0.0006 in scientific notation is 6 × 10-4
<h2>Step by step Explanation:</h2>
All numbers in scientific notation or standard form are written in the form
m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.
To convert 0.0006 into scientific notation, follow these steps:
Move the decimal 4 times to right in the number so that the resulting number, m = 6, is greater than or equal to 1 but less than 10
Since we moved the decimal to the right the exponent n is negative
n = -4
Write in the scientific notation form m × 10n
= 6 × 10-4
Therefore,
6 × 10-4 is the scientific notation form of 0.0006 number and 6e-4 is the scientific e-notation form for 0.0006
<h2>HOPE IT HELPS ☺️</h2>
speed of light=wavelength*frequency
3*10^8m/s=wavelength*1.20*10^13
wavelength=3*10^8/1.2*10^13
=10^-5/0.4=2.5*10^-5m
Answer:
The structure is given in attached file.
Explanation:
Explanation
2-bromocyclopentamine (Figure attached) is a synthetic compound which is synthesized by substitution reaction of cyclopentamine and hydrobromide. Its molecular formula and molecular mass are C5H10NBr and 164.05 mol/g respectively. It is a very reactive compound so it doesn’t available in pure form, it is present in market as a mixture of 2-bromocyclopentamine and Hydrobromide.
Properties
:
Its boiling point is 115 0C
Its melting point is – 75 oC
It is highly flammable
It is highly toxic
It is irritant
It is corrosive in nature
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
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