Answer:
Explanation:
Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid
number of moles of cyclopropane = 1.84 g / 42.08 g/mol = 0.044 mol
number of mole of water = 1010000 g / 18.02 g/mol = 56048.83 mol
mole fraction of cyclopropane = mole of cyclopropane / ( mole of water + mole of cyclopropane) = 0.044 mol / 56048.87 = 7.85 × 10⁻⁷
P cyclopropane = k henry × mole fraction of cyclopropane = 4273 atm × 7.85 × 10⁻⁷ = 0.00335 atm
b) number of cyclopropane molecules present in each cubic centimeter = ( 0.044 mol / 1010) × 6.022 × 10 ²³ / 1000 cm³ = 2.623 × 10 ¹⁶ molecules in cm³
Because im pretty sure the molecules would just fall apart like salt which is ionic
it would be b hope i helped
Answer:
CuCO₃ is the formula of the substance that will precipitate out first
the concentration of carbonate ion when this precipitation first begins = 1.212 × 10⁻¹¹ M
Explanation:
![Pb(NO_3)_2 \ + \ Na_2CO_3 -----> PbCO_3 \ + \ 2NaNO\\\\ksp \ of \ PbCO_3 = 1.6*10^{-13} \\\\ \\Cu (CH_3COO)_2 \ + \ Na_2CO_3 ------> CuCO_3 \ + \ Na(C_2H_3O_2)\\\\ksp \ of CuCO_3 = 1.3*10^{-10}\\\\As \ ksp \ of CuCO_3 > PbCO_3\\\\Then \ \ CuCO_3 \ will \ precipitate \ out \ first](https://tex.z-dn.net/?f=Pb%28NO_3%29_2%20%5C%20%2B%20%5C%20Na_2CO_3%20-----%3E%20PbCO_3%20%5C%20%2B%20%5C%202NaNO%5C%5C%5C%5Cksp%20%5C%20of%20%5C%20PbCO_3%20%3D%20%201.6%2A10%5E%7B-13%7D%20%5C%5C%5C%5C%20%5C%5CCu%20%28CH_3COO%29_2%20%5C%20%2B%20%5C%20Na_2CO_3%20------%3E%20CuCO_3%20%5C%20%2B%20%5C%20Na%28C_2H_3O_2%29%5C%5C%5C%5Cksp%20%5C%20of%20%20CuCO_3%20%3D%201.3%2A10%5E%7B-10%7D%5C%5C%5C%5CAs%20%5C%20ksp%20%5C%20of%20CuCO_3%20%3E%20PbCO_3%5C%5C%5C%5CThen%20%5C%20%5C%20CuCO_3%20%5C%20will%20%5C%20precipitate%20%5C%20out%20%5C%20first)
![PbCO_3 ----> Pb^{2+}_{(aq)} + CO^{2-}_{3(aq)}](https://tex.z-dn.net/?f=PbCO_3%20----%3E%20%20%20Pb%5E%7B2%2B%7D_%7B%28aq%29%7D%20%2B%20CO%5E%7B2-%7D_%7B3%28aq%29%7D)
will be ![[Pb(NO_3)_2] = 1.32 *10^{-2} \ M](https://tex.z-dn.net/?f=%5BPb%28NO_3%29_2%5D%20%3D%201.32%20%2A10%5E%7B-2%7D%20%5C%20M)
![ksp = [Pb^{2+}][CO^{2-}_3]](https://tex.z-dn.net/?f=ksp%20%3D%20%20%5BPb%5E%7B2%2B%7D%5D%5BCO%5E%7B2-%7D_3%5D)
![1.6*10^{-13} = (1.32*10^{-2})(CO^{2-}_3)\\\\(CO^{2-}_3) = \frac{1.6*10^{-13}}{(1.32*10^{-2})}\\\\(CO^{2-}_3) = 1.212 *10^{-11} \ \ M](https://tex.z-dn.net/?f=1.6%2A10%5E%7B-13%7D%20%3D%20%281.32%2A10%5E%7B-2%7D%29%28CO%5E%7B2-%7D_3%29%5C%5C%5C%5C%28CO%5E%7B2-%7D_3%29%20%3D%20%5Cfrac%7B1.6%2A10%5E%7B-13%7D%7D%7B%281.32%2A10%5E%7B-2%7D%29%7D%5C%5C%5C%5C%28CO%5E%7B2-%7D_3%29%20%3D%201.212%20%2A10%5E%7B-11%7D%20%5C%20%5C%20M)
Answer:
Explanation:
Materials used for sacrificial anodes must be relatively pure active metals.
Or we can say any metal with more negative potential that that of iron ( +0.59) is used for sacrificial anode for iron
so from the above the following can be used as sacrificial anode for iron
Mg = +1.06
Al = +0.96
Zn = +0.7
The following will not be used as sacrificial anode for iron because their negative reduction potential is less than that of iron ( +0.59)
Ag = -0.76
Cu = -0.44
Pb = 0
Sn = +0.34
Au = -3.05