(3.45 x 10^20 molecules of sulfur dioxide ) x

= 5.73x10^-4 mole of sulfur dioxide
Answer:
Both are only physical changes.
Explanation:
Nothing about the objects chemically change, and they remain the same.
Their changes arent due to heating/cooling.
Answer:
f. Sn^4+
c. second
e. Al^3+
d. third
Explanation:
This question comes from a quantitative analysis showing the flowchart of a common scheme for identifying cations.
Now, from the separation scheme, Let's assume that Sn⁴⁺ & Al³⁺ were given; Then, Yes, the separation will work.
However, there will be occurrence of precipitation after the 1st step1.
So, the <u>Sn⁴⁺</u> cation will precipitate after the <u>second </u>step. Then the <u>Al³⁺</u> cation will precipitate after the <u>third</u> step.
Answer:
Precipitation Reactions
They contain two aqueous reactants, one aqueous product, and one solid product. In this reaction, two soluble products, Pb(NO3)2 and KI, combine to form one soluble product, KNO3, and one insoluble product, PbI2. This is a precipitation reaction, and PbI2 is the precipitate.
Answer : The current passing between the electrodes is, 
Explanation :
First we have to calculate the charge of sodium ion.

where,
q = charge of sodium ion
n = number of sodium ion = 
e = charge on electron = 
Now put all the given values in the above formula, we get:

Now we have to calculate the charge of chlorine ion.

where,
q' = charge of chlorine ion
n = number of chlorine ion = 
e = charge on electron = 
Now put all the given values in the above formula, we get:

Now we have to calculate the current passing between the electrodes.



Thus, the current passing between the electrodes is, 