So we have Barium nitrate with a solubility of 8.7g in 100g water at 20°C.
using that relation
i.e.
8.7g (barium nitrate) =100g (water)
1g barium nitrate = 100/8.7 g water
27g barium nitrate = (100/ 8.7 ) × 27
= 310.34 g
therefore,
you need 310.34g of water is in the jar.
Answer:
3
three half-filled orbitals each capable of forming a single covalent Bond and an additional lone - pair of electrons
Those elements with similar properties are in the same column.
Example:
sample density of gasoline, 20 g of weigth into 5 <span>mL
Answer:
D = m / V
D = 20 g / 5 mL
D = 4 g/mL</span>
