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:
It lies in in the <u>visible</u><u> </u><u>region</u><u>.</u>
Frequency: <u>is</u><u> </u><u>7</u><u>.</u><u>4</u><u>0</u><u>7</u><u> </u><u>×</u><u> </u><u>1</u><u>0</u><u>^</u><u>-</u><u>1</u><u>4</u><u> </u><u>Hz</u>
Explanation:
V is speed of light.
f is frequency
lambda is the wavelength
Answer:
None
Explanation:
Cl₂ is above Br₂ in the activity series.
Bromine will not displace chlorine from its salts.
The reaction will not occur.
I don’t know I don’t know I don’t know