄(aq)
The complete ionic equation includes all the ions and the molecular species.
2 H⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.
2 H⁺(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq)
Identify the precipitate or lack thereof for the following:CaCl2(aq) + K2CO3(aq) --> CaCO3 + 2KClA) CaCO3B) KClC) No precipitateFeCl2(aq) + (NH4)2SO4(aq) --> FeSO4 + 2NH4ClA) FeSO4B) NH4ClC) No precipitate
Answer:
A) CaCO₃
C) No precipitate
Explanation:
To answer these questions we need to consider the solubility rules.
Identify the precipitate or lack thereof for the following:
CaCl₂(aq) + K₂CO₃(aq) ⇒ CaCO₃(s) + 2 KCl(aq)
Group II carbonates are insoluble. Thus, CaCO₃ is insoluble.Salts containing Group I cations are soluble. Thus, KCl is soluble.FeCl₂(aq) + (NH₄)₂SO₄(aq) ⇒ FeSO₄(aq) + 2 NH₄Cl(aq)
Most sulfates are soluble. Thus, FeSO₄ is soluble.Salts containing the ammonium ion are soluble. Thus, NH₄Cl is soluble.A gas obeys the ideal-gas equation of state P V = N k T , where N = n NA is the number of molecules in the volume V at pressure P and temperature T and n is the number of gm-moles of the gas. Calculate the volume occupied by 1 gm·mole of the gas at atmospheric pressure and a temperature of 313.7 K . Avogadro’s constant is 6.02214 × 1023 mol−1 , Boltzmann’s constant is 1.38065 × 10−23 N · m/K , and 1 atm = 1.013 × 105 N/m2 . Answer in units of L.
Answer:
The volume occupied is 25.7 L
Explanation:
Let's replace all the data in the formula
P . V = N . k . T
N = nNA
1 gm.mole . 6.02x10²³
k = Boltzmann's contant
T = T° in K
1 atm = 1.013 × 10⁵ N/m2
1.013 × 10⁵ N/m2 . Volume = 6.02x10²³ . 1.38065 × 10⁻²³ N · m/K . 313.7K
Volume = (6.02x10²³ . 1.38065 × 10⁻²³ N · m/K . 313.7K) / 1.013 × 10⁵ m2/N
Volume = 2607.32 N.m / 1.013 × 10⁵ m2/N = 0.0257 m³
1 dm³ = 1 L
1m³ = 1000 dm³
25.7 L
The volume of 1 gm·mole of gas at atmospheric pressure and a temperature of 313.7 K is approximately 25.7 Liters, as calculated using the Ideal Gas Law PV=nRT, with n=1, P=1 atm, T=313.7K, and R=0.08206 L·atm/mol·K.
Explanation:To calculate the volume occupied by 1 gm·mole of the gas at atmospheric pressure and a temperature of 313.7 K, you'll want to use the Ideal Gas Law (PV=nRT), where P is the pressure of a gas, V is the volume it occupies, n is the number of moles of the gas, T is the temperature, and R is the universal gas constant. We'll use the given values: n=1 gm·mole, P=1 atm, T=313.7K, and also utilize the known universal gas constant for these units R=0.08206 L·atm/mol·K.
The ideal gas equation PV=nRT can be rearranged to solve for the volume of the gas: V=nRT/P. Substituting the values, V=(1 gm·mole * 0.08206 L·atm/mol·K * 313.7K) / 1 atm. By performing the calculation, we find that the volume V is approximately 25.7 Liters.
Learn more about Ideal Gas Law here:https://brainly.com/question/30458409
#SPJ3
A mixture of carbon dioxide and hydrogen gases contains carbon dioxide at a partial pressure of 401 mm Hg and hydrogen at a partial pressure of 224 mm Hg. What is the mole fraction of each gas in the mixture?
Answer:
[tex]X_{H_2}=\0.3584[/tex]
[tex]X_{CO_2}=\0.6416[/tex]
Explanation:
According to the Dalton's law of partial pressure, the total pressure of the gaseous mixture is equal to the sum of the pressure of the individual gases.
Partial pressure of carbon dioxide = 401 mmHg
Partial pressure of hydrogen gas = 224 mmHg
Total pressure,P = sum of the partial pressure of the gases = 401 + 224 mmHg = 625 mmHg
Also, the partial pressure of the gas is equal to the product of the mole fraction and total pressure.
