Answer:
The test solution is acidified using a few drops of dilute nitric acid, and then a few drops of silver nitrate solution are added. Different coloured silver halide precipitates form, depending on the halide ions present: ... iodide ions give a yellow precipitate of silver iodide.
There's two sig figs: the 1 and the zero. The 10^-2 contributes no sig figs.
Answer:
0.28M
Explanation:
From the question given, we obtained the following information:
C1 = 1.1 M
V1 = 125mL
V2 = 500mL
C2 =?
Using the dilution formula C1V1 = C2V2, we can easily find the molarity of the diluted solution as follows:
C1V1 = C2V2
1.1 x 125 = C2 x 500
Divide both side by 500
C2 = (1.1 x 125)/500 =
C2 = 0.28M
Answer:
Explanation
Let, Wavelength = W,
Given, frequency (f) = 50 Hz, wave velocity (V) = 342 m/s
We know, V = f × W
=> 342 = 50 × W
=> W = 342/50 metre
=> W = 6.84 metre
