Pb(NO3)2(aq) + KCl(aq) ---- → KNO3(aq) + PbC12(s)
1 answer:
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It is really easy if you look up an electron configuration table (one that looks like a periodic table) and then just go down the rows left to right, top to bottom, and just stop when you get to where the element is on the table.
Answer:
The wavelength the student should use is 700 nm.
Explanation:
Attached below you can find the diagram I found for this question elsewhere.
Because the idea is to minimize the interference of the Co⁺²(aq) species, we should <u>choose a wavelength in which its absorbance is minimum</u>.
At 400 nm Co⁺²(aq) shows no absorbance, however neither does Cu⁺²(aq). While at 700 nm Co⁺²(aq) shows no absorbance and Cu⁺²(aq) does.
Answer:
There are 0.93 g of glucose in 100 mL of the final solution
Explanation:
In the first solution, the concentration of glucose (in g/L) is:
15.5 g / 0.100 L = 155 g/L
Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.
- 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)
The concentration of the second solution is:
So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:
1 L --------- 9.3 g
0.1 L--------- Xg
Xg = 9.3 g * 0.1 L / 1 L = 0.93 g