The information that the third quantum number of an electron gives is the direction the electron in spinning. That is option A.
<h3>What are quantum numbers?</h3>
Quantum number are those numbers that are used to specify the properties of the atomic orbitals and the electrons in those orbitals.
The types of quantum numbers include the following:
The third quantum number is shows the direction of the electron while spinning through specifying its angular momentum.
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Answer:
3853 g
Step-by-step explanation:
M_r: 107.87
16Ag + S₈ ⟶ 8Ag₂S; ΔH°f = -31.8 kJ·mol⁻¹
1. Calculate the moles of Ag₂S
Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S
2. Calculate the moles of Ag
Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag
3. Calculate the mass of Ag
Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag
You must react 3853 g of Ag to produce 567.9 kJ of heat
There are 1.48 × 10²⁵ molecules of zinc oxide in a 2 kg sample. Details about number of molecules can be found below.
<h3>How to calculate number of molecules?</h3>
The number of molecules of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.
According to this question, there are 2000g of ZnO in a sample. Zinc oxide has a molar mass of 81.38 g/mol.
no of moles = 2000g ÷ 81.38g/mol
no of moles = 24.57mol
number of molecules = 24.57 × 6.02 × 10²³
number of molecules = 147.95 × 10²³
Therefore, there are 1.48 × 10²⁵ molecules of zinc oxide in a 2 kg sample.
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Answer:

Explanation:
A reaction is spontaneous at all temperatures by the following combinations:
=> A negative enthalpy change (
)
=> A positive entropy change (
)
See the attached file for more better understanding!
Answer:
0.51
Explanation:
Given the Nernst equation;
E= E° - 0.0592/n logQ
E= 355 mV or 0.355 V
E° = 0.34 - 0= 0.34 V
n= 2(two electrons were transferred in the process)
Equation of the reaction;
H2(g) + Cu^2+(aq) -----> 2H^+(aq) + Cu(s)
Substituting values;
0.355 = 0.34 - 0.0592/2 log([H^+]/1)
0.355 - 0.34 = - 0.0296 log [H^+]
0.015/-0.0296 = log [H^+]
Antilog (-0.5068) = [H^+]
[H^+] = 0.311 M
pH = -log[H^+]
pH= - log(0.311 M)
pH = 0.51