So,
[tex]P_{CO_2}=X_{CO_2}\times P[/tex]
[tex]X_{CO_2}=\frac{P_{CO_2}}{P}[/tex]
[tex]X_{CO_2}=\frac{401\ mmHg}{625\ mmHg}[/tex]
[tex]X_{CO_2}=\0.6416[/tex]
Similarly,
[tex]X_{H_2}=\frac{P_{H_2}}{P}[/tex]
[tex]X_{H_2}=\frac{224\ mmHg}{625\ mmHg}[/tex]
[tex]X_{H_2}=\0.3584[/tex]
A chemist prepares a solution of mercury(II) iodide HgI2 by measuring out 0.0067μmol of mercury(II) iodide into a 350.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's mercury(II) iodide solution. Round your answer to 2 significant digits.
Final answer:
The concentration of the chemist's mercury(II) iodide solution is 0.019 μmol/L after rounding to two significant digits.
Explanation:
The concentration of the chemist's mercury(II) iodide solution can be calculated using the formula for molarity, which is the number of moles of solute divided by the volume of the solution in liters.
In this case, we are given 0.0067μmol of HgI2 and a volume of 0.350 L:
Molarity (μM) = amount of solute (in micromoles) / volume of solution (in liters)Molarity (μM) = 0.0067μmol / 0.350 L = 0.01914μmol/L
After rounding to two significant digits, the concentration of the solution is 0.019 μM.
WILL MARK BRAINLIEST
What pressure (in atm and in bars) is exerted by a column of methanol (CH3OH) 305 m high? The density of methanol is 0.787 g/cm3.
Answer
23.52 atm pressure is exerted by a column of methanol
Explanation:
formula for pressure
Pressure = force / area
And we know that Force F= mass x acceleration (ma)
And mass = volume x density
And volume = cross sectional area x height?
so that...
P = F/A
Putting value of F
P = (mg)/A
Putting value for mass
P = (ρ V) g / A
Putting value for volume
P = ρ (A h) g / A
Canceling A
P = ρgh
Now add the value
now watch the units...
P = (0.787g/cm³) x 9.80 m/s² x 305m... units= g x m² / (cm³ s²)
and if you recall for pressure unit Pressure ...
1 Pa = 1 N / m² = (1 kg m / s²) / m²
so know
[g x m² / (cm³ s²)] x (1 kg / 1000g) x (100 cm / m)³
converts the units to (1 kg m² / s²) / m
= (1 kg m / s²) / m²
Addition of all values
P = (0.787g/cm³) x 9.80 m/s² x 305m x (1 kg / 1000g) x (100 cm / m)³ x (1 atm / 101325 Pa)
Pressure= 23.52 atm
Give two examples (each) of strong electrolyte, weak electrolyte, and nonelectrolyte. 2. Predict the products for reactions below. Which of the following reaction(s) produce a precipitate? A) LiOH + Na2S B) (NH4)2SO4 + LiCl C) Sr(C2H3O2)2 + Na2SO4 D) KNO3 + NaOH E) None of the above solution pairs will produce a precipitate. 3. What are the spectator ions in the precipitation reaction you chose above? 4. Write the molecular, complete ionic, and net ionic equations for the reactions in Q2.
Answer:
1. Strong electrolytes: HCl and NaOH.
Weak electrolytes: CH₃COOH and NH₃.
Nonelectrolytes: (NH₂)₂CO (urea) and CH₃OH (methanol).
2. C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]
3. The spectator ions are Na⁺ and C₂H₃O₂⁻.
4. Written in explanation section.
Explanation:
An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.A characteristic of strong electrolytes is that the solute is assumed to be 100 percent dissociated into ions in solution, therefore they are great conductors of electricity. Eg: HCl and NaOH.
On the other hand, weak electrolytes are not completely dissociated in solution, therefore they are poor conductors of electricity. Eg: CH₃COOH and NH₃.
(By dissociation we mean the breaking up of the compound into cations and anions).
A nonelectrolyte does not conduct electricity when dissolved in water. Eg: (NH₂)₂CO (urea) and CH₃OH (methanol).
2. The products formed on each reaction are (always remember to balance the equations):
A) [tex]2 LiOH (aq) + Na₂S (aq) → Li₂S (aq) + 2NaOH (aq)[/tex]
B) [tex](NH₄)₂SO4 (aq) +2 LiCl (aq) → 2NH₄Cl (aq) + Li₂SO₄ (aq)[/tex]
C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]
D) [tex]D) KNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + KOH (aq) [/tex]
The reaction C will produce SrSO₄, a white color precipitate.
