When determining the amount of oxidant present by titration, you can use iodine/starch as an indicator. First, the oxidant, like hypochlorite, oxidizes Choose... When starch and iodine are both present, the solution is Choose... During the titration, a titrant like thiosulfate reduces the

The question is incomplete, the complete question is;

When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.

First, the oxidant, like hypochlorite, oxidizes

Choose...

neutral iodine into iodide ion

iodide ion into neutral iodine

iodate polyatomic ion into iodide ion

When starch and iodine are both present, the solution is

Choose...

blue-black

brownish yellow

clear

During the titration, the titrant, like thiosulfate, reduces the

Choose...

iodide ion into iodate polyatomic ion

neutral iodine into iodide ion

iodide ion into neutral iodine

When the iodine has completely reacted at the endpoint of the titration, the solution should become

Choose...

clear

blue-black

brownish yellow

1. iodide ion into neutral iodine

2. blue-black

3. neutral iodine into iodide ion

4. clear

Explanation:

Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;

ClO-(aq) + 2H+(aq) + 2I-(aq) > 6 I2(l) + Cl- (aq)+ H2O(l)

When iodine is added, the colour of the starch solution immediately changes to blue-black.

A reduction reaction occurs when the titrant, thiosulfate is added as follows;

I2 + 2S2O32- → 2I- + S4O62-

The solution at end point is found to become clear again.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

they it depends on if one believes about radioactivity, they make a geiger counter beep but maybe it was going to beep anyway electronics do wierd things all the time or it was programmed to beep near certain metals.