Answer:
The original concentration of ethanol was 18 % (v/v)
Explanation:
For the analysis of a sample, mostly dilutions are made of the original concentrated sample. The analysis results obtained from the diluted sample are then calculated for the concentrated samples.
According to the given data, the wine sample was diluted 75 times. This means that the initial concentration of ethanol was 75 times more than the final concentration i.e. 0.24 % (v/v). So mathematically:
original concentration (v/v)= final concentration (v/v) x times diluted
original concentration (v/v)= 0.24 % x 75
original concentration (v/v) = 18 %
The best answer between the two choices would be the first option TRUE because the scientific method is used to do more advance research and investigation on things.
Answer:
See Explanation
Explanation:
Given that;
N/No = (1/2)^t/t1/2
Where;
No = amount of radioactive isotope originally present
N = A mount of radioactive isotope present at time t
t = time taken
t1/2 = half life
N/1000=(1/2)^3/6
N/1000=(1/2)^0.5
N = (1/2)^0.5 * 1000
N= 707 unstable nuclei
Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;
Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1
Hence;
A=Aoe^-kt
Where;
A = Activity after a time t
Ao = initial activity
k = decay constant
t = time taken
A = Aoe^-3 *0.1155
A=Aoe^-0.3465
Answer:
1. 
molecules of CO₂
2. 10⁴ molecules of H₂O
3. 8.75×10³² molecules of C₆H₁₂O₆
Explanation:
1. molecules of CO₂
2. 10⁴ molecules of H₂O
3. 8.75×10³² molecules of C₆H₁₂O₆