3. When ionic compounds dissolve in water, they break apart into their component cations and anions. To be more realistic, the equations should show the dissociation of dissolved ionic compounds into ions. This is called a ionic equation which shows dissolved species as free ions. To see whether a precipitate might form from this solution, we first combine the cation and anion from different compounds, and refer to the solubility rules. The spectator ions are ions that are not involved in the overall reaction.
Therefore, for the equation chosen above:
[tex]Sr²⁺ + 2C₂H₃O₂⁻ + 2Na⁺ + SO₄²⁻ → SrSO₄ + 2Na⁺ + 2C₂H₃O₂⁻[/tex]
Because spectator ions appear on both sides of an equation, they can be eliminated from the ionic equation.
[tex]Sr²⁺ + SO₄²⁻ → SrSO₄[/tex]
Finally, we end up with the net ionic equation, which shows only the species that actually take part in the reaction.
In this reaction, the spectator ions are Na⁺ and C₂H₃O₂⁻.
4. Molecular equations:
A) [tex]2 LiOH (aq) + Na₂S (aq) → Li₂S (aq) + 2NaOH (aq)[/tex]
B) [tex](NH₄)₂SO4 (aq) +2 LiCl (aq) → 2NH₄Cl (aq) + Li₂SO₄ (aq)[/tex]
C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]
D) [tex]D) KNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + KOH (aq) [/tex]
Complete ionic equations:
A) [tex]2Li⁺ + OH⁻ + 2Na⁺ + S²⁻ → 2Li⁺ + S²⁻ + 2Na⁺ +OH⁻[/tex]
B) [tex]2NH₄⁺ + SO₄²⁻ + 2Li⁺ + 2Cl⁻ → 2Li⁺ +S²⁻ +2NH₄⁺ +2Cl⁻ + 2Li⁺ + SO₄²⁻[/tex]
C) [tex]Sr²⁺ + 2C₂H₃O₂⁻ + 2Na⁺ + SO₄²⁻ → SrSO₄ + 2Na⁺ + 2C₂H₃O₂⁻[/tex]
D) [tex]K⁺ + NO₃⁻ + Na⁺ +OH⁻ → Na⁺ + NO₃⁻ + K⁺ + OH⁻[/tex]
If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, no precipitation reaction occurs. The only net equation can be written for reaction C):
C) [tex]Sr²⁺ + SO₄²⁻ → SrSO₄[/tex]
Estrone, which contains only carbon, hydrogen, and oxygen, is a female sexual hormone that occurs in the urine of pregnant women. Combustion analysis of a 1.893-g sample of estrone produced 5.543g of CO2 and 1.388g H2O. The molar mass of estrone is 270.36g/mol .Find the molecular formula for estrone.Express your answer as a chemical formula.
Answer:
The formula of Estrone = [tex]C_{18}H_{22}O_2[/tex]
Explanation:
Mass of water obtained = 1.388 g
Molar mass of water = 18 g/mol
Moles of [tex]H_2O[/tex] = 1.388 g /18 g/mol = 0.07711 moles
2 moles of hydrogen atoms are present in 1 mole of water. So,
Moles of H = 2 x 0.07711 = 0.1542 moles
Molar mass of H atom = 1.008 g/mol
Mass of H in molecule = 0.1542 x 1.008 = 0.1555 g
Mass of carbon dioxide obtained = 5.543 g
Molar mass of carbon dioxide = 44.01 g/mol
Moles of [tex]CO_2[/tex] = 5.543 g /44.01 g/mol = 0.126 moles
1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,
Moles of C = 0.126 moles
Molar mass of C atom = 12.0107 g/mol
Mass of C in molecule = 0.126 x 12.0107 = 1.5133 g
Given that the Estrone only contains hydrogen, oxygen and carbon. So,
Mass of O in the sample = Total mass - Mass of C - Mass of H
Mass of the sample = 1.893 g
Mass of O in sample = 1.893 - 1.5133 - 0.1555 = 0.2242 g
Molar mass of O = 15.999 g/mol
Moles of O = 0.2242 / 15.999 = 0.01401 moles
Taking the simplest ratio for H, O and C as:
0.1542 : 0.01401 : 0.126
= 11 :1 : 9
The empirical formula is = [tex]C_9H_{11}O[/tex]
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 9×12 + 11×1 + 16= 135 g/mol
Given, Molar mass = 270.36 g/mol
So,
Molecular mass = n × Empirical mass
270.36 = n × 135
⇒ n = 2
The formula of Estrone = [tex]C_{18}H_{22}O_2[/tex]
Final answer:
To find the molecular formula of estrone, calculate the empirical formula first by finding the moles of carbon and hydrogen. Then, determine the mole ratio between carbon and hydrogen and round to the nearest whole number to get the empirical formula. Finally, divide the molar mass of estrone by the molar mass of the empirical formula to find the ratio, and round to the nearest whole number to get the molecular formula.
