2Na+MGF2=2NAF+MG
I hope that's right :)
I believe the answer is a balance or mechanical scale
Answer:
CCl4- tetrahedral bond angle 109°
PF3 - trigonal pyramidal bond angles less than 109°
OF2- Bent with bond angle much less than 109°
I3 - linear with bond angles = 180°
A molecule with two double bonds and no lone pairs - linear molecule with bond angle =180°
Explanation:
Valence shell electron-pair repulsion theory (VSEPR theory) helps us to predict the molecular shape, including bond angles around a central atom, of a molecule by examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement which tends to minimize repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom are either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far apart from each other as possible.
Lone pairs and multiple bonds are known to cause more repulsion than single bonds and bond pairs. Hence the presence of lone pairs or multiple bonds tend to distort the molecular geometry geometry away from that predicted on the basis of VSEPR theory. For instance CCl4 is tetrahedral with no lone pair and four regions of electron density around the central atom. This is the expected geometry. However OF2 also has four regions of electron density but has a bent structure. The molecule has four regions of electron density but two of them are lone pairs causing more repulsion. Hence the observed bond angle is less than 109°.
Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium
Heat gained by water = 100 × 4.186 × 30.5
= 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5
= 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules
This problem could be solved through the Graham’s law of
effusion (also known as law of diffusion). This law states that the ratio of
the effusion rate of the first gas and effusion rate of the second gas is
equivalent to the square root of the ratio of its molar mass. Thus the answer
would be 0.1098.
