The answer would be C.
The salt and the water have both undergone physical changes.
Hope this helps!
This problem is easily solvable because radioactivity equations are common and well-established. The pseudo-first reaction is written below:
A = A₀(1/2)^(t/h)
where
A is the final amount
A₀ is the original amount
t is the time
h is the half life
5,000 = A₀(1/2)^(24,000/6,000)
Solving for A₀,
<em>A₀ = 80,000 atoms</em>
Answer:
See Explanation
Explanation:
For SF6;
Since;
1.25 g of S corresponds to 4.44g of F
1 g of sulphur corresponds to 1 * 4.44/1.25 = 3.55
For SF4;
Since;
1.88 g of S corresponds to 4.44g of F
1 g of sulphur corresponds to 1 * 4.44/ 1.88 = 2.36
Hence;
Mass of oxygen per gram of sulphur in SF6/Mass of oxygen per gram of sulphur in SF4
=
3.55/2.36 = 1.5
Hence the law of multiple proportion is obeyed here.
The sample must be sufficiently soluble (fig. 2) to yield an NMR spectrum. For 1H and 1H observed NMR, it is recommended to dissolve between 2 and 10 mg in between 0.6 and 1 mL of solvent so that the sample depth is at least 4.5 cm in the tube (fig. 3).
Answer: 0.745 g of 
will be produced from 1.08 g of sodium sulfate
Explanation:
To calculate the moles :
is the limiting reagent as it limits the formation of product and 
is the excess reagent.
According to stoichiometry :
3 moles of produce = 3 moles of
Thus 0.0076 moles of will require=
of
Mass of 
Thus 0.745 g of will be produced from 1.08 g of sodium sulfate