Explanation:
To find the molecular formula for estrone, we need to determine the empirical formula first. The empirical formula represents the simplest whole-number ratio of the elements in a compound. We can use the information from the combustion analysis to calculate the empirical formula.
First, we need to determine the moles of carbon and hydrogen in the estrone sample.
Moles of carbon = mass of CO₂ / molar mass of CO₂
Moles of carbon = 5.543 g / 44.01 g/mol = 0.1259 mol
Moles of hydrogen = mass of H₂O / molar mass of H₂O
Moles of hydrogen = 1.388 g / 18.015 g/mol = 0.0770 mol
Next, we need to find the mole ratio between carbon and hydrogen by dividing the moles of each element by the smallest mole value.
Dividing both moles of carbon and hydrogen by 0.0770 mol, we get the ratio 1.6367 : 1.
Finally, we round the ratio to the nearest whole number to obtain the empirical formula. In this case, the empirical formula is C2H2O.
To find the molecular formula, we need to know the molar mass of estrone. The molar mass of estrone is given as 270.36 g/mol. Dividing the molar mass of estrone by the molar mass of the empirical formula (42.08 g/mol), we get a ratio of approximately 6.42.
Rounding the ratio to the nearest whole number, we find that the molecular formula for estrone is C6H6O3.
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.
Estimate ΔG°rxn for the following reaction at 387 K. HCN (g) + 2 H2 (g) → CH3NH2 (g) ΔH° = −158.0 kJ; ΔS° = −219.9
Answer:
ΔG°rxn = -72.9 kJ
Explanation:
Let's consider the following reaction.
HCN(g) + 2 H₂(g) → CH₃NH₂(g)
We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:
ΔG°rxn = ΔH° - T.ΔS°
where,
ΔH° is the standard enthalpy of the reaction
T is the absolute temperature
ΔS° is the standard entropy of the reaction
ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)
ΔG°rxn = -72.9 kJ
The estimated ΔG°rxn for the given reaction HCN(g) + 2 H2(g) → CH3NH2(g) at 387K, calculated using the equation ΔG = ΔH - TΔS and the given values ΔH° = −158.0 kJ and ΔS° = −219.9 J/K, is -72.86 kJ.
Explanation:To estimate ΔG ° rxn for the given reaction HCN(g) + 2 H2(g) → CH3NH2(g) at 387K, we need to use the equation ΔG = ΔH - TΔS. Given that ΔH° = −158.0 kJ and ΔS° = −219.9 J/K (remember to convert kJ to J, so ΔH is -158000 J), we substitute the values into the equation: ΔG = (-158000 J) - (387K * -219.9 J/K) = -158 kJ + 85.14 kJ = -72.86 kJ. So, the estimated ΔG°rxn for the reaction under these conditions is -72.86 kJ.
Learn more about Thermodynamics here:brainly.com/question/35546325
#SPJ3
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.
A flask is charged with 3.00 atm of dinitrogen tetroxide gas and 2.00 atm of nitrogen dioxide gas at 25ºC and allowed to reach equilibrium.
N2O4(g)-->2 NO2(g) Kp = 0.316
Which of the following best describes the system within the flask once equilibrium has been established?
The rate of the decomposition of N2O4(g) is equal to the rate of formation of NO2(g).
The partial pressure of N2O4(g) is equal to the partial pressure of NO2(g).
The rate of the decomposition of N2O4(g) is greater than the rate of formation of NO2(g).
The rate of the decomposition of N2O4(g) is less than the rate of formation of NO2(g).
Answer:
The correct answer is The rate of decomposition of N2O4 is equal to the rate of formation of NO2.
Explanation:
According to theory of reaction kinetics we all know that at equilibrium the amount of reactant is equal to the amount of product.
In other word it can be stated that the rate of the decomposition of N2O4 is equal to the rate of the formation of NO2.
Final answer:
At equilibrium, the partial pressure of N2O4 is equal to the partial pressure of NO2 in the flask.
Explanation:
Once equilibrium has been established, the partial pressure of N2O4(g) is equal to the partial pressure of NO2(g) in the flask. This is because the equilibrium constant (Kp) expression for the decomposition of N2O4 is (PNO2)^2 / PN2O4, which means the partial pressures of the two gases are directly related. Therefore, answer choice B, 'The partial pressure of N2O4(g) is equal to the partial pressure of NO2(g),' best describes the system within the flask at equilibrium.
When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydrofluoric acid results. Write the net ionic equation for the reaction. (Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)
Answer:
F⁻(aq) + H⁺(aq) ⇄ HF(aq)
Explanation:
When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydrofluoric acid results. The corresponding molecular equation is:
KF(aq) + HCl(aq) ⇄ KCl(aq) + HF(aq)
The full ionic equation includes all the ions and the molecular species. HF is a weak acid so it exists mainly in the molecular form.
K⁺(aq) + F⁻(aq) + H⁺(aq) + Cl⁻(aq) ⇄ K⁺(aq) + Cl⁻(aq) + HF(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.
F⁻(aq) + H⁺(aq) ⇄ HF(aq)
What is the percentage composition of each element in hydrogen peroxide, H2O2?
A)7.01% H and 92.99% O
B)7.22% H and 92.78% O
C)6.32% H and 93.68% O
D)5.88% H and 94.12% O
Answer:
The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).
Explanation:
Step 1: Data given
Molar mass of H = 1.0 g/mol
Molar mass of O = 16 g/mol
Molar mass of H2O2 = 2*1.0 + 2*16 = 34.0 g/mol
Step 2: Calculate % hydrogen
% Hydrogen = ((2*1.0) / 34.0) * 100 %
% hydrogen = 5.88 %
Step 3: Calculate % oxygen
% Oxygen = ((2*16)/34)
% oxygen = 94.12 %
We can control this by the following equation
100 % - 5.88 % = 94.12 %
The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).
The percentage composition of hydrogen peroxide (H₂O₂) is D) 5.88% H and 94.12% O.
The calculation can be represented as follows:
1) Molar masses:
H = 1.008 g/mol O = 15.999 g/mol2) Molar mass of [tex]H_2O_2:[/tex]
[tex]$\ce{H_2O_2} = (2 \times 1.008) + (2 \times 15.999) = 34.014$ g/mol[/tex]3) Percentage of hydrogen:
[tex]\[ \% \ce{H} = \frac{\text{mass of H}}{\text{total mass}} \times 100\% = \frac{2 \times 1.008}{34.014} \times 100\% \approx 5.88\% \][/tex]4) Percentage of oxygen:
[tex]\[ \% \ce{O} = 100\% - \% \ce{H} = 100\% - 5.88\% = 94.12\% \][/tex]Therefore, the percentage composition of [tex]$\ce{H_2O_2}$[/tex] is approximately 5.88% H and 94.12% O.
Comparing with the given options, we can see that the closest match is D) 5.88% H and 94.12% O
When 15.5 g of an organic compound known to be 70.58 % C , 5.9 % H , and 23.50 % O by mass is dissolved in 718.1 g of cyclohexane, the freezing point is 3.29 ∘ C . The normal freezing point of cyclohexane is 6.59 ∘ C . What is the molecular formula for the organic compound?Assume that the organic compound is a molecular solid and does not ionize in water. Kf values for various solvents are given here.
Answer:
The molecular formula for the organic compound is C₈H₈O₂
Explanation:
70.58 % C , 5.9 % H , and 23.50 % O
This is a percent by mass, so:
in 100 g of compound I have 70.58 g of C, 5.9 g of H and 23.5 g of O.
We need to apply the formula for freezing point depression, the colligative property.
ΔT = Kf . m
Where ΔT means T°F (pure solvent) − T°F (solution)
Kf, the cryoscopic constant and m is molality (mol of solute in 1kg of solvent); Kf for cyclohexane is 20.8 °C/m
6.59°C - 3.29°C = 20.8 °C/m . m
3.3°C = 20.8 °C/m . m
3.3°C /20.8 m/°C = molal
0.158 m/kg
In 1kg of solvent I have 0.158 moles, but I have 718.1 g of cyclohexane
1000 g _____ 0.158 moles of solute
718.1 g _____ 0.114 moles of solute
To find out the molar mass of my compound I have to divide
mass/mol → 15.5 g /0.114 m = 136.04 g/m
Now I have to work with the percent and think this rules of three:
100 g of compound has ____ 70.58 g of C __5.9 g of H __23.5 g of O
136.04 g of compound has ___ 96.02 g of C __8.02 g of H __ 32 g of O
Molar mass of C = 12 g/m
Molar mass of H = 1 g/m
Molar mass of O = 16 g/m
g / molar mass = moles
C: 96 g / 12 g/m = 8
H: 8 g / 1 g/m =8
O: 32 g/ 16 g/m = 2
The molecular formula for the organic compound is C₈H₈O₂
To find the molecular formula of the organic compound, calculate the empirical formula first, which is CH₃O. Then calculate the molar mass of the compound, which is 29g/mol. Divide the molar mass of the compound by the molar mass of the empirical formula to find the molecular formula ratio. The molecular formula is CH₃O.
