Answer:
1.13 × 10⁶ g
Explanation:
Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.
Al³⁺ + 3 e⁻ → Al
We can establish the following relations:
- 1 Ampere = 1 Coulomb / second
- The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons circulate
- The molar mass of Al is 26.98 g/mol.
The mass of aluminum produced under these conditions is:

Answer:
133.74 L
Explanation:
First we <u>convert the given pressures and temperatures into atm and K</u>, respectively:
- 750.0 Torr ⇒ 750/760 = 0.9868 atm
- 20°C ⇒ 20+273.16 = 293.16 K
- 40°C ⇒ 40+273.16 = 313.16 K
Then we<u> use the PV=nRT formula to calculate the number of moles of helium in the balloon</u>, using<em> the data of when it was on the ground</em>:
- 0.9868 atm * 8.50 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293.16 K
Then, knowing the value of n, we <u>use PV=nRT once again, this time to calculate V</u> using <em>the data of when the balloon was high up:</em>
- 0.550 atm * V = 2.866 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 313.16 K
Answer:
Antimony, Sb, an atom contains 51 protons inside the nucleus and 51 electrons outside the nucleus.
Explanation:
Option “C” is correct because protons exist inside the nucleus and electrons exist outside the nucleus. Moreover, Sb has the number of protons 51 that is found in the nucleus and it has 51 electrons that exist outside the nucleus. However, the nucleus contains protons and neutrons while electrons exist outside of the nucleus. All the given options do not follow such condition therefore, all the statements are incorrect accept option 3rd.
Answer:
Electronegativities generally decrease from top to bottom within a group due to the larger atomic size.
Of the main group elements, fluorine has the highest electronegativity (EN= 4.0) and cesium the lowest (EN=0.79).
Explanation: