Answer:
a) IUPAC Names:
1) (<em>trans</em>)-but-2-ene
2) (<em>cis</em>)-but-2-ene
3) but-1-ene
b) Balance Equation:
C₄H₁₀O + H₃PO₄ → C₄H₈ + H₂O + H₃PO₄
As H₃PO₄ is catalyst and remains unchanged so we can also write as,
C₄H₁₀O → C₄H₈ + H₂O
c) Rule:
When more than one alkene products are possible then the one thermodynamically stable is favored. Thermodynamically more substituted alkenes are stable. Furthermore, trans alkenes are more stable than cis alkenes. Hence, in our case the major product is trans alkene followed by cis. The minor alkene is the 1-butene as it is less substituted.
d) C is not Geometrical Isomer:
For any alkene to demonstrate geometrical isomerism it is important that there must be two different geminal substituents attached to both carbon atoms. In 1-butene one carbon has same geminal substituents (i.e H atoms). Hence, it can not give geometrical isomers.
The amount ( in moles of excess reactant that is left is 0.206 moles
Explanation
FeS(s) + 2HCl (aq) → FeCl2 (s) + H2S (g)
- by use of mole ratio of FeS: HCl which is 1:2 this means that 0.223 mole of FeS reacted completely with 0.223 x 2/1 =0.446 moles 0f FeCl2.
- HCl was in excess because 0.446 moles of HCl reacted and initially there was 0.652 moles.
- Therefore the amount that was left
= 0.652- 0.446 =0.206 moles
Answer: so the answer would likely be
Explanation:
Assuming the temperature is 20° C,
100 g of water dissolves 34 g,
200 g of water will dissolve: 200 x 34 / 100
= 68 g
The third option is correct.
<u>Answer:</u> The a solution having pH = 7 has 100 times more hydroxide ions than in a solution having pH = 5
<u>Explanation:</u>
pH is defined as the negative logarithm of hydrogen ion concentration present in the solution.
Mathematically,
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution.
Mathematically,
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
Putting values in above equation, we get:
![7=-\log[H^+]](https://tex.z-dn.net/?f=7%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-7}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-7%7DM)
Putting values in above equation, we get:
![5=-\log[H^+]](https://tex.z-dn.net/?f=5%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-5%7DM)
Taking the ratio of hydrogen ion for both the pH:
![\frac{[H^+]_{pH=5}}{[H^+]_{pH=7}}=\frac{10^{-5}}{10^{-7}}\\\\\frac{[H^+]_{pH=5}}{[H^+]_{pH=7}}=10^2](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%5E%2B%5D_%7BpH%3D5%7D%7D%7B%5BH%5E%2B%5D_%7BpH%3D7%7D%7D%3D%5Cfrac%7B10%5E%7B-5%7D%7D%7B10%5E%7B-7%7D%7D%5C%5C%5C%5C%5Cfrac%7B%5BH%5E%2B%5D_%7BpH%3D5%7D%7D%7B%5BH%5E%2B%5D_%7BpH%3D7%7D%7D%3D10%5E2)
![[H^+]_{pH=5}=100\times [H^+]_{pH=7}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_%7BpH%3D5%7D%3D100%5Ctimes%20%5BH%5E%2B%5D_%7BpH%3D7%7D)
To calculate the pOH, we use the equation:
pOH = 14 - 7 = 7
Putting values in above equation, we get:
![7=-\log[OH^-]](https://tex.z-dn.net/?f=7%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-7}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-7%7DM)
pOH = 14 - 5 = 9
Putting values in above equation, we get:
![9=-\log[OH^-]](https://tex.z-dn.net/?f=9%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-9}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-9%7DM)
Taking the ratio of hydrogen ion for both the pH:
![\frac{[OH^-]_{pH=7}}{[OH^-]_{pH=5}}=\frac{10^{-7}}{10^{-9}}\\\\\frac{[OH^-]_{pH=7}}{[OH^-]_{pH=5}}=10^2](https://tex.z-dn.net/?f=%5Cfrac%7B%5BOH%5E-%5D_%7BpH%3D7%7D%7D%7B%5BOH%5E-%5D_%7BpH%3D5%7D%7D%3D%5Cfrac%7B10%5E%7B-7%7D%7D%7B10%5E%7B-9%7D%7D%5C%5C%5C%5C%5Cfrac%7B%5BOH%5E-%5D_%7BpH%3D7%7D%7D%7B%5BOH%5E-%5D_%7BpH%3D5%7D%7D%3D10%5E2)
![[OH^-]_{pH=7}=100\times [OH^-]_{pH=5}](https://tex.z-dn.net/?f=%5BOH%5E-%5D_%7BpH%3D7%7D%3D100%5Ctimes%20%5BOH%5E-%5D_%7BpH%3D5%7D)
Hence, the a solution having pH = 7 has 100 times more hydroxide ions than in a solution having pH = 5
