Answer:
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The gases that get released form bubbles in the solution
The answer to the problem is 4.5 kilometers. you can solve this problem by cross multiplying
Answer:-
Carbon
[He] 2s2 2p2
1s2 2s2 2p2.
potassium
[Ar] 4s1.
1s2 2s2 2p6 3s2 3p6 4s1
Explanation:-
For writing the short form of the electronic configuration we look for the nearest noble gas with atomic number less than the element in question. We subtract the atomic number of that noble gas from the atomic number of the element in question.
The extra electrons we then assign normally starting with using the row after the noble gas ends. We write the name of that noble gas in [brackets] and then write the electronic configuration.
For carbon with Z = 6 the nearest noble gas is Helium. It has the atomic number 2. Subtracting 6 – 2 we get 4 electrons. Helium lies in 1st row. Starting with 2, we get 2s2 2p2.
So the short term electronic configuration is [He] 2s2 2p2
Similarly, for potassium with Z = 19 the nearest noble gas is Argon. It has the atomic number 18. Subtracting 19-18 we get 1 electron. Argon lies in 3rd row. Starting with 4, we get 4s1.
So the short electronic configuration is
[Ar] 4s1.
For long term electronic configuration we must write the electronic configuration of the noble gas as well.
So for Carbon it is 1s2 2s2 2p2.
For potassium it is 1s2 2s2 2p6 3s2 3p6 4s1
Answer:
2370.0 contains 4 significant digits and Option (c) is correct .
1.20\times 10^{-3}\ contains\ three\ significant\ digit.
Option (b) is correct .
Step-by-step explanation:
Rules for finding significant digit .
1 : Non-zero digits are always significant.
2: Any zeros between two significant digits are significant .
3: Trailing zeros in the decimal number is also significant.
As the number given be 2,370.0.
= \frac{23700}{10}
Simplify the above
= 2370
Thus by using the rule given above.
2370.0 contains 4 significant digits.
Option (c) is correct .
As the number given be 0.00120 .
= \frac{120}{100000}
Simplify the above
= \frac{1.20}{1000}
= 1.20\times 10^{-3}
Thus by using the rule given above.
1.20\times 10^{-3}\ contains\ three\ significant\ digit.
Option (b) is correct .
