Answer:
C₁₁H₁₂NO₄
Explanation:
In order to determine the empirical formula of doxycycline, we need to follow a series of steps.
Step 1: Determine the centesimal composition
C: 59.5 mg/100 mg × 100% = 59.5%
H: 5.40 mg/100 mg × 100% = 5.40%
N: 6.30 mg/100 mg × 100% = 6.30%
O: 28.8 mg/100 mg × 100% = 28.8%
Step 2: Divide each percentage by the atomic mass of the element
C: 59.5 /12.0 = 4.96
H: 5.40/1.00 = 5.40
N: 6.30/14.0 = 0.450
O: 28.8/16.0 = 1.80
Step 3: Divide all the numbers by the smallest one
C: 4.96/0.450 = 11
H: 5.40/0.450 = 12
N: 0.450/0.450 = 1
O: 1.80/0.450 = 4
The empirical formula of doxycycline is C₁₁H₁₂NO₄
Answer:
The compound contains lauryl sulfate and ammoium ions. Lauryl sulfate contains lauric acid (in black and white) , the fatty acid formed by the covalent bonds between C-C attached to hydrogens, and sulfate ions attached to lauric acid (in red) with C-S covalent bond. Sulfer is attached to oxygen by covalent bonds. In Ammonium ions, N is surrounded by four hydrogen atoms.
When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:
MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-
The initial concentrations are as follows:
[Mg2+] = .220(<span> 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
</span>[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-
Answer:
25 grams
Explanation:
You strat off with 400 grams of your substance. By day 14, half has dcayed and you only have 200 grams left. By day 28, there are 100 grams of the substance. On day 42, there are 50 grams left. Finally, on day 56, the substance has been through four half-lives and 25 grams remain.
