Elements are used to create bonds and are classified as particles.
Given :
Three block of same mass name A , B and C .
Sides of block A , B and C is 3.0 cm , 5.0 cm and 10.0 cm .
To Find :
Which block has the higher density .
Solution :
We know , density
is given by :
......1 )
Here , V is volume .
Now , volume V for cube is given :
.......2 )
( Here , a is the side of cube )
Now ,form equation 1 we can see if mass remains constant then density decrease with increase in volume.
Therefore , cube with minimum side will have higher density , in this case it is 3 cm block .
Hence , this is the required solution .
A. An element is the answer
<span>aqueous is the colorful one</span>
NaBr < H3O+1 = OH-1 < Na^+1 = Br^-1 < H2O
<span>Least is NaBr (100% dissolved so no NaBr remains, only Na^+1 and Br^-1 </span>
<span>H2O yields 10^-7 M H3O^+1 and 10^-7 M OH^-1 (Kw = 1x10^-14 = [H3O+][OH-] </span>
<span>Na^+1 and Br^-1 will bothe be 0.1 M </span>
<span>H2O is slightly less that 1000 g / L in a 0.1 M NaBr solution, so its concentration is about 55.5 M</span>