Answer is: 6.022·10²² molecules of glucose.
c(glucose) = 100 mM.
c(glucose) = 100 · 10⁻³ mol/L.
c(glucose) = 0.1 mol/L; concentration of glucose solution.
V(glucose) = 1 L; volume of glucose solution.
n(glucose) = c(glucose) · V(glucose).
n(glucose) = 0.1 mol/L · 1 L.
n(glucose) = 0.1 mol; amount of substance.
N(glucose) = n(glucose) · Na (Avogadro constant).
N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.
N(glucose) = 6.022·10²².
The model would look something like the image below.
There would be a <em>central nucleus</em> containing <em>20 protons</em> and <em>20 neutrons</em>.
Surrounding the nucleus would be four concentric rings (energy levels) containing <em>20 electron</em>s.
Going out from the nucleus, the number of electrons in each ring would
be <em>2, 8, 8, 2</em>.
Answer: the empirical formula is C3H4O3
Explanation:Please see attachment for explanation
Answer:
The answer to your question is 7160 cm
Explanation:
Data
diameter = 1 mm
length = ?
amount of gold = 1 mol
density = 17 g/cm³
Process
1.- Get the atomic mass of gold
Atomic mass = 197 g
then, 197g ------------ 1 mol
2.- Calculate the volume of this wire
density = mass/volume
volume = mass/density
volume = 197/17
volume = 5.7 cm³
3.- Calculate the length of the wire
Volume = πr²h
solve for h
h = volume /πr²
radius = 0.05 cm
substitution
h = 5.7/(3.14 x 0.05²)
h = 5.7 / 0.0025
h = 7159.2 cm ≈ 7160 cm
Answer:
The freezing point will be 
Explanation:
The depression in freezing point is a colligative property.
It is related to molality as:

Where
Kf= 
the molality is calculated as:




Depression in freezing point = 
The new freezing point = 