06ce mc Chem

2008-04-27 9:28 pm
10.
Solution X is prepared by mixing 100.0 cm3 of 2.0 M Na2SO4 with 50.0 cm3 of 1.0 M NaNO3. What is the concentration of Na+ ions in X ?
A. 1.5 M
B. 1.7 M
C. 3.0 M
D. 3.3 M


Another Q is CE MC 33 2006
Why the answer is D??
i don't know why gas bubbles are formed on the copper surface but not on the zinc surface ?!
Why copper can react with dilute sulphuric acid to form bubbles ??
更新1:

if u know one of them , just answer it!! thanks a lot p.s. please answer with steps

回答 (1)

2008-04-28 12:07 am
✔ 最佳答案
10.
The answer is C.

No. of moles of Na2SO4 = MV = 2 x (100/1000) = 0.2 mol
1 mole of Na2SO4 gives 2 moles of Na+ ions.
Hence, no. of moles of Na+ ions from Na2SO4 = 0.2 x 2 = 0.4 mol

No. of moles of NaNO3 = MV = 1 x (50/1000) = 0.05 mol
1 mole of NaNO3 gives 1 mole of Na+ ions.
Hence, no. of moles of Na+ ions from NaNO3 = 0.05 mol

Total number of moles of Na+ ions = 0.4 + 0.05 = 0.45 mol
Volume of solution X = 100 + 50 = 150 cm3 = 0.15 dm3
Concentration of Na+ = mol/V = 0.45/0.15 = 3.0 M

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33.
The hydrogen gas does not come from the reaction between copper and dilute sulphuric acid. Actually, there is no reaction between copper and sulphuric acid.
I do not have the question, but I guess what the question is. If my guess is incorrect, please inform me via email.

I guess that the zinc and the copper are connected with a metal wire, or made contact directly. This is a cell, or can be treated by using the principle of chemical cells.

Zinc is more reactive than copper (or zinc is at a higher position in electrochemical series than copper). Therefore, zinc is preferentially oxidized to zinc ions and electrons are released.
Zn(s) → Zn2+(aq) + 2e-

Electrons flow from zinc to copper. At the surface of copper, hydrogen ions in the solution gain electrons and are thus oxidized to give hydrogen gas.
2H+(aq) + 2e- → H2(g)


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