LD50 is defined as the lethal dose 50% which describes the amount of material required to kill 50% of the testing population. It is given in units of mg of chemical per kg of bodyweight of the recipient.
Comparing hydrogen peroxide and acetic acid, we see that peroxide has a lower LD50 of 900 mg/kg, with acetic acid having LD50 = 3310 mg/kg. When comparing LD50 values, the smaller value will be the more toxic compound. What this means is that in this case, a smaller amount of peroxide is required to kill 50% of the testing population compared to acetic acid.
Therefore, 3% hydrogen peroxide is more hazardous to consume.
Answer:
your answer is hawks
Answer:
0.56L
Explanation:
This question requires the Ideal Gas Law: 
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.
Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:
Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means 
and 
Lastly, we must calculate the number of moles of 
there are. Given 0.80g of , we will need to convert with the molar mass of . Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: 
Thus, 
Isolating V in the Ideal Gas Law:
...substituting the known values, and simplifying...
So, 0.80g of would occupy 0.56L at STP.
Answer:
89.88 g
Explanation:
Atomic Mass of Ar: 39.948
Mass = moles * AM
Replacing moles = 2.25 and AM = 39.948 you get the mass of Ar:
Mass = 2.25 * 39.948
Mass = 89.88 g
Answer:
gde
Explanation:
We are attempting to synthesize 1-butyne from 1-chlorobutane. Since 1-chlorobutane is a primary alkyl halide, 1-butene is formed when 1-chlorobutane is reacted with a bulky base such as t -BuOK or t -BuOH in presence of strong heat. This is an E2 reaction.
Secondly, the 1-butene is reacted with bromine in carbon tetrachloride. The vicinal dihalide (1,2-dibromobutane) is formed. This can now undergo further elimination reactions in the presence of sodamide and strong heat to yield 1-butyne which is the desired product. These reactions involve the elimination of the first HBr molecule to give an alkenyl bromide. A second elimination now gives the terminal alkyne.
