2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
Answer:
The atomic number that should be here, 57, is located at the bottom of the table in the row called the Lanthanides. Directly below the space in Row 6, in Row 7, is another empty space, which is filled by a row called the Actinides, also seen at the bottom of the chart.
Explanation:
hope this helps!
Density = Mass / Volume
V = 1.00 * 4.00 * 2.50 = 10 cm³
22.57 g/cm³ = Mass / 10 cm³
M = 22.57 g/cm³ * 10 cm³
M = 225.7 g
Answer: The mass of the block of osmium is 225.7 g.
It's a base because it increases the concentration of hydroxid ions
