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.
To solve this question you need to calculate the number of the gas molecule. The calculation would be:
PV=nRT
n=PV/RT
n= 1 atm * 40 L/ (0.082 L atm mol-1K-<span>1 * 298.15K)
</span>n= 1.636 moles
The volume at bottom of the lake would be:
PV=nRT
V= nRT/P
V= (1.636 mol * 277.15K* 0.082 L atm mol-1K-1 )/ 11 atm= <span>3.38 L</span>
0.000735 in scientific notation is 7.35 x 10^-4
Gain or lose.
The exchange of electrons in chemical bonding seeks to fulfill the octet rule. There are some exceptions, such as with hydrogen and helium, whose valence shells have a capacity of two electrons.
When Iodine-131 emits a β particle will produce Xe-131
<h3>Further explanation
</h3>
Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,
- alpha α particles ₂He⁴
- beta β ₋₁e⁰ particles
- gamma particles ₀γ⁰
- positron particles ₁e⁰
- neutron ₀n¹
So for reaction Iodine-131 :
