The empirical formula is N₂O₅.
The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.
The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of N:O</em>.
I like to summarize the calculations in a table.
<u>Element</u> <u>Moles</u> <u>Ratio¹ </u> <u> ×2² </u> <u>Integers</u>³
N 1.85 1 2 2
O 4.63 2.503 5.005 5
¹To get the molar ratio, you divide each number of moles by the smallest number (1.85).
²Multiply these values by a number (2) that makes the numbers in the ratio close to integers.
³Round off the number in the ratio to integers (2 and 5).
The empirical formula is N₂O₅.
In standard notation, the following number would be 86618.5.
Answer:
92.49 %
Explanation:
We first calculate the number of moles n of AgBr in 0.7127 g
n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g
n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol
Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and
From n = m/M
m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol
m = 0.0038 mol × 79.904 g/mol = 0.3036 g
% Br in compound = m₁/m₂ × 100%
m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)
m₂ = mass of compound = 0.3283 g
% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %
Fungi and bacteria release nitrogen stored in dead tissue through a decomposition process called B) ammonification.
