Standard equation would be N2(g)+3H2(g)==>2NH3(g), so through stoichiometry, (4 mol N2)(2mol NH3/1 mol N2), assuming excess H2, would yield 8 moles of NH3.
Answer:- 1840 g.
Solution:- We have been given with 3.35 moles of and asked to calculate it's mass.
To convert the moles to grams we multiply the moles by the molar mass of the compound. Molar mass of the compound is the sum of atomic masses of all the atoms present in it.
molar mass of = atomic mass of Hg + 2(atomic mass of I) + 6(atomic mass of O)
= 200.59+2(126.90)+6(16.00)
= 200.59+253.80+96.00
= 550.39 gram per mol
Let's multiply the given moles by the molar mass:

= 1843.8 g
Since, there are three sig figs in the given moles of compound, we need to round the calculated my to three sig figs also. So, on rounding off to three sig figs the mass becomes 1840 g.
To minimize the sharp pH shift that occurs when a strong acid is added to a solution, IT IS PRACTICAL TO ADD A WEAK BASE.
When a strong acid is added to a solution, it usually brings about a sharp change in the pH of the concerned solution. To avoid this, one can add a weak base to the solution first. The weak base will serves as a buffer for the strong acid and prevents the solution from experiencing sharp pH variations.
Answer:
The water in the hydrate (referred to as "water of hydration") can be removed by heating the hydrate. When all hydrating water is removed, the material is said to be anhydrous and is referred to as an anhydrate.
Explanation:
I will need a picture if the periodic table