It is limited by the depth divers and equipment can go.
Answer:
The correct answer is B the tertiary halides reacts faster than primary halides.
Explanation:
During SN2 reaction the nucleophile attack the alkyl halide from the opposite side resulting in the formation of transition state in which a bond is not completely broken or a new bond is not completely formed.
After a certain period of time the nucleophile attach with the substrate by substituting the existing nuclophile.
An increase in the bulkiness in the alkyl halide the SN2 reaction rate of that alkyl halide decreases.This phenomenon is called steric hindrance.
So from that point of view the that statement tertiary halides reacts faster that secondary halide is not correct.
To be able to calculate the number of moles for this problem, for simplicity, we assume that it is an ideal gas. We use the equation PV = nRT. We do as follows:
PV = nRT
n = PV / RT
n = 1(100000) / 0.08206 (27 + 273.15)
n = 4060.04 mol
Answer: 107 grams
Explanation: got it right on the test
The molarity of the acetic acid solution is calculated as below
calculate the moles of KOH used
moles =molarity x volume
= 20.9 x0.267 = 5.58 moles
write the reaction equation
KOH + CH3COOH = CH3COOK + H2O
since the reacting ratio between KOH to CH3COOH is 1:1 the moles of CH3COOH is also = 5.58 moles
molarity of CH3COOH = moles/volume
=5.58 /29.1 =0.192 M
