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
Answer: The correct answer is E.
Explanation:
The path for the ejaculation of the sperm starts in the testicles, more exactly in the seminiferous tubules, then the path continues trough the epididymis for reaching vas deferens. Finally, the path is completed by the sperm passing through the ejaculatory duct for being released by the urethra.
Therefore the correct answer is E.
If you were to use the formula to find density which is D=M/V and plug in your numbers M which stands for mass and and plug in your numbers for V which what’s for Volume and should be a number in mL and then divide them you should get your answer. So 43.5g divided by 55.0mL and your density would = 0.79g/mL
Answer:
Please find the solution in the attachment file.
Explanation:
