Which of the following required Bohr's model of the atom to need modification ? A. Energies of electrons are quantized. B. Quantized electron energies are responsible for emission spectra lines. C. An electron's energy increases the farther it moves from the nucleus. D. Electrons do not follow circular orbits around the nucleus..

Answers

Answer 1

Answer:

Electrons do not follow circular orbits around the nucleus

Explanation:

Bohr's model of the atom is a combination of elements of quantum theory and classical physics in approaching the problem of the hydrogen atom. According to Neils Bohr, stationary states exist in which the energy of the electron is constant. These stationary states were referred to as circular orbits which encompasses the nucleus of the atom. Each orbit is characterized by a principal quantum number (n). Energy is absorbed or emitted when an electron transits between stationary states in the atom.

Sommerfeld improved on Bohr's proposal by postulating that instead of considering the electron in circular orbits, electrons actually orbited around the nucleus in elliptical orbits, this became a significant improvement on Bohr's model of the atom until the wave mechanical model of Erwin Schrödinger was proposed.

Answer 2

Answer:

Answer:

Electrons do not follow circular orbits around the nucleus

Explanation:

Related Questions

Calculate the individual percent recoveries of benzoic acid, naphthalene and 3-nitroaniline if you were able to collect 9.75 g of benzoic acid, 6.41 g of naphthalene, and 7.71 g of 3-nitroaniline from a set of extractions. The starting mass of the mixture was 26.24 g. (0.6 pt)
Describe the range of radii of most atoms in nanometers (nm)
To completely neutralize a 0.325 g sample of pure aspirin, 15.50 mL of a sodium hydroxide solution is added. If 16.25 mL of the same sodium hydroxide solution must be added to an aspirin tablet sample during a titration to reach the endpoint, calculate the mass of aspirin in the tableA. 0.310 g B. 0.288 g C. 0.392 g D. 0.450 g E. 0.341 g
Consider the reaction, Cl2 + H2S => 2 HCl + S, which is found to be first order in Cl2. Which step of the proposed mechanism must be slow in order to agree with this rate law? Cl2 => 2 Cl Cl + H2S => HCl + HS Cl + HS => HCl + S A. only 3 B. only 2 C. 1 D. Either 2 or 3
100.0 g of liquid copper (molar mass 63.546 g/mol; melting point 1358 K; density 8.02 g/mL) is placed in a rigid container of volume 10.0 L at temperature 1508 K. The container is placed in an evacuated chamber and a small hole of area 3.23 mm2 is made in the upper container wall. After 2.00 hours, the mass of copper in the container has decreased by 1 0.0168 g. Assuming the mass loss is due to effusion, calculate the vapor pressure of liquid copper at 1508 K. Hint: because the liquid constantly evaporates, the pressure inside the container is constant

A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2 (aq) and Mg(s), producing Cd(s) and Mg2 (aq). Write half-reactions for the anode and cathode and then write a balanced cell reaction. Please include the states of matter in the equations.

Answers

Answer:

Cd²⁺(aq) + 2 e⁻ → Cd(s)

Mg(s) → Mg²⁺(aq) + 2 e⁻

Cd²⁺(aq) + Mg(s) → Cd(s) + Mg²⁺(aq)

Explanation:

A voltaic cell is composed of two half-reactions:

Reduction (cathode): Cd²⁺(aq) + 2 e⁻ → Cd(s)

Oxidation (anode): Mg(s) → Mg²⁺(aq) + 2 e⁻

Cd²⁺ reduces to Cd (the oxidation number decreases from 2 to 0), whereas Mg oxidizes to Mg²⁺ (the oxidation number increases from 0 to 2). We can get the overall cell reaction by adding both half-reactions:

Cd²⁺(aq) + 2 e⁻ + Mg(s) → Cd(s) + Mg²⁺(aq) + 2 e⁻

Cd²⁺(aq) + Mg(s) → Cd(s) + Mg²⁺(aq)

QUESTION 2: The molar mass for this compound is 1.11g / m * o * l The molecular formula for this compound is

Answers

I need the percentage again dude

A chemist titrates 80.0mL of a 0.3184M pyridine C5H5N solution with 0.5397M HBr solution at 25°C . Calculate the pH at equivalence. The pKb of pyridine is 8.77.

