The answer would be 425.599 because 1 ATM is 760 mmHg.
Answer:
1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵
Explanation:
1) Ca:
Its atomic number is 20. So it has 20 protons and 20 electrons.
Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.
So, you have two more electrons (20 - 18 = 2) to distribute.
Those two electrons go the the orbital 4s.
Finally, the electron configuration is [Ar] 4s².
2) Pm
The atomic number of Pm is 61, so it has 61 protons and 61 electrons.
Pm is in the row (period) 6. So, the noble gas before Pm is Xe.
The atomic number of Xe is 54.
Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.
The resultant distribution for Pm is: [Xe]6s² 4f⁵.
<span>Start with the number of grams of each element, given in the problem.Convert the mass of each element to moles using the molar mass from the periodic table.Divide each mole value by the smallest number of moles calculated.<span>Round to the nearest whole number.</span></span>
Answer:
14 mol e⁻
Explanation:
Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese
8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)
Step 2: Calculate the moles corresponding to 110 g of manganese
The molar mass of Mn is 55 g/mol.
110 g × 1 mol/55 g = 2 mol
Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn
According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.
2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻
Answer:
7?
Explanation:
Its somewhat hard to comprehend the question, but if the way I read it was right, its 7.
