A10.00 g sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 g baso4 (m- 233.4). which barium salt is it? 2. (a) bacl2 (m- 208.2) (b) ba(o2ch)2 (m- 227.3) (c) ba(no3)2 (m 261.3) (d) babr (m- 297.1)

The salt is barium chloride.

Explanation:

Moles of barium sulfate =

According to reaction, 1 mol of barium sulfate is produced from 1 mol of .

Then 0.0480 moles will be produced from:

of .

Mass of used = 10.00 g

Moles of =\frac{10.00 g}{\text{Molar mass}}[/tex]

Molar mass of = 208.33 g/mol

The nearest answer to our answer is .

The correct answer barium chloride with molar mass of 208.2 g/mol.

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i  

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.