Thomson in 1897 was the first to suggest that one of the fundamental units was more than 1,000 times smaller than an atom, suggesting the subatomic particle now known as the electron. Thomson discovered this through his explorations on the properties of cathode rays.
The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal
How to determine the mole of glucose
Mass of glucose = 49 g
Molar mass of glucose = 180.2 g/mol
Mole of glucose = ?
Mole = mass / molar mass
Mole of glucose = 49 / 180.2
Mole of glucose = 0.272 mole
How to determine the energy released
C₆H₁₂O₆ →2C₂H₆O + 2CO₂ ΔH = -16 kcal/mol
From the balanced equation above,
1 mole of glucose released -16 kcal of energy
Therefore,
0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal
Thus, -4.4 Kcal were released from the reaction
Learn more about stoichiometry:
brainly.com/question/14735801
#SPJ1
Answer:
Answer
The limitations are-
1. Privacy of the individuals involved in the research process.
2. Physical and psychological risks should be minimized
3. The subjects should be chosen equitably
4. Only reasonable exposure to risks is admissible.
Explanation
The privacy of the individuals involved in the research must be taken into consideration. This will ensure safety of patient data and information. The risk of physical and physiological well being of the person must be taken into consideration in such a research. In addition to that, the subject must be made to understand every procedure and the risks involved before testing. Moreover, only minimal exposure to risks is allowed and must have been previously tested in animals to avoid deaths.
To find the atomic mass, multiply the individual atomic masses with their relative abundances and add them up.
10.012*0.1991 + 11.009*0.8009 = 10.810 amu (5 s.f.)
Element X is Boron.