Explanation:To determine the molecular formula of the organic compound, we need to calculate the empirical formula first.
The percentages of carbon, hydrogen, and oxygen in the compound represent the mass ratios of these elements. Assume we have 100g of the compound, which means we have 70.58g of carbon, 5.9g of hydrogen, and 23.50g of oxygen. Now we convert these masses to moles using the molar masses: 12g/mol for carbon, 1g/mol for hydrogen, and 16g/mol for oxygen.
Next, we divide the moles of each element by the smallest number of moles to get the simplest mole ratio. The empirical formula of the compound is CH3O.
To find the molecular formula, we need to know the molar mass of the compound. This can be calculated by adding the atomic masses of the elements in the empirical formula: 12g/mol for carbon, 1g/mol for hydrogen, 16g/mol for oxygen. The sum is 29g/mol.
Since the molar mass of the compound is 29g/mol and 15.5g of the compound is dissolved in 718.1g of cyclohexane, we can calculate the moles of the compound: 15.5g / 29g/mol = 0.534moles.
The molecular formula is found by multiplying the empirical formula by an integer value to match the molar mass. To do this, divide the molar mass of the compound by the molar mass of the empirical formula: 29g/mol / 29g/mol = 1.
Therefore, the molecular formula of the organic compound is CH3O.
Learn more about Molecular formula determination here:https://brainly.com/question/13397333
#SPJ11
Consider an AB3 molecule in which A and B differ in elec-tronegativity. You are told that the molecule has an over-all dipole moment of zero. Which of the following could be the molecular geometry of the molecule? (a) Trigonal pyramidal (b) Trigonal planar (c) T-shaped (d) Tetrahedral (e) More than one of the above
Answer:
(b) Trigonal planar
Explanation:
The molecular geometry is the one that stabilizes better the bonds and the free electron pairs. If the molecule is nonpolar (overall dipole moment zero), so, there's no free electron pairs at the central atom. So, the molecule has the central atom A surrounded by three atoms of B, which is the trigonal planar geometry.
In an AB3 molecule, where A and B differ in electronegativity, for the molecule to have an overall dipole moment of zero, the molecular shape must be calculated in a way that the polar bonds sum up to zero. The correct molecular shape in this case is Trigonal Planar. In this geometry, all three B atoms are symmetrically distributed in a plane around the atom A (120° apart), which leads to cancellation of dipole moments.
Explanation:In order for a molecule with disparity in electronegativity (polar bonds) to have an overall dipole moment of zero, the molecular structure needs to be in such a way that the dipole vectors cancel each other out. This typically happens when the molecule is symmetric. In the case of an AB3 molecule where A and B differ in electronegativity, the geometrical structures under consideration that could meet this condition are: Trigonal pyramidal, Trigonal planar, T-shaped, and Tetrahedral.
Upon reviewing these, the correct answer is (b) Trigonal planar. This is because this geometry allows for all three B atoms to be distributed in a plane around the A atom symmetrically (120° apart), allowing the dipole moments from the polar bonds to cancel each other out, which results in an overall zero dipole moment. The other geometries do not allow such cancellation, hence they can't be the correct answer.
To understand better, trying to draw the shapes out or use molecular 3D models can help visualise the planar shape and how the dipole vectors cancel each other out.
Learn more about Molecular Geometry here:https://brainly.com/question/30185738
#SPJ3
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.
The complex [Zn(NH3)2Cl2]2+ does not exhibit cis-trans isomerism. The geometry of this complex must be ________.
a. tetrahedral
b. trigonal bipyramidal
c. square planar
d. octahedral
e. either tetrahedral or square planar
Answer:
A. tetrahedral
Explanation:
The only complex ion geometries that can exhibit cis-trans isomerim are the octahedral and square planar.With the above information in mind, the correct answer would then have to be either a) or b), however the complex [Zn(NH₃)₂Cl₂]²⁺ has 4 ligands. This means that the correct answer is a), because the trigonal bipyramidal geometry has 5 ligands.
