Answer:
Rb: [Kr] 5s
Step-by-step explanation:
Rb is element 37, the first element in Period 5.
It has one valence electron, so its valence electron configuration is 5s.
The noble gas configuration uses the symbol of the previous noble gas as a shortcut for the electron configurations of the inner electrons.
The preceding noble gas is Kr, so the electron configuration is Rb: [Kr] 5s.
Yes, electron follows the same path when it absorb and loses energy.
Yes, when an electron moves from a higher orbit to a lower orbit it always follow the same path as it moves from a lower orbit to a higher orbit. When electron absorb energy it has the power to move from lower orbit to higher orbit or energy level.
While on the other hand, when an electron loses that energy, it comes back to its original position from which it moves earlier when it absorb energy so we conclude that electron follows the same path when it absorb and loses energy.
Learn more: brainly.com/question/24962163
Since the exponent is negative, you move the decimal (2.0) to the left two spots leaving you with .02
Answer:
Mass = 88.12 g
Explanation:
Given data:
Mass of iron oxide = 126 g
Mass of iron formed = ?
Solution:
Chemical equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Number of moles of iron oxide:
Number of moles = mass/molar mass
Number of moles = 126 g/ 159.69 g/mol
Number of moles = 0.789 mol
Now we will compare the moles of iron with iron oxide.
Fe₂O₃ : Fe
1 : 2
0.789 : 2/1×0.789 = 1.578 mol
Mass of iron:
Mass = number of moles ×molar mass
Mass = 1.578 mol × 55.84 g/mol
Mass = 88.12 g
Answer:
D. You can no longer tell if your original sample contained Na+ or K+.
Explanation:
Group reagents are the reagents which are used to separate a whole group of cations from the other groups of cations.
For example, for group 1 , the group reagent is HCl and for group 5, the group reagent is ammonium carbonate.
<u>When sodium chloride and potassium carbonate is used, it means we are contaminating our sample with foreign cations which may interfere in the process of detection. Also, we will not be able to decipher whether the sample has these ions in the initial or not as we have added them.</u>
