When 2 moles burn 6 moles of O2 is produced !
so for 16 moles , (6/2)×16= 48 moles will be produced !
so answer is A , 48 moles !
Answer:
2.2%
Explanation:
Percentage error,
You apply the formula,
[(Estimated value - Actual value)/Actual value] × 100%
; [(43.26 - 42.32)/42.32] × 100
; (0.94/42.32) × 100
; 0.022 × 100
Percent error = 2.2%
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.
Answer:
Answers are in the explanation
Explanation:
Ksp of CdF₂ is:
CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)
Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²
When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:
6.44x10⁻³ = [X] [2X]²
6.44x10⁻³ = 4X³
X = 0.1172M
<h3>[F⁻] = 0.2344M</h3><h3 />
Ksp of LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]
When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:
1.84x10⁻³ = [X] [X]
1.84x10⁻³ = X²
X = 0.0429
<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
Answer:
pH after the addition of 10 ml NaOH = 4.81
pH after the addition of 20.1 ml NaOH = 8.76
pH after the addition of 25 ml NaOH = 8.78
Explanation:
(1)
Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles = 2 x 10⁻³ moles,
Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles
CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O
Initial conc. 2 x 10⁻³ 1 x 10⁻³ 0
Equilibrium 1 x 10⁻³ 0 1 x 10⁻³
Final volume = 20 + 10 = 30 ml = 0.03 lit
So final concentration of Acid = 
Final concentration of conjugate base [CH₃CH₂CH₂COONa]
Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .
Using Henderson Hasselbalch equation to find the pH
![pH=pK_{a}+log\frac{[conjugate base]}{[acid]} \\\\=-log(1.54X10^{-5} )+log\frac{0.03}{0.03} \\\\=4.81](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%2Blog%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D%20%20%5C%5C%5C%5C%3D-log%281.54X10%5E%7B-5%7D%20%29%2Blog%5Cfrac%7B0.03%7D%7B0.03%7D%20%5C%5C%5C%5C%3D4.81)
