Question

calculate the total no.of ions present in the 3.01×10^23 ions of Al^+3​

Answer 1

Answer:

Solution

Explanation:

Correct option is

C

0.042kg

6.022×10

23

Mg

2+

ions in 1 mole of MgCO

3

3.01×10

23

CO

3

2−

ions in 1 mole of MgCO

3

=

6.022×10

23

3.01×10

23

=0.5mole

So, mass of sample (MgCO

3

)=(24+12+48)×0.5=84×

10

0.5

g=42g=0.042Kg