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:
A. Ba(NO₃)₂(aq) + Na₂CO₃(aq)
Explanation:
- The reaction between aqueous Ba(NO₃)₂ and Na₂CO₃ is a precipitation reaction.
- The complete equation for the reaction is;
Ba(NO₃)₂(aq) + Na₂CO₃(aq) → BaCO₃(s) + NaNO₃(aq)
- Barium carbonate is the precipitate formed during the reaction.
- All nitrates are soluble in water and all salts of sodium are soluble in water, therefore, Ba(NO₃)₂, Na₂CO3, and NaNO₃ are soluble in water.
- Precipitation reaction involves reaction of soluble salts to form a precipitate as one of the product.
Answer:
13598 J
Explanation:
Q = m × c × ∆T
Where;
Q = amount of energy (J)
m = mass (grams)
c = specific heat capacity
∆T = change in temperature
m = 65g, specific heat capacity of water = 4.184J/g°C, initial temperature= 100°C, final temperature = 150°C
Q = 65 × 4.184 × (150 - 100)
Q = 271.96 × 50
Q = 13598 J
Hence, 13598 J of energy is required to boil 65 grams of 100°C water and then heat the steam to 150°C.
Answer:
the answer is C hope I helped
