Then answer would be D. Answer D is correct because you would need to use a better solvent to see the ink separate on the chromatography paper. Hope that helps. :)
N2 + 3H2 ----> 2NH3
<span>you can see 3 moles H2 reacts to form 2 moles NH3 </span>
<span>Therefore moles NH3 = 2 / 3 x moles H2 </span>
<span>= 2/3 x 12.0 mol </span>
<span>= 8.00 mol NH3 hope this help</span>
Answer:
C. Its oxidation number increases.
Explanation:
- <em><u>Oxidation is defined as the loss of electrons by an atom while reduction is the gain of electrons by an atom</u></em>.
- Atoms of elements have an oxidation number of Zero in their elemental state.
- When an atom looses electrons it undergoes oxidation and its oxidation number increases.
- For example, <em><u>an atom of sodium (Na) at its elemental state has an oxidation number of 0. When the sodium atom looses an electrons it becomes a cation, Na+, with an oxidation number of +1 , the loss of electron shows an increase in oxidation number from 0 to +1.</u></em>
Answer:
Kb for CH₃NH₂ (methylamine) is 4.4 × 10⁻⁴
Hope that helps.
Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.