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Answer:
![CH_3 CH_2 CH_2 CH_2 CH_2 CH_2 CHO + 2[Ag(NH_3)_2]+ 3 H_2 O](https://tex.z-dn.net/?f=CH_3%20CH_2%20CH_2%20CH_2%20CH_2%20CH_2%20CHO%20%2B%202%5BAg%28NH_3%29_2%5D%2B%20%203%20H_2%20O)
= 
Explanation:
Tollens test is carried out to perceive difference between aldehydes and ketones on the basis of their capability to oxidized easily.
when Tollens react with aldehyde (heptanal) , a silver mirror is form on inner side of container.
The reaction between tollens and heptanal is given as
![CH_3 CH_2 CH_2 CH_2 CH_2 CH_2 CHO + 2[Ag(NH_3)_2]+ 3 H_2 O](https://tex.z-dn.net/?f=CH_3%20CH_2%20CH_2%20CH_2%20CH_2%20CH_2%20CHO%20%2B%202%5BAg%28NH_3%29_2%5D%2B%203%20H_2%20O)
= 
Answer:
3.50*10^-11 mol3 dm-9
Explanation:
A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.
Ag2C2O4 --> 2Ag+ + C2O4 2-
So Ksp = [Ag+]^2 * [C2O42-]
In 1 L, 2.06*10^-4 mol of silver oxalate dissolve, giving, the same number of mol of oxalate ions, and twice the number of mol (4.12*10^-4) of silver ions.
So Ksp = (4.12*10^-4)^2 * (2.06*10^-4)
= 3.50*10^-11 mol3 dm-9
The solution is here,
given mass of copper wire(m)=25 g
no. of moles in given mass of Cu=0.393
a) no. of atoms= no. of moles×avogardo's number
=0.393×6.023×10^23=2.36 ×10^23 atoms
b) 25 g of Cu wire has 0.393 moles
so 200 g of Cu wire has 0.393/25 × 200=3.144 mol.
no. of atoms= 3.144×2.023×10^23=1.89×10^24 atoms.
c) 25 g of Cu wire has 2.36×10^23 atoms
so 5 g of Cu wire has (2.36×10^23)/25 × 5
=4.72×10^22 atoms
( b can also be solve like c)