Fructose-1-P is hydrolyzed according to: Fructose-1-P + H2O → Fructose + Pi If a 0.2 M aqueous solution of Froctose-1-P is allowed to reach equilibrium, its final concentration is 6.52 × 10-5 M.
What is the standard free energy of Froctose-1-P hydrolysis?
Answer:
[tex]\Delta G^{\circ}=-15902 J/mol[/tex]
Explanation:
In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):
[tex]\Delta G^{\circ}=-R*T*ln(K_{eq})[/tex]
Being Keq:
[tex]K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}[/tex]
Initial conditions:
[tex][Fructose-1-P]=0.2M[/tex]
[tex][Fructose]=0M[/tex]
[tex][Pi]=0M[/tex]
Equilibrium conditions:
[tex][Fructose-1-P]=6.52*10^{-5}M[/tex]
[tex][Fructose]=0.2M-6.52*10^{-5}M[/tex]
[tex][Pi]=0.2M-6.52*10^{-5}M[/tex]
[tex]K_{eq}=\frac{(0.2M-6.52*10^{-5}M)*(0.2M-6.52*10^{-5}M)}{6.52*10^{-5}M}[/tex]
[tex]K_{eq}=613.1[/tex]
Free-energy for T=298K (standard):
[tex]\Delta G^{\circ}=-8.314\frac{J}{mol*K}*298K*ln(613.1)[/tex]
[tex]\Delta G^{\circ}=-15902 J/mol[/tex]
The decomposition reaction of N2O5 in carbon tetrachloride is 2N2O5−→−4NO2+O2. The rate law is first order in N2O5. At 64 °C the rate constant is 4.82×10−3s−1. (a) Write the rate law for the reaction. (b) What is the rate of reaction when [N2O5]=0.0240M? (c) What happens to the rate when the concentration of N2O5 is doubled to 0.0480 M? (d) What happens to the rate when the concentration of N2O5 is halved to 0.0120 M?
Answer:
(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]
(b) rate = 1.16 x 10⁻⁴ M/s
(c) rate = 2.32 x 10⁻⁴ M/s
(d) rate = 5.80 x 10⁻⁵ M/s
Explanation:
We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so
(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]
(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s
(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s
(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to
1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s
Answer:
a) r = 4.82x10⁻³*[N2O5]
b) 1.16x10⁻⁴ M/s
c) The rate is doubled too (2.32x10⁻⁴ M/s)
d) The rate is halved too (5.78x10⁻⁴ M/s)
Explanation:
a) The rate law of a generic reaction (A + B → C + D) can be expressed by:
r = k*[A]ᵃ*[B]ᵇ
Where k is the rate constant, [X] is the concentration of the compound X, and a and b are the coefficients of the reaction (which can be different from the ones of the chemical equation).
In this case, there is only one reactant, and the reaction is first order, which means that a = 1. So, the rate law is:
r = k*[N2O5]
r = 4.82x10⁻³*[N2O5]
b) Substituing the value of the concentration in the rate law:
r = 4.82x10⁻³*0.0240
r = 1.16x10⁻⁴ M/s
c) When [N2O5] = 0.0480 M,
r = 4.82x10⁻³*0.0480
r = 2.32x10⁻⁴ M/s
So, the rate is doubled too.
d) When [N2O5] = 0.0120 M,
r = 4.82x10⁻³*0.0120
r = 5.78x10⁻⁴ M/s
So, the rate is halved too.
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.
The following reaction is found to be at equilibrium at 25 celcius: 2SO3--->O2 + 2SO2 + 198kJ/mol.
What is the expression for the equilibrium constant, Kc?
A.) [SO3]^2/[O2][SO2]^2
B.) 2[SO3]/[O2]2[SO2]
C.) [O2][SO2]^2/[SO3]^2
D.) [O2]2[SO2]/2[SO3]
Final answer:
The expression for the equilibrium constant (Kc) for the reaction 2SO₃ --> O₂ + 2SO₂ + 198 kJ/mol is C.) [O₂][SO₂]^2/[SO₃]^2, according to the law of equilibrium.
Explanation:
The question involves understanding how to write the expression for the equilibrium constant (Kc) for a given chemical reaction. Given the reaction 2SO₃ --> O₂ + 2SO₂ + 198 kJ/mol, the correct expression for Kc reflects the concentration of products over reactants, raised to the power of their stoichiometric coefficients in the balanced equation.
The correct expression for Kc is therefore C. [O₂][SO₂]^2/[SO₃]^2. This follows from the equilibrium law, which states that Kc is calculated by taking the concentration of the products, [O₂] and [SO₂] squared (because the coefficient of SO₂ is 2), over the concentration of the reactants, [SO₃] squared (because the coefficient of SO₃ is also 2).
Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How many milliliters of 5.50 M HCl (aq) are required to react with 6.25 g Zn (s) ?
Answer:
34.7mL
Explanation:
First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.
So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn
All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.
So now we need to go to HCl!
To do that we multiply by the molar coefficients in the chemical equation:
[tex]\frac{0.0956g Zn}{1 } (\frac{2 mol HCl}{1molZn})[/tex]
This leaves us with 2(0.0956) = 0.1912 mol HCl
Now we use the relationship M= moles / volume , to calculate our volume
Rearranging we get that V = moles / M
Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl
V= 0.0347 L
To change this to milliliters we multiply by 1000 so:
34.7 mL
The statement that "the lowest energy configuration for an atom is the one having the maximum number of unpaired electrons allowed by the Pauli principle in a particular set of degenerate orbitals" is known as: A. the quantum model B. Hund's rule C. the aufbau principle D. Heisenberg uncertainty principle E. the Pauli exclusion principle
Answer:
B. Hund's rule
Explanation:
Hund's rule -
According to Hund's rule ,
As, the electrons are negatively charged and hence , like poles repel , so they repel each other in order to stabilizes themselves and hence , the electron firstly occupies a vacant orbital , before actually pairing up , so as to reduce repulsion .
The rule of maximum multiplicity states that ,
The term with the lowest energy is the one with the highest value of the spin multiplicity .
hence , the statement given in the question is about Hund's rule .
Answer:
its actually A
Explanation:
Calculate the number of free electrons per cubic meter for silver, assuming that there are 1.3 free electrons per silver atom. The electrical conductivity and density for Ag are 6.8 × 107(Ω.m)–1and 10.5 g/cm3, respectively.
Answer : The number of free electrons per cubic meter for silver are [tex]7.62\times 10^{28}m^{-3}[/tex]
Explanation :
To calculate the number of free electrons per cubic meter for silver by using the following equation:
[tex]n=1.3\times N_{Ag}[/tex]
[tex]n=1.3\times \frac{\rho_{Ag}\times N_A}{A_{Ag}}[/tex]
where,
[tex]\rho_{Ag}[/tex] = density of Ag = [tex]10.5g/cm^3[/tex]
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}atoms/mol[/tex]
[tex]A_{Ag}[/tex] = 107.87 g/mol
Now put all the given values in the above formula, we get:
[tex]n=1.3\times \frac{\rho_{Ag}\times N_A}{A_{Ag}}[/tex]
[tex]n=1.3\times \frac{(10.5g/cm^3)\times (6.022\times 10^{23}atoms/mol)}{107.87g/mol}[/tex]
[tex]n=7.62\times 10^{22}cm^{-3}[/tex]
[tex]n=7.62\times 10^{28}m^{-3}[/tex]
conversion used : [tex]1cm^{-3}=10^6m^{-3}[/tex]
Therefore, the number of free electrons per cubic meter for silver are [tex]7.62\times 10^{28}m^{-3}[/tex]
Final answer:
To calculate the number of free electrons per cubic meter for silver, we can use the density and molar mass of silver to determine the number of silver atoms per cubic meter. Since there is one free electron per silver atom, we can then calculate the number of free electrons per cubic meter.
Explanation:
To calculate the number of free electrons per cubic meter for silver, we first need to calculate the number of silver atoms per cubic meter. We can do this by using the density of silver, which is given as 10.5 g/cm³. Converting this to kg/m³ gives us a density of 10500 kg/m³. Next, we need to find the molar mass of silver, which is 107.87 g/mol. Using Avogadro's number (6.02 × 10^23 atoms/mol), we can calculate the number of silver atoms per cubic meter.
n = (10500 kg/m³) / (107.87 g/mol * (1 kg / 1000 g) * (6.02 × 10^23 atoms/mol))
n ≈ 9.48 x 10^28 atoms/m³
Since there is one free electron per silver atom, we can conclude that there are approximately 9.48 x 10^28 free electrons per cubic meter of silver.