<u>Answer:</u> The formula of the compound formed between rubidium and fluorine is RbF
<u>Explanation:</u>
Ionic bond is defined as the bond which is formed by complete transfer of electrons from one atom to another atom.
The atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom. This bond is usually formed between a metal and a non-metal.
Rubidium is the 37th element of the periodic table having electronic configuration of 
This will loose 1 electron to form 
ion
Fluoride is the 9th element of the periodic table having electronic configuration of 
This will gain 1 electron to form 
ion
To form 
compound, 1 rubidium ion is needed to neutralize the charge on fluoride ion
The formation of the given compounds is shown in the image below.
Answer is because
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First you find the molar mass of the formula.
3C: 36.03g + 8H:8.08g + N: 14.01g + 5O: 80g + P: 30.97g = 169.09g
Then do the mole ratio (304.3g)x(1mole/169.09g) = 1.800 moles
Answer:
Un enlace químico es una atracción duradera entre átomos, iones o moléculas que permite la formación de compuestos químicos.
Explanation:
1.506 × 10²³ carbon atoms
Explanation:
First we calculate the number of moles of calcium carbonate (CaCO₃) contained in 25 g.
number of moles = mass / molecular weight
number of moles of CaCO₃ = 25 / 100 = 0.25 moles
Now we make use of Avogadro's number which tell us that we have 6.022 × 10²³ molecules (or atoms) in one mole of substance.
Now in our case:
if in 1 mole of CaCO₃ we have 1 × 6.022 × 10²³ carbon atoms
then in 0.25 moles of CaCO₃ we have X carbon atoms
X = (0.25 × 1 × 6.022 × 10²³) / 1
X = 1.506 × 10²³ carbon atoms
Learn more:
Avogadro's number
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