Potassium will loss it's electron because it have a character of electropositively therefore it's electron will transfer to flouren which will make stable and complete of it's electro configuration. Now potassium will have negative charge due to loss it's outer most electron on it's orbit and flourine will take negatively charge too because of complete it's outer most shell to be completed.(But we have to know that at first flourine was had a positive charge because of not completed of it's outer most shell)
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The reason why you don’t see solid or liquid oxygen is because oxygen is a natural occurring gas. Gas can not be solid or liquid because there is too much energy to have volume/shape. That’s why you see oxygen tanks, as the tank keeps the gas contained and that is why you can’t see it in the air outside or in something like H20 (water).
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Answer : (C) "Higher frequencies have larger spaces between lines".
Explanation:
In Young's experiment, the condition for constructive interference is given by :
.........(1)
n is order or number of lines observed
d is distance between slits
is the angle between the path and the line from screen to the slits.
We also know that, 
or
where,
c is the speed of light
is frequency
is wavelength
So, equation (1) turns into
So,
or
Higher frequencies have larger spaces between line.
So, correct option is (C).
Answer:
Animals tend to use carbohydrates primarily for short-term energy storage, while lipids are used more for long-term energy storage. Carbohydrates are stored as glycogen in animals while lipids are stored as fats (in plants carbohydrates are stored as cellulose and lipids as oils)
Explanation:
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Answer:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
Explanation:
We identify the reactants:
CuBr₂ and Pb(CH₃COO)₂
The products will be: Cu(CH₃COO)₂ and PbBr₂
You may know these information:
Salts from acetate are soluble.
Bromide can make solid salts with these cations: Ag⁺, Pb²⁺, Hg₂²⁺, Cu⁺
PbBr₂ is formed, so this will be our precipitate
The equation is:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
