1,3-cyclopentanediol has 3 possible stereoisomers. Two of which are enantiomers and are therefore, optically active, while one is optically inactive. The enantiomers are the <em>trans-</em> isomers which are (1R,3R)-1,3-Cyclopentanediol, and (1S,3S)-1,3- Cyclopentanediol.
The stereoisomer that is not optically active is the <em>cis- </em>isomer which is (1S,3R)-1,3-Cyclopentanediol and is shown in the illustration below.
A physical property does not change the substance.
Solubility would be the answer since all of the rest are changing the substance. They all deal with bonds except solubility.
Answer: D. Solubility
Answer:
Moles of Hydrogen produced is 5 moles
Explanation:
The balanced Chemical equation for reaction between zinc and sulfuric acid is :
![Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)](https://tex.z-dn.net/?f=Zn%28s%29%20%2B%20H_%7B2%7DSO_%7B4%7D%28aq%29%20%5Crightarrow%20ZnSO_%7B4%7D%28aq%29%20%2B%20H_%7B2%7D%28g%29)
This equation tells that ; when 1 mole of Zn react with 1 mole of sulfuric acid, it produces 1 mole of zinc sulfate and 1 mole of hydrogen.
Since sulfuric acid is in excess so Zinc is the limiting reagent
(Limiting reagent : Substance which get consumed when the reaction completes, limiting reagent helps in predicting the amount of products formed)
Limiting reagent (Zn) will decide the amount of Hydrogen produced
![Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)](https://tex.z-dn.net/?f=Zn%28s%29%20%2B%20H_%7B2%7DSO_%7B4%7D%28aq%29%20%5Crightarrow%20ZnSO_%7B4%7D%28aq%29%20%2B%20H_%7B2%7D%28g%29)
![1\ mole\ zinc\rightarrow 1\ mole\ H_{2}](https://tex.z-dn.net/?f=1%5C%20mole%5C%20zinc%5Crightarrow%201%5C%20mole%5C%20H_%7B2%7D)
So,
![5\ mole\ zinc\rightarrow 5\ mole\ H_{2}](https://tex.z-dn.net/?f=5%5C%20mole%5C%20zinc%5Crightarrow%205%5C%20mole%5C%20H_%7B2%7D)
Hence moles of Hydrogen produced is 5 moles
The concentration of OH- ( symbol: [OH-] ), is equal to 10^-pOH (ten to the pOH'th power). The pOH equals 14 minus the pH, because the pH + the pOH = 14. So the pOH is 14-13= 1. Now the concentration of OH- is 10^-1 (= 1) moles/Litre
<span>NaOH (s) --> Na+ (aq) + OH- (aq) </span>
<span>1. : 1. : 1 </span>
<span>So by dissolving one mole of NaOH, you get one mole of Na+ and one mole of OH-. Meaning that the molarity (number of solved NaOH in one Litre) of NaOH is 1 mole/Litre, because the ratio is 1:1. This means, in ten litres of water there are also ten moles of NaOH. And the weight of one mole of NaOH is 40.00 grams (look it up in literature). So in ten litres solution with a pH of 13, there are 40.00*10 = 400 = 4 * 10^2 grams of NaOH dissolved</span>
Molar mass CaCl₂ = 111.0 g/mol
number of moles:
n = mass of solute / molar mass
n = 85.3 / 111.0
n = 0.7684 moles of CaCl₂
M = n / V
0.788 M = <span>0.7684 / V
</span>
V = 0.7684 / 0.788
V = 0.97512 L
hope this helps!