Explanation:
1g = 1000mg
154000mg = 154g
No of days she can drink = 154÷ 11 = 14days
<span>There is five main area of study in Chemistry, these are:
Analytical, this focusses on experimental equipment and methods used in chemistry (e.g., NMR, Spectroscopic methods, etc.)
Biochemistry - focuses on the chemistry of compounds and processes in living things (e.g., amino acids, proteins, DNA, cellular respiration, Krebs cycle, etc.)
Organic - focuses on the chemistry on most carbon-based molecules found in living things (e.g., hydrocarbons, alcohols, carbolic acids. Amines, ester, etc.)
Inorganic - (focuses on all elements other than carbon (e.g., fluorine, silicon, xenon, etc.)
Physical - focuses on the basic structure and energetic son atoms and molecules (e.g., subatomic structure, is nice and covalent bonding, thermodynamics, reactions, etc.)</span>
Answer:
Conduct more trials
Explanation:
Theoretical Probability can be defined as what someone is expecting to happen
Experimental Probability on the other hand, is defined as what actually happens.
Probability is usually calculated in the same way for experimental probability and that of theoretical probability. You divide the total number of possible ways in which a particular outcome can happen, by the total number of outcomes itself.
In Experimental probability, the more times a probability is tried, it gets closer and even more closer to theoretical probability.
So, for the question, Jamie should improve the number of tries more, so as to get his experimental probability results to be closer to the theoretical probability result.
Answer:
a) 
b) 
Explanation:
Equation of reaction:
Initial pressure 3 1 0
Pressure change 2P 1P 2P
Total pressure = (3-2P) + (1-P) + (2P)
Total Pressure = 3.75 atm
(3-2P) + (1-P) + (2P) = 3.75
4 - P = 3.75
P = 4 - 3.75
P = 0.25 atm
Let us calculate the pressure of each of the components of the reaction:
Pressure of XO2 = 3 - 2P = 3 - 2(0.25)
Pressure of XO2 =2.5 atm
Pressure of O2 = 1 - P = 1 -0.25
Pressure of O2 = 0.75 atm
Pressure of XO3 = 2P = 2 * 0.25
Pressure of XO3 = 0.5 atm
From the reaction, equilibrium constant can be calculated using the formula:
![K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BPXO_%7B3%7D%5D%20%5E%7B2%7D%20%7D%7B%5BPXO_%7B2%7D%5D%20%5E%7B2%7D%5BPO_%7B2%7D%5D%20%7D)
Standard free energy:
b) value of k−1 at 27 °C, i.e. 300K
Answer:
The answer to your question is all the formulas in bold has the same empirical formula
Explanation:
Data
Empirical formula CH₂O
Process
To solve this problem factor the subscripts of each formula and compare the result with the empirical formula given.
a) C₂H₄O₂ factor 2 2(CH₂O)
b) C₃H₆O₃ factor 3 3(CH₂O)
c) CH₂O₂ this formula can not be simplified
d) C₅H₁₀O₅ factor 5 5(CH₂O)
e) C₆H₁₂O₆ factor 6 6(CH₂O)
