Given :
Moles of Na : 1.06
Moles of C : 0.528
Moles of O : 1.59
To Find :
The empirical formula of the compound.
Solution :
Dividing moles of each atom with the smallest one i.e 0.528 .
So,
Na : 1.06/0.528 = 2.007 ≈ 2
C : 0.528/0.528 = 1
O : 1.59/0.528 = 3.011 ≈ 3
Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.
So, empirical formula is
.
Hence, this is the required solution.
This is given by Avogagro number: 1 mol = 6.02*10^23 particles
Then you can do whichever to these two relations, because they are equivalent:
- 1mol / 6.02*10^23 representative particles, and
- 6.02*10^23 representative particle /1 mol
Only the second option of the question includes one of the valid conversion factors. Then, the conversion factor of the second option is the right answer
Answer:
option D is correct
D. This solution is a good buffer.
Explanation:
TRIS (HOCH
)
CNH
if TRIS is react with HCL it will form salt
(HOCH
)
CNH
+ HCL ⇆ (HOCH
)
NH
CL
Let the reference volume is 100
Mole of TRIS is = 100 × 0.2 = 20
Mole of HCL is = 100 × 0.1 = 10
In the reaction all of the HCL will Consumed,10 moles of the salt will form
and 10 mole of TRIS will left
hence , Final product will be salt +TRIS(9 base)
H = Pk
+ log (base/ acid)
8.3 + log(10/10)
8.3
Answer:
2.5 mol Cl2 will react with 5 mol Na.
Explanation:
Answer:
The correct would be C i think :)
Explanation:
Stay postivie :)