I used coulomb's law to find the force of attraction between the ionic atoms in BeS and BeO.
For BeS, F = -6.85x10^29 J/m
For BeO, F = -9.078x10^29 J/m
Which has the stronger force of attraction? BeO? What does the "-" sign mean?
To determine which substance has a stronger force of attraction, we need to compare the magnitudes of the forces. In this case, we have:
For BeS, F = -6.85x10^29 J/m
For BeO, F = -9.078x10^29 J/m
The "-" sign in front of the numbers indicates that the forces are attractive forces. In Coulomb's law, the force between two charges of opposite signs is attractive, while the force between two charges of the same sign is repulsive.
Comparing the magnitudes, we can see that the force of attraction for BeO (-9.078x10^29 J/m) is larger than the force for BeS (-6.85x10^29 J/m). Therefore, BeO has a stronger force of attraction between its ionic atoms compared to BeS.
To determine which compound has the stronger force of attraction, we need to compare the magnitudes of the forces. In this case, we have:
For BeS, F = -6.85x10^29 J/m
For BeO, F = -9.078x10^29 J/m
Since both forces are negative, the "-" sign indicates that the force of attraction is an attractive force. In other words, it means that the ionic atoms are being pulled towards each other.
To compare the magnitudes, we can ignore the negative sign.
Comparing the magnitudes:
|F(BeS)| = 6.85x10^29 J/m
|F(BeO)| = 9.078x10^29 J/m
Since 9.078x10^29 J/m is greater than 6.85x10^29 J/m, the force of attraction in BeO is stronger than in BeS.