Answer:
B,C,D
Explanation:
The quantity of CCl4 produced is contingent on the amount of CH4 used in a 1:1 ratio. Given that there are twice as many moles of Cl2 compared to CH4, some Cl2 will remain unreacted. To fully utilize all Cl2, additional CH4 must be introduced into the reaction.
1) To express 0.89% m/v, it equals 0.89 grams of NaCl per 100 ml of solution.
This corresponds to 8.9 grams of NaCl in 1000 ml of solution, or 8.9 grams in 1 liter.
2) Molarity is represented as M = moles of solute / liters of solution.
Thus, we need to determine the moles in 8.9 grams of NaCl.
3) The molar mass of NaCl is calculated as 23.0 g/mol + 35.5 g/mol = 58.5 g/mol.
4) Therefore, the number of moles of NaCl calculates as mass / molar mass = 8.9 g / 58.5 g/mol = 0.152 moles.
5) Consequently, M = 0.152 moles of NaCl / 1 liter of solution = 0.152 M.
Answer: 0.152 M
A secondary alkyl halide would be characterized by having a carbon atom connected to two other carbon atoms, with bromine attached to that carbon.
Therefore, bromo-hexane qualifies as a 2-degree or secondary alkyl halide
Initially, we calculate the moles of gas using the ideal gas law:
PV = nRT
n = PV / RT
n = (1.4 * 226.4) / (0.082 *(27 + 273.15))
n = 12.88
Next, we apply the given percentages to estimate the moles of helium:
Moles of helium = 0.655 * 12.88
Moles of helium = 8.44
We then use the formula:
Mass = moles * molar mass
Mass of helium = 8.44 * 4
Mass of helium = 33.76 grams.