Answers

Answer:pH = 2.96

Explanation:

C5H5N + HBr --------------> C5H5N+ + Br-

millimoles of pyridine = 80 x 0.3184 =25.472mM

25.472 millimoles of HBr must be added to reach equivalence point.

25.472 = V x 0.5397

V =25.472/0.5397= 47.197 mL HBr

total volume = 80 + 47.197= 127.196 mL

Concentration of [C5H5N+] = no of moles / volume=

25.472/ 127.196= 0.20M

so,

pOH = 1/2 [pKw + pKa + log C]

pKb = 8.77

pOH = 1/2 [14 + 8.77 + log 0.20]

pOH = 11.0355

pH = 14 - 11.0355

pH = 2.96

Calculate the molarity of each solution.a. 0.38 mol of lino3 in 6.14 l of solution
b. 72.8 g c2h6o in 2.34 l of solution
c. 12.87 mg ki in 112.4 ml of solution

Answers

Q1)
molarity is defined as the number of moles of solute in 1 L solution
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M

Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M

Q3)

Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M

The molarities of the given solutions: (a). 0.38 mol of LiNO₃ in 6.14 L of solution has a molarity of 0.062 M. (b). 72.8 g of C₂H₆O in 2.34 L of solution has a molarity of 0.675 M. (c). 12.87 mg of KI in 112.4 mL of solution has a molarity of 0.000688 M.

To calculate the molarity (M) of a solution, you can use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

a. 0.38 moles of LiNO₃ in 6.14 L of solution:

Molarity (M) = 0.38 moles / 6.14 L = 0.062 M

b. 72.8 grams of C₂H₆O (ethyl alcohol) in 2.34 L of solution:

First, you need to convert grams to moles using the molar mass of C₂H₆O.

Molar mass of C₂H₆O = 2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.08 g/mol

Now, calculate moles of C₂H₆O:

moles = 72.8 g / 46.08 g/mol = 1.58 moles

Molarity (M) = 1.58 moles / 2.34 L = 0.675 M

c. 12.87 mg of KI in 112.4 mL of solution:

First, convert milligrams to grams (1 g = 1000 mg):

12.87 mg = 12.87 g (since 12.87 mg / 1000 = 0.01287 g)

Now, convert mL to liters (1 L = 1000 mL):

112.4 mL = 0.1124 L

Calculate moles of KI:

Molar mass of KI = 39.10 g/mol (for K) + 126.90 g/mol (for I) = 166.00 g/mol

moles = 0.01287 g / 166.00 g/mol = 7.75 × 10⁻⁵ moles

Molarity (M) = (7.75 × 10⁻⁵ moles) / 0.1124 L = 0.000688 M

So, the molarities of the solutions are as follows:

a. 0.062 M

b. 0.675 M

c. 0.000688 M

To know more about moles:

brainly.com/question/34302357

#SPJ3

A 0.7 ft diameter hole forms in a tank containing butane at 19 atmg and 76 degrees Fahrenheit. Determine the maximum possible mass flow rate through this leak in lb m / min, if the external pressure is 1 atm.

Answers

Answer:

Q = 3,534.4 lbm/s = 212,062 lbm/min

Explanation:

Mass flowrate of discharge or leakage mass flowrate (Q) is given as

Q = AC₀√(2ρgP)

A = Cross sectional Area of leakage = (πD²/4) = (π×0.7²)/4

A = 0.385 ft²

C₀ = discharge coefficient = 0.98 (For maximum discharge flow rate, the flow is turbulent with discharge coefficient within 1% of 0.98)

ρ = density of butane at 76°F = 35.771 lbm/ft³

g = acceleration due to gravity = 32.2 lbm.ft/lbf.s²

P = Gauge Pressure in the tank = (absolute pressure) - (external pressure) = 19 - 1 = 18 atm = 38091.9 lbf/ft²

Q = AC₀√(2ρgP)

Q = (0.385)(0.98)√(2×35.771×32.2×38091.9)

Q = 3,534.4 lbm/s = 212,062 lbm/min

Hope this Helps!!!

50 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M sodium acetate (the conjugate base). The Ka of acetic acid is approximately 1. 74 X 10 -5. What is the pH of the resulting solution?