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
B-it allows the filament to heat up and glow
Answer:
i can't see the image so sorry
Explanation:
i couldnt see the imagee
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)
Answer:
As heat is applied to liquid water, the molecules move faster and the temperature again increases. During the phase change from liquid to gas, the added heat is stored in the molecules as, once again, potential energy, and the temperature remains constant.
Explanation:
