Answer: B) 2 (as indicated by electron distribution shown), but taking into account the real properties of this element, 4,7,8 also occur (see below).
Explanation:
This is the electron complement/atomic number of ruthenium, which actually has the structure [Kr] 5s1 4d7
Nevertheless, Ru does not form Ru(I) compounds and few Ru(II) compounds (RuCl2, RuBr2, RuI2). It also forms Ru(III)Cl3 and a larger number of Ru(IV) compounds, e.g. RuO2, RuS2. It also forms RuO4
When electrons are filling energy levels, the lowest energy sublevels are occupied first. This is Hund's rule.
Hund's rules state that:
Every orbital in a sublevel has to be singularly occupied before any other orbital is able to be doubly occupied.
All of the electrons in single occupied orbitals have to have the same spin to maximize the total spin.
The total number of atoms in 7.10g of chlorine is 1.204 × 10²³atoms.
HOW TO CALCULATE NUMBER OF ATOMS:
- The number of atoms in a substance can be calculated by multiplying the number of moles in that substance by Avogadro's number as follows:
- no. of atoms = no. of moles × 6.02 × 10²³ mol-¹
- The number of moles in 7.10g of Cl is calculated as follows:
no. of moles = mass ÷ molar mass
no. of moles = 7.10g ÷ 35.5g/mol
no. of moles = 0.2mol
no of atoms = 0.2mol × 6.02 × 10²³
no. of atoms = 1.204 × 10²³atoms.
- Therefore, the total number of atoms in 7.10g of chlorine is 1.204 × 10²³atoms.
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Answer:
Sodium bicarbonate
Explanation:
Sodium bicarbonate ( NaHCO₃ ) -
Sodium bicarbonate , according to the IUPAC nomenclature , its name is sodium hydrogen carbonate ,and in common terms also refereed to as baking soda .
It is a white crystalline solid , it is basic in nature .
<u>The cation and anion of this salt are the sodium ion ( Na⁺) and the anion bicarbonate anion (HCO³⁻) .</u>
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Answer:
About 7.9 L.
Explanation:
We can utilize the ideal gas law. Recall that:
Because the amount of carbon dioxide does not change, we can rearrange to formula to:
Because the right-hand side stays constant, we have that:
Hence substitute initial values and known final values:
Therefore, the final volume is about 7.9 L.
