The standard notation of 3.004 × 10⁻³ is 0.003004
To express enormous figures which are in standard notation like 1,300,000 or exceedingly minuscule quantities like 0.0000000000045, scientists use scientific notation. Scientific notation, also referred to as exponential form, is one of the earliest mathematical techniques. It is well regarded by practitioners. People use scientific notation to handle situations where numbers are too large or too small to be calculated easily. Scientists, engineers, and mathematicians all use this technique. where as standard notation is way writing numbers in normal decimal form. The scientific method id most convenient denoting numbers as the numbers will be either too big or too small.
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The STP is Standard temperature and pressure.
Standard temperature is defined as zero degrees Celsius (0° C)
Standard pressure supports 760 millimeters in a mercurial barometer (760 mmHg)
we have that
760 mmHg= 1 atm
therefore
the answer is 0 deg C and 1 atm.
Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
Try sugar. I hope this helps!!!!!
I believe the answer is (B) as molten Kevlar when spun into fiber, its polymers have a crystalline arrangement.
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