Answer:
The correct answer is
e. NH3(aq) + H+(aq) --> NH4+(aq)
Explanation:
To solve this, we write out the indidual ionization reation for aqueous ammonia and nitric acid thus
For aqueous ammonia we have
NH₃(aq) + H₂O(l) ↔ NH₄⁺(aq) + OH⁻(aq)
Aqueous ammonia is a weak base and therefore undergoes partial ionization hence the reversible reaction sign
As the level of ionization will not be more than 5% OH⁻ cannot represebt the weak base
For nitric acid we have
HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)
a strong acid like nitric acid undergoes conplete ionization in the solution
The total equation is NH₃(aq) + HNO₃(aq) → NH₄NO₃(aq)
The sum of the ionic equation is
NH₃(aq) + H⁺(aq) + NO₃⁻(aq) → NH₄⁺(aq) + NO₃⁻(aq)
The ionic equation is
NH₃(aq) + H⁺(aq) → NH₄⁺(aq)
Ni(OH)2+H2SO4=NiSO4+2H2O
Double Replacement Reaction
Answer: 1.10x10²³ atoms of C
110202600000000000000000 atoms C
Explanation:The solution process is shown below.
0.183 mole C x 6.022x10²³ atoms C / 1 mole C
= 1.10x10²³ atoms C
or 110202600000000000000000 atoms C
To find the empirical formula you would first need to find the moles of each element:
58.8g/ 12.0g = 4.9 mol C
9.9g/ 1.0g = 9.9 mol H
31.4g/ 16.0g = 1.96 O
Then you divide by the smallest number of moles of each:
4.9/1.96 = 2.5
9.9/1.96 = 6
1.96/1.96 = 1
Since there is 2.5, you find the least number that makes each moles a whole number which is 2.
So the empirical formula is C5H12O2.
Um try B? i think that would be correct. either B or C
