New answers tagged

2

The oxidation number of hydrogen is mostly $1$. Unless it is in elemental state, i.e. as $\ce{H2}$, here the oxidation number of $\ce{N}$ is $-3$ and hydrogen is $+1$. Hydrogen can also take up $-1$ depending on the atom it is attached to. What you really need to lookout for is the electronegativity of the atom attached to it.


2

Okay first, it's the oxygen in the air that causes oxidation, also dissolved oxygen in water, but if you let your tobacco get wet enough for the latter to be significant you will be getting more rot than curing ... Anyway, I don't think a sealed jar would be a good idea. Ammonia build-up, wet rot, etc would tend to be an issue. You could build a proper ...


0

OK, I have a different opinion on the curing of tobacco step from my colleagues. Water in contact with vegetation (likely rich in transition metals) with dissolved O2 and CO2 likely produces (especially in direct or diffused sunlight) powerful radicals that slowly removes/decompose organics. This is part of the science cited in the advanced oxidation process ...


3

If you didn't understand what Poutnik has sain in his comment, this is how it should be done: You have realized your parent compound is $\ce{H4P2O7}$ (adding two $\ce{H+}$). Correct? If so, concentrate on following hypothetical reaction: $$\ce{2H3PO4 -> H4P2O7 + H2O} \tag1$$ I believe you know what $\ce{H3PO4}$ is looks like. Now, draw two $\ce{H3PO4}$ ...


1

In general, gallium can have +1 and +3 oxidation states, though +1 is a very rare condition. In its turn arsenic possible oxidation states are -3, +3 and +5. As it is stated in Wikipedia, gallium oxidation state in gallium arsenide is +3. Once again, arsenic has -3 oxidation state in arsenides or intermetallic compounds. Therefore, gallium arsenide oxidation ...


1

As the formula of the substance is known, you should first state the formula of the ions produced when the substance is dissolved into water. Here, $\ce{K_4Fe(CN)_6}$ gets dissolved in water and produced $4$ ions $\ce{K^+}$ so that the $4$ corresponding negative charges must be fixed on the remaining anion, which has the formula $\ce{[Fe(CN)_6]^{4-}}$ with $...


1

$$\ce{2 Na3PO4 + 3 BaCl2 -> 6 NaCl + Ba3(PO4)2}$$ Now this is clearly evident that dividing whole equation by 2 we get 1 mole of $\ce{Na3PO4}$ and at the right hand side 3 moles of $\ce{NaCl}.$ Therefore, 1 mole of phosphate ion is replaced by 3 moles of $\ce{Cl-}$ ion. Hence $n\text{-factor} = 3.$ In reactions like this where there's no change in ...


Top 50 recent answers are